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H: Do you need to solder the center point in this type of BGA chip?
I have the following chip. I'm not sure if its called a BGA chip or not. I need this chip replaced and I'm able to remove it using a hot air station, but I would like to know if I require soldering the center pad that is underneath the chip. Is this point usually used for important data that needs to be read or is just mainly for securing the chip with glue?
AI: This style of package is QFN (Quad Flat No-leads package). The specific instructions about connecting the pad would be found in the datasheet. (@brhans have mentioned this already.)
The central pad is usually connected to GND, although GND is sometimes also wired to one of the pins on the periphery.
The pad may also also serve for heat sinking the IC to the PCB. For some ICs the heat sinking aspect is important (power ICs, for example). For other ICs, it's not important. Solder provides a better thermal path than lack thereof. |
H: Schottky diode part of a laser power supply enable circuit
In this circuit from Jim Williams' AN90 the enable line pulls down the base of Q1 through Q2, through a Schottky diode (1n5712). Why is this Schottky diode there? I imagine it's a Schottky diode so it has a diode drop smaller than Q1's, so it's possible to turn off Q1... but why does it need to be there at all? It seems as if you could short over it and prevent Q1 from conducting just fine. It must be there for a reason! He even seems to omit it in an earlier figure (http://cds.linear.com/docs/en/application-note/an90f.pdf Fig. 2)
AI: Q2 also pulls down the left side input voltage, which sets the laser current, through the right-facing 1N4148 diode. The Schottky diode keeps these two functions separate.
Without the Schottky, a low voltage output from A1 could pull the current control input low, limiting the input voltage to some value that depends on the laser's forward voltage. There would also be a positive feedback path which could lead to oscillation. |
H: LMR16030 voltage regulator circuit
I am trying to set up a circuit to use the LMR16030 buck regulator to step down 60V to 9V. I bought the Powergood version and I'm not sure how to connect the power good pin.
As you can see in the datasheet on page 3:
http://www.ti.com/lit/ds/symlink/lmr16030.pdf
"PGOOD pin for Power Good version, open drain output for power-good flag, use a 10 kΩ to
100 kΩ pull-up resistor to logic rail or other DC voltage no higher than 7 V."
I'm not sure how to apply this as I do not have a DC voltage at 7V in my circuit.
I based my circuit on the application section of the datasheet page 19 and encountered a few problems with the Powergood pin open. I first forgot to connect the GND and obtained the desired voltage, but oscillating, as my arduino was turning on and off rapidly. When I then noticed my error, I connected the GND pin properly and it instantly shorted with a big spark at the GND pin. Could this be due to the fact that I left the PGOOD pin opened?
Thank you for your help!
AI: Could this be due to the fact that I left the PGOOD pin opened?
No. That is an optional output. Leaving it unconnected won't cause a problem.
When I then noticed my error, I connected the GND pin properly and it instantly shorted with a big spark at the GND pin.
Until you have enough experience to know when to break it, a good rule of thumb is: Never make changes to a circuit with power still applied.
Even if you are changing to the correct connections, doing so with power applied can cause partial (sometimes hidden) or fatal damage to components. So please, despite the temptation to fix the connection problem quickly when you notice it, power off (and wait for any capacitors to discharge) first! :-)
The spark suggests a large potential difference between the ground at the output (which you connected at the time you saw the spark?) and the ground of your regulator. Although I can't say that this definitely caused damage, it won't have done any good to anything to equalise the potentials so quickly.
Regarding the PGOOD pin:
I'm not sure how to apply this as I do not have a DC voltage at 7V in my circuit.
That topic is answered in the datasheet, section 7.3.9:
Voltage seen by the PGOOD pin should never exceed 7 V. A resistor divider pair can be used to divide the voltage down from a higher potential.
Therefore if you were actually going to use that PGOOD signal, you could use a resistor divider from the 9V output (once your regulator is supplying a stable output!) to generate whatever pull-up voltage you require. However if you are leaving that pin unconnected, it sounds like perhaps you aren't using that signal, in which case you simply don't use it.
I first forgot to connect the GND and obtained the desired voltage, but oscillating, as my arduino was turning on and off rapidly.
Personally, if I'm designing a power supply, I don't use something valuable connected to the regulator's output while I'm testing, in case of unexpected problems which might cause damaging output voltages. Instead of using your Arduino, you may want to consider making a "disposable" load e.g. car light bulbs, or high-wattage resistors. There are also robust electronic loads available.
Finally, I wanted to point out that this IC requires special care regarding the PCB layout (see datasheet section 10) and component choice (inductor, diode and capacitors, see datasheet section 8). If you are trying to make this using some kind of breadboard arrangement, you may not be successful - even if it seems to work, it may not be stable, and/or may emit lots of EMI :-(
Update: As discussed in later comments, TI offer evaluation module LMR16030PEVM for this IC, which provides a "known working" PCB, and was a successful solution in this case. |
H: minimum spacing between hobby circuit board edge and DIP IC
Since I don't run a labratory and I have a standard printer capable of printing 600 DPI, I'm trying to make a small circuit board but I'm wondering, whats the closest I should place the pins of a DIP IC to the edge of a circuit board?
My normal track settings are 0.3mm thickness, 0.3mm minimum spacing between everything.
I am at a hobbyist/amateurish level so I don't want to screw up my circuit only because an IC is too far to the edge.
AI: Depending on your etching setup, 0.3mm may be too small, and you might find you're etching away part of your tracks or you might find shorts between them. Home-etched boards generally require a lot more space to prevent these problems, I would probably shoot for around 0.5mm minimum trace width/spacing.
For my professionally made boards I generally aim for 20 mil (~0.5mm) minimum distance between any copper or pads and the board edge. For hand-etched boards I would probably aim for about twice that (~1mm) minimum from the board edge. This will minimize the chances of problems.
It does very much rely on your experience and confidence level, though. I suggest giving it plenty of space to begin with, and as you make more boards and you find the tolerance is pretty tight, you can reduce the clearance and trace widths as needed. It's entirely up to you, but the numbers I offered should at least provide a jumping off point. |
H: What do these deprecated markings mean?
I found this Moog Prodigy synthesizer schematic that I'm trying to figure out how it works in order to learn more about how electronics work.
What do the circled symbol at the top of the schematic mean ?
What does "N.C" mean ? An example of it is located at the far right.
What does the "*" next to R2 mean ?
What does the "1%" next to R5 mean ?
What does -9V and -12V mean ?
What does the arrow next to R14 mean ?
AI: The item in the top left is a 2-pole switch that has 3 positions (32', 16', 8'). In the top right, NC means no connection. The 1% is the tolerance of resistor R5. R14 is a potentiometer (variable resistor). The arrow is the wiper arm of R14. The -12V is the voltage powering that part of the circuit. The voltage (-9.0V) at the output of U2B is either for purposes of troubleshooting or for an adjustment. The meaning of the star above R2 is probably explained in a note somewhere else on the schematic. |
H: Supernode with 2 voltage sources and 1 current source: node voltage analysis
I'm struggling with defining a super node in this case. My initial thought was to create a super node with V1 and V2, but I'm not sure how to handle the current source. Doing KCL in this configuration would mean that the current source cancels itself out and becomes irrelevant. Is that correct?
The other option is to also include the current source in the super node, but I'm not sure how I would do that.
Note that the goal of the problem is to find the voltages at each node.
simulate this circuit – Schematic created using CircuitLab
AI: As Photon points out and you already seemed to know, the arrangement means that the current source has no impact. But I wanted to add a redrawn schematic to help emphasize the fact. Ideal voltage sources have zero impedance. Looking at the redrawn schematic, it is pretty easy to see which path the current source's current magnitude will take: through \$V_1\$ and \$V_2\$. And that means it doesn't affect the voltages at any of the nodes.
This schematic also emphasizes that there really is only one unknown node, \$V_x\$. The others are determined from that. The only additional problems are working out the currents in the two voltage sources.
simulate this circuit – Schematic created using CircuitLab
Using nodal analysis this works out to:
$$\begin{align*}
\frac{V_x}{R_3} + I_{V_2} &= 3\:\textrm{A} + \frac{0\:\textrm{V}}{R_3} ~~~\therefore~~~~ V_x=R_3\cdot\left( 3\:\textrm{A}-I_{V_2}\right)\\
\\
\frac{V_x+9\:\textrm{V}}{R_1} + I_{V_1} &= I_{V_2} + \frac{0\:\textrm{V}}{R_1} ~~~\therefore~~~~ V_x=R_1\cdot\left(I_{V_2}-I_{V_1}\right)-9\:\textrm{V}\\
\\
\frac{V_x+30\:\textrm{V}}{R_2} + 3\:\textrm{A} &= I_{V_1} + \frac{0\:\textrm{V}}{R_2} ~~~\therefore~~~~ V_x=R_2\cdot\left(I_{V_1}- 3\:\textrm{A}\right)-30\:\textrm{V}
\end{align*}$$
The above is three equations in three unknowns which works out to \$I_{V_1}=4.3\:\textrm{A}\$ and \$I_{V_2}=4.05\:\textrm{A}\$ and \$V_x=-10.5\:\textrm{V}\$.
If I had to guess what would happen when you remove \$I_1\$ from the circuit, I'd guess that \$I_{V_1}=1.3\:\textrm{A}\$ and \$I_{V_2}=1.05\:\textrm{A}\$ and \$V_x=-10.5\:\textrm{V}\$. Which would show that all of the current from \$I_1\$ goes through \$V_1\$ and \$V_2\$. (And a solution would show that this is, in fact, the correct result.) |
H: Why are there no BGA chips with triangular tessellation of circular pads (a "hexagonal grid")?
Ball grid arrays are advantageous integrated circuit packages when a high interconnect density and/or low parasitic inductance is paramount. However, they all use a rectangular grid.
A triangular tiling would allow π⁄√12 or 90.69% of the footprint to be reserved for the solder balls and the surrounding clearance, while the ubiquitous square tiling only allows π/4 or 78.54% of the footprint to be used.
Triangular tiling would theoretically allow either reducing the chip footprint by 13.4% or increasing the ball size and/or clearance while maintaining the same footprint.
The choice seems obvious, yet I have never seen such a package. What are the reasons for this? Would signal routing become too difficult, would manufacturability of the board somehow suffer, would this make adhesive underfill impractical or is the concept patented by someone?
AI: Unless you use via-in-pad, which costs more, you need room to put routing vias in between the pads, like this |
H: Why is VGA connector has only one pin for red color data but is composed of 8 bits?
VGA connectors have 15 pins, three of these are used for color data of red, blue and green. Based on this project, each color has a width of 8 bits. However if under the scheme of 8-bit color, each color will have 8 bits of data. (Upon further inspection, 8-bit color means 3 bits red, 3 bits green and 2 bits blue) How can you have multiple bits of data in one pin?
AI: A VGA signal cable uses analog signals for R, G, and B. H-sync and V-sync are... digital sync pulses.. but the color channels need to be converted into a digital signal through a sampling process to be fed into an LCD.
Traditionally these signals were fed into a CRT with a red, green, and blue cathode ray, and so these signals were simply modulated into the beam intensity of each. |
H: 2.4 ghz RF radio and antenna - nrf24 underneath a 7-segment display
I have this tiny nrf24 module that I want to use underneath a 7-segment display.
How much will this affect the range and the quality of the radio signal?
I have done a lot of research and testing on this other similar nrf24 module. Mostly about modifying the antenna to get a better signal/range, but unfortunately, none of those mods really do anything. Since then I went back, read more and found out that the antenna design in these nrf24 module are called Meandered Inverted-F Antenna (MIFA), which means it is the already a good design and trying to modify it is a bad idea? After done my own homework, I am able to get twice the range than before, and I am pretty happy about it, but longer is always better.
AI: These nrf24 modules use the 2.4 GHz ISM band. The wavelength of such a signal is around 12 cm. Antennas get useful when longer than one quarter of that so in this case 3 cm. That is in the order of the length of that meander. You adding wires to it actually makes the antenna perform worse !
At 2.5 GHz you cannot do much with standard wires. You could disconnect the meander on the module (there is probably a resistor or capacitor feeding the signal to the antenna) and use an antenna for Wifi instead. Note that these will have a small coax cable to feed the signal to the antenna. These are quite small and hard to solder properly. Some experience is required ! Then mount that Wifi antenna somewhere where it has some space around it.
It is hard to predict what range you will get when mounting the module between other modules. Anything containing electrically conductive structures will harm the effectiveness of the antenna. It is always a good idea to give the antenna some space to "breathe". |
H: capacitive proximity sensor - how does it work
I know 3d printers such as the i3 use a Capacitive proximity sensor to try compensate for the uneven surface of the print bed. for example this one seems to be popular
https://www.amazon.co.uk/gp/product/B00542U3M4/ref=oh_aui_detailpage_o00_s00?ie=UTF8&psc=1
If am not mistaken a capacitor is formed with two plates and dialetric in this case glass.
Does this sensor work by forming a capacitor between the bed and the sensor i.e. the bed is one plate of the capacitor then the glass is the insulator and then the sensor itself is the other plate.
AI: No, the bed is not the second plate.
Those sensors contain both plates of the capacitor, but they are arranged so that the area in front of the sensor is part of the dielectric volume.
If some material is brought into that area, its dielectric constant and its distance to the sensor changes the capacitance. This allows to detect almost any type of material (your idea would work with conductors only), though calibration is needed to adjust the sensor to the actual material.
When the object has a large, flat surface, the sensor can be used to measure the exact distance as well. |
H: Fourier serie coefficients of real function
Let's take a periodic wave function defined by f(t) = (1 - t) with a period of 1.
The Fourier coefficients can be calculated:
$$
\omega_0 = 2\pi/T=2\pi \\
F(n) = \int_{0}^{1}(1 - t)e^{-jn 2\pi t}dt=-\frac{j}{2\pi n} \\
F(0) = 1/2
$$
There is a property of Fourier series which states:
If f(t) is real and can be developed into a Fourier series: F(n) is purely imaginary and odd if and only if f(t) is odd.
My question is, why is the F(n) expression I get is purely imaginary if my f(t) function isn't odd (nor even in fact).
I might just be blind, I might not see something, in either case I would appreciate you point it to me, but I get that:
f(t) is not odd, nor even (\$f(-0.3) = 0.3, f(0.3) = 0.7\$)
F(n) is purely imaginary as \$ F(n) = -j/2\pi n \$ (no real part)
So does this example contradicts the stated property of Fourier transforms, or is it due to something I don't see? I expected to get a complex F(n), with both a real and an imaginary part. Might turn out to be a dumb question, oh well...
Thanks
AI: Let me rewrite that statement for you, I believe I know where your misunderstanding lies:
\$F(n) \space \forall \space (n \in \mathbb{Z}) \$ is purely imaginary and odd if and only if f(t) is odd.
In plain English: \$F(n)\$ (for all n in integers including 0) is purely imaginary and odd [...]. In your example, \$F(0)\$ is not imaginary. Therefore the statement remains correct.
Now think of a new waveform that looks exactly the same as your original "sawtooth" function, but that is now odd: \$g(t) = f(t) - F(0) = f(t) - 1/2\$. If you redo the exercise, you will find \$G(0) = 0\$, confirming the statement. |
H: is soldering mobile processors something that i could do at home?
I want to connect a mobile processor such as a Samsung Exynos or Qualcomm Snapdragon, which both have their connectors on the bottom to a PCB but im not sure if this is something i can do on my own or needs a pick and place machine, to be honest im not entirely sure what the method of soldering used is called. Can you please inform me if a socket is available for these also? Sorry if this came across as a dumb question, thanks for the help :)
AI: It is not possible to solder components with contacts on the bottom (called Ball Grid Array, or BGA) using a soldering iron. You would need either a reflow oven, or a BGA rework station.
Some people do have reflow ovens at home, or have converted toaster ovens to do the job, but even then it's a bit tricky and easy to mess up.
So the answer to your question is yes, you can do this at home if you have the right tools, but if you only have a soldering iron then you can't. |
H: Has anyone used partsim and if they have, where is the potentiometer?
Thanks you all for reading. Lately I have been trying to use www.partsim.com to model a vintage synthesizer in hopes of building a real one with real hardware. However it's really expensive to do so and I don't know anything about electronics so I don't want to waste my time and energy if I'm not good enough to do the project. Was wondering if anyone has used it and if they know if there is a potentiometer object and a 2 pole - 3 way switch object.
AI: There is no potentiometer in partsim.
Why not just use two resistors and change the resistance as required?
Partsim is a simulator rather than a schematic package, so using the resistance values rather than an adjustable potentiometer should be fine, (to my knowledge tracks are not simulated).
Same again for the switch, just build two circuits, one switch and one unswitched, (again to my knowledge the sim does not model switching transients).
Alternatively, email digikey and ask them to generate a potentiometer for the sim, (support@partsim.com.) |
H: Understanding microcontroller ADC input pins (atmega328)
If I set up the ADC on the ATmega328 to read from a certain ADC pin, what state is the ADC pin put in to?
Since I don't explicitly enable any pull ups, is the pin floating/high voltage/low voltage?
AI: Since I don't explicitly enable any pull ups, is the pin floating/high voltage/low voltage
The pin state is whatever voltage you are feeding into that pin. If you are feeding in 3V then the pin reads the binary equivalent of 3V. If you feed in 2V then the pin reads the binary equivalent of 2V.
If you don't connect anything to the pin then it is floating.
When working with ADC there is no concept of HIGH or LOW, only discrete quanta of voltages (\$\frac{1}{1024}V_{ref}/LSB\$). |
H: What happens with a diode and a battery in this circuit?
What would happen if I have a battery and a diode in a circuit with the diode trying to force the negative part of the battery to the positive part? Just wondering...
simulate this circuit – Schematic created using CircuitLab
AI: You would have a somewhat fuzzy-worded short circuit. Depending on the exact realization,
if you put the diode in the direction shown in your picture, it will not conduct, and nothing will happen.
Earlier answers (before you included the circuit):
if you have a battery with a voltage lower than the voltage at which the diode starts conduction very little would happen
the diode could explode, or fail less spectacularly, probably open
some wire could melt
the battery could start to boil, leak, explode, or just fail
a moderate current could flow, limited by the resistances in the circuit, until the battery is empty
If you are referreing to an ideal battery and diode: you would have created a circuit that violates the requirements for being analyzable. |
H: Help with converting scheme
I have this scheme and I can't make it look normal in order to count total resistancce
AI: So it doesn't look like you want us to solve it for you which is the reason I am helping. If you had wanted it solved for you, you would have provided values.
my guess is perhaps you just want to look at the circuit in a different layout, in which case:
simulate this circuit – Schematic created using CircuitLab
looking at it this way may be what you are looking for. |
H: Is it possible to import a netlist from Multisim into Kicad?
So at the moment I'm using two different programs for modeling circuits. Multisim and KiCad. I use Multisim because it allows me to do SPICE simulations, something that KiCad doesn't allow me to do. I then need to route the circuit on a PCB, something that Multisim by itself can not do. That's where KiCad comes in. I've used KiCad before to route circuits on a PCB and it's worked out well for me. The problem is the netlist that Multisim outputs doesn't work when I try to import it in KiCad. What I think is going on is that the .net file type used by KiCad runs a different format than the one in Multisim. Is there a way to convert these two files so they are compatible?
AI: Have you tried exporting to previous multisim formats? I dont use it but, perhaps, it has the feature to export to other versions, or even other formats, like the gerber format (this can be read with gerbview).
I personally what to do its to simulate in gEDA thats a full suite to simulate (ngspice) and make the pcb doing in pcb and other things.
See ya! |
H: Passing Voltage above threshold
I am trying to design a electronic circuit which will only pass voltage between a threshold, like a dual clipper.
The input is a varying DC input and the system (would) allow output only above 3.3V.
Initially I tried to design a negative clipper with bias voltage but then the bias voltage powers up the output when the input voltage goes below the bias voltage, which would be detrimental as I would need to recharge the source.
Please suggest alternate approaches.
Edit 1: The source is a low impedance input ranging in 10 - 30 ohm.
The load is a boost circuit which boosts the voltage to steady 5V.
The voltage tolerance is not an issue because the load can withstand voltage from 2.5 V - 5.5 V.
I would like the bias voltage to not be forwarded to the output, thus when the input falls below threshold, the output falls to zero.
AI: simulate this circuit – Schematic created using CircuitLab
Adjust R ratios to suit supply voltage , and 1 diode drop below threshold
Output equals input with very little offset. |
H: Crowbar circuit with TL431 help
So I'm trying to create a OVP using TL431 an a triac, i have checked the datasheet but since i suck at math i don't really get how to set the ref voltage..
datasheet: http://www.ti.com/lit/ds/symlink/tl431b-q1.pdf
My example circuit:
I'm having a 19v power supply, and i want to make it so that if the supply for some reason should reach over 20v, it blows the fuse.. Any tips? Thanks!
AI: That might work, but try flipping the triac so MT2 is grounded and MT1 connects to the input voltage, and connect the 'cathode' of the TL431 ONLY to the gate.
Calculate the gate resistors as @Tony suggests, so for about 19.5V you could use 2.49K and 16.9K, standard E48 values.
This circuit won't work with an SCR because SCRs don't generally come in the opposite polarity.
Edit: Like the below
simulate this circuit – Schematic created using CircuitLab
When the voltage at the REF input of the TL431 exceeds ~2.5V the gate current increases until the triac triggers and shorts out the supply, blowing the fuse. The triac is operated in quadrant III (gate and MT2 both negative wrt MT1), which is always acceptable. Below the trigger voltage the TL431 conducts < 1mA which should not trigger most triacs. If it is a problem (or just to be sure), you can connect a 100 ohm resistor from MT1 to gate. |
H: Capacitor output differentiation issue
I am an electronics student and today we were looking at the voltage output of a charged capacitor against time, which can be modelled in relation to R, C and e (I forget exactly how) but our teacher then differentiated it to find the gradient and it confused me because it was not how I thought differentiation worked
I thought differentiation meant that ax^n went to axn^(n-1), but this clearly isn't what happened, so if someone could explain this it would be greatly appreciated!
AI: In this case 'e' is not a variable it is a constant: eulers number, ~2.718.... more commonly known as the base of natural log, ln().
The derivation rules for an exponential function such as a^x (where a is constant) state that the derivative for:
$$\frac{d}{dx}a^x = a^x \ln(a)$$
In your case, we have another constant to worry about, the 2 in front of the x. In which case, we copy down the constant to in front of the base (not bring it down as you were thinking) so:
$$\frac{d}{dx}a^{bx} = b a^{bx} \ln(a)$$
Where b is another constant.
So, there is one more point remaining, \$\ln(e) = 1\$. Relplacing \$a\$ with \$e\$ means that:
$$be^{bx}\ln(e) = b e^{bx}$$
by simply replacing b with 2, we get what wolfram alfa told you for the derivative of \$e^{2x}\$.
The main point: What you were thinking about differentiation is when the variable is the base of the exponent, not when the variable is in the exponent itself. |
H: MOSFET switch circuit design feedback
I would like to have some feedback on this circuit.
V1: 0,6A Solenoid
V1_LED: Arbitrary small LED
V1_IN: Input signal from Arduino (3.3V, 20mA)
Should I change something?
Is the IRLML2502 suitable? Is there a better choice?
Edit:
Link to MOSFET:
http://www.infineon.com/dgdl/irlml2502.pdf?fileId=5546d462533600a401535667f44d2602
AI: I would:
Ditch Q1/R7/R8/D2
Drive Q2 direct from the Arduino (maybe with a small resistor to limit inrush current, though not so large as to slow down the switching needlessly)
Use a pull-down to GND on the gate
Choose a MOSFET with a low enough \$R_{DSON}\$ at 3.3V
Make sure I chose a MOSFET that could happily dissipate enough heat for the current through it.
simulate this circuit – Schematic created using CircuitLab
(Depending on thresholds you may want to move the 10K pull-down resistor to the left of the gate resistor, but it generally doesn't really make any difference. It's only there to keep the FET turned off when the GPIO pin is floating at boot-up.)
Looking at the datasheet it looks like the MOSFET you have chosen may do the job. At 2.5V \$V_{GS}\$ the \$R_{DSON}\$ is (typically) 0.05Ω. With 0.6A through it you have 0.018W, which is much less than the 1.25W limit at 25°C. |
H: How to safely measure a 480VAC waveform with a DAQ?
I have a LabJack U6 Data Acquisition unit and would like to measure 12 (4 sets of 3 phase) 480VAC lines simultaneously at 600 samples/second. This would be used to evaluate a 480VAC 3phase, generator and utility system that has intermittent brown-out and black-out issues. The voltage transducers in the existing hardware are not responding fast enough to determine which system or load is causing the issue. I have looked at other voltage transducers, but it seems they all average and have a relatively slow refresh rate (~250ms).
My initial thoughts on this would be to use a voltage divider of 100:1 to get the voltage lower, feed that to a linear optocoupler, then feed the isolated side of the optocoupler into the DAQ. Am I reinventing the wheel here?
I would prefer to find a manufactured device for this, but have had no luck. I'm a mechanical engineer so maybe I just don't know the correct vocabulary.
AI: Use an error amplifier similar to the ADuM3190. Those provide isolation and decent feedback for current sense/line sense.
Use the device as a standard amplifier for your divider and feed the output into the A/D of your microprocessor (if available). The sample rate will likely be limited by your AD/DAQ. |
H: Resistors soldered 1cm off board
I've got a 8 year old pair of powered studio monitors that have recently started hissing for about 5-10 minutes after being powered on from being completely off. The culprit was some bad caps that I am replacing. While I was at it, I noticed that there are two large resistors coming from a bridge rectifier that are positioned in such a way that they are 1cm vertical from the board. I was wondering, is there any reason for resistors, or other components for that matter, to be soldered in such a manner?
AI: The resistors are spaced away from the PC Board so as to allow better cooling.
One side benefit of doing this is that heat damage to the PCB is reduced or eliminated.
It is common to see discolored circuit board under and around resistors that run hot. Spacing the resistor away from the board both helps keep the resistor cool as well as minimizing board discoloration and damage. |
H: Applying KVL to DC bias calculation
I'm trying to use Kirchhoff's voltage law to find the bias point of a bjt amplifier circuit, as shown below:
I begin with switching off the signal and opening all capacitors
simulate this circuit – Schematic created using CircuitLab
The transistors are identical, both with beta of 150.
I then assume Q1 and Q2 are in forward active mode, and arrive at the following:
Vc - VBE2 + r4* ix = 0 ---> (1)
Vc - VBE2 -iy* R6 - VBE2 - R3* ix = 0 --- >(2)
ix = 151*iy ---> (3)
Solving the above in Wolfram, I find that both ix and iy are negative. What do I do then? Does it mean Q1 is not forward active, and all the assumptions are wrong?
I hate to ask a "please check my calculations" sort of question, but I've verified the steps again and again for hours and am getting nowhere.
Please enlighten me!
AI: simulate this circuit – Schematic created using CircuitLab
Try again with KVL with same schematic re-arranged, so that you do not confuse Ie1 with Ie2 both being the same Ix. |
H: Will this PCB trace GSM antenna be affected by EMI?
I am designing a PCB in which there are 3 SIM800 GSM modules sharing a (commercial) external antenna. The external antenna is connected to the PCB via an antenna connector, and from that connector, the antenna signal is “shared” via PCB traces running to the GSM_ANT port of the GSM modules (please see photo)
The reason I prefer this design is that it is both cheap and convenient.
But without in-depth knowledge on antenna design and EMI, my question is Could this be a bad design? Or should there be scenarios that this design does not work reliably?
Just to provide more info, the PCB will be housed inside a steel enclosure, and its main PSU is an AC-DC adapter (good brand) which is already sealed for EMI. The adapter is also inside the steel enclosure. The steel enclosure will be deployed in a residential area where there are no large machineries around.
Thanks in advance!
Dave
UPDATES:
Following some of your answers, I update the question as follows:
Below is the old design, which uses 3 separate antenna connectors for the 3 GSM modules. However, these 3 antenna cables are combined (soldered) into one (with seal) outside the PCB and go to the external antenna.
The 3 GSM modules are supposed to be sending signal one at a time.
This old design has been tested and worked (though I suspect that signal strength are just around 8/10 of what it should be - good enough for the application though)
AI: I think a more reasonable way would be to use an external three way splitter and the old design to connect your three GSM modules to one antenna.
You don't seem to have the knowledge to do RF PCB designs (not knocking you, I don't either) so it would be safer for you to use an external device that "Just Works."
Three-way splitter:
The old design may have been more advanced than you think - proper lengths of coax joined correctly can form an RF splitter. See "Wilkinson power divider" on the Wikipedia power divider page. For the uninformed eye, this would look like somebody just lashed a bunch of coax cables in parallel and hoped for the best - when in reality the lengths and connections were carefully selected to make a proper splitter that isolates the three sources from each other.
Wilkinson divider:
You can also make a Wilkinson divider on your circuit board, though that may take too much space.
Microstrip Wilkinson divider: |
H: LEDs and voltage optimisation
I am looking for some advice.
I understand that LED drivers transform a supply from AC to DC as well as the amount of volts from input to output, for this reason i believe they are classed as inductive loads.( these are not inductive, these are non-linear resistive loads)
Will there be an increase in efficiency using LEDS on a site with voltage optimisation and if so what would the average increase in efficiency be. Just so you know there will be a lot of 22w LED tubes.
I want to say they voltage optimisation will increase effiency but im not sure if it will be worth it considering the buy back time and the fact there are minimal inductive loads on site.
Thank you in advance for any replies
Liam
ps. £60,000 for voltage optimiser
2000 LED lamps on site average 22w
AI: "Voltage optimisation" - essentially reducing mains voltage so that it is at a lower voltage than average, is almost certain to be a waste of money and have negative rate of return in your application.
ie In most applications VO is expensive snake oil. In a very few it may be worthwhile BUT you need to know why and design the system well. In almost all cases, don't even think about it.
In your application they are almost certainly of no value because an LED tube's inbuilt power supply is designed to operate the LEDs at a designed operating point and designed power level. To do this the supply takes the available mains voltage and draws whatever energy is needed to operate the LEDs correctly. If the mains voltage is increased the current drawn will decrease proportionately. Similarly, if the mains voltage is decreased the current drawn will increase proportionately. In each case the input power will be essentially the same.
I say "essentially" as the efficiency of the mains AC to LED drive converter will probably vary somewhat as mains voltage varies - but this will vary with design and possibly amongst inits of the same model, and the worst case differences are liable to be small.
It is possible but unlikely that the tubes that you are buying have a significantly greater efficiency at some input voltages than others. In this unlikely event the optimum voltage may be at the low end or high end or at the average expeccted mains voltage. This would be due NOT for the reasons that voltage optimisers are claimed to work for (lower Vin = lower power draw) which can apply in SOME special cases, but due to manufacturing and design issues that the voltage optimiser maker is unaware oif and unable to design for.
If you really care you could take say 10 tubes and operate then from a variac from Vmin_allowed to Vmax_allowed in say 10% steps and see if there is any useful change in efficiency for a single tube and, if so, if this is consistent across the samples tested. It is extremelt likely that this will prove that a voltage optimiser serves no useful purpose here.
VO's and PFCs
Voltage Optimisers have some overlap with PFCs = but usually not much.
Simplistic mains AC to LED drive circuits tend to draw high current at mains voltage peaks and lower current as voltage drops in a cycle and then probably no power for much of the ranage around zero crossing.
This leads to large current peaks at voltage peaks and no current for a large part of the cycle. This can be "deconstructed" as a current waveform rich in harmonics - the end effect is effectively low power factor and violation of regulatory standards aimed at exactly this issue as electronic loads increase. To meet standards modern LED drivers incorporate waveform correction circuitry - also able to be seen as power factor correction circuitry, which converts the current drain to essentially mains frequency sinusoidal with less than an allowed amount of distortion.
A "good" voltage optimiser may well deal with peaky loads as a part of its voltage shifting task. Failure to do so would make them unsuited to some roles due to regulatory non-compliance. This is no guarantee that they actually will behave as desired. If the load already draws essentially sinusoidal current the VO serves no purpose in that area and probably little or none on any other.
Related:
VO - Wikipedia
Jeff Howell - fergetaboutit
Some real tests - Bzzzzzzt |
H: Intel i7 Core CPU numbers on the front side
What is the numer at the fourth row on the front side of Intel i7 CPU? (pictured)
I saw two Intel CPUs in the store described as "Core i7-6700" but have slightly different prices (both OEMs). Only the difference I found is the numbers on the front side.
P.S. I'm already guessing it is kind of a serial number but not sure.
AI: That is the Finished Process Order serial number, and is essentially a lot number. It is used by Intel to determine warranty information.
Source 1
Source 2 |
H: Technical term for smooth scrolling digital volume encoder
I am making a bluetooth volume control from a machined aluminium billet.
I cannot find a technical name for the digital volume encoder that would provide that smooth rotation.
Or am I approaching this wrong and should be looking at smooth analog pots and then convert the signal to digital?
AI: What you're looking for is called a "detentless rotary encoder". They often come with quadrature output, so you may need to decode it in order to provide discrete up/down steps. |
H: Why gmbs is not zero when B and S are tied together?
I tried to run this simple circuit for DC operating point. The transistor is in saturation with Vth = 800mV.
What makes me confused is about gmbs.
It is not zero but also very large compared to gm.
Why gmbs is not zero when B and S are tied together?
Thank you.
AI: The value \$g_{mbs}\$ is a small-signal parameter. It is a linearization around a given operating point.
It answers the question how much the dependent variable would change if the independent variable is varied. Even though \$V_{BS}=0\$ we could change this voltage by a small amount \$v_{bs}\$ and consequently the drain current would change as well. The amount is given by \$v_{bs} g_{mbs}\$.
Likewise the \$g_m\$ of a transistor is not zero even if the gate is connected to a fixed voltage source.
Make a transient simulation of the following example and try to understand what's going on. I am sure the value of gmbs will suddenly make sense to you.
Update:
The reason why gmbs is not equal to zero lies in the fact that backgate voltage acts through the backgate-effect.
The drain current in saturation is given by the following equation
$$
I_D = \frac{K'}{2} \frac{W}{L} \left(V_{GS} - V_T(V_{BS})\right)^2
$$
where VT is a function of VBS!
Without going into the details of the derivation this finally results in
$$
g_{mbs} = g_m \frac{\gamma}{2\sqrt{2\Phi - V_{BS}}} = g_m \cdot \eta
$$
So, gmbs is proportional to gm!
Even if \$V_{BS}\$ is zero a small change will alter the threshold voltage an therefore the drain current changes as well. |
H: Algorithm for testing EEPRM
What are the steps or the algorithms to verify that an EEPROM is working ?
Right now I just write 10 bytes and I read 10 bytes from 10 addresses, and that's naïve approach. Is there an algorithm or an approach to verify that the eeprom is correct ? or also the problem for flash memory testing ?
AI: For an initial test, you could use the walking ones approach.
Your flash may or may not require a complete buffer to be written.
The walking ones test is well known and is one of the fastest tests to check proper operation of a memory device.
Update in response to comment.
I implemented this specific test to determine the presence or absence of a flash device in the early 1990s, so yes, it works with flash.
[Update for full buffer write]
Some older (and many serial access) devices require that a complete write buffer be filled; this simply (internally) increments a counter so you can programme the data word using the above algorithm and simply fill the rest of the buffer with 0xFF for every other byte - that way only one word actually gets truly programmed (flash devices are 0xFF on a byte basis when not programmed). |
H: How do I assemble this using the following circuit?
I'm a third year Computer Science student and have decided to create a pulse sensor for my third year project. I'm mainly interested creating software to visualise results but I thought it would be cool to attempt to actually build the device also.
I'm attempting to follow this guide but having difficulty reading this schematic:
So far, this is what I've come up with...
1) Connect 1 leg of the LDR to 6v power and the other leg to the base to base (B) of Q1.
2) Connect C of Q1 to a 1k resistor and then connect that to my 6v (Unsure how I'd connect this to my 6v if the 6v is already connected to leg of LDR)
3) Connect Q1 (B) to R3 and then to ground.
From here I'm a little lost, would appreciate some advice.
I'm also a little confused to how I connect multiple things to ground/power. Will each connection have its own wire running to ground or do I need to create some sort of T like junction from the existing power/ground wire?
I've tried to give this a go in advance so I'm not just poaching for the answer but would appreciate some help.
Thanks in advance.
AI: Physical Connections do not have to look exactly same as logical connections shown in logic diagram as long as they match "logically"
The LDR is a photon current source that changes its effective resistance from >1M in dark to <<10K in bright light. ( they often have a dynamic range of 3-4 decades max, where as a Panasonic 5mm Light sensor has a dynamic range of 4-5 decades and is faster and FET buffered. (also cheap)
The LDR has a large area and thus more current can be captured but also large C, so the RC equivalent response time depends on light intensity and equivalent R with a fixed internal C, As stated by @FakeMoustache this causes slow response at high R values ( or light pulse turn off )
Since the supply shown here is 6V and the Vbe threshold is 0.6V, a 10:1 voltage divider is needed to activate Q1 , thus trimpot of 10k+ for R1 determines this divide ratio is my suggestion.
i.e. if R1 is 5K and LDR is 50K it is almost low enough to turn on LED. If R1 is too high then ambient ceiling light will trigger it without a colour filter. Usually IR emitters and IR detectors work better and faster with high gain Op Amps with a black optical daylight filter. But for ambient daylight the LDR is fine or an unfiltered (optical) Silicon PD (photo diode) is fine, but since smaller, it needs higher gain OpAmp but also,is faster since low C due to small size.
Since this is your 1st DIY, try to experiment and watch for false triggers from ceiling lights, with high R1.or too low R1 with light source too far away.
If you want a better light sensor design look up the Panasonic Light sensor 5mm at Digkey |
H: Any better way to determine source of light by analyzing the electromagnectic spectrum of the light
Project overview
I am tasked to develop a microprocessor based device that when shown a light upon, can determine the source of the light (Natural light, Flourescent Bulbs, LED Bulbs, Incandescent Bulbs, Flame - Forest fire). At this stage, only the visible light is considered.
From my research, the only way to differentiate the source of light is by analyzing the emission spectrum and matching it closely to known values. Example:
Solutions considered
Measuring RGB composition ratio of the light
I have considered going this route as it doesn't seem too complicated, smaller device, can be integrated easily into the bigger project as a forest fire detector and even suggested by my supervisor.
But I have doubts that this would be very accurate as some light sources might have close values (the intensity is what is being measured on a ball park wavelength).
The sensor I am looking at currently is the Hamamatsu’s S10917-35GT RGB color sensor, sensitive to only the required wavelengths.
Building a high resolution spectrograph with a diffraction grating film
This route is much more complicated and requires external processing of the images to determine the light source. Basically, you build a spectrograph with a diffraction grating film and a high resolution camera. The image is processed with a computer software to plot the emission spectrum graph and you can analyse the graph to determine the light source. Development guide is here
Unfortunately, this is not very convenient as we would prefer the device's primary objective to function on it's own without any networking.
So, the question
Is there any drawback on my first solution?
Is there a better solution? Preferably can fit on a standalone device?
This would probably be far fetched but is there a sensor out there that can analyse a light emission and provide intensity values on a range of chosen wavelengths? Or atleast something that would help me build a device that does such.
AI: You really are looking for someone who's already solved this, I suppose. But I don't know of any project, myself. So all I can offer are some thoughts to consider.
On spectrometers:
For a spectrometer device, a DVD-RW (don't use DVD-R, as it will absorb substantial bands in the red region) provides 1350 \$\frac{\textrm{lines}}{\textrm{mm}}\$, so that is very cheap and readily available.
Small megapixel digital cameras are also cheap. An array could also be used, but these days it seems an entire 2D camera is cheaper and more available. So I wouldn't bother with an array.
Using a DVD-RW you can actually separate the yellow spectral lines of
mercury at 577 nm and 579 nm. (Not with a CD, though.) I've done this, myself, using a DVD-RW and a mercury-argon lamp.
Wavelength calibration is cheap. Just get a mercury-argon lamp. You'll get the argon lines in the first minute or so, then the mercury lines will dominate later. From the combination of them, you can easily calibrate your camera pixels vs wavelength. Hg-Ar lamps used for calibration used to cost me about $8, but I expect they are more expensive now.
Intensity calibration is expensive. You need a standard lamp, traceable to NIST standards, and these have to be recalibrated after 100 hrs use, or so. They are cheap bulbs, uncalibrated. But the calibration process costs real money. Then you have to set up a proper optical arrangement, too. But this is the only way to figure out just how each of your pixels respond to each of the wavelengths they are being hit with. Frankly, I'd try and avoid any of this and hope I didn't need it or could just apply a basic templated approximation of a standard lamp and not waste money on actual calibration, hoping that what I got was good enough. Or just not bother at all and use a rigged up equation and figure, "oh, well," and see how it goes. Chances are, you can make this step go away and still get useful results if you just think carefully.
You probably can consider going from 450 nm to 750 nm, but you cannot hope to exceed an octave with a single grating. You may want some kind of filter involved so that you don't get mixed up spectral energies on the same pixels. Or just don't worry about it and do some experimenting.
Optical baffling will be desired to avoid getting extraneous light where it isn't wanted.
Tony just reminded me... you'll need a narrow slit -- about as narrow as you can make it. I prefer the use of two old-style razor blades that can be adjusted. One fixed, one movable. But for the card stock paper box, I just used an exacto blade 'very carefully' to create a narrow and uniformly narrow slit.
I've done all this using a sheet of paper (card stock) that I print out and then cut, fold tabs, use Elmer's glue, and create a box with baffles made essentially out of paper. The baffling uses special dark flocking to help absorb and block wayward light. The DVD slides in at the correct angle and a small camera is then placed at the exit. I've used this with my own eye to observe different lighting in the house and it works PERFECTLY well, in my opinion. I have no trouble differentiating between incandescent, fluorescent, and LED lighting sources. And the sun, for that matter. I tried a DVD-R and immediately saw a huge missing band in the red, which is why I'm telling you that you need DVD-RW if you care about that region.
I could publish some plans for all this, I suppose. Location of slit, angle of DVD, etc. While my box design uses the entire DVD-RW (because I wanted to be able to drop in other DVD media and/or a CD (at a different angle so I'd made two different insertion slots for that purpose), only a tiny part of the DVD-RW surface is actually involved (if baffled correctly.) So I also liked using the entire DVD-RW for that reason, as well, because cutting the DVD into pieces would stress it and I didn't want to do that, either.
Just by way of a little info, the box used a 70 mm vs 40 mm tilt for the DVD (1350 \$\frac{\textrm{lines}}{\textrm{mm}}\$) and 50 mm vs 40 mm for a CD (625 \$\frac{\textrm{lines}}{\textrm{mm}}\$.) The slit was positioned on the 40 mm face, positioned about 10 mm from one edge in either CD or DVD case.
On RGB:
The RGB sensor you mentioned has, as I expected to see, very wide acceptance of wavelengths in each of the three sensors. LEDs tend to have very wide response ranges (they emit and receive over a wide range of wavelengths.) That sensor has modestly overlapping responses. How well all that will work for you, would be a matter of experimentation, I think. You could apply some computer code, instead, using your curves and the response functions of the sensor to see if it would be serviceable. But I'm not going to even try and write it for you. Perhaps the best thing would be for you to knuckle down and buy the sensor and do some testing with it. It may be just fine for your needs. But I can't tell you yes or no, from a quick scan of it. I also haven't tried to do this with RGB, so that's another reason I can't promise anything here and you'll have to just try it for yourself.
I liked Eugene's comment about frequency, too. Incandescent bulbs (and I've tested this using a very sensitive instrument -- with tens of microKelvin resolution and hundreds of microKelvin accuracy traceable to NIST standards, as I work on such things) will vary about 3% of their amplitude during the AC cycling at 60 Hz. (Would be different with 50 Hz.) Fluorescents operate at mains frequencies and also at high frequencies (both are manufactured and used.) But their emissions are through phosphors, which often have fast response times. (Some phosphors are slow, order of millisecond taus due to depending upon forbidden triplet to singlet transitions. But many of them are quite fast -- microsecond taus.) You may have to do some experimenting here. But I think this could be fruitful, because you can design electronic circuits for very narrow bands if you want to. You'd have to worry about conditioning the signal so that you don't saturate the amplifier chain. But that's doable. I haven't looked at the frequencies used in modern LED bulbs, though. And I'll leave it to you to google up details there. All that said, I think Eugene's point has merit worth examining, as well.
Personally? I'd go with the DVD-RW because I have a lot of experience with doing that, know that I can do it easily, quickly, and cheaply, and because I think I could avoid the intensity calibration step to get where you need to go. The cameras are dirt cheap and so is the Hg-Ar lamp for wavelength calibration, periodically. It's almost no work at all. Plus, I already have walked around the house checking out different light sources with a hand-held card-stock box with no electronics at all and was perfectly able to see the differences in various light sources, by eye. So I know I can get there from here.
EDIT: A couple of images from an old fluorescent bulb. One of them across the spectrum and the other zoomed up a bit. Pretty cool separation of the mercury doublet there!
I specialized in binning LEDs for Siemen's OSRAM division years ago, as a contractor. So this stuff comes partly from that experience. We first used expensive spectrophotometers, but switched to Ocean Optics some time later (much cheaper.) But in the meantime I had a lot of fun with DVDs and CDs, used with all that fancy calibration equipment laying about. (Including disappearing filament calibrators, which I forgot to mention above.) Spent a LOT of my time studying human response reports prior to and since the CIE 1931 standard and the later ones in the 1960s. Also really enjoyed Edwin Land's work in the late 1970's and early 1980's -- very interesting stuff. |
H: Questions and confusions on transmission line theory and lumped element model
I want to model a simple signal transmission system by considering it as travelling through a transmission line.
Here is the system:
The original signal generated by the transducer is 0 to 10V step function like signal with very low 3Hz fundamental frequency.
So the signal has high harmonics which matters since it is squarewave-like step function signal.
This is a general question which is related to my previous question: Passive low-pass filtering question for a transducer output
Basically I have confusion when to use which model. Transmission line theory or lumped model...
So I want to model this in LTspice as a Lumped LC Model or something in that sort to see phenomena like ringing.
To achieve an approximate model I will write about each sections from left to the right of the circuit:
Questions when trying to use lumped model:
1-) Rload in the above illustration represents a scope or very high input impedance of a measurng instrument which is known.
Rout in the above figure is the output of a transducer which is unknown. How to measure that if it is not a function generator with a known output impedance?
I will use the following method to find the output resistance:
http://www.qsl.net/w/w2aew//youtube/How_to_measure_output_impedance.pdf
Do I really need the output impedance or can I take it roughly some hundreds of ohms?
2-) Coaxial-cable in the above illustration is a standard 50 Ohm intrinsic impedance BNC cable. I know its length and lets say it is 20 meters long.
Since we use lumped element model we will not use 50 Ohm right? And capacitance and the inductance will vary with length?
In other words, how can I model this cable in LTspice?
A question when trying to use impedance matching in transmission line theory:
Secondly let's forget about the LTspice and lumped model and just assume we only want to achieve impedance matching.
Here is the model considering impedances:
So now we have a transducer output impedance which is Rout=Zout, coaxial cable impedance which is 50 Ohm, and we have a load Rload=Zload which is lets say 100M.
So in this case to achieve impedance matching throughout the line I need a 50 Ohm resistor in parallel with Rload just before Rload to make the Rload 50 Ohm.
And I also need to know the Rout and I need to add a series or parallel resistor to it such that its equivalent or Thevenin would be 50 Ohm.
Is this method/model right? If it is than I will have the following issue:
The 50 Ohm parallel resistor will load the transducer and I will have more error right?
It seems to me for sending data this might not be problem but in this case the signal's voltage level is important.
What would you suggest in this case?
AI: Coaxial-cable in the above illustration is a standard 50 Ohm intrinsic impedance BNC cable. I know its length and lets say it is 20 meters long. Since we use lumped element model we will not use 50 Ohm right? And capacitance and the inductance will vary with length? In other words, how can I model this cable in LTspice?
First, when we talk about transmission lines, we talk about characteristic impedance. "Intrinsic impedance" is not a term that has any specific meaning in the area of transmission lines.
A lumped element model of a transmission line with 50 ohms characteristic impedance does not involve a 50 ohm resistive element in series. Characteristic impedance describes the ratio between voltage and current in the travelling wave that can propagate along the line. It doesn't cause any power loss like a series resistance would.
It might involve a series of capacitive and inductive elements in a pi or T section arrangement. A pi-section model of a lossless unbalanced line would look like this:
simulate this circuit – Schematic created using CircuitLab
C1 and C6 would have half the value of C3, C4, and C5, because the intermediate capacitors actually each represent the shunt legs of two pi sections in parallel. The total capacitance should add up to the line's capacitance per unit length times its length. The total inducance should add up to the line's inductance per unit length times its length.
Obviously this model will fail when the frequency gets too high, because the first and last capacitance elements will effectively short out signals approaching in the forward and reverse directions. By increasing the number of sections you can reduce the capacitance per section in the model, and so increase the frequency where this issue occurs.
So now we have a transducer output impedance which is Rout=Zout, coaxial cable impedance which is 50 Ohm, and we have a load Rload=Zload which is lets say 100M.
The load impedance is not particularly realistic. Typical scope inputs are 1 or 10 Megohms. Scopes designed for measuring reasonably high frequencies will usually have an option to program an input impedance of 50 ohms.
So in this case to achieve impedance matching throughout the line I need a 50 Ohm resistor in parallel with Rload just before Rload to make the Rload 50 Ohm.
Yes, if your scope doesn't have a 50 ohm input impedance option, and reflections become an issue, you can add a 50-ohm parallel resistance at the input to reduce these reflections. It will also reduce the signal seen by the scope.
And I also need to know the Rout and I need to add a series or parallel resistor to it such that its equivalent or Thevenin would be 50 Ohm.
You don't strictly need to match both ends of the transmission line. If you match one end very well it will eliminate reflections that reach that end, so you won't see ringing from multiple reflections.
The 50 Ohm parallel resistor will load the transducer and I will have more error right? It seems to me for sending data this might not be problem but in this case the signal's voltage level is important. What would you suggest in this case?
You could either provide a buffer amplifier at the transducer to produce a signal with low output impedance.
You could move the scope closer to the transducer so that the line can be shorter and impedance matching not needed.
You could provide an RC filter at the transducer output to reduce the edge transition speed so that less high frequency signal is present and impedance matching is not needed. |
H: Transistor Switch Maximum Permissible Resistance
I have been learning from a series called "The Great Courses". This is a question that modifies a given example with new numbers. No matter what I do, I cannot reach the correct answer, and the book gives no explanation.
The image is a little messy but all given info is there. Is there something I am doing wrong?
AI: As a Rule of thumb all Transistors are rated at Ic/Ib=10 even if the Vce(sat) is reasonably low at a ratio of 20 or special devices with a ratio of 50, but NEVER use hFE for a switch. Rule of thumb is use 10% of hFE or 10 which ever is higher unless pushing the Ic max limits.
Thus using stead state Tungsten Lamp resistance which drops to <15% R_cold when hot ( ignore that for now)
Neglect Vce(sat)= 0.7V which you used ok (close as in ballpark close...)
R_lamp=(24)/500mA = 48 Ω
Neglect Vbe drop for now = 0.7 (ballpark)
choose R = 10 * R_load = 480 Ω
Checking
Ib= (24-0.7)/480 Ω = 55 mA
so Ic/Ib= 500mA/55= 9.1
So add 10% ~ 20% and use 520 Ω ~ 570 Ω for cool operating transistor
500mA *0.7V = 350mW @ 0.250'C/mW for a TO-92 will be 88'C rise above 25'C room which is too hot, so choose a transistor with Vce(sat) =0.3V @ 500mA =150mW
- RULE OF THUMB using a Common Emitter SWITCH
- (linear load biased from same supply as load.
Make Rb/Rc=10 |
H: Confusion about the rated voltage of a Permanent magnet synchronous motor
Confusion about the rated voltage of a Permanent magnet synchronous motor.
I’m currently trying to do a motor control experiment and I’m really confused by a term called “rated voltage” from the motor specifications.
Below is my Simulation scheme for the speed control of a PMSM (with sinusoidal back-emf).
Under experiment circumstance, it would be something like this.
The outputs of the inverter are Vao, Vbo, and Vco. The line-to-neutral voltages would be
VaN=(1/3)(2*Vao-Vbo-Vco)
VbN=(1/3)(2*Vbo-Vao-Vco)
VcN=(1/3)(2*Vco-Vao-Vbo)
And the line-to-line voltage would be
Vab=Vao-Vbo
Vbc=Vbo-Vco
Vca=Vco-Vao
My question is, when we mean rated voltage of a PMSM. Does this value refer to the RMS value of line-to-line voltage, the RMS value of the line-to-neutral voltage, or simply the DC bus voltage of the inverter?
I want to clarify this confusion because I know that to drive a motor, the DC bus voltage for an inverter should be 1.414 times larger than the AC RMS line voltage of the Motor.
Assume that the rated voltage of a PMSM is 24V. If this rated voltage means the RMS value of the line-to-line voltage, then the required DC bus voltage (Vdc) should simply be 1.414*24=34V. If this rated voltage mean the RMS value of the line-to-neutral voltage, for a three-phase PMSM, the line-to-line voltage will be 1.732*24=41.5V. And the required DC bus voltage for the inverter will be 59V. If this rated voltage refer to the DC bus voltage of the inverter, then the DC bus voltage for the inverter will simply be 24V. And the line-to-line voltage of this PMSM under rated condition will be 24/1.414 = 17V.
Can anyone clarify this for me?
Thanks.
@ Bruce Abbott
Based on the datasheet I provided in the comment. I picked the part number DN42040S24-026 and used the following motor specs:
Stator phase resistance: 0.75ohm;
Lds & Lqs: 0.0021H;
Flux linkage established by PM: 0.00477564V.s;
Inertia: 2.4e-06;
Pole pairs:4.
I have no idea about calculating the viscous damping based on the datasheet, so I assumed the viscous damping to be 0.0001N.m.s. I tried to simulate the motor’s performance at rated condition in Simulink. I initially supplied the motor with 800/3Hz three phase sinusoidal voltage inputs as (let’s call this group 1)
Vao=24*1.414*cos[2*pi*(800/3)*t];
Vbo=24*1.414*cos[2*pi*(800/3)*t-2/3*pi];
Vco=24*1.414*cos[2*pi*(800/3)*t-4/3*pi].
The motor speed is around -5000rpm and have large amplitude oscillations. But when I change the supply voltage to (let’s call this group 2)
Vao=24*1.414*cos[2*pi*(800/3)*t];
Vbo=24*1.414*cos[2*pi*(800/3)*t+2/3*pi];
Vco=24*1.414*cos[2*pi*(800/3)*t+4/3*pi].
The motor speed reaches a steady-state of -4000rpm. I’m confused here because I think I should have the same speed response with either group. Also, I don’t understand why I get a negative speed response at -4000rpm rather than +4000rpm. I tried to switch the phase sequences of group 2, but I get the same speed response as group 1: -5000rpm with large oscillations. I do not know what the problem is here. Any suggestions? Thank you.
AI: It depends on whether the motor is designed to be operated from an AC or DC source. If the motor is rated for 24V DC then rms doesn't apply, because rms and peak voltage are the same for DC. If it is supposed to work off AC power (either directly or through an inverter) then that AC power would normally be measured in rms.
A PMSM designed for DC operation a usually called a brushless DC motor (BLDC) while one designed for AC operation is normally described as a brushless AC motor. But these terms are not exclusive. BLDC could refer to any motor that doesn't have brushes and runs from a DC source (either directly or through a controller) - including variable reluctance, stepper, and voice coil motors - while brushless AC includes induction and hysteresis motors that don't have permanent magnets.
Most low voltage 'BLDC' motors are intended to be driven with '6 step' commutation, which produces a trapezoid drive waveform. These motors often have non-sinusoidal back-emf because they are designed to work best with trapezoid drive. Low voltage brushless motors described as 'PMSM' are usually intended to be driven with sine waves.
However both types of motor can be driven with either waveform, the only difference being that on trapezoid drive a PMSM will produce higher than rated speed with greater torque ripple, while on sine wave drive a 'trapezoid' BLDC motor will have lower than expected performance. So it is possible to describe the same motor as both BLDC and PMSM, and even rate it differently depending whether it is driven with 'AC' (sine waves) or 'DC' (6 step commutation). |
H: bi-directional half-duplex communication between two AT89C2051's micros without serial port
I'm trying to do a direct two-way communication between two AT89C2051 microcontrollers with only two GPIO pins so that I have other GPIO lines free for other tasks.
P3.7 is the data wire, and P3.2 (INT0) is the control wire, and they are externally pulled high to VCC (of 5 volts) through 10K resistors.
One microcontroller has an 18.432Mhz crystal connected to it so I can connect it to my computer at a clean 9600bps baud. The other microcontroller is connected to a 20Mhz crystal (which is the highest speed lowest-priced compact crystal I ordered from the internet) for fast memory and wireless communication.
Initially I was going to reserve 4 GPIO pins for data communication but I wouldn't have any free for other tasks such as memory and LED manipulation.
This is my idea behind the two wires.
The sender loads P3.7 with the one bit and then sends P3.2 low then waits a bit for remote processing. This repeats until the whole byte is sent.
When the receiver receives data, the interrupt is supposed to be executed (from the remote sender making P3.2 low) and the byte received from P3.7 is stored in a temporary location until all bytes are received which a counter tracks. Once all are received, a flag is set.
The sending seems to be ok, but the receiving always stalls.
Do I need to adjust my timings or is there a better approach of doing half-duplex bi-directional communication with only two wires connected to the same pins (with pull-up resistors) on both micros?
Here is my code for reference. I have to remove the first line of code after programming the first chip to ensure the correct code is on each chip.
UNIT equ 1h ;1 = 8051 to interface to PC, 0 = 8051 to interface to radio.
RDATS EQU 40h ;remote data temp
RDAT EQU 42h
RDATCT EQU 41h
RDI EQU 0h ;remote-data-interrupt bit
CT EQU 1h ;carry temp - receive
CTS EQU 2h ;carry temp - send
org 0h
ljmp main
;INT 0 natural address is here (03h)
ljmp intZ
org 002Bh
;called
intZ:
clr EX0
ifdef UNIT
clr P3.4 ;This is supposed to execute on 8051 connected to PC but does not.
endif
mov CT,C
push ACC
mov C,P3.7
mov A,RDATS
RLC A
mov RDATS,A
inc RDATCT
mov A,RDATCT
cjne A,#8h,no8
setb RDI
mov RDAT,RDATS
mov RDATCT,#0h
no8:
jnb P3.2,$
pop ACC
mov C,CT
ifdef UNIT
setb P3.4
endif
setb EX0
reti
sendrdat:
clr EX0
mov CTS,C
mov R7,#8h
push ACC
sendrmore:
clr C
rlc A
mov P3.7,C
clr P3.2
nop
nop
setb P3.2
djnz R7,sendrmore
pop ACC
mov C,CTS
setb EX0
ret
serialin:
clr RI
jnb RI,$
mov A,SBUF
ret
serialout:
clr TI
mov SBUF,A
jnb TI,$
ret
pserial:
rewr:
clr A
movc A,@A+DPTR
jz exl
acall serialout
inc DPTR
ajmp rewr
exl:
ret
main:
mov RDAT,#0h
mov RDATS,#0h
mov RDATCT,#0h
clr RDI
mov P3,#0FFh
mov P1,#0FFh
mov SP,#50h ;skip most ram so SP doesnt be wrecked
mov scon,#50h
mov tmod,#020h
mov 087h,#80h ;pcon.7
mov TH1,#0F6h ;9600 baud on 18.432Mhz
setb EX0
setb EA
setb TR1
IFDEF UNIT
ajmp PCmode
ELSE
ajmp RADmode
ENDIF
;branch
PCmode:
clr RI
jnb RI,$
mov DPTR,#ptest1
acall pserial
;send 0AAh to remote 8051
mov A,#0AAh
acall sendrdat
mov DPTR,#ptest2
acall pserial
; code stalls here forever.
; It should only stall until remote 8051 calls this one.
rwait:
nop ;no overloading processor
jnb RDI,rwait
;RDAT = remote byte received
mov A,RDAT
push ACC
mov DPTR,#ptestr
acall pserial
pop ACC
acall serialout
mov A,#0Dh
acall serialout
mov A,#0Ah
acall serialout
mov DPTR,#ptestdone
acall pserial
sjmp $
RADmode:
;radio mode... only obey commands at this time
nop
jnb RDI,RADmode
clr RDI
mov A,RDAT
cjne A,#0AAh,nosig ;AAh=get signature = 41h (A)
mov A,#041h
acall sendrdat
nosig:
ajmp RADmode
sjmp $
ptest1:
db 'Sending data to unit 2...',0Dh,0Ah,00h
ptest2:
db 'Receiving data from unit 2...',0Dh,0Ah,00h
ptestr:
db 'Received: ',00h
ptestdone:
db 'This is a test that worked!',0Dh,0Ah,00h
END
AI: I want to enable as much multitasking as possible
That just means that you will be using interrupts in some fashion.
The alternative, salting your code with short snippets of identical polling code placed strategically to simulate periodic events is as ugly as it gets. So you don't ever do it. I've seen it done before. And it's terrible and will make your code for all practical purposes unmaintainable.
until all bytes are received which a counter tracks
half-duplex bi-directional communication
These suggest to me that you actually may want a master/slave relationship, but where either can become the master for the period of the communication. It can be done by having only one master, ever. But the way you write suggests you would either like a protocol without a specific master (such as the asynchronous protocol used in RS-232, for example) or else a means by which either can become a master for the purposes of a burst of communication.
18.432Mhz crystal
20Mhz crystal
With two different processors, even if they are running at the so-called same clock rate, will start out with some phase difference which also drifts around relative to each other. Nothing is perfect in life. But with two different clock rates, one thing I also look at is the GCD. In this case, that's 32000. This means one option might be to use a divisor of 576 on your \$18.432\:\textrm{Mhz}\$ micro and a divisor of 625 on your \$20.000\:\textrm{Mhz}\$. That would get the bit timing, if you used a protocol where that mattered, into the similar ballparks. You'd still have phase error and drifting. But if you chose to use an asynchronous RS-232 like protocol, you could probably manage that after a fashion at a bit rate of \$32\:\textrm{kHz}\$.
But I also think Dave is right, too. Using something like \$\textrm{I}^2\textrm{C}\$ has its advantages over dinking around with too much over-sampling (you may need 2X over-sampling or more, regardless, since this is software; but this varies a bit with the hardware interrupt modes you can use, too) and getting your bit timings zeroed in.
The tradeoffs are now something like this:
Use multi-master \$\textrm{I}^2\textrm{C}\$. This takes a little extra effort in the code for master and slave to achieve the multi-master parts. But there is a standard for this too and plenty of example code to examine, I'm sure. The limitation here is that your communications are half-duplex using two lines. But you've already accepted this. And arbitration issues might have you scrambling for those GCD divisors I mentioned above.
Use a copy of the asynchronous format (start bit, N data bits, and let's say two stop bits) used by RS-232 and commonly found in UART hardware. The advantage here is that you get full-duplex communications. But you've indicated you don't need that. You'd probably want to use those GCD divisors here, too, to get the bit timing close enough to each other.
Use \$\textrm{I}^2\textrm{C}\$ in simple master-slave mode. Slaves can not push any data to the master. Instead, the master would have to ask the slave if there was something to send back. But you could also go this route. It would spare you arbitration issues, too.
In all cases you will have interrupts involved and either, (A) use state machines; or, (B) use an operating system with threads (cooperative ones, if not pre-emptive.)
It doesn't take much, not even on the 8051, to pony up a simple cooperative set of threads. (Yes, I've done it on a SiLabs C8051F061 and using SDCC and not Keil.) It really cleans up code and makes it very maintainable, because it avoids the chain of state transitions.
But a clearly documented set of state transitions isn't bad, either. Most with any experience using state machines can follow the clear documentation and see which code performs which step. So I think either approach would be fine.
they are externally pulled high to VCC (of 5 volts) through 10K resistors.
So this is good as \$\textrm{I}^2\textrm{C}\$ wants that and asynchronous protocols can work with that.
Why don't you do some googling on \$\textrm{I}^2\textrm{C}\$ state machines, read through them until you feel you understand them, and also read through the documentatation available from this \$\textrm{I}^2\textrm{C}\$ documentation that is available to you. Similarly, consider an asynchronous protocol, as well.
One thing in your favor is that you do have the ability to get similar bit times on each device. This doesn't fix phase differences or drift. But a little bit of over-sampling can address that. Or you can just leave them free-run and work through the issues of that process.
Chances are, there are many different examples of code readily available on the web to cover you here, too. I'd recommend doing some searching there, as well. You are the one who needs this. So you should be doing the up-front work here. Not asking us to do that for you.
That said, if you get yourself educated about these methods and can narrow down your questions to something specific, I'm sure you'll get some wise advice. But right now, you are just throwing code against a wall to see what sticks. And I think we'd like you to do your own work getting up to speed on existing methods which are tried and true, before you start asking anyone to come up with something new for you or something based only on your own (apparently quite limited) thinking about this. If you are experienced, you can indeed draft up your own protocol and make it work -- first time, too, out of the box. But you aren't experienced. So you should first work from the benefit of others. |
H: Adding Slab to a Capacitor
I was solving a question in an Electromagnetics book and need some help: two parallel plates were connected to a 100V battery. The battery was removed and then a slab of dielectric material was placed between the plates touching the upper one but leaving 0.25 mm of free space in the bottom. It is required to find electric field intensity and some other quantities after inserting the slab.
The idea of the question is know what remains constant before and after inserting the slab. It makes sense to assume that the charge would remain the same since it has no way to go after the battery is removed. However, after I solved the question with this assumption, I found that the total energy stored in the capacitor after inserting the slab is less than the total energy before inserting it. I thought the energy would also be conserved but apparently it is not. Can anybody give a physical explanation why the total energy of the system is less? Where did this energy go?
More details:
Plates area: 2x10^2 m2
Distance between the plates: 1.25 mm
Thickness of dielectric slab: 1 mm
Relative permittivity of the dielectric slab: 5
Solution: before inserting the slab:
Total capacitance = 1.42e-10 F
Voltage = 100 v
Energy (W) = .5cv^2 = 7.08e-7 Joules
surface charge density = electric flux density = 1.42e-8 c/m2
After inserting the slab:
in free space region: C=7.08e-10 F, E = 80,000 v/m, v=20V, W = 1.42e-7 Joule
in dielectric slab: C=8.85e-10 F, E=16,000 V/m, V=16 V, W=1.13e-7 Joule
Total W = 2.55e-7 Joules compared to 7.08e-7 before inserting the slab. Where did the 4.53e-7 Joule go?
AI: As the slab is brought to the plates, it will experience an attractive force that will tend to pull it into the gap. If you do not restrict it, the slab will accelerate.
In an ideal environment, the energy loss in the capacitor will be balanced by the increase in kinetic energy of the slab. Over time, this energy will be lost to friction or drag.
Once the slab is at rest inside the plates, work must be done to remove it against the electric field. This work will raise the energy stored in the capacitor. |
H: Why can't I find Phase to phase voltage like this
I know that that in 3phase ac,
Vll = sqrt(3) * Vln
I got confused about the sqrt(3) part so I draw phaser diagram and the resultant vector will give me the phase -phase voltage
But it only gives me 230V , why?
Can anyone explain how it got sqrt(3) in equation??
AI: What you are calculating (the black line) isn't actually \$V_{AB}\$, it is rather the vector sum of \$V_{AN}\$ and \$V_{BN}\$ which is actually just \$-V_{CN}\$.
\$V_{AB}\$ is the voltage of \$V_A\$ referenced to \$V_B\$. To calculate that, you have to subtract one from the other. So the vector sum is actually:
$$V_{AB} = V_{AN} - V_{BN}$$
Doing the calculation, we get:
$$\begin{align}\\
V_{AB} = V_{AN} - V_{BN} &= 230\angle0^\circ - 230\angle120^\circ\\
&=398\angle30^\circ\\
&=(230\sqrt{3})\angle30^\circ\\
\end{align}$$
Now lets look at it in a vector diagram:
Notice how the calculation you did (left) and the correct form (right) differ. You can see from the diagram that the line drawn for \$V_{AB}\$ is actually equal to the vector that takes us from the point \$V_B\$ to \$V_N\$ (-\$V_{BN}\$), and then from \$V_N\$ to \$V_A\$ (\$V_{AN}\$).
If you want to calculate that it is exactly \$\sqrt{3}\$, we can do a bit of trigonometry on the newly formed triangle:
From that we can see that:
$$\frac{V_{AB}}{2} = 230\times\sin{60}$$
We know that \$\sin(60) = \frac{\sqrt{3}}{2}\$, so we can say directly that:
$$V_{AB} = 2\times230\times\frac{\sqrt{3}}{2} = 230\sqrt{3} = V_{AN}\sqrt{3}$$ |
H: Hardware sources of entropy on an FPGA
I'm building an FPGA feed handler and one of the problems is I want to do Monte-Carlo simulations which require a high-quality entropy source, i.e. a pseudo-random LSR implementation won't do. I've read that one can try using metastability as a source of randomness, does anyone do this, is this a viable option?
Also, since FPGAs have external clock inputs, can those be leveraged by some hardware "true RNG" solution? Do such solutions exist on the market?
AI: Metastability is really not a viable option in a modern FPGA technology because the metastable timing window is tiny - many orders of magnitude smaller than the setup/hold timing uncertainty window, which is dominated by things like clock skew, routing delays and variations with voltage and temperature.
Unfortunately there's a lot of confusion about metastability and this timing window is sometimes called the metastable timing window, as other (sometimes unexpected) sources of uncertainty in an output due to clock timings are loosely lumped together and incorrectly called metastability.
While this wider window generates uncertainty in the output, it's highly correlated with the above causes, not entropic (outside the tiny true metastable window).
If you need the mathematical and practical details, search the Usenet comp.arch.fpga newsgroup for "metastability" and "Peter Alfke".
TL/DR : look elsewhere for entropy : an avalanche diode, (i.e. a zener well above 6V, say 12V) amplified, sliced to logic levels, fed to an input pin would be one good choice. |
H: Would this be a legal VHDL architecture?
I want to design a basic BCD to XS3 converter using VHDL. I am fairly new to
conditional and behavioral architecture implementation, but I was wondering if this would be a legal program, especially my assign statements
entity bcd2xs3 is
port (bcd : in std_logic_vector(3 downto 0);
xs3 : out std_logic_vector(3 downto 0));
end bcd2xs3;
architecture behav of bcd2xs3 is
begin
case bcd is
when X"0"|X"1"|X"2"|X"3"|X"4"|X"5"|X"6"|X"7"|X"8"|X"9" => xs3 <= std_logic_vector(unsigned(integer(unsigned(bcd))+3));
when others => xs3 <= X"0";
end case;
end behav;
AI: I prefer this solution
library ieee;
use ieee.std_logic_1164.all;
use ieee.numeric_std.all;
entity bcd2xs3 is
port (bcd : in std_logic_vector(3 downto 0);
xs3 : out std_logic_vector(3 downto 0));
end bcd2xs3;
architecture behav of bcd2xs3 is
signal bcd_int: integer;
begin
bcd_int <= to_integer(unsigned(bcd));
process (bcd_int )
begin
case bcd_int is
when 0 to 9 =>
xs3 <= std_logic_vector(to_unsigned(bcd_int + 3), bcd'length);
when others =>
xs3 <= X"0";
end case;
end process;
end architecture behav; |
H: How to add part in eagle board that doesn't have corresponded in the schematic "jumpers"?
I am using Eagle to design single side board and I want to use smd jumpers to connect between some points.
So basicly I don't want to add those smd jumpers as a component to the circuit in the schematic, I just want to be able to add them whenever I need without jumping every time to the schematic window and add them.
Just add them from the pcb window directly.
I have tried to search this but couldn't find something useful.
AI: To figure it out you need to understand how EAGLE works. Each device you find in library consists of three components - package, or board footprint with pads (through hole or SMD), symbol with pins, and correspondence between package and pins.
You can add package to board without adding symbol to schematics, but you will not be able to connect to its pads properly, and DRC will give you "overlapping" errors.
The right way is to add jumper to schematics, connect its pads respectively, and wire pads at the board level with tracks. All other ways is hacking, which may lead to
your mistakes, and tool will not be able to advise you what is wrong;
errors in DRC, and you may not be able to find the issue quickly.
EAGLE is a powerful tool, but please refrain misusing it. You will spend much more time figuring out why EAGLE gives you errors, or why your board does not work properly. Just follow rules and design board the way EAGLE supports.
Thus the answer is
You must add jumper to schematic, and wire it as needed at both schematic and board levels. |
H: connect headphones to TV RCA output?
My TV has two RCA female connectors for audio output. I assume it is the line-out. No female jack connector specfic for headphones.
I want to connect usual headphones that has a male jack connector. I've buy an adapter "two male RCA => one stereo jack" (Y-cable) and everything seems ok, just volume a few low.
However, after read wikipedia:
https://en.wikipedia.org/wiki/Line_out_(signal)
it seems this is really a bad idea, headphones has not the valid input impedance, and that can cause low sound, damage my TV, and an economical world crisis.
Any hint?
Addendum:
According to samsung (see here), the solution is the Y-cable. It seems they are not worried about impedances (and, in photo, it seems they are not worried about what is RCA or jack ?).
AI: That Wiki page sums it up pretty clearly. Except for the world crisis part, you mentioned.
Why not use a commercially offered amplifier designed for the purpose? These days, there are fewer available amplifiers (in the US, anyway) which are designed only for this purpose (you usually have to buy more than just an amplifier), but they exist. They are rare these days and usually only supplied to "tube" people, now, it seems. I kind of hate the fact that the market for a simple stereo amplifier has dwindled down to such a small market. But it has. That said, for example, something like this will do the job quite well and it does exactly and only the job you want done.
I don't know your limitations, though.
You could also consider the idea of making something. If so, would you prefer a kit amplifier, an IC solution, or discrete schematic? If you are posting here, perhaps you are comfortable approaching one of those?
Regardless, you do some some kind of active amplification. It doesn't have to be complicated. And you'd have to decide if you want the two RCA sides combined into a mono-output or if you wanted stereo retained. But it will need a source of power, whether battery or otherwise.
Interesting catch from Samsung. I still wouldn't go that way. It doesn't seem correct to me. You already mentioned that the audio seemed muted to you. I think that tells you what you need to know, here..
You might consider something like a TDA2822, for example. It's a stereo IC and will operate off of low voltages. But it lacks an intrinsic volume control and may be a little over-powered for your needs. The TDA7053A is also a stereo amplifier, with a built in volume control, but it is bridge-tied and headphones usually have a common ground. So when reading these, you need to be aware that you probably want single-ended operation. Search around for something still closer than those two I just found. They are usually very easy to use. (EDIT: Just saw the TPA6021A4 show up on a search, which at first glance looks closer. I'm not going to read the datasheet fully for now, though. But you might check it out. They are available at the usual suspects, Digikey, Arrow, and so on, for about $2 each. It operates both as single-ended and bridge-tied. Also, TI has an evaluation module that retails for about $50 called the TPA6021A4EVM. Here is the user's guide for it. Look at Figure 3 for a headphone connection that may work for you.)
The value for \$V_{pk}=\sqrt{2 P R}\$, plus a little for whatever overhead is required. That's \$V_{pk}\$ and not \$V_{pp}\$ so if using a single DC supply keep in mind that you'd need twice that value and more for a DC supply rail. But I think the power you require for headphones is very modest, so it's likely that whatever voltage supply you use, so long as you have a way to control the volume, will work. |
H: Boolean Algebra with Demorgans Law
I have to prove that:
!(a(b + z(x + !a))) = !a + !b(!z + !x)
I am assuming that I have to use DeMorgans Rule so this is as far as I'm getting with:
!a + (!b!z + (!xa))
I'm not sure where to go from here. More specifically I'm not sure how to get rid of the a in (!xa).
AI: !(a(b + z(x + !a)))
= !(ab + azx + az!a) [distribution]
= !(ab + azx) [a./a = 0]
= !(a.(b+zx)) [factoring]
= !a + !(b+zx) [DeMorgan]
= !a + !b.!(zx) [DeMorgan]
= !a + !b.(!z + !x) [DeMorgan]
There are not many rules to learn in Boolean logic.
Get used to them : There is DeMorgan, factoring, things like a+/a.b = a+b, etc... |
H: What kind of transistor do I use for switching high voltage with a microcontroller?
I am working on a project with Nixie tubes which require a high voltage supply. I have some "indicator dot" tubes that require 90V DC, and I want to turn them on/off using a microcontroller control line.
I am thinking a transistor is the right tool for this, but I am not familiar enough with the different types to know what will work (BJT vs MOSFET, NPN vs PNP, high-side vs low-side switching).
simulate this circuit – Schematic created using CircuitLab
Please forgive this schematic. It is probably wrong; it is merely indicative of what I want to accomplish.. that is the question I am asking. What is the right way to do this, and what type of transistor should I use? Specific part numbers would also be appreciated.
To be clear, I don't need to amplify any voltages, I already have the voltage supply I need for the 90V line.
AI: NPN is correct transistor type BUT load MUST be in collector.
That way low voltage at Vbe drives the base, and the collector switches HV.
The transistor must be rated for usefully more than 90 V.
See www.digikey.com selection guide.
MPSA42 was the olde time favourite at about 300V rating.
With load in the collector the circuit inverts - ie Vin high = on.
Whether this suits depends on your need.
This can be altered to Vin high = off with additional components.
simulate this circuit – Schematic created using CircuitLab |
H: How do you apply the negative feedback laws when there are two feedback networks?
I have the following circuit, which has a series-shunt feedback in the form of r3 and r7, and a shunt-series feedback in the form and r4 and r6.
How do I apply the usual steps in analysing negative feedback amplifiers, then? For example, to find the closed loop gain, I very much want to use Aol/(bAol + 1), but I don't understand what b I should use as there is a current-sampling, current-feedback and a voltage-sampling, voltage-feedback circuit.
What about the other negative feedback theorems that relate to bandwidth and input/output impedance? How do I find them?
AI: Let me start things out by looking only at the DC operating point. You'll see some problems that may be present, then. This circuit's behavior has some overly strong dependencies on \$\beta\$.
simulate this circuit – Schematic created using CircuitLab
I get the following:
$$\begin{align*}
I_{B_1} &= \frac{V_{E_2}-V_{BE_1}}{R_6 + \left(\beta_1+1\right)\cdot R_3} \\
\\
I_{C_1} &= \beta\cdot I_{B_1}\\
\\
V_{C_1} &= V_{B_2} = V_{CC} - R_2\cdot I_{C_1} \\
\\
V_{E_2} &= V_{B_2} - V_{BE_2}
\end{align*}$$
At this point, it's possible to solve for \$V_{E_2}\$:
$$\begin{align*}
V_{E_2} &= V_{B_2} - V_{BE_2} \\
\\
&= V_{CC} - R_2\cdot I_{C_1} - V_{BE_2} \\
\\
&= V_{CC} - R_2\cdot \beta_1\cdot I_{B_1} - V_{BE_2} \\
\\
&= V_{CC} - R_2\cdot \beta_1\cdot \frac{V_{E_2}-V_{BE_1}}{R_6 + \left(\beta_1+1\right)\cdot R_3} - V_{BE_2} \\
\therefore \\
V_{E_2} &= \frac{\left(V_{CC}-V_{BE_2}\right)\cdot\left(R_6 + \left(\beta_1+1\right)R_3\right)+\beta_1 R_2 V_{BE_1}}{R_6 + \beta_1 R_2 + \left(\beta_1+1\right)R_3}
\end{align*}$$
Assuming \$\beta_1=100\$, \$\beta_2=100\$, \$V_{BE_1}=680\:\textrm{mV}\$, and \$V_{BE_2}=700\:\textrm{mV}\$, then \$V_{E_2}\approx 2.236\:\textrm{V}\$. From there, we find that \$I_{C_1}\approx 1.5\:\textrm{mA}\$, \$I_{C_2}\approx 4.7\:\textrm{mA}\$, \$V_{C_1}\approx 3\:\textrm{V}\$, \$V_{CE_1}\approx 2.1\:\textrm{V}\$, \$V_{C_2}\approx 3\:\textrm{V}\$, and \$V_{CE_2}\approx 700\:\textrm{mV}\$. The first stage has a little bit of working room; perhaps \$V_{pk}=1-1.5\:\textrm{V}\$ with a positive going input swing at \$V_{IN}\$. But the second stage is already crossing over into saturation and has no room to work, at all, for a negative going input swing at \$V_{IN}\$. I consider this to be an essentially useless state of affairs.
Assuming \$\beta_1=200\$, \$\beta_2=200\$, \$V_{BE_1}=680\:\textrm{mV}\$, and \$V_{BE_2}=700\:\textrm{mV}\$, then \$V_{E_2}\approx 1.931\:\textrm{V}\$. From there, we find that \$I_{C_1}\approx 1.57\:\textrm{mA}\$, \$I_{C_2}\approx 4.1\:\textrm{mA}\$, \$V_{C_1}\approx 2.7\:\textrm{V}\$, \$V_{CE_1}\approx 1.8\:\textrm{V}\$, \$V_{C_2}\approx 3.9\:\textrm{V}\$, and \$V_{CE_2}\approx 2\:\textrm{V}\$. The first stage still has a little bit of working room, though less; perhaps \$V_{pk}=1\:\textrm{V}\$ with a positive going input swing at \$V_{IN}\$. And luckily now, the second stage has a little room left for a negative going input swing at \$V_{IN}\$, too. But again on the order of \$V_{pk}=1\:\textrm{V}\$, but as the gain is about 3 this means that the first stage can't present more than about \$300\:\textrm{mV}\$ in that direction. So this means the first stage is thus actually limited to an output swing of \$\pm 300\textrm{mV}\$ or an input swing of about \$\pm 35\textrm{mV}\$ (before we take into account the negative feedback from \$R_6\$.) This isn't entirely useless.
The above exposes some questions about the DC operating point that was chosen and its sensitivity to \$\beta\$. Depending on the actual value, the circuit goes from limited utility to useless.
Now, it's clear that \$R_6\$ is providing DC bias for stage 1. But it is also providing negative feedback, since \$Q_2\$'s emitter follows the signal at \$Q_1\$'s collector which itself is an inverted copy, with gain, of the input signal. (Ignoring limitations noted above about the DC operating point.)
You may want to compute this local negative feedback's \$B_1\$ value (a value from 0 to 1, we expect to apply in your open-loop to closed-loop equation) before taking on what \$R_7\$ then does. Roughly speaking, you want to compute this as:
$$B_1=\frac{\textrm{d}V_{B_1}}{\textrm{d}V_{E_2}}$$
But to get that, you need to take into account what is happening at the input node. You actually have three sources and impedances all vying for it. The input signal through its impedance, the emitter resistance via \$\beta_1\$ to ground, and \$V_{E_2}\$ through \$R_6\$. Let's add that to the schematic now:
simulate this circuit
The equation to compute the voltage there, just cobbled up as a DC equation for now and ignoring the impedance of the \$10\:\mu\textrm{F}\$ input capacitor, would be:
$$\begin{align*}
V_{B_1}&=\frac{V_s \beta_1 R_3 R_6+V_{E_2}\beta_1 R_3\left(R_s+R_1\right)}{\beta_1 R_3 R_6 + \beta R_3 \left(R_s+R_1\right) + R_6 \left(R_s+R_1\right)} \\
\\
&=\frac{V_s R_6+V_{E_2}\left(R_s+R_1\right)}{R_6 + R_s+R_1 + \cfrac{R_6 \left(R_s+R_1\right)}{\beta_1 R_3}}
\end{align*}$$
Holding \$V_s\$ constant, we can now compute \$B_1\$ as:
$$B_1=\frac{R_s+R_1}{R_6 + R_s+R_1 + \cfrac{R_6 \left(R_s+R_1\right)}{\beta_1 R_3}}$$
For \$\beta_1=100\$, this works out to \$B_1\approx .188\$. For \$\beta_1=200\$, this works out to \$B_1\approx .204\$.
Given your equation that takes the open-loop gain for the first stage as \$A\approx \frac{4.7\:\textrm{k}\Omega}{4.7\:\textrm{k}\Omega} \approx 8.3 \$, then it follows that for \$\beta_1=100\$ that \$A_{closed}\approx 3.2\$ and for \$\beta_1=200\$ that \$A_{closed}\approx 3\$. So the closed loop gain on the first stage is about 3, taking into account \$R_6\$ (which serves both as a biasing resistor and as a negative feedback.)
However, that figure still doesn't take into account the fact that the input source is, itself, loaded down by the first stage's input impedance. That input impedance works out to about:
$$\begin{align*}\frac{\left(\beta_1+1\right) R_3 R_6}{R_6 + \left(\beta_1+1\right) R_3} \approx 33\:\textrm{k}\Omega
\end{align*}$$
Taking that into account with the input impedance, you'd get about:
$$\begin{align*}\frac{33\:\textrm{k}\Omega}{33\:\textrm{k}\Omega + 13.5\:\textrm{k}\Omega} \approx .7
\end{align*}$$
So from input signal to output of stage 1, including the input loading and negative feedback from \$R_6\$, I figure a total gain at the output of stage 1 of about 2.1.
I'm leaving the last step for you, where you include the impact of \$R_7\$. |
H: Bridge rectifier peak voltage at load
Assuming the peak voltage at \$V_1\$, \$V_{1p}=9.25V\$ and ideal silicon diodes (\$0.7V\$ drop in forward bias) and the DC output in \$V_0\$ is \$5V\$. I want to find the \$V_{op}\$, peak voltage at \$V_0\$. Given that the ripple voltage, \$V_r=0.2V\$ Then:
\$V_{DC}=V_{op}-\frac{V_r}{2}\$
And we have \$V_{op}=5.1V\$
I can understand why I couldn't apply KVL and write \$V_{op}=V_{1p}-1.4\$. Shouldn't this be the \$V_{op}\$?
If I apply KVL with \$V_{op}=5.1V\$ the equation doesn't hold true. What's going on? Excuse me because I this question may look similiar to another one I posted a while ago. This exercise is confusing me.
AI: I can't tell if this is homework, or not. But if you actually are trying to figure out how to design a bridge rectifier system that yields a desired output and ripple specification, and you have complete freedom to build a transformer for it rather than being forced to buy one from a list of readily available units, then the process is something like this:
Specifications:
\$V_{OUT_{REGULATED}}=5\:\textrm{V}\$ (the final regulated voltage)
\$V_{DROP_{REGULATOR}}=1.5\:\textrm{V}\$ (the minimum overhead used by the linear regulator)
\$I_{MAX}=1\:\textrm{A}\$ (the maximum load current to be supported)
\$V_{RIPPLE}= 500\:\textrm{mV}\$ (the ripple voltage at the input of the linear regulator)
\$f=60\:\textrm{Hz}\$ (the operating frequency of the mains supply to the transformer)
From that you can work out the rest. For example, let's use the above values to get:
$$\begin{align*}
t_{charge} = \frac{\textrm{sin}^{-1}\left(\frac{V_{OUT_{REGULATED}}+V_{DROP_{REGULATOR}}}{V_{OUT_{REGULATED}}+V_{DROP_{REGULATOR}}+V_{RIPPLE}}\right)}{4 \pi f}&\approx 1.6\:\textrm{ms} \\
\\
C = \frac{I_{MAX}}{V_{RIPPLE}}\left(\frac{1}{2 f}-t_{charge}\right)&= 13.5\:\textrm{mF}\\
\\
I_{DIODE_{MAX}} = \frac{\sqrt{2}\cdot V_{RIPPLE}\cdot C}{t_{charge}}&\approx 6\:\textrm{A}
\end{align*}$$
At this point, even though the capacitor isn't a standard value, you can make an estimate for the voltage drop of your rectifier diode. I generally take this to be about \$800\:\textrm{mV}\$ when supplying \$1\:\textrm{A}\$, so in this case I'd guess about:
$$V_{DIODE}=800\:\textrm{mV}+120\:\textrm{mV}\cdot \textrm{ln}\frac{6\:\textrm{A}}{1\:\textrm{A}}\approx 1\:\textrm{V}$$
Now, you can estimate the transformer's RMS value as:
$$V_{RMS} = \frac{V_{OUT_{REGULATED}}+V_{DROP_{REGULATOR}}+V_{RIPPLE}+2 V_{DIODE}}{\sqrt{2}}\approx 6.3\:\textrm{V}$$
Luckily, that happens to be a standard transformer value.
But note that this is under load. The capacitor chosen matters here.
I'm going to stop at this point and let you clarify your question some more. Perhaps the above process isn't what you wanted, at all. But it gives you an idea of some of the factors involved. Perhaps someone else understands you better and can provide a better response. But for now, I'm still not sure what you are doing. |
H: Best way to hand-solder this TQFP (SMD) component
So I'm working on a small project right now that requires a ATmega32U4RC. I'm trying to test/prototype with it, but it's a surface-mount component, so I can't just place it in my breadboard. So I found this, which is perfect for what I want.
The problem is that I haven't soldered in quite some time. I've also never soldered a component with such tiny leads.
Needless to say, I ended up with this:
Yea, I know it's bad :(
The solder-wick didn't really seem to want to do it's job (anyone know why this is?), so I couldn't really clean it up.
Are there any tips for doing this (tools that might help me, etc.)? Is there an easier way to do this than hand-soldering?
AI: Learn to use "drag soldering" technique. Basically, instead of trying to solder each pin individually, you just get a bead of solder on the tip and roll it across the pins. As you hit each one, it'll leave just enough solder on the pin to attach it.
A video will do a lot to help explain here:
https://www.youtube.com/watch?v=wUyetZ5RtPs
There's a few things that help make this work:
Apply flux - lots of it - before you start soldering. Yes, I know the solder you're using probably has a flux core, but that's not enough here. (If you use paste flux, the paste can also help keep the part in place.)
Using the right kind of tip helps a lot. A small tip is not what you want here! I use a Hakko "hoof" tip (T18-C2); there's probably equivalents for most other good irons. |
H: Why negative feedback (common source degeneration) increases linearity range?
From this online lecture (page 190-20), I see that the negative feedback (Rs in common source degeneration) helps increases linearity range.
I am wondering if anybody know an explanation for this intuitively or mathematically.
Another question is what is the difference between the left and right hand pictures in the image below?
For small signal, they are exactly the same.
Thank you.
AI: Suppose you have a nonlinear transfer function \$\frac{v_O}{v_{IN}} = a(v_{IN})\$
If you add negative feedback 1/m such that:
\$ \frac{v_O}{v_{IN}} = a(v_{IN})(v_{IN} - \frac{ v_O}{m})\$
so
\$ \frac{v_O}{v_{IN}} = \frac {a(v_{IN})}{1+\frac{a(v_{IN})}{m}}\$
For \$\frac{a(v_{IN})}{m} \gg 1, \frac{v_O}{v_{IN}} \approx m \$ (linear gain of m)
For \$\frac{a(v_{IN})}{m} \ll 1, \frac{v_O}{v_{IN}} \approx a(v_{IN}) \$ (nonlinear with negligible feedback)
As you can see, the closed loop gain must be reduced significantly from the open-loop gain: \$m \ll a(v_{IN}) \$ for there to be a large reduction in nonlinearity. |
H: How can earth fault meter work through inactive IGBTs
I have a setup with a Bender IRDH375 connected to the DC bus of an IGBT based inverter. If I connect a 50kOhm resistor to earth while running the inverter, it detects it well. That part is reasonably ok to understand. However while not running it still detects about 200-400kOhms to ground. How is this possible?
Schematic of setup illustrated below. Only 1 of 3 phases are illustrated for brevety.
simulate this circuit – Schematic created using CircuitLab
Shouldn't the IGBT be almost fully blocking while "OFF"?
Update:
If I take off R2, the earth meter will show almost maximum insulation (10MOhms). Same if the inverter is running or not.
AI: It starts with diagnosing the situation. The dector is very sensitive so you can use the instrument for finding the problem.
Disconnect the incomming DC and inverter connections from the AGH150W and read the effects on the IRD375. If that is wrong then the problem is limited to Bender units.
2.When the first step is ok connect the DC source to the AGH150W and read again.
If OK then disconnect the DC source again and connect the inverter and read again.
The last step is to connect both the DC source and the inverter without running and read again.
4 Then run the inverter and read again.
You should be able te find the error by cutting the system in little pieces
From the action taken by OP it has become clear that there is no insulation error but a remaining resistance (not open) between the in and output side of the inverter.
This being the case the only way to make the unit working is to lower the thresholt resistance of the unit just below the measured resistance value. This can be done between 1kOhm and 10MOhm. |
H: Relationship between voltage/current & LED strips?
I recently bought a WS2812 LED strip (5 meters, 300 LEDs)
When the LEDs are driven at full brightness and each color lit, they draw 60mA each--or 18 amps for all 300!
This led to a problem, which led to a question.
A friend suggested that I not attempt to power the whole strip at once, because even if I could, 18 amps would probably burn it up. So, the other option is to splice in power leads every so often, which is recommended by the manufacturer.
I took a look at my power supply and realized that while it can supply between 0-24vdc, it can only supply 5 amps.
Is there no difference between 12v at 5 amps, and 5v at 5 amps other than power? I understand that Power = amps * volts, so 12v at 5 amps is 60 watts, and 5v at 5 amps is only 25 watts. My question is this: shouldn't the limiting factor be the wattage, not the current? If my power supply can supply 24v at 5 amps (it can), this is 120 watts. In order to supply 120 watts at 5 volts, I would be hitting 24 amps.
AI: You are partially correct. All three: power, current, and voltage are limiting factors for your power supply. If your power supply is labeled 0-24VDC @ 5A max, then it means it can handle these max ratings:
Max Voltage: 24V
Max Current: 5A
Max Power: 120W
So you say your LED's draw 18A total. This is only half of the equation. You also need to know what voltage the strip needs to be supplied. If the voltage that the strip requires is less than 1/18 of 120 (6.67V), then congrats, your power supply can handle your LED strip. If the required strip voltage is > 6.67V, then your total power draw will be > 120W which is no good. However, you can't just willy nilly plug the LED strip into the power supply because remember, your power supply can only handle 5A. The easiest way to overcome this problem is through the use of a power transformer to turn 24V into your required voltage.
simulate this circuit – Schematic created using CircuitLab |
H: Why microcontroller takes many clock cycles to start up with PLL clock source?
I was looking over the ATTiny85 datasheet and noticed on page 26 that with a PLL clock source, the fastest startup time is 14CK + 1K (1024) CK + 4 ms. Am I misinterpreting what they mean by 1K CK, or is it that the PLL needs time to set up? Compared to other clock sources it seems to take many more cycles.
Thanks
AI: Deeper interpretation: The PLL is actually producing clock cycles during that whole time. The problem is that until it achieves "lock", the clock cycles may at times be too short to allow the CPU to operate correctly, as the VCO control voltage swings both above and below the target value before settling down.
So what this specification is really telling you is that this is how long it takes before the clock frequency being produced by the PLL is guaranteed to be within the range required by the rest of the chip.
This specification is based not on how fast the VCO itself is running, but rather on the bandwidth of the PLL's internal feedback. There's a tradeoff between fast start-up (wide bandwidth) and low jitter (narrow bandwidth). Some chips actually give you the ability to select different bandwidths for different applications. |
H: How are RS 485 transceivers implemented?
I would like to know how transceivers (which are used to send and receive differential signals) are implemented. AFAIK the receiver the transmitter (TX) could be something similar to an H bridge. I already took a look at datasheets of some popular transceivers but I could not find any details about the inner working.
EDIT : the term transceiver might be a little to large, by transceiver I mean the ones used for RS 485 (eg : for USB or DMX-512)
AI: Assuming the receiver side of a transceiver is boring to you (just a comparator with some hysteresis):
TI's DS9638 (datasheet of the SNLS389D) actually has an equivalent circuit for a single transmitter circuit:
To illustrate what's happening a bit better:
In green, the identical output stages. Look at the left:
Q15 has the job of pulling the output low as soon as the base of Q6 sees super-threshold voltage
Q18+Q19+Q17 form a current amplifier
rest of the combination of Q16's collector and Q18 base invert
The whole functionality of the center part is to make sure Q5 and Q6 see the inverse, but at very tightly controlled thresholds. |
H: Digital gain control with a digital pot?
I want to make an audio amplifier with digital volume control. I figure that by adjusting the feed back path that I won't change the input impedance and won't affect the frequency response. (other than the gain bandwidth product)
Does the following setup look reasonable?
Would a digital pot be good for this application?
Are there any drawbacks of digital pots that i'm not aware of that would make them undesirable for this application?
simulate this circuit – Schematic created using CircuitLab
AI: I can see two things that are not optimal in that design. For the discussion I will use a random digital potentiometer I found from this article: The DS1881. Most are similar in construction.
Wiper resistance
A digital potentiometer has a pretty high wiper resistance, especially compared to a mechanical potentiometer. Not only is it high, possibly in the order of hundreds of ohms, but it is also not very stable or well-defined. The DS1881 lists it as 160 Ω typical, but up to 250 Ω maximum. It can vary with the signal voltage and such.
It is not a problem if it is in series with a high impedance input, for example the input of an op-amp. Then there is negligible current through it, and thus, a negligible voltage drop. In your case, the wiper is in parallel with a section of the potentiometer:
simulate this circuit – Schematic created using CircuitLab
Since some amount of current will flow here, it will affect the signal. How bad? Not much, it is somewhat swamped by the 10 kΩ series resistance and the rest of the potentiometer section, but that leads to the next problem:
Absolute resistance variation
A digital potentiometer has a well-defined and trimmed ratio, but its absolute resistance can vary a lot:
PARAMETER
SYMBOL
CONDITIONS
MIN
TYP
MAX
UNITS
End-to-End Resistance
REE
+25 °C
45
kΩ
End-to-End Resistance Tolerance
+25 °C
-20
+20
%
Ratiometric Temperature Coefficient
30
ppm/°C
End-to-End Resistance Temperature Coefficient
750
ppm/°C
Wiper Resistance
RW
160
250
Ω
Note the ±20% resistance tolerance, and the 750 ppm/°C temperature coefficient.
Since you want to use it as a rheostat, you are hit by this. If it matters, that is up to you. It will really only make the gain fluctuate a bit by temperature and so on. A digital potentiometer should really be used as a voltage divider, so you can take advantage of the close matching between the steps.
The article I linked to, from Maxim, has a few audio circuits you may want to take a look at, with pros and cons. |
H: Determine the current i through the 4 ohm resistor
Due to different frequencies, i applied superposition by shorting the right source
and applied voltage division rule to get the voltage across the resistance.
Vr=3 ∠0° * 4/ (4 + 6j) = 1.666∠-56°, and Ir=Vr/R
But the answer is not correct (for the right source also). What did I do wrong?
AI: When you short the right source you end up with this circuit:
simulate this circuit – Schematic created using CircuitLab
So in order to obtain the voltage across the resistor you have to take into account the second inductor.
$$
Vr=3∠0°*\frac{R1||L2}{L1 + R1||L2}
$$
R1||L2 means the parallel of Resistor 1 and Inductor 2
$$
R1||L2 = \frac{R1· j \omega L2}{R+ j \omega L2}
$$
Also you have to apply superposition shorting the left source and then adding the two results. |
H: Triangular wave generator
As part of a college project I have to design a working Class-D audio amplifier.
One of the first challenges is to generate a triangular wave at a frequency arround 100kHz.
In order to do this I have in mind generating first a square wave (using a 555 timer in an astable configuration) and then integrate it. Then I encounter some problems:
The device has to be single powered (5V taken from a USB plug)
Becouse of that limitation I cannot use a simple Inverter Integration using an operational amplifier.
Any ideas on how to solve my problem?
Thanks in advance, and sorry for my bad english :/
AI: A class-D audio amplifier does not necessarily need a triangular wave generator, a good alternative could be a self-oscillating class-D amplifier (-> Google).
In case you have to use a carrier-based class-D you could use the following configuration for the carrier generation (link)
A 100kHz carrier frequency might be a little bit too low, I suggest to increase it to 200 or even 300kHz. |
H: Specification of DC motor
I have a motor, like to use it for a project. I don't have the specification details of it(volt, Amps, RPM). Can someone help me in this regard?
Image source/Item description
AI: This motor is manufactured by Mabuchi. The part number 'RK-370SD-4045' actually tells you quite lot about it. Referring to the chart below, 'RK' means it has a round (circular) case and carbon brushes. 370 is Mabuchi's designation for motor size and number of armature poles. '4035' means 35 turns of 0.4mm diameter wire on each armature pole. 'SD' means it has anisotropic Ferrite magnets.
Unfortunately Mabuchi only publishes specs for a small selection of their motors, not every variation. However if you have the specs for one particular wind (wire size and number of turns) it is possible to calculate the specs for other winds. Specs for the RK-370SD-2870 are:-
Voltage: range 4.5~9.6V, nominal 7.2V
No-Load: 16500rpm, 0.34A (at 7.2V)
At Stall: 36.3 mNm torque, 8.77A (at 7.2V)
Your motor is identical except for having half as many turns (35 vs 70) of wire which has double the cross-sectional area (0.4mm2 vs 0.28mm2) and therefore 1/4 the resistance and 4 times higher stall current (at the same voltage). With half as many turns the torque constant would be halved, but with 4 times higher stall current the stall torque would be doubled. No-load rpm should be twice as high at the same voltage, as should no-load current (probably a bit more than double due to increased windage and eddy current loss at the higher rpm).
However since apart from the windings both motors have identical construction, their maximum power and rpm ratings should be the same. For the same performance (rpm, torque, power and efficiency) your motor would need half the voltage. So we can estimate the specs of your motor as:-
Voltage: range 2.25~4.8V, nominal 3.6V
No-load: 16500rpm, 0.68A (at 3.6V)
Stall: 36.3mNm torque, 17.5A (at 3.6V)
From these numbers we can derive the motor constants Kv (rpm/Volt), Kt (nM/A), and Rm (resistance) which are used to calculate performance. Mabuchi's performance graph for the RK-370SD-2870 is shown below. Your motor should perform similarly on 3.6V, but with double the current draw. I have marked the changes for your motor in red.
Note that these calculations (including Mabuchi's own performance graph!) are only theoretical, based on a simple DC motor model. In practice there may be differences due to factors such as brush resistance, timing advance (may be forward, neutral, or reverse), applied voltage etc. To get the actual performance on a particular voltage you will have to test a real motor. |
H: Finding value of Resistance for Maximum Power
I am badly stuck in finding value for R in this circuit for maximum power transfer in the network,
It is my assignment question and I am badly confused.
Please help ,
Thanks.
AI: For maximum power transfer you are expected to not only know how to set up equivalent circuits, but also how to find maximums and minimums by setting a derivative to zero.
The first step, of course, is to develop the equivalent minimum circuit:
simulate this circuit – Schematic created using CircuitLab
You should have been able to follow the above steps. If not, then you have some serious study to do and you won't yet be able to solve these problems until you can perform the above steps on your own.
Now that you have the final circuit, you can work out the following equation:
$$\begin{align*}
I &= \frac{V}{R+R_3} \\
\\
P_3 &= I^2 R_3 = \frac{V^2 R_3}{\left(R+R_3\right)^2}
\end{align*}$$
Now, in the above equation for \$P_3\$, \$V\$ and \$R\$ are known constants. \$R_3\$ is the variable whose value you want to find in order to maximize \$P_3\$. This is the point where you are supposed to know from calculus courses (or reading) that you take the derivative, set that new expression to zero as an equation (to find where the slope is zero), and then solve. This will give you either a minimum or a maximum.
$$\begin{align*}
\frac{\textrm{d}P_3}{\textrm{d} R_3}&=V^2\frac{R-R_3}{\left(R+R_3\right)^3} \\
\\
\textrm{so,} \\
\\
V^2\frac{R-R_3}{\left(R+R_3\right)^3}&= 0
\end{align*}$$
I'll leave the last step of solving that answer to you. But that's all that is left. |
H: Set the threshold voltage of CMOS inverter to VDD/2 for both rising and falling edge: possible?
Using 0.35u technology (VDD= 3.3v, Vt=0.7, Tox = 0.7 nm), I am trying to set the threshold voltage of an inverter to VDD/2.
If I set the width ratio of PMOS/NMOS to 5 (means the Width of PMOS would be 5 times larger than the NMOS) then the threshold voltage at rising edge of input shows a HI skew, but is around VDD/2 at the falling edge.
If I reduce the width ratio to 1 or 1.25 then at the rising edge of input I see a threshold voltage of around VDD/2, but at the falling edge I can see a LO skew.
I tried playing with Width Ratio to get a threshold voltage of around VDD/2 for both rising and falling input edges, but it seems it is impossible. I even looked at the specs of standard inverter cell coming from the foundry and it shows the same behavior.
It seems it is impossible to set the threshold voltage of CMOS inverter to VDD/2 for both rising and falling edge. Am I correct?
AI: No, it is not impossible to set the threshold voltage to be right at half of the supply voltage. Just to be clear, the threshold voltage is defined as the DC input voltage at which the output voltage is equal to VDD/2. It does not necessarily correspond to an exact delay between rise and fall however.
You can easily set the threshold voltage as accurately as you wish by iteration using something like Newtons Method to size one FET vs. the other (or analysis if the FET properties are handy).
But, this may not necessarily create a gate which has equal rise and fall delays or output slopes (although increasing the PMOS size relative to the NMOS, for example, will both increase the threshold voltage and reduce the rise delay).
Usually the most important characteristic is the output delay and slopes for each load, unless you are trying to use the inverter as a comparator or something.
If you want a balanced gate, what I could recommend is to size the transistors for equal slope at the same load. This generally equates to equal delay as well in a single stage logic gate. |
H: Is it a good idea to connect aluminum AC mains wires by soldering
Provided I have an appropriate flux for aluminum and I'm able to solder the wires will this connection be durable?
I'm not sure if it will corrode or not considering aluminum and the solder (60-40 solder) are different metals.
AI: Provided I have an appropriate flux for aluminum and I'm able to solder the wires will this connection be durable?
That's nearly impossible. You can weld aluminium under protective atmosphere (inert gas welding), but you cannot solder it with normal solder. Or at least, I've never seen anyone succeed (and by god, I tried to use alu sheet metal as ground planes). I've heard it's possible, but you'll need a special solder, which seems to be very expensive.
will this connection be durable?
If you used the right solder and technique, probably, yes. But since that would be your first alu solder joints: No. Good soldering takes practice, even on much simpler materials like copper. Don't expect good results on your first try :)
Also, other aluminium connections won't be durable: read the comments, it creeps, and that's not a good property for a wire that you want to screw in or stick in to some locking connector.
But: don't use aluminium wires. There's a reason they aren't used anymore today, and it's very simple, and physically inevitable:
Aluminium doesn't like being bent as much as e.g. copper does. You don't want your wire to slightly break when installing it, because that breaking point will have a higher resistance, and get hot, and burn your house down.
Aluminium generally has a much higher resistance than copper. You just convert your electrical energy to heat. Using copper-clad aluminium wire does make sense in some cases, but that cable needs to be thicker than a copper cable carrying the same current (and usually isn't even cheaper, in the end, for installation cables). You find these types of cables in free-hanging high-voltage supply lines – they aren't good for installation, because you can't bend them as much as copper cable (see 1.), and because they are bigger than their copper alternatives.
In other words, don't use aluminium cabling. |
H: Can I apply nodal analysis if the wire doesn't have a resistance?
simulate this circuit – Schematic created using CircuitLab
How do I apply it in the wire containing D4 and D3 if it doesn't have a resistance? The way I learned is: \$\frac{V-V_{D4}-V_{D3}}{R}\$ but in this case there is no resistance across \$V\$ and ground. Should I put \$R=0\$?
AI: Yes, you can. That is essentially how SPICE works.
If you're doing back of the envelope calculations, you'll likely be modelling D4 and D3 as voltage sources when they are in zener operation and forward bias operation respectively. In that case you cannot apply nodal analysis (nodal analysis doesn't apply to circuits with voltage sources) but you can apply modified nodal analysis.
The way I learned is: \$\frac{V-V_{D4}-V_{D3}}{R}\$ but in this case there is no resistance across \$V\$ and ground. Should I put \$R=0\$?
What you should have learned is to write the equations for KCL at each node:
$$i_1 + i_2 + i_3 = 0$$
assuming \$i_1\$, \$i_2\$, and \$i_3\$ are the current of the three branches connected to node v, defined so they all flow in to the node. Then you need to figure out how to express these three currents in terms of the voltage at node v and the other nodes of the circuits. If the circuit elements were all resistors, you might get an equation like the one you used. But if the elements are diodes (neglecting zener or avalanche operation), you'd need to use the Shockley equation:
$$ I_D = I_s \exp\left(\frac{qV_D}{nRT}-1\right)$$
where \$I_D\$ is the current through a diode and \$V_D\$ is the voltage across the diode (something you'd be able to write as \$v_x - v_y\$ where \$x\$ and \$y\$ are two nodes in your circuit), and \$I_s\$ and \$n\$ are characteristics of the diode. To model the zener diode you'll need an even more detailed model that includes zener behavior. |
H: Fried motherboard resistors, what to do?
A motherboard failed under load recently, and here is the worry:
I am waiting for a new mainboard. What must i do about it? is it a very easy repair? I am mostly confused about identifying the resistors.
AI: These a filtering caps on one of many of platform voltage rails. Capacitors blow up only due to overvoltage, or too much of ripples/ringing. Which means that some active components (power MOSFETs) are likely blown as well, causing this, and likely some other damage. These caps can be high-value (47uF-100uF) low-voltage (2.5V-4V) caps. Without proper schematics for this board it is not possible to repair it. And the schematics will be impossible to obtain. So throw the board away, or keep it for re-use its components for DIY projects. |
H: Use external reference for frequency counter
I recently bought an HP 5316A frequency counter. It has the ability to use an "external reference". Does that mean it ignores its internal clock and uses the external signal in lieu of the clock?
Is it complicated to provide the external reference signal? What is involved?
AI: Yes, the counter will use external clock as its time base. The instrument itself has an internal reference clock, but it has certain limited accuracy. If someone has a better quality "atomic clock" (as rubidium standard or else), one can use it to improve the counter parameters significantly.
What is involved? An external clock generator and BNC coaxial cable. |
H: MAT12 matched pair transistor substrate connection
On Analog Devices' MAT12 data sheet, there is a note (see on the top right, note #2) saying that the substrate, which is connected to the case, is usually connected to the most negative circuit potential. Since there is no pin connected to the substrate (or to the case), how do I connect the substrate to the most negative circuit potential?
AI: From the datasheet:
Substrate is connected to case on TO-78 package.
Now, here is a TO-78:
As you can see, the case is actually metal.
If you necessarily want to connect the substrate, the recommended way is to do it using a heat sink. I'm not sure if you actually gain anything by doing that. It could be feasible to solder a small wire or pin to the tab, but I've never done it, and I would be afraid that it could damage the matching. As Tom points out in the comments:
basically if you use a metal heat sink with no insulation, don't connect that heat sink to anything but the most negative circuit potential
I suspect that the only reason they mention that you can connect the substrate is because you sometimes have to, due to the heat sink. Not because they think it is warranted. If it gave any performance benefits, I'm sure they would point it out. |
H: 12v battery reading 10.7v not supplying 5v load
I have a solar charge controller connected to a solar panel (30w), also connected to the charge controller are a 12v battery (4.5Ah) and a raspberry pi Zero (5v, draws approx. 200mA).
This evening the charge controller is showing the battery voltage as 10.7v.
I turn on the load switch to supply power to the Raspberry Pi but as soon as it draws current the controller switches power off again.
So my question is, should my 12v battery be able to supply enough power to the raspberry pi (5v & 200mA ) when it is only charged to 10.7v?
If not, why not?
I would have thought 10.7v being far higher than 5v so it should be well capable.
Battery details:
12volt/4.5Ah
Constant voltage charge(25C)
cycle use: 14.5 - 14.9V
Standby use: 13.6 -13.8V
Initial current: 1.35A
Charge controller details:
Anself: https://www.amazon.com/Anself-Controller-Regulator-Overload-Protection/dp/B01A0D18MU?SubscriptionId=AKIAILSHYYTFIVPWUY6Q&tag=duckduckgo-ffab-20&linkCode=xm2&camp=2025&creative=165953&creativeASIN=B01A0D18MU
Schema:
30w solar panel----charge_controller/Battery(USB interface)---raspberry pi
AI: Consider that the lead acid battery is 100% resting at 12.5V and near 0% @ 11.5V and charging should be in the range of 13.5 to 14V.
Your "mileage will vary" with flooded vs SLA from battery chemistry differences slightly and deep discharge marine batteries.
Below 11.5V starts to rise in series impedance quickly so any load drops the voltage more easily than in the normal charge voltage range. 12.0V is midpoint or around 50% SoC .
All it takes is one bad cell to spoil the pack.
If you think it is fully charged and it drops to 10.7 then my guess is one cell is self shorted partially and should be 1.5V higher or 12.2V
Take readings after adequate settling time with some loading following charger off. |
H: How to measure the input impedance of a summing amplifier?
I have a summing amplifier set up like so, and I need to come up with a test to find the input impedance, I know because its a 741 IC that it will be in the region of 250k - 10 mil ohms, but don't know how to test it and explain the test results
AI: Anything wrong with the age-old trick for measuring the input impedance of a "black box"? Create a voltage divider of a resistor and the input to be measured, selecting a type of resistor or variable resistor that will behave as a pure resistance at the frequency involved. One then substitutes resistors, or adjusts the variable resistor, until the signal amplitudes across the resistor and the input are the same. The DC resistance of the resistor should then be the same as the input's impedance at that frequency, no? (If a variable resistor is used, it would be removed from the circuit and measured with an ohmmeter.) Of course, this is a simplification. A simple calculation would avail with a ratio of amplitudes other than 1:1. |
H: Random signal lost or extreme weak signal with remote controller on ISM frequency (433Mhz)
There is a task given to me that I have to investigate a problematic RF remote controller against its weakness and make a suggestion report how it can be improved.
This RF remote controller is used with two hands like PS console remotes. While it is functioning normally, in some cases, signal is not correctly or completely received by the receiver.
Edit: The receiver is under the bed controlling motors.
Here are my findings, considerations and modifications;
1. Hands or human body on the remote controller are causing attenuation over the transmitted signals significantly at some directional orientation. (Observed with RF spectrum analyzer in dBm). At those moments, sometimes data is not properly received by the receiver but not always, unpredictable.
2. Antenna is on the 4-layers PCB. This anteanna is working OK when there is no obstacle. See image.
3. RF circuitry and micro-controller are all together and no shielding on the RF side. See image.
4. I have cut out on-board antenna and added a smd female RF connector. This way I experimented these steps;
a. with teflon brown cable and connected a 17.3 cm wire.
Antenna is inside the box: More or less same dBm measured as the on-board one. Data lost is not checked yet since it is not always the case.
b. with teflon brown cable and connected a 433Mhz whip antenna. This is shorter, most likely it has a spiral version of 17.3 cm antenna inside.
Antenna is outside the box: 30% more dBm measured than the on-board antenna does. And most importantly, hands and human body factors are quite less effective over the signals.
5. My point of view, the RF circuit part standalone is looking OK to me in terms of proper matching and harmonic suppression.
So, After those info and the image, what mistakes are you seeing (if there are)?
How can this RF circuitry be made more effective by all the RF design technics considered?
I hope that I could explain what I am up to.
AI: How much is 30% more dB?
Ricean fading or (reflection cancellation) will be your dominant cause of dropout even if RSSI or SNR shows adequate but marginal signal, so reflective surfaces and people interfering with reflections can cause dropouts when a reflected signal is near equal amplitude but out of phase with the main signal path.
Keeping the whip antenna at right angles to the direction of propagation give the most signals. To see if you are in a Ricean Fading zone with weak signals, behind the Tx, move the unit on a table top slightly over a +/- 1/2 wavelength zone in small increments to FIND a dead spot and then you know you will get dropouts with random orientations at that distance or path loss. Try to get the RSSI or AGC signal out of the Rx to display on a scope to get better sensitivity to Path loss, interference and Ricean Fading Effects.
Then consider the best gain/dispersion antenna and location for the fixed receiver with an improved ground plane for the antenna. |
H: Jacob's Ladder and Component Identification
What would be the component in the bottom left of the picture?
AI: It looks like an option for 120/240 primary windings. My guess is it is connected in series for 240Vac operation.
The arc generates ozone, UV and other nasty EMI, not for newbies. |
H: Use superposition to find voltage V1
I first open circuited the 50 mA source and used Ohm's law to find \$V_1\$
$$V_1= 20 ∠0° \frac{j50 \times (-j25+40)}{j25 +40}$$
but the answer is not correct. Also, how do I find \$V_1\$ when the 20 mA source is opened?
AI: (a) The circuit across the active current source is a parallel circuit of C and a series circuit of R and L.
Note that components values are given as admittances (not as impedances).So you can't calculate \$V = IZ\$ (\$Z=\$impedance) butyou have to use \$V = \frac{I}{Y}\$ (\$Y=\$admittance).
The total admittance is \$Y=Y_C + Y_{RL}\$ (parallel admittance are calculated like series impedances or resistances: just add them) and
\$Y_{RL}=\frac{Y_R Y_{L}}{Y_R + Y_{L}}\$ (series admittance are calculated like parallel impedances or resistances: product divided by sum)
If you follow the steps you get exactly the given solution.
(b) Now the circuit across the active current source is a parallel circuit of L and a series circuit of R and C.
You get V1 by
using the current divider formula to get \$I_L\$ and \$I_{RC}\$
\$V1 = \frac{I_C}{Y_C}\$ where \$I_C = I_{RC}\$ (because R and C are in series). |
H: How can I tell if my building plumbing provides an adequate ground?
I work in an old building with two-wire wiring, i.e. no separate ground. Is there some minimum-equipment-purchase way to tell if a wire soldered to a building plumbing pipe is an adequate ground? I have a standard hand-held multi-meter.
AI: http://homeguides.sfgate.com/plumbing-system-electrical-grounding-94939.html
This author above suggests plumbing can supplement and earth ground rod, but not be a primary connection.
Building codes in your area determine the ultimate answer.
Ask relevant city authorities for reference to building codes for commercial ground methods or consult an electrician.
Certainly deep water lines are better than no ground for equipment that needs to protect you from stray leakage from dust and humidity on metal shell electrical tools.
I might measure the AC voltage from the plumbing to Neutral and see how much voltage drop there is. ( must be < 5% line) ) ... this is grounded outside at the transformer but current can cause a voltage drop. |
H: Confusions on FFT of a square-wave in theory and in scope and simulation
Below is an ideal square-wave in time domain, and its harmonic components in freq. domain:
As you see above a square-wave is composed only of its odd harmonics as spikes nothing in between.
And below is an oscilloscope's FFT of a square-wave:
Here we see two things are different. First of all the above FFT is not composed of spikes but widened curves. Secondly it is continuous not discrete.
Maybe scope may involve some noise and the non-zero rise time of the square-wave might effect the FFT results.
So instead of scope lets look at what LTspice shows the FFT of a square-wave(pulse-train in this case) with the following setup:
Here is the FFT:
This doesn't look good either. So I set the rise times much shorter than LTspice's default as follows:
Now the FFT became better:
Now I can see one of the problem when viewing FFT of a square-wave or a pulse in a scope the rise and fall times are never zero. The other problem is there is some noise involved.
Here is my question:
Here is what I don't understand.. At the beginning I provided a spectrum of an ideal square-wave which was discrete odd harmonics as spikes. But both in scope and in LTspice the FFT is continuous.
I'm confused at this point. Let me give an example. In my last plot above, the pulse frequency is 100Hz. So I would expect it is composed of its odd harmonic sinusoids of 300Hz, 500Hz, 700Hz,.. so forth and so on. I wouldn't expect it would have 130Hz component for instance or 102Hz. In fact according to the last FFT plot there is component at 102Hz and it is even greater than 300Hz component.
Any idea which one represents reality here? What am I knowing wrong?
AI: Welcome to the real world!
The "mathematically perfect" transform you show at the top, with the "discrete" harmonics is generated assuming that the rise and fall times of the waveform are zero, and that you're doing a continuous transform — no discrete sampling in the time domain. It assumes that the integrations are over an infinite span — or equivalently, they're done over an exact whole number of cycles.
When you generate a waveform in a simulator, or feed a real waveform into an oscilloscope, these assumptions no longer hold. The rise and fall times have nonzero values, and the signal is sampled in the time domain. Most importantly, the sampling does not normally encompass an exact number of whole cycles, nor is it precisely synchronized to the waveform.
Violating these assumptions causes the peaks you see in the transform output to "spread out". This does not mean that there's any energy at, say 102 Hz, but only that the limitations of the algorithm means that it can't distinguish between whether or not there's energy there, given the limitations of the set of samples it is working with. |
H: Comparator input offset voltage
This is an example about input offset of a comparator in Analog Integrated Circuit Design by Tony Chan Carusone (page 414).
Could anyone explain why the plot in Fig 10.1 is a normal distribution?
AI: There are two parameters in a Normal Distribution, the mean and σ, the standard deviation.
The mean here is +3.0 mV and the standard deviation is shown for the probability of a logic 1 output. The inverse curve would be the probability of a logic "0" output.
This Normal distribution curve of a transition or the "Bell curve" would indicate the probability than any threshold transition would occur.
The Cumulative Function of the Normal distribution is the integral of the area under the curve across the range of inputs was what was actually shown.
ref Wiki |
H: Connecting the linear regulator directly to AC line
There are several out of the shelf solutions: LR8 from Microchip (formaly known as Supertex product) and NCP785 from ON-Semi. The last one is able to handle up to 700 volts at the input which is impressive.
Both components is very attractive to be used in my application. However, LR8 does not have enough voltage strength (I need to withstand 400 VAC = 625 VDC) and NCP785 is not easy to buy in my country (Russia).
One of the solutions is to put an external transistor like this:
However I think that it should be possible to add an external transistor to linear regulator (LP2981 in my case) without additional components (zener, resistors...) - this would be perfect as I need to reduce PCB size.
From my early engineering career I remember that the external transistor can be used to increase the current of the regulator, something like this:
Is it possible to do same thing to input voltage?
AI: As far as I can see you need to rectify the incomming phase (diode + capacitor). Feeding from an alternating source is not possible.
The basic model for increasing the current of a regulator should be a little different. Here the corresponding image with the principle.
Realizing regulator with high voltage input is diffent but also feasable.
I did not find a regulator but you can build one yourself as shown in the schematic.
Ofcourse you can modify the parts etc. according to your needs and parts available. |
H: LDR not detecting light from LED when finger placed between the two
I'm currently working on a project which uses an LED and a photoresistor to work out my BPM.
I have a schematic which I've built into a prototype which all works fine and when I run the serial monitor in Arduino I'm getting output values.
The problem is, when I place a finger in between the LED and photoresistor, for some reason the readings are going to 0 and I'm unsure why?
I've encased with led and photoresistor to avoid any other light but for some reason I'm not getting any values.
Here is the schematic that i'm following:
You'll have to excuse my choice of encasement:
Unsure of what reasons which could be preventing the LDR of outputting.
AI: The situation is simple. If you make it impossible for light to enter the LDR then the resistance of that device is increasing as you can see from the added picture.
Removing the light with your finger makes the resistance go up. The result is that the base of Q1 does not receive enough current anymore to make it conduct. The collector of Q1 goes up and Q2 starts to conduct in full. So the led D! starts to glow.
The next step is to understand what happens when the led is putting it's light on the LDR.
Looking at the schematic again you can see that if Q1 starts to conduct complete Q2 will be cut of and the led goes off. However that is not possible because then there is no light falling on the LDR anymore.
The result will be a balance whereby the resistance of the LDR lowers enough with the light from the LED to keep Q1 and Q2 conducting so much that the light from the led is sufficiant to keep the LDR at the resistance needed.
Now if you start playing with this combination then you are able to change the balance and send the resulting signal to the analog input of arduino. |
H: how to improve RF meter?
On https://www.rfsafe.com/product/diy-rf-meter-led-powered-rf-detector-diodes/ , there is a kit which contains:
nine 1SS86 Schottky diodes and
one LED,
which uses to detect mobile phone and microwave radiation.
It states:
range of application: mobile phone GSM signal. It is useless for PHS, fixed-line telephone,CDMA mobile phone
How can I improve the circuit, in order to detect also cordless phone radiation, like the dect 6.0 cordless phone, which, actually, operates on 1.9GHz?
BTW:when they explains how this circuit works, they using the word germanium diodes, while refering to the 1SS86 diodes. As far as i understand,they are not germanium diodes,they are silicon diodes (silicon Schottky diodes).
Am I right?
AI: These people are scammers, and do not deserve your money
How can I improve the circuit, in order to detect also cordless phone radiation, like the dect 6.0 cordless phone, which, actually, operates on 1.9GHz?
With this circuit, you can only get exactly one band. I'd suspect this circuit was meant for GSM in the 900 MHz band (they don't mention – could be 800 MHz or 1800 MHz, too, one reason more to stay away from that product).
The circuit itself is kind of funny: it's two things in one:
a loop antenna (hence the circular layout)
a voltage multiplying diode cascade, which I can only guess works by exploiting parasitic capacitance.
This device will not work. Even if we assume the LED gets this bright with but 0.5 mA (it will not), a single 1SS86 would have a forward voltage of about 0.18 V; nine of them in series would have a forward voltage of 1.62 V. Then add the forward voltage of the LED (something around 2V). Without a very large antenna, you cannot get peak-to-peak voltages that high from an EM field as weak as legal for GSM.
I tried to watch the video, but it was so full of bullsh*t that I couldn't stand to finish it. Don't buy anything from these people. If you put your cell phone in something that tries to block electromagnetic radiation, you will be subject to more energy, not less.
One thing that struck me was: in the assembly instructions, there was no external antenna whatsoever, whereas in the video where they demonstrate this thing lights up, the have about 15cm of wire attached to the device. That's a scam. Not only doesn't the device work on its own, it's almost certain they can now use whatever secondary transmitter (and by the length of that cable, maybe a 433MHz analog FM radio transmitter that can send 100s of watts – much much much more than cell phones) to light up the LED. |
H: How to control a device that need to be updated over internet?
I know, the question is very vague, but I'll try to explain better.
I'm working in a school project and I got interested in continuing this project after school. It's a automatic aquarium. I want to control everything that happens in a aquarium, so I used an Arduino Uno connected to some sensors and some activators - sorry, I don´t know the right terms.
Everything works ok.
But... I got the idea to control this system remotely, connecting it to the internet through a wifi connection using an wifi shield. Ok, seems possible to do, but I also want to change all "program" remotely. I mean, the user, by a site or app, would be able to select a fish and the Arduino would change all the environment to better grow that specie. And I imagine it requires a "reprogramming" of Arduino and its seems impossible to be done by wifi connection.
So... The question is:
Can I achieve it using an Arduino? What version of Arduino? And How?
If its impossible to achieve it with an Arduino, how could I achieve it?
AI: There are high chances you don't need to reflash remotely. Just write a generic control algorithm supporting all target sensors and actuators, and let the user download configuration parameters which could be stored in EEPROM. Then update those parameters remotely.
If you insist on remote reflashing, you should also design a bootloader which takes a hand if the last reflashing attempt was unsuccessful. Otherwise your system will be easy to "brick". |
H: Should I use a current limiter on a battery powered system
I'm designing a small battery powered system, actually I'm battery powering a module which already has all the switching regulators it needs. This provides me with a wide input range of 5-10V. I was thinking of connecting the batteries directly to the module instead of using a regulator because I don't want to pay an efficiency penalty, especially in low power mode. I was worried about inrush current though when the batteries are first turned on. I know the internal resistance of the batteries should limit it some but should I add some kind of current limiting or soft-start circuit to protect the batteries from damage or overheating?
That'd be easy to do with a regulator, not sure what I'd use without one. Maybe a high side switch with some time based control. Should I even be concerned about this? Target batteries are LiOn that won't be recharged in the device itself.
AI: You haven't said anything about the current draw of your system, how much inrush you expect, nor the type of battery, so the question can't be answered directly. However, it is unlikely that inrush current from occasionally turning something on is going to matter to the battery or the power supply input circuitry. If you are just powering a few 10s of mA of ordinary electronics, then you are overthinking this. |
H: Addressing issue in flash M29W800DB
I wanted to loop through the addresses in that flash M29W800D, so I started a counter from 0 to 0xFFFFE, then I try to write an unsigned int to an address, but it fails. When I made the loop counter to increment each even addresses, it worked.
but why ? what's so special with that chip that doesn't address all the addresses ?
here is a link to the datasheet http://www.datasheetcatalog.com/datasheets_pdf/M/2/9/W/M29W800D.shtml
I'm using blackfin 533 with the chip.
I'm using 16 bit mode with the flash
AI: Most probably you set chip into 8-bit mode (/BYTE is low), and try to access it using 16-bit data bus. |
H: Supercompilation for electronic design
I'm reading Advanced FPGA Design and the first example, one of simply raising a number to the 3rd power, quickly diverges into various quite clever optimizations to make the valuation faster.
Now, personally, I'd argue that such minutae is best left to the computer, so my question is this: is supercompilation (synthetically combining program elements until an optimal algorithm is found) used when appled to HDL?
TBH, I always assumed that that's exactly what synthesis is: taking HDL code and making sure it runs in optimal time. But the book hints at that not being the case, and altering code to shorten the evaluation path actually makes a difference.
AI: Is supercompilation (synthetically combining program elements until an optimal algorithm is found) used when appled to HDL?
Not in the sense of the sort of optimizations described in that book.
Generally speaking, an HDL synthesizer is free to make any optimizations which do not affect the externally visible behavior of the design. They can make changes which improve internal timing, such as refactoring logical expressions, changing internal signals, replicating registers, and even sometimes moving logic across internal registers, but they cannot make any changes which cause a change in externally visible behavior. In particular, any change which would cause a module to take more or fewer clocks to return a result, such as introducing or removing pipeline stages, is strictly forbidden.
In short: HDL synthesizers can make adjustments to a design to allow it to run at a higher clock frequency, but they cannot make changes which make it return a result in more or fewer clock cycles. |
H: interfacing buttons with python/c++
As a preface: I have 0 experience in interfacing with peripherals other than keyboard and mouse, but am willing to learn, just don't know where to start.
I would like to write a software in python (if C/C++ would be needed or is the easier way to go I can interface that) running on a PC, that can be manipulated by two external buttons (it will be an experimentel setup for small children, so we need big easily identifiable buttons, that are easy to push, so the keyboard/mouse is out of the question). The easiest would be to buy two USB buttons, that can be easily accessed, but I've been told these do not exist, and that I would have to build an interface myself (microcontrollers and stuff). If this is not true that would be great, if it is, I would like to invest my time into something that is not horribly hard and long to learn to use, and both the knowledge and equipment easily reusable if other issues like this come up. What would you recommend?
Thanks in advance:)
AI: People who told you they don't exist sorta lied.
What you will need:
The cheapest mouse you can find. Really, the cheapest.
Your Big Easy To Use Whatever You Need buttons. Grab an SPDT type.
Some fancy tripolar cable to connect them, grab the size and color you like
A soldering iron, and somebody able to use it
First of all, crack open your mouse. You will see something like this:
By Job at English Wikipedia - Transferred from en.wikipedia to Commons by Ansumang using CommonsHelper., GFDL, Link
See those tiny little guys on the right, with red little thingies on top? These are microswitches, something like this:
By Benjamin D. Esham / Wikimedia Commons, CC BY-SA 3.0 us, Link
without the metal thing on top. The red thingies correspond to the black piece of plastic on the top left of the switch case.
You now need to identify the terminals of the switch. Odds are that they are labeled as C or COM, NO and NC, as COMmon, Normally Opened, Normally Closed.
Now you need to start soldering. Your Buttons will have three similarly labeled terminals: you need to remove the switches and solder your cables where the switches used to be, then solder the cables to the terminals on your buttons. Really, it's easier said than done.
Finally, enjoy making your rig nice and all so that a young human fancies using it and is not endangered by anything.
And... Bingo. You're done.
Now, this is not really useful to you, in the sense that you do not learn much. But this solution has many pros: it is dirty cheap, easy for you to interface to, no worries about safety for the youngsters, great flexibility, you name it. |
H: Selection of DC Motor Drivers
Is there different Motor Drivers available for Permanent Magnet DC Motor and Wound Rotor DC Motor .Specifically for Series and Parallel wound DC Motor .
AI: Yes. A brushless Permanent magnet DC motor requires a controller that switches the polarity of the voltage applied to the windings similarly to supplying AC to the windings.
A DC motor with a commutator (and wound rotor) requires variable-voltage DC applied to the armature. The field is usually supplied by a separate fixed or variable-voltage DC supply rather than being series or parallel connected. The field can also be a permanent magnet with the same type of controller less the separate power supply for the field.
If you search PMDC motor control and armature voltage DC motor control you should be able to find detailed explanations of each type. There are a lot of details and variations to the designs, a lot of features, and a lot of aspects of the their of operation. There is much more than can be covered in this question and answer format. |
H: Schematic Numbering Scheme
I need help determining what are some helpful bits of information to include in a new schematic number scheme. For example:
Schematic ID
Revision (Major, Minor, Fix)
etc.
Its not critical, but I would like to define some good info to include in the numbering system before I commit to it for this new project.
Thanks
AI: I've consulted for several medical device companies. They have a common theme in revision numbering.
Revisions for released documents were letters: A then B then C etc.
Revisions for unreleased R&D versions were numbers: 01 then 02 then 03 etc.
Then eventually the package would get released and become A.
That way, it was easy to tell released documents from unreleased ones.
I was also asked to maintain a revision history for schematics. It was a table on the title sheet of the schematic. It had 4 columns: rev number, summary of changes, who, when.
[update in response to the comment by @DerStrom8]
Good point. I do the PRELIMINARY stamp too. The color is carefully chosen to be eye-catching, but not alarming. |
H: Use of diodes and resistors in a push-pull amplifier
What is the function of the two resistors (R4 and R5) and two diodes (D1 and D2) in the circuit below?
AI: The diodes keep the bases of the transistors 1.4V apart. This reduces crossover distortion. The resistors are to provide a bias current for the diodes.
So the diodes keep the transistors "close" to being on. If they weren't there, as the input voltage went between ~0.7 and ~-0.7V the transistors would be in cutoff and the output voltage would be zero. This causes the output to have distortion around the zero crossing. Keeping them biased on like this is called class AB operation.
Feedback can be useful to help minimize distortion, but class AB operation helps as well. Another technique often used is a Vbe multiplier instead of the two diodes. The Vbe multiplier allows more control over how much bias you are providing the output transistors. (Google it for more info.)
Often the diodes are mounted on the same heatsink as the transistors (or sometimes in the same package as the transistors) in order to track the change in forward voltage/Vbe with temperature. |
H: power for the load in push-pull amplifier
I wanted to ask:
what is the power needed for the load resistor?
Is there any problem? and what might be the solution for it?
I really appreciate you help, thank you!
AI: 2) The circuit in your other question is the improved version of the one in this question.
Let's take it as reference:
If you look at the circuit carefully, voltage gain is 1. But since the output impedance seen from the common-emitter point is about a few miliOhms, this circuit can act as a power amplifier.
1) The power needed for the load depends on the input, of course. But be careful, VCC and VEE are limiting values for the output signal before clipping. Applying maximum 8Vpp = 4Vp = 2.83Vrms as input can be a good choice.
Suppose we apply 8Vpp = 2.83Vrms as input. So the output signal at the common-emitter point will be Vin = 8Vpp = 2.83Vrms. Since R9 and RL form a voltage divider, the voltage across the load will be 2.83 x 100 / 112 = ~2.5Vrms. So the power dissipated by the load will be 2.5²/100 = ~63mW. |
H: Capacitor charging circuit review needed
I recently posted a thread about using a coin cell to power an electronics project that features an SD card (here). The problem at hand is that the SD card will pull high current during read/write operations, much higher than a coin cell can handle. However, the fact that the SD card pulls very little current outside of read/write operations leads me to believe that I can use a capacitor charge/discharge circuit for supplying higher power to SD cards when it is needed. I'm hoping I can get some verification regarding my design.
After doing some rough hand calculations, I've calculated that I will need to write to my SD card every 2 seconds. The amount of data that will need to be written is 512 bytes. Referencing a previous post I made, the user jonk provided this picture:
I will be using an 8Mhz clock, along with a 3.3V power supply, thus I calculated an SD card write speed of 4 Mb/s. Using this clock rate on my 512 bytes, that comes out to a write time of 0.000128 seconds. Because I assume there is overhead involved, I rounded this up to a 'hopefully generous' 1ms.
So my calculations suggest that I can charge the capacitor for up to 2 seconds, and then can discharge it at a rate such that an appropriate voltage is sustained for at least 1ms.
Referring to my circuit design, I am using a 3.3V coin cell (CR2032) whose voltage is regulated by a 120mA switched cap regulator. This regulator acts as the main power source for other components that can consume 6mA current continuous. Note that I am open to selecting a different coin cell that can provide a higher continuous current. Also note that I realize that the 6mA by itself is higher than the rated continuous current of a CR2032 coin cell, however, I am willing to lose cell capacity for higher current. While I'm ok with 6mA, I am not okay with an additional 30mA pulsating SD load.
I've selected a 3.3K resistor to limit the capacitor charging current to 1mA max, and I've select a 180uF capacitor given that it seems to charge to 3.3V in the allotted 2 seconds.
I've estimated that the SD card will pull 30mA during write operations, the 180uF capacitor appears to be able to hold enough charge to sustain a 30mA current for 1ms. I did a quick simulation in SPICE and after 1ms of capacitor discharge, the voltage dropped to 2.9V which is still acceptable for powering an SD card. I modeled my SD card as a 100 ohm resistor that would switch on every 2 seconds for a duration of 1ms.
So my questions:
How much overhead is involved when writing to an SD card (FAT32 file system)? Would my estimate of a 1ms write time cover all the overhead required to write to a text file on an SD card?
For my application, is it fair to model an SD card as a resistor that switches on periodically?
Does it appear that I may be overlooking anything?
AI: I think you've jumped far too quickly into the design stage of the electronics portion that you now do understand better and have given short shrift (by waving hands in the air) to studying the SD card specification. You should shift gears and set aside the easier part (which is the electronics that simply flows out of the information you still need to collect) and delve more deeply into the SD card portion of this.
Study the specification. Look, for example, at 4.6.2.2. What times do you see there? Have you read 4.4 and looked at going into energy saving modes for the SD card? If you are doing a write, will you also need to do a read prior to that? (Are you partially modifying a block?)
If you are planning on using the Arduino library without understanding the code that was written for it and how it relates to the SD card specification, first hand, then you need a way to make measurements with the library so that you can observe what it actually does do and what the power requirements look like.
In the best of circumstances, you should at least read the SD card specification for the relevant parts of what you are trying to do. Not necessarily because you are going to write code -- it seems you want to use the library code -- but because you need to understand the specification itself to know what your library must contend with. This allows you to anticipate circumstances that 1000 measurements of actual tests won't tell you about. You can test all day long with the library and never trip up on some detail. So you'll just go and assume, figuring that since you haven't seen a problem, there won't be a problem later on.
This is the point of reading the SD card specification, as a matter of independently informing yourself and not just blindly using a library. Sure, the author probably read the specification and does know these things. But when they create the library, they don't document everything they've learned. They just write the library. You have the job, separately, of knowing the specification and the reach of it in cases where you can't actually test. (It's impossible for you to go get every SD card ever manufactured. But if you read the specification, you can make sure that you cover all the conforming cards and situations, at least.)
There's no escaping your education here. You can focus on what you care about: Write, Power Saving Mode, Read (perhaps), and anything else you can think up that you might do (or you think the library actually will do.) But to design the electronics which is responsible for maintaining sufficient power, you need to know just how long the worst case might be. And you need to know that, firmly, in your own head and to to exactly why you know what you know. You could be able to rattle it off almost as fast as you can talk.
Once you have a full handle on that part, and assuming your library does its job as it should, then you can make some estimates. But really, the electronics part of this is almost trivial. Falls out so easy. The difficulty is all the rest here and making sure you know what the worst case might be. With that, the electronics design just falls out.
For example, from Section 4.6.2.2:
For a Standard Capacity SD Memory Card, the times after which a
timeout condition for write operations occurs are (card independent)
either 100 times longer than the typical program times for these
operations given below or 250 ms (the lower of the two).
and,
While the card should try to maintain that busy indication of write
operation does not exceed 250ms in the case of SDXC card, if the card
is not possible to maintain operations with 250ms busy, the card can
indicate write busy up to 500ms including single and multiple block
write in the following scenarios:
a) The last busy in any write operation up to 500ms including single and multiple block write.
b) When multiple block write is stopped by CMD12, the busy from the response of CMD12 is up to 500ms.
c) When multiple block write is stopped by CMD23, the busy after the last data block is up to 500ms.
d) Busy indication at block gap in multiple block write is up to 250ms except a following case. When the card executes consecutive two
blocks write (2*512Bytes) and it spans across the physical block
boundary, the busy after the each block can be indicated up to 500ms.
Especially regardless of the above definition, a speed class writing
mode specified by CMD20 shall keep write busy up to 250ms in any case
until the end of speed class write is indicated.
Or this, as one of their final notes:
It is strongly recommended for hosts to implement more than 500ms
timeout value even if the card indicates the 250ms maximum busy
length.
What do you make of all the above, for example?
Here's another interesting note (to me) that seems to relate a corner case to your desire for 512 byte writes. From Section 4.4:
For example, in the case that a host with 512 Bytes of data buffer
would like to transfer data to a card with 1 KByte write blocks. So,
to preserve a continuous data transfer, from the card's point of view,
the clock to the card shall be stopped after the first 512 Bytes. Then
the host will fill its internal buffer with another 512 Bytes. After
the second half of the write block is ready in the host, it will
continue the data transfer to the card by re-starting the clock
supply. In such a way, the card does not recognize any interruptions
in the data transfer.
From this, I have learned that not all SD cards use 512 byte blocks. That's interesting. It makes me wonder if the Arduino library supports this mode of lower power operation. And it makes me wonder if you should perhaps consider writing your own code, as this could actually be helpful when performing writes and wanting to lower overall power consumption.
What do you think?
Just searching the web for others who have wanted to keep times short when writing to an SD card, I did find the following comment you might want to consider -- not as fact, but as motivation for you to work harder in understanding what's ahead of you.
The comment was (found here):
Reading the specifications, it should "typically" take 0.5 ms to write
a block, with a worst case of 240 ms. What I find is a rather steady
100 ms. If I can't reduce this, it would burn a lot of power and
require a re-design for bigger batteries.
The writer of this, above, did appear to have read the specification, as they noted the 240 ms worst case (which is, as we can now read, actually 250 ms.) So that adds some credibility to the question and to the note about what they are actually finding, in practice, as well.
It's one data point, of course. But it should provide some motivation for you to really knuckle down and understand what the SD card specification allows. |
H: Issues with using higher capacitance than required
http://www.digikey.ca/product-detail/en/stmicroelectronics/LD1117V33/497-1491-5-ND/586012
I am planning on using this regulator for a home project. I noticed on the datasheet that a minimum 10uF capacitance can be used. On the design diagram, they also have a 10uF listed.
What is the downfall of using a 22uF capacitor instead of the 10uF recommended capacitor? Is there a downfall at all? It is just a minimum recommending but I can go higher if I wanted to?
Thank you!
AI: Generally, none. As long as you're above the minimum capacitance for stability, and the capacitor is relatively close to the regulator output (within a few cm), then you'll be alright. If anything, adding more capacitance will generally reduce the output noise.
The issue with "too much capacitance" comes in when you start getting into extreme capacitance values, in the range of Farads (not tens of microFarads like the cap you're talking about, but 1F). In that case, those capacitors take a lot of current for a long time to charge, which can cause a substantial amount of losses across your regulator, if it doesn't mistake the current draw as a short-circuit and shut itself off first. |
H: How to Add SRA to a MIPS One Cycle CPU
I'm trying to add the MIPS command SRA to the following one cycle CPU:
If I knew how many bits needed to be shifted I could easily replace the lower bits with 0's, but that's never a known till the instruction is being executed. What would be a way to implement this so it doesn't matter how far the bits need to be shifted?
AI: Use a single cycle rotator + masking (for shift operation). The rotator is like a coupled 32:32 mux (32 parallel 32:1 muxes with the same select signals and inputs, except the inputs are shifted by 1-bit for each subsequent 32:1 mux). The masking unit is a 32-bit thermometer decoder.
If you want this to work in a single cycle you may need to spend some time at the rotator (since it depends on both data and selects). If you will not ever need the rotation (as opposed to just shifting), you can save a lot of time on the critical path by hard wiring the rotator's second (and any subsequent stages) stage LSB (which in a rotator scheme would be the MSB of the previous stage) to either '0' or '1'; whichever you wish to shift in at the LSB.
The critical path will almost certainly be through the selects to the rotator (since it must both reach the first stage for the data as well as fan out to 32 gates).
The masking unit (aka 32 bit thermometer decoder) can be implemented requiring only the select signals and can be as simple as an AND with each bit (before or after the rotator; the ideal timing depends on the arrival times of the selects vs. the data signals).
For example, see the barrel shifter which relies completely on combinational logic (and thus able to be completed in a single cycle, assuming the logic is fast enough). |
H: DC motor control ULN2803
I'm trying to do some basic DC motor (pump) control through an Adafruit Feather Huzzah (ESP8266) and an ULN2803 Darlington transistor array; I've wired it up as follows:
And my code goes a little something like this:
void setup() {
Serial.begin(9600);
pinMode(14, OUTPUT);
analogWrite(14, 0);
}
void loop() {
if (Serial.available())
{
int speed = Serial.parseInt();
if (speed >= 0 && speed <= 1023)
{
analogWrite(14, speed);
Serial.print("PWM set to ");
Serial.println(speed);
}
}
}
Unfortunately, unless analogWrite() is set to 1023, the motor only humms and does not rotate.
What I've done to troubleshoot:
I've used a (cheap) voltmeter to validate that the voltage going to
the motor is variable as I adjust the analogWrite() value.
Also, I've replaced the motor with a LED strip and it fades properly as I adjust the analogWrite() value.
Finally, I've hooked the motor up to a desktop power supply, and was
able to validate that as I reduce the voltage from 12V to 0V, the
motor slows down.
What am I missing?
AI: @WesleyLee got me on the right track; it looks like the starting voltage has to be high enough to get the motor spinning (at least initially).
The sketch below will get the motor running at a very low voltage:
void setup() {
Serial.begin(9600);
pinMode(14, OUTPUT);
}
void loop() {
analogWrite(14, 1023);
delay(15);
analogWrite(14, 200);
delay(3000);
}
It's not a beautiful solution but I suppose it works for my application. That's what you get for buying cheap pumps I guess. |
H: Measuring small capacitor parallel to big inductance (Network Analyser)
So what I have at hand is a cheap network analyser that can present the measured impedance in form of the values of different equivalent circuit topologies, e.g. R-L series or RL parallel.
What I want to do is measure the electrical parameters of a simple PCB antenna structure, which is typically modeled as as capacitor (mainly inter-winding capacitance) parallel to ( R + L).
So I did the "SOL" calibration and measured S11. Concerning R and L, I chose the available Network Analyser Representation of the equivalent circuit called "RL" (I have no C//(R+L) representation available)
This measurement shows the values for both, R-L series and R-L parallel equivalent circuits and I get reasonable values so no problem here.
But how can I determine the value of the capacitor? Is it possible at all? I thought of writing down the formula for the real and imaginary parts of the impedance of the topology I want to model (i.e. C//(R+L)) and solve it using the measured values of S11.
Besides of being inconvenient I think I would get a wrong result since any physically capacitance present in the structure is already indirectly included in the values i get for R-L (by having a slightly smaller L than without cap), right?
(Of course the inclusion of the capacitor in the presented R-L values is possible only for a specific range of frequencies because above some frequency the capacitor will dominate and "get visible" in the measurements, but that does not help either...)
AI: Set up your network analyzer to plot the response as magnitude of impedance (\$\left|Z_L\right|\$), with log scale on the y axis.
Assuming 1. your network analyzer can reach a high enough frequency, and 2. the connection from the reference plane to the termination you're trying to measure is short, you should see \$|Z_L|\$ increasing from low frequencies up to some peak and then decreasing after the peak.
The equation for the part of the curve where it's increasing is approximately \$Z_L(f)=j2\pi{}fL\$. The equation for the part of the curve where it's decreasing is approximately \$Z_L(f)=\frac{1}{j2\pi{}fC}\$. By picking out a couple of points on the appropriate parts of the curve, you should easily be able to estimate both \$L\$ and \$C\$.
If you already know \$L\$ you can get \$C\$ really easily knowing the resonance frequency (the frequency where the graph reaches its peak) is \$f_R=\frac{1}{2\pi\sqrt{LC}}\$. |
H: Laser for Engraving & Cutting
Can I use 500mw 808nm Infrared IR Laser Diode for engraving and cutting of wood, plastic and metal?ASER
https://www.banggood.com/Focusable-500mw-808nm-Infrared-IR-Laser-Diode-Dot-Module-12V-TTL-Fan-Cooling-p-1227542.html?rmmds=search&cur_warehouse=CN
AI: No. 500mW at IR wavelengths cannot cut Metal. Reflection issue aside, it cannot produce the heat needed in the time needed to cut through it. The metal will absorb the heat and dissipate it away too quickly. At best, if you angle it, and paint it black or use tape, you may, may, lightly etch it.
That laser may cut balsa wood, very very not dense material, with multiple passes.
Plastic depends on the plastic type, how deep, etc. Keep in mind that some plastic has chlorine in it, like PVC, and it will damage the lens.
In short, a 500 mW laser is best used to engrave, and can only really cut paper. I struggled to cut poster board, and most cardboard, even the 1/16 inch stuff was too much to cut. I myself own one of these lasers. |
H: Input of boost converter or charge pump in the energy harvest
I read some paper about the energy harvest, and lots of paper said the induced voltage from, for example, the thermo-electric harvesting could be as low as tens of millivolts in a realistic case, so their input of boost or charge pump is less than 100 mV.
But in some paper, I found the author will set the input from 30 mV to 300 mV, some paper even set the input to 3V. Why? If the induced voltage is only tens of millivolts, then 100 mV should be unless, because you cannot have this high voltage, higher than 100 mV, to feed the boost or charge pump, then why do some authors still let the voltage be so high? Maybe induced voltage can be higher than 100 mV in some other kind of harvest. If so,can anyone give me an example?
Paper example 1(input 30mV~300mV):http://ieeexplore.ieee.org/document/6419064/
Paper example 2 (input 450mV~3V):http://ieeexplore.ieee.org/document/8234200/
AI: The voltage range for energy harvesting output is as diverse as there are different forms of energy harvesting. Each harvesting type will have it's own energy storage method, boot method and boost method.
You could be talking about RF energy harvesting, thermal difference harvesting, kinetic energy harvesting or solar harvesting. To each it's own electronics. Study each in detail and you'll understand better. |
H: Daisy chaining a large number (20) of TLC5940 in serial. Bolstering signals?
Working with shift registers (tlc5940 and Library) on Sparkfun breakout boards driving individual LEDs.
The boards are designed to connect VPRG, GSCLOCK, BLANK, XLATCH, SIN/OUT, SCLK,VCC, GND in serial. Everything works perfectly on Board #1-10. However, I need quite a few more boards/chips (up to 28) and the signals get crazy from Board 11 onwards.
I was hoping to find some options to increase the signal strengths to and past board #11, all while keeping the timing correct.. Any help with possible solutions/specific schematics is GREATLY appreciated.
The TLC breakout boards are spaced every three inches. LEDs, chips and Arduino all are powered from a 5V regulated, 40amp source (MeanWell). Power is getting to last board and it's Leds. The connections/soldering are solid and not shorting. Each led and board has been tested individually.
I understand that this is a tall order, impedance, connections, conductance, the board, etc are in play here but I very much need to keep the boards in this lengthy configuration (necessary for art installation). Even if this boost doesn’t get all 28 chips working, I’d like to understand more how the various clock, latch, data signals can work through an extended serial bus and what I can do to clean up, bolster and push them further.
Chip Datasheet: https://www.sparkfun.com/datasheets/Components/General/tlc5940.pdf
Spark fun Board: https://www.sparkfun.com/products/10616
EDIT: ROUND TWO
Hoping the break from this project doesn't bury it, but I've got a few more questions based on trying to use the above suggestions.
I've redesigned the art installation so that I could get the chip/boards centralized and next to each other/Arduino rather than separated out to the 4 grids of LEDs. I am going to run a much, much thicker solid straight line for connections, ground and power (which I will increase to 6v). Main question I have is about the difference between a Buffer Amp and the clock driver. Since the TLC5940 boards are designed to run in serial, and since I need so many (approx30), It's been suggested that I run a 74HC7014 Non-invert amp every three-ish boards. I can do that, but should I drive all of the signals from the same Buffer? Data, VPRG, Blank, and both clock signals (GSclock, Sclk)?
Or should I push both clock signals in parallel from a "robust" clock driver (any suggestions as to which one?) and the rest of the signals in serial through the boards and Buffers?
I was also wondering how and if I need to do anything to terminate any of the signals/ground? Would it help keep everything working as supposed? Right now, all signals/power and ground are just in a long line, from the Arduino straight through all boards and drivers (It does bend in half, though, to keep it compact) Is there anything that should be done after the last board? Or can they just end?
Could use a little more suggestion on the schematic before I solder this up and try it out. @oldfart ?
AI: First:
I noticed that each board has a 5V regulator for the LED driver. You state that your power supply is 5V too. This will mean that the regulator does nothing but cause a light voltage drop. The TLC5940 will work on a voltage slightly UNDER 5V. It has 3..5V operating range so that is no problem for the chip.
If your LEDs can stand it use a higher VCC voltage. It would be better if you e.g. use 6V and then you can loose 1 volt over the wire and the regulator.
Possible solution:
As it seems to work on 11 boards I would suggest you make a small 'signal repeater' board. Just a set of buffers which take the input signals and send them out again. Make e.g. five or six of those boards and insert them at regular intervals. (Do not think: it worked with 11 so 28/11 = ~3 boards. You will need a safety margin). I had a quick look and the 74HC7014 looks good: six non-inverting buffer with Schmidt trigger input. Probably good up to 1MHz at 5V. Maybe you can find the 14-DIP version which is easy to solder on a breadboard.
Last:
Next time you embark on an electronic project this size talk to an experienced electronics engineer. Preferable an old fart like me. We know the pitfalls for constructions like this.
Your connection diagram looks better except for one detail: You are feeding the HCT clock driver boards from the VCC which is 6V. But your expansion boards have a regulator which make them 5V. Try to find a 5V signal on one of the the expansion board adjacent to the clock driver and connect that to the HCT board instead of VCC. The alternative would have been to add a 5V regulator to each HCT buffer board.
You should buffer all the signals. That way they all get more or less the same delay and thus as a group there is little change. In this case it works as all signals go in the same direction. There is no 'return' channel.
You do not need to terminate the power/ground. As to terminating the serial signals. It is not bad idea but as the system worked before with ~10-ish boards I would first try without. You can always add them.
It would be nice to have a scope picture of the signals at the begin and end. |
H: Using a capacitor instead of a coil for wireless charging?
This might sound like a stupid question, but I'm kinda interested in a detailed explanation.
So while inductive charging is rather ubiquitous, things like electric toothbrushes and Qi chargers come to mind, as well as to a lesser extent, wireless charging pads for cars, the are all inductive charging systems, basically forming half of a transformer each (the charger is one half, the charged device the other half).
But would it be technically possible to make such a charging setup, using a large capacitor?
I.e. having an AC source connected to a large plate, forming one half of a large capacitor. The charged device would then form the other half of the capacitor.
I understand the distances would have to be rather small, but in cases like Qi chargers, the're right on top of each other anyway. Also, the dielectric properties of the materials between the two conductive plates, would obviously play a role.
But from a purely engineering perspective, why is it a bad idea? One problem I can come up with, is using very high frequencies. The charger side of the capacitor would essentially be an antenna, with the sizes of a smart phone, that would mean frequencies in the 500MHz-3GHz range. I assume it'd be simply too complicated to use a signal generator at these frequencies to send sizeable amounts of power?
AI: First point: A transformer has a built-in return path, so you have a complete circuit. That is to say, it has two wires on each side, so current can flow out through one and back through the other. To make an equivalent device with capacitors, it would be need to have two of them, one for out and one for return.
Next lets put some rough numbers on. For a phone, you have an area of about 5000mm^2 for each capacitor. If you put the capacitor plates near the surface, then you might get a gap of 1mm and the plastic case could be engineered with a dielectric constant of say 10. That works out about 0.4nF per capacitor.
Now lets think about transferring power through that. At 1kHz the impedance would be 400Mohm. At 1MHz, it would be 400kohm, and 1GHz it would be 400ohm. That sort of high frequency causes lots of headaches around avoiding interference with other devices. And at any real power, not microwaving the user. And even if you go all the way up to 1GHz, then you still have to deal with a relatively large impedance. To get a watt flowing through the capacitors will need 30V.
So the engineering problems are probably not insurmountable, but it is still much easier (and therefore cheaper) to use an inductive method. |
H: Difference between push-pull and totem-pole
I can't understand/see the difference between a push-pull and a totem-pole output.
Push-pull:
Totem-pole:
Do these circuits have a transistor in the input for something like "double amplification?"
AI: You'll notice in the push pull stage that you need a PNP and an NPN transistor, whereas the totem pole driver uses only NPN transistors. This is useful because NPN type transistors are usually easier to make, and support higher current for a given size than PNP type transistors.
To address your question on "Double amplification", a push pull driver doesn't necessarily give more amplification, but is used because it is more efficient than a single transistor amplifier because (ideally) only one transistor is on at a time, and all the current goes through the load. |
H: How to improve design of naive MW AM transmitter for lab demo
I'm trying to prepare small demo for school lecture on radio history.
sorry for confusion, MW here means medium wave, not megawatt :)
And I want to create small and very simple transmitter which could be "heard" with old factory built receiver. There are two points in this demo:
to show that AM is really simple kind of modulation and explain construction as best as I could (my pupils have some basic understanding of physics and electronics);
to show that old receivers are not "broken" (as I was asked several times), but rather that there are almost no stations broadcasting.
I'm working from the idea expressed in the picture below (it refers to "Electronics World, August 2004" - but I couldn't find original) - simply to have logic generator on desired frequency (here 555 timer) driving antenna with its push-pull output:
I'm not sure that author of this design really succeeded as he explains in making it work for distances up to 25 meters. It's said "antenna is just 6-10 feet of wire".
I've used spare Arduino to drive pin with 1.5 MHz (most 555s I've seen are not working well at the frequencies above 100-200 kHz) and apply some trivial modulation. Attaching 2-3 feet telescopic antenna for test I've found that the signal is "heard" with portable radio in a wide range of frequencies (say, from 1.8 to 0.8 MHz - I believe because output is not sine wave) and only in very close range.
I've googled a bit and found the "rule of thumb" that if your antenna is short (compared to wavelength) it should be loaded with series inductance (and with capacitance if long). I've tried several which I had at hand and found the best result is with 22uHn. Now it is heard only when receiver is precisely tuned and at about 1-1.5 meters.
But I'm curious to improve it to range of 10 meters (at least it would be good to have range larger than antenna).
So I have questions:
what can be done (besides trying longer antenna) to improve range (if this is possible at all - I'm not sure I've ever seen working MW transmitter of small scale);
is there any guide to load antenna with inductance (choosing proper value etc.)
can some different type of output for driving antenna make it work better (e.g. transistor with inductance in collector?) - are there any general recommendations
may I benefit from trying loop antenna? As I understood from wikipedia, small antennas for medium/long waves are not good for transmitters...
Thanks in advance!
AI: I think you are using the 555 as an FM modulator. This describes the function of pin 5 (the pin you are applying your signal into): -
Control Voltage, This pin controls the timing of the 555 by overriding
the 2/3Vcc level of the voltage divider network. By applying a voltage
to this pin the width of the output signal can be varied independently
of the RC timing network. When not used it is connected to ground via
a 10nF capacitor to eliminate any noise.
Hence what you get is FM and not AM but, with virtually any form of modulation on pin 5 you will get something detectable on an AM radio tuned to one of the harmonics. However, if you want to demonstrate something discernible and believable to your pupils, then use AM for sure. You can make a simple AM modulator from an oscillator, a diode and an LC filter and it can be made to work very well. Applying an amplifier at the output can boost signals considerably but may be illegal in your area of the world.
what can be done (besides trying longer antenna) to improve range (if
this is possible at all - I'm not sure I've ever seen working MW
transmitter of small scale)
A single monopole antenna needs a decent ground for it to work so that is one point.
is there any guide to load antenna with inductance (choosing proper
value etc.)
For a short monopole (i.e. electrically short compared to a quarter wavelength), then the antenna looks "capacitive" so please look at this graph and use the same reactance inductor in series: -
can some different type of output for driving antenna make it work
better (e.g. transistor with inductance in collector?)
There is no great substitute for making a "short" monopole longer and using an earth plane. Using an inductor to cancel the capacitance gets a third of the job done - the radiating resistive impedance that remains after the cancellation is still very low (just a few ohms for a short antenna) so another improvement is to use a step-down transformer to project the low radiation resistance of 1 or 2 ohms up to something manageable with low power semiconductor outputs.
may I benefit from trying loop antenna?
At circa 1 MHz the loop is virtually purely inductive hence it generates a magnetic field and very little electric field. Given that the impedance of free space is 377 ohms, it is preferable to generate an electric field strength to magnetic field strength ratio of 377:1.
Small loop antennas at this frequency therefore only produce a magnetic field and this decays with distance squared (rapidly becoming a distance-cubed at about one diameter from the loop).
Compare this with a proper EM wave - both fields sustain each other and the decay in strength is proportional only to distance. The is the real beauty of an EM wave.
So, going to a magnetic loop is not going to give a benefit. They work greate at picking up the magnetic content of an EM transmission but, when they are small relative to wavelength they are crappy as transmitters. |
H: Creating Light Sensor
So the situation is like this, I have a project with a budget of $25. I have to build a robot that is capable of connecting a magnet to a tin can autonomously. I also have a vex kit which I can use for free. I need to create a system so that the robot can find the tin can and drive to connect the magnet from up to 3m away with no obstacles between the robot and the tin can. We are provided with an IR beacon that flashes as 10Hz but previous people who have done this experiment have told me that the IR light is almost impossible to detect without aimlessly driving around the arena for a while. We are allowed to use a different detection method if we would prefer.
I want to use visible light and a phototransistor to detect when I am looking at the tin can. I would wrap the tin can with retroreflective tape and have a LED ring light on the robot. The LED light would bounce back and go through a magnifying glass to focus the light on the phototransistor. However, the Vex kit only provides 5V out and the LED light has 15 LEDs which I think makes it require 10.5V to function at full power. Is there a cheap way to power the light without a car battery or is there a way I can cut down on other costs, like removing the need for the tape/magnifying glass?
AI: A large difficulty to be overcome is the variability of background illumination.
Another problem is one of dynamic range of detected light intensity.
Variable background: As the robot points to different locations, it detects more light (perhaps when pointing toward a window) or less light (perhaps while pointing to a dark corner). Detected light from a beacon or from a retro-reflector may be only a little greater than this background illumination. What you want to detect is the difference between background and background + light source.
Modulation of the light source is a method of overcoming the variable-background problem. The light source is rapidly turned on and off, perhaps 1000 times-per-second. A reasonably fast light detector (a photo transistor) is used to detect the changes from on-to-off. The alternating amplitude of its photo current can be passed through a band-pass filter centered at 1000 Hz to yield an AC signal whose amplitude is proportional to detected light from only the light source, not from the variable background.
simulate this circuit – Schematic created using CircuitLab
Dynamic range: Detected light amplitude may vary greatly with distance. At ten feet distance, detected amplitude will be small, even when optimally pointed at the source. At a few inches distance, detected amplitude will be a great deal larger. A photodetector must cope with this large signal range, yet still be able to measure small changes in amplitude. A variable-gain amplifier can help: high gain at large distance, low gain at short distance. |
H: Getting a Constant Value From ADC
I run a 10bit ADC from which I get a number which varies in-between +/-5.
Ex:- if ADC value is 512 then it may vary from any number in between 517 and 507.
I do not want to use any filter mechanism as it is computationally intensive.
I just want to get a pick a fixed number from last 10 measurements which has more frequency.
Please help me in writing a function for above.
Microcontroller- infineonTLE
IDE- Keil(c)
AI: This is a very simple 1-stage IIR filter:
float state = 0;
float filter(float sample)
{
state = (1 * sample + 0.2 * state)/1.2;
return state;
}
You can tweak the response by changing the coefficients before sample and state. Don't forget to adjust the coefficient sum, too.
For more knobs to tweak, you can remember the previous, pre-previous etc. states, too.
float states[3] = { 0, 0, 0 };
float filter(float sample)
{
float latest;
latest = (1 * sample + 0.2 * states[0] + 0.1 * states[1] + 0.05 * states[2])/1.35;
states[2] = states[1];
states[1] = states[0];
states[0] = latest;
return latest;
} |
H: Why 2 zero-crossing detection circuits in heater?
I reverse engineered the power board of an electric heater. I think, I understand most aspects of this circuit. One thing is not entirely clear to me though: The circuit seems to have two distinct ways to detect the zero-crossing point which is needed to control the Triac:
Via the opto-isolator (U1) that is directly driven by the primary
Via the voltage divider R7/R8 that taps the secondary of the transformer
Both signals feed directly into a microcontroller (PIC16F886). Both pins (RA5/RB0) on the microcontroller can be configured to be inputs to the built-in ADC.
What could be the reason for having those 2 separate circuits?
AI: What could be the reason for having those 2 separate circuits?
The opto U1 produces a square wave output in phase with the incoming AC and is probably used as a zero crossing detector by the MCU. If the AC level drops it will still produce a square wave i.e. it is a digital detector of the AC power voltage and will hardly vary its signal even if the incoming supply dropped by a half.
D6, R7 and R8 in combination with the unspecified filter capacitor C4 measure the peak voltage of the secondary and that is probably used by the MCU to determine what the line AC level is. Knowing the line level you could, for instance, alter the triac duty cycle so that on low AC voltages it stayed conducting for longer.
Short story is: heater power level control irrespective of AC supply voltages. |
H: Adjustable brightness for IR remote signals
I'm using and Arduino, an IR Led, and a transistor to for sending RC5 signals (36khz).
For the next step I need to be able to adjust the brightness of the signal, basically changing the range. The value could be set by an analog signal, spi or i2c or anything basic.
I thought about using a digital potentiometer, but they are not taking enough current for the LED - should be up to about 250mA (for a short time).
Searched the web but really not sure how to approach this. How do I create a circuit to adjust brightness for IR remote signals?
AI: The carrier pulses contain harmonics and the RC5 spec is rated at 25%~33%. Since the receiver has a BPF and AGC, attenuation of the fundamental will have a direct impact on path length. The question asks how to do this by analog methods, but in reality, any method that controls the fundamental Tx amplitude is identical including reducing the the PWM duty cycle(d.c.) of carrier pulses.
The Q of Rx BPF determines the bandwidth f/Q and approx. limitation for smallest d.c. for steady carrier.
Using fourier spectral analysis, I measured duty cycle ratio vs fundamental , f1 below, using a peak pulse at 0dB;
d.c. (~) f1 [dB]
1/2 . . . +2.1 ( Note: f1 pk is greater than squarewave pk)
1/3 . . . +0.9
1/4 . . . -0.9
1/8 . . . -6.4
1/12. . . -9.5
1/20. . .-13.7
1/32. . .-17.4
1/80. . .-26.0
These pulses will be filtered by the Rx BPF and added to the baseline noise to give a SNR and path loss for threshold of errors est. 15~20dB which in turn controls range.
Conclusion
Define range reduction, get Friss loss then choose duty cycle(d.c.) of carrier pulses for that loss on f1 from above table.
no need to add any circuit. |
H: Lithium Ion Battery Pack - Looking to calculate watt-hours from V1 to V2
I've got a 2S4P Lithium Ion 18650 pack (that's 2 sets of 4 cells in parallel where each set is in series).
These batteries can be changed to typically around 4.1 or 4.2v and can technically be discharged low (2.5v), however the wider this range, the less life you can our of the cells. I plan on charging my cells to 4.2v and providing a voltage cutoff at 3.0v.
What I'd like to understand is how I get capacity (in watt-hours I believe?) out of the voltage range I've selected.
Let's assume each cell has a rated life of 2000 mAh.
AI: Alrighty, so there will be the answer that we can extract from the information on the datasheet/discharge curve chart. That will give you a fairly good idea of what to expect. Or if you want more that than you can run tests on your specific cells to find out. But I think you are just needing a more of a ballpark answer. Ballpark is the only thing that is really of any value, unless you know exactly what temperature the cells will be at all the time as well as some other variables.
So, li-ion cells have a non-linear discharge curve. If we look up the discharge curve for your particular cells it will look something like this one.
Now you can understand why you would need to know the current draw if you wanted more than a ballpark guess; the capacity in W-H changes with load because different loads cause different voltage sags.
So anyways, lets assume we are talking about 2A draw only. Look at the 2A curve (the black line with a negative slope)
You are saying you wanted to set the cutoff at 3v. Well since we are ballparking, we can notice that the discharge curve for the region you are interested in is close enough to linear that you could just easily calculate the area under the curve without an integral.
If you aren't sure if you need the capacity in WH or just AH well ah is easily found by just finding the capacity that relates to the cutoff voltage you want in the discharge curve.
If you do want WH though:
If we say the voltage range is 4.1 to 3.1, and the capacity is 3.2ah well then you get a rectangle with a triangle on top of it that you can find the area of and know your Watt Hours. WH = (3.2*3.1) + 1/2(3.2* (4.1-3.1)) |
H: Implementing a Mealy Machine in Verilog
I have a piece of code that I don't understand:
always_ff @(posedge CLK)
state <= RST || ~A[0]===1'bx || ~A[1]===1'bx ? 0 : nextstate;
A is the input. What is this piece of code exactly doing? What is ~A[0]===1'bx or ~A[1]===1'bx?
My guess is that this code is checking if either one of the inputs aren't valid (1'bx). If one of the inputs aren't valid, we jump back to the starting state 0.
But why the negation in ~A[0]===1'bx?
AI: I assume this is a shortcut for:
A[0]===1'bx || A[0]===1'bz
If A[0] is 1'bx, ~A[0] will still be 1'bx.
If A[0] is 1'bz, ~A[0] will make it 1'bx.
I have to assume this is in a test bench because that code can not be synthesized.
A more meaningful way to write the code is:
state <= RST || $isunknown(A[1:0]) ? 0 : nextstate; |
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