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H: PCB electroplating then etching
How can I protect the through holes copper which is electroplated from being etched away by the ferric chloride? Is there a resistive solution or it`s just the conductive ink.
AI: The same photoresist that keep the copper from being etched away should generally keep the hole plating from being etched away. If not, they can be painted explicitly. |
H: How to calculate right current to charge 2 cell series 8.4V 4000mAh battery lipo?
I have 4000mAh 2 cell series 8.4V lipo battery
I need make a custom charger I am thinking of using LT3650-8.4 2A charging Ic
if I set 2A
how long will it take to charge this fully /if timer is disabled timer?
is it (1st cell 4000mAh + 2nd cell 4000mAh) / 2000mA = 4h ?
if I set 1A
how long will it take to charge this fully /if timer is disabled timer?
is it (1st cell 4000mAh + 2nd cell 4000mAh) / 1000mA = 8h ?
What is the maximum capacity of 2 cell battery this IC can charge properly? without a timer?
as I know we have to manage the full charge cycle (5.6-8.4) within 4h at least, so to charge it within 4h with the set current to 2A as in the figure, how long will it take to fully charge a 4000mAH 2 cell battery?
I think everybody has misunderstood the question I am asking. now everybody should be able to understand the question
AI: Summary:
Charging 2 x 4Ah cells in series with 2A current will charge them to ABOUT
time = Ah/Charge_rate x 90% = 4/2 x 0.90 = 1.8 hours
At that stage they will be at 8.4V and ABOUT 90% of capacity.
Beyond that, if you wish to continue charging you must hold voltage at 4.2V/cell maximum and allow current to decrease under control of battery chemistry.
Reduction of current to about C/5 will take around an additional 2 hours. Stopping at V=4.2V/cell without CC taper charging greatly increases battery life in return for a small reduction in per cycle capacity.
__________________________________
LOOK at the LT3650 data sheet that Andrew cited.
READ what it says.
The graph on the 1st page shows the theoretical and practical result of CCCV LiIon charging. You are dangerously short on theoretical knowledge to build your own charger.
If charging current flows in series through multiple identical cells it will charge all of them at the same rate and in the same time. The voltage needed will increase with the cell count. You can only charge a cell at CC (constant current) until it reaches Vmax (typically 4.2V/celll).
Then you MUST either cease charging or hold Vcell constant at Vmax and allow the current to "taper"(reduce) under control of the cell chemistry. If you hold I at 2A then after a while Vcell will rise above Vmax and the cell will be damaged or destroyed.
IF the cell is rated at 1C = 4A charge max then you can charge at C/2 = 2A until cell reaches Vmax. Cell capacity will be about 90% of max. If there are 2 cells in series the time taken will be the same but the required voltage will be double. |
H: How do you make custom THT (through-hole) connectors?
I have a project that requires a connector that is no longer manufactured which would interface with an old device. For prototyping, I have been salvaging this connector from old peripherals that used it, but I need a way to get these connectors in bulk that doesn't rely on using other devices for parts. Specifically, I need the THT (through-hole) version of this connector.
Is there a way to either make the connector myself or to get them manufactured? The connector plug is like any other USB, HDMI, DVI, or similar THT plug, it is just of a completely custom design.
AI: Certain manufacturers offer small runs and custom connector services.
Check Samtec, but be ready for tooling and MoQ charges. |
H: Noisy RS485 signal with MAX13487E and Raspberry Pi
I am trying to use the MAX13487E RS485 transreceiver with a Raspberry Pi 3 B+. I chose this chip specifically because it does not require a RTS signal, which is complicated to setup on the PI. My schematic is the following:
The MAX13487E is a 5V chip so I stepped down the RX signal to 3.3V, but I directly connected the TX line to DI, which I think should be fine because the datasheet states that the input high voltage of DI is 2V.
I am receiving data correctly on the PI, but I can't send anything without the other device rejecting frames because of a noisy signal. Here is a scope capture of the TX line in yellow and the RS485 B line in green. This capture was done without connecting the device to the bus.
And here is a capture of the differential signal (pink curve is A - B):
Should I be feeding a 5V signal to DI or is something else wrong with my circuit?
AI: The block diagram of the first page of the datasheet for the MAX13487E indicates that the driver can shut itself down based on the conditions on the line. It can do this even if you have manually disabled the receiver by forcing RE LO (i.e. turning off the receiver doesn't mean forcing on the transmitter) since you can see RE is unable to influence the driver enable.
This probably exists so that you don't have two devices fighting each other trying to drive the line. That means that if your idle line conditions are wrong, the MAX13487E will think a device is transmitting and back off by shutting down its transmitter.
I think your 10K resistors are too large relative to your 120 Ohm and are causing your idle voltage to not be >200mV which is confusing your auto-detection circuitry and making it think another device on the line is transmitting so it reacts by shutting itself down spuriously during transmission.
I don't use auto-detection transceivers so I don't know for sure. Let me know how it works out. |
H: Can we use ULN2003A to power up another module
I have one requirement wherein, I need to power up one of the modules occasionally. I do not have relay in my controller board, but I have ULN2003A in my board. The module consumes around 400mA. So can I use ULN2003A to drive this external module? In case I can drive, how the connection will be? In case of relays, we give power to one side of the relay and the other side is connected to the collector of ULN.
AI: The ULN2003A is a 7 channel low-side driver that is rated for 500mA per channel, but drops ~1V at 400mA.
If your module will work with ~1V less than your supply voltage and its ground pin being raised by 1V isn't a problem then it may work (though the ~0.4 W power loss in the ULN2003A could make it very hot).
However if another device has to communicate with the module then the raised ground potential probably will be a problem, so it would be better to use the ULN2003A to switch a high-side power transistor, like this:-
simulate this circuit – Schematic created using CircuitLab
The high-side switching device can either be a PNP bipolar transistor (as shown here) or a P-channel MOSFET designed for 'logic level' Gate drive (eg. IRLML5203). Either type should be rated for several Amps to ensure low voltage drop at 400mA. |
H: 74HC14 waveform question
I am trying to get more familiar with using a 74HC14, as an experiment I have an ldr set up in a voltage divider configuration and the center tap goes to the input of the 74HC14 input 1. I have all of the other inputs tied to ground. Everything is soldered together on a perf board. I have a .1uF, .01uF, .001uF, .0001uF cap on the power pin (probably overkill). Input voltage is 5v. The voltage divider and the IC are both powered off the same rail
The IC triggers as expected when light is shined on/removed from LDR. Below is the waveform I am getting for the falling edge, it is very consistent from test to test.
What I don't understand is why the signal is dropping to -2.4 or so when triggered then staying below 0.
The rising edge has an overshoot to about 6v but is smoother. Any information/guidance would be appreciated.
https://drive.google.com/open?id=1oG6Eh2Gwj9bf4IXGBJ9i5LoJoLvwybN6
AI: When trace and wires are inductive at the rate up to 10nH/cm interact with the scope probe coax and standard ground leads, any measurements with rise times faster than what the 20MHz filter allows will have impedance mismatch faulty signals, both due to the board design and the scope probe ground lead.
Either apply the 20MHz DSO filter at all times or learn how to capture proper signals and learn the effects of the [robe tip capacitance. It is no secret that Nch low side drivers are much lower impedance than Pch High side drivers and breadboards without a ground plane closer than trace width the chances of matched impedance on traces is zero. Thus overshoot depends on the degree of how much lower the driver impedance is.
If you want any chance of getting a clean signal; the ground plane, IC decoupling cap and output must all be within 1cm be probed as below Also any long traces or loads examined for reflection effects.
Recall 80% Tr = 0.35/f (-3dB) so Tr=0.35/20MHz =17.5ns
This means any rise times < 18 ns common to all 74HCxx CMOS will be inaccurate on a scope unless you follow this test method... or ignore it and apply the 20MHz DSO filter switch.
Getting rid of noises after RC low pass filter stage |
H: Adding Ethernet ports to my SoC
I'd like to add multiple (>5) Ethernet ports to a board I am developing. Linux on the CPU has to see each port individually, be able to route the traffic among ports, and also add traffic on its own (coming from other sources, such as an LTE modem and a video source).
So far I've been unable to find a SoC satisfying all my other requirements and that has at least an Ethernet port, so I wanted to add additional Ethernet ports on the PCIe port. However, I an not even sure what I should look for: I've searched for an Ethernet switch IC and I found the VSC7514 by Microsemi which, according to an ELC2018 presentation, allows the ports to be seen as individual interfaces by Linux. However, the presenter says that only control traffic passes on the PCIe (whereas I have to send data to the switch too).
What should I look for if I want to add more Ethernet interfaces to my board? For a multiport Ethernet transceiver IC?
Thanks in advance for any help! :)
AI: If you're talking PCIe, then any PCIe NIC chip should work. Not sure if you can get quad port chips, or if you would need two dual port chips. But it should be doable. Switch chips are also an option, but bear in mind that a switch chip does not send all traffic through to the SoC, the SoC is simply another port on the switch. You can set up VLANs and the like, but if you send everything to the SoC for routing, then the SoC could end up being a bottleneck - both the link bandwidth of the SoC port as well as the processing power of the SoC can limit the overall rate. You could have the same problem with multiple PCIe NICs as well. |
H: Transfer function of a classic controller
I want to calculate the transfer function and the gain, zeros and poles of the transfer function, considering that: $$R_1=R_2=R_3$$ $$C_1=C$$ I want to know if this circuit is a P, or PI, or PD, or PID controller or a lag/lead compensator.
To get to the transfer function, I think AMPOP entries must be matched. I think that is a lead controller, but I'm not sure. How can I solve?
My work so far:
$$\cfrac{v_0-v_i}{R+\bigg(\cfrac{1}{R}+\cfrac{1}{Z_C}\bigg)^{-1}}=\cfrac{v_i}{R}\Leftrightarrow \cfrac{v_0-v_i}{R+\bigg(\cfrac{1}{R}+sC\bigg)^{-1}}=\cfrac{v_i}{R}\Leftrightarrow \cfrac{v_0-v_i}{R+\bigg(\cfrac{1+sCR}{R}\bigg)^{-1}}=\cfrac{v_i}{R}\Leftrightarrow \cfrac{v_0-v_i}{R+\cfrac{R}{1+sCR}}=\cfrac{v_i}{R}\Leftrightarrow \cfrac{v_0}{R+\cfrac{R}{1+sCR}}=\cfrac{v_i}{R}+\cfrac{v_i}{R+\cfrac{R}{1+sCR}}\Leftrightarrow v_0\Bigg(\cfrac{1}{R+\cfrac{R}{1+sCR}}\Bigg)=v_i\Bigg(\cfrac{1}{R}+\frac{1}{R+\cfrac{R}{1+sCR}}\Bigg)\Leftrightarrow \cfrac{v_0}{v_i}=\cfrac{\cfrac{1}{R}+\cfrac{1}{R+\cfrac{R}{1+sCR}}}{\cfrac{1}{R+\cfrac{R}{1+sCR}}}=\cfrac{\cfrac{1}{R}}{\cfrac{1}{R+\cfrac{R}{1+sCR}}}+1\Leftrightarrow \cfrac{v_0}{v_i}=\cfrac{1}{R}\times \big(R+\cfrac{R}{1+sCR}\big)+1=2+\cfrac{1}{1+sCR}=\cfrac{2+s2CR+1}{1+sCR}=\cfrac{3+s2CR}{1+sCR}$$
poles:
$$1+sCR=0\Leftrightarrow s=-\frac{1}{CR}$$
zeros:
$$3+s2CR=0\Leftrightarrow s=-\frac{3}{2CR}$$
AI: Well, that is more related to circuit theory than control. You probably know that:
the impedance of a capacitor can be written as \$ Z_C = \frac{1}{sC}\$,
the amp op will behave as if having a virtual ground between the two inputs,
that amp op configuration is an non-inverting amplifier.
So write down the KCL equations of the circuit, or just use the gain formula for non-inverting amplifiers (fixed in the current edit):
$$ v_i = \frac{R_3}{R_3 + R_1 + R_2||Z_C} v_o.$$
The transfer function will be
$$ \frac{v_i}{v_o} = H^{-1}(s) = \frac{R_3}{R_3 + R_1 + (1/R_2 +sC)^{-1}},$$
$$ H^{-1}(s) = \frac{R}{2R + (1/R +sC)^{-1}} = \frac{R(1/R +sC)}{2R(1/R +sC) + 1}, $$
$$ H^{-1}(s) = \frac{(1 +sRC)}{2(1 +sRC) + 1} = \frac{(1 +sRC)}{(3 +s2RC)},$$
$$ H(s) = \frac{(3 +s2RC)}{(1 +sRC)} = k_1 \frac{\frac{3}{2RC} +s}{\frac{1}{RC} +s} = k \frac{1 + \frac{2RC}{3}s}{1 + RCs},$$
Where \$k_1\$ is some gain, and \$ k = H(0)\$. This transfer function has a pole and a zero, and it is not of the form
$$ PID(s) = k_p + k_i/s + k_d s,$$
since it has no purely derivative term (although in real systems they will never will due to sensor error amplification), nor it has an ideal integrator. In fact, it seems to be some compensator.
\$ \frac{1}{CR} > \frac{3}{2CR} \Leftrightarrow \$ lead compensator,
\$ \frac{1}{CR} < \frac{3}{2CR} \Leftrightarrow \$ lag compensator.
The gain will be \$k = H(0) = 3\$.
Obs:
\$R_2||Z_C\$ is the equivalent of \$R_2\$ and \$Z_C\$ in parallel. |
H: What is the difference between pulling up multiple digital IC pins to Vdd with a single resistor vs. one resister per pin?
I am not very good at understanding current, so I was hoping someone could help calrify the difference between pulling up multiple digital IC pins to Vdd with either a single resistor OR one resistor per pin?
if you are connecting multiple pins to the same resistor, would you just use a higher value resistor? Ie. instead of 4 10k resistors connected to Vdd, pull-up everything with one 40k resistor? Or perhaps the other way around?
Example:
VERSUS
AI: Assuming DC loading isn't a factor here (generally not with modern CMOS devices), one advantage of the first (individual resistors) is that it makes it easier to use one of the unused inputs just by connecting a wire/signal to that pin. If the second option were being used, you'd have to isolate the desired pin from the shared pullup trace first.
Another reason for individual resistors is that it makes routing easier on the PWB.
Many times you'll see a hybrid, where you might have one pull up resistor for a given IC. Or one resistor for a common signal, such as an enable for a bunch of differential line drivers. |
H: I2C clarification
This question is mainly to check if I understood i2c protocol and hardware correctly. Because pin modes confusing me a little, ia m jsut a beginner.
I2C SDA and SCL lines are pulled high with pull up resistor. In my case I use 4k resistor.
Master pulls SDA line to ground and starts clocking on SCL line after that, and thats called start condition.
Master send out an adress to slave and r/w bit.
Here is what confuses me.
Master reconfigures the pin SDA in to input mode to recieve ACK bit while slave reconfigures its input pin to output pin and pulls SDA line to ground to send and ACK bit.
And then same repeats with the rest of the bytes untill stop condition is met.
The question is, is pin actually reconfigured from output to input on the master to read the ACK bit or there is some other mechanisms that i am not aware of?
From what I read on AVR forums I can do the following:
In case my SDA pin is on PB0, to make it high I would switch PB0 to input without internal pull up (leaves it floating). To make pin go low I would set it to output low. While I am input mode I could also read the input from the same pin.
Please correct me if i am wrong.
AI: I²C has two states on the SCL/SDA lines. A weak high level, and a strong low level. To achieve that by bit-banging, set the output register to low, and leave it alone later.
Only ever switch the SCL/SDA data direction register between input/weak high level and output/strong low level.
For SCL you have to do the same as for SDA because the slave may choose to pull the SCL line low for clock stretching. You have to release clock, then check if the slave also released it. Not supporting clock stretching properly is a bug even found in professional I²C hosts implemented in hardware.
Most AVR controllers have an I²C host/device controller built in hardware, and it's one of the few which is implemented correctly. You don't have to bitbang I²C on those controllers. Check the datasheet. |
H: Designing an air gap transformer Misunderstanding
Suppose my core is actually with a design which make it saturated and the voltage on the secondary is too high.
So I have to decrease the secondary voltage:
For this I can increase the number of turn on the primary side N1. Nevertheless I will increase the magnetic excitation H which is equal with some appoximation to H = N1*I1/l. It will then saturated even "higher" my core. So I can't go this way.
I can lowering N2 : Unfortunately N2 is at its minimum.
Last solution to reduce the secondary voltage according to the Faraday's Law, reducing the variation of the magnetic flux over the time. (Consider that we cannot change the duty cycle).
How can I decrease the variation of the magnetic flux? Here is what I do not understand:
By decreasing N1, I decrease the excitation H1, so the magnetic flux. (cf B-H curves)
Nevertheless if I consider the Faraday's law, as the primary voltage is constant and the duty cycle fix, If I decrease N1, I increase the variation of the magnetic flux.
Where is my error?
AI: For this I can increase the number of turn on the primary side N1. Nevertheless I will increase the magnetic excitation H which is equal with some appoximation to H = N1*I1/l. It will then saturated even "higher" my core.
That would be entirely true if the magnetization current remained the same but, it doesn’t...
Increasing the number of primary turns increases the primary magnetization inductance. This, in turn naturally reduces the primary magnetization current because more inductance means a higher reactance at the excitation frequency (presumably 50 or 60 Hz or some other fixed value).
And, if you look at the formula for inductance, you’ll find it’s proportional to turns squared hence, if you were to double the number of turns, you would get 4 times the inductance and, for a given primary voltage and frequency, the magnetization current would reduce by four.
Another example - if turns increase by 10% then inductance increases by 21% and the current that causes saturation reduces by 21%.
Below is the formula for a solenoid but the same applies for a transformer winding (where \$\mu_0\$ is increased by the relative permeability of the core material and \$\ell\$ is the mean length around the core).
So, if you double the turns (for example), the overall effect on ampere turns (and H field) is that it halves. This is because current has quartered but turns have only doubled.
This means that the effects of saturation are reduced.
You can also introduce an air gap to reduce the effective magnetic permeability of the core. This also reduces inductance (by the amount the permeability is reduced) and, this is “corrected” by more turns but, remembering that inductance is proportional to turns squared, there is still a net benefit on reducing saturation.
Suppose my core is actually with a design which make it saturated and
the voltage on the secondary is too high.
The only way you’ll get too high a secondary voltage is because the turns ratio is incorrect. Saturation does not make the secondary voltage increase.
By decreasing N1, I decrease the excitation H1, so the magnetic flux.
Decreasing the primary turns by (say) 2, lowers inductance by 4 and increases magnetization current four times hence ampere turns (and H field) increases by 2 and you get more saturation. |
H: Will beeping dc motors cause damage to the motors?
I have noticed from a drone (of a friend of mine) it has a feature that lets its motors beep so when it gets lost it can be easily found. I have some searching and this seems to be done by alternating turning forward and backward quickly enough for the motors not be able to spin.
I would hopefully like to exploit this characteristic as maybe a feedback to the user for example a button is pressed and the motors would beep, this would mean that I do not need a piezo speaker.
But will this cause some damage to the motor or motor driver? My intuition is telling me it does, but on some commercial products (drones) or high performance robots (the one used in competitions) I often hear these beeps.
AI: Summary: Safe enough unless you try really really hard to design it to be destructive.
It MAY be possible to damage the motor this way if you tried hard enough, but that applies to any use of the motor, and it will be very easy to implement such a system without harming the motor. Controllers usually have inbuilt hardware or software mediated protection systems which prevent one applying excessive levels of stupidity. Even without this it would be hard to destroy a system at sensible power levels.
Issues may include
Causing excessive thermal dissipation by applying high energy levels without the cooling usually afforded by the propellor downdraft
Operating the motor at zero revs is its potential highest torque point - if you were going to break something this will be a likely region to do it. But, any system not designed to easily withstand peak torque, even repeatedly applied, is not a well designed one.
Operating a motor in reverse applies reverse torque to the propellor to shaft coupling. Some systems use appropriate sense left or right hand threads to lock the propellors in position. Even almost zero revs reverse torque will tend to unlock the propellor - BUT it will be locked again by the next half cycle of 'sound' drive. and properly locked again when normal motion is applied.
Bonus:
For "extra points" you may be able to make it talk.
"Help me. Help me. I'm over here ...".
(I've heard of road surfaces designed to make a car's suspension "talk" - decipherable "speech".) |
H: Connecting two RCD's serially
I want to make a electric leakage protection circuit for a water heater in case of heater gets corrupted in time. We use rezistance 220V - 1500 watt. We grounded the water (because rezistance have coating outside and doesnt convey electric to water) but when there is electricity in water, RCD at office cuts off electricity for all devices. We only want heater to turn off. So, we bought another RCD and connected it to water heater, then we plug water heater to electiricty. It works but two RCD triggered at the same time. We only want one of the RCDs that connected to the heater to trigger. I can't connect different RCD in fuse board because system must be portable and this safety system must be included to main circuit.
Is it possible to do this with RCDs? Or can we add something after RCD of water heater to delay the current, so RCD of office wont trigger?
AI: If you can incorporate an isolation transformer into your build then this will prevent the main fuse board RCD tripping. The isolation transformer fits between your local RCD and the incoming AC power feed from the fuse board.
Alternatively, if you can get an RCD that can be easily dismantled and modified, you could make it more sensitive by looping the through going live and neutral wires through the internal magnetic toroid 2 or 3 times. |
H: How I could use a 2v LED to indicate line continuity on a 220VAC line
I have a 2v-2.5v DC LED and I want to use it to indicate if the line fuse is working.
ON fuse is ok
OFF fuse blown up
What is the proper way to connect it since obviously a direct connection would fry it instantly (and using a dedicated AC-DC adapter just for a led would be crazy)?
AI: Well if you want to keep it really simple, you could achieve an inverted logic(led off when fuse is okay and on when fuse is blown), you could simply connect the led circuit in parallel to your fuse.
Your led circuit would be simple, your LED in series with a higher reverse voltage diode(something like 1n40007 could work) and a resistor. Your led, the diode and led would all be in series and would typically stay off if the potential difference across the fuse is 0v(fuse is intact). The potential difference would rise to line voltage when the fuse is blown and hence your led would light up. You may find a bit of flicker in led due to line voltage frequency but It shouldnt be a major challenge.
If you certainly want the led to be on while your fuse is intact, you would have to wire up the led circuit between live and neutral. The led, diode and resistor would be connected between live(live after fuse) and neutral. From an execution perspective, solution 1 would be simpler to wire up and execute.
Hope this helps. |
H: What does the arrow going through the battery symbol represent?
I am currently studying Practical Electronics for Inventors, Fourth Edition, by Scherz and Monk. Chapter 2.5 Resistance, Resistivity, and Conductivity presents the following image (without description) when discussing the concept of resistance:
I have the following questions:
What does the \$ A \$ symbol pointing to the left-hand-side of the Ohmic material represent?
What does the arrow going through the battery symbol represent?
I would appreciate it if people would please take the time to clarify this.
AI: A is the cross sectional area of the resistor.
The arrow through the voltage source indicates its voltage is variable. The device is likely not actually a battery because a variable battery is not an easily realizable device.
An arrow through a component is a common indicator for a variable value; for example you'll see variable resistors and variable capacitors indicated this way also. |
H: Relation between diodes and laser diodes
I am currently studying Practical Electronics for Inventors, Fourth Edition, by Scherz and Monk. Chapter 2.5 Resistance, Resistivity, and Conductivity presents makes the following claim when discussing the concept of resistance:
Ohm's law can be applied only to ohmic materials -- materials whose resistance remains constant over a range of voltages. Nonohmic materials, on the other hand, do not follow this pattern; they do not obey Ohm's law. For example, a diode is a device that allows current to pass easily when the voltage is positive, but prevents current flow (creates a high resistance) when the voltage is negative.
I'm wondering what, if any, relationship there is between this explanation of diodes and the laser diode? The explanation provided is obviously very basic, but how does it, if at all, contribute to the functioning of laser diodes?
I would greatly appreciate it if people would please take the time to clarify this.
AI: I'm wondering what, if any, relationship there is between this explanation of diodes and the laser diode?
If you read the Wikipedia page that you linked to, you'll see that a "laser diode is like an LED".
A semiconductor junction diode works because the arrangement of the P-type materials and N-type materials allow current to flow in just one direction.
An LED (which is a kind of semiconductor junction diode) emits light because of some special quantum mechanical properties of the materials used such that when it conducts, instead of the energy dissipated by the diode drop times current generating heat, at least some of it generates photons.
A laser diode (which is a kind of LED) undergoes Light Amplification by Stimulated Emission of Radiation (LASER), because as it conducts in leaves a bunch of electrons in an excited state (called a population inversion), and when a photon smacks into such an electron the electron falls to its ground state and generates another photon in the same direction and phase as the first -- which is what the laser effect is all about.
You could possibly use a laser diode as an "ordinary" diode, but it would be unhappy. They're quite specialized devices -- but under the hood, there's a PN junction, and many of the same quantum-mechanical processes that make a plain old 1N4148 conduct current in one direction but not the other are present in a laser diode. |
H: Finding relay datasheet: HC1-AC240V LR42758 30111H AP311998 3A30VDC 10A 1/3HP 125,250VAC
I have a dickens of a time finding a datasheet for a spare relay that I want to reuse. Here are the images of it:
The text from the side there is:
HC1-AC240V LR42758 30111H AP311998 3A30VDC 10A 1/3HP 125,250VAC
The pins:
The other side showing "Aromat" (but I cannot tell if that is the manufacturer of the coil inside the relay, or the entire relay):
I've web-searched laboriously and came up short.
One datasheet looks close: https://www.alliedelec.com/m/d/2454bde37e51c885d9084c84a437bef1.pdf but I doubt that is the exact one because the breakout for the part number has many more dashes than "HC1-AC240V".
I speculate that the "LR" implies it is a Latching Relay but not sure. I get a lot of search results for "LR42758"
Is there a online search engine somewhere whereby I can put in this information and get back datasheets specifically for it? Or am I just going to have to chuck it and start with something new (they are not that expensive, just hate throw something useful away).
AI: Aromat is Matsushita is Panasonic. That's an obsolete item, we were using them back in the early 90s. There are, however, compatible products out there. The LRxxx number is a file number for the safety approvals and won't give you any more information about that specific device (it would cover many different part numbers).
Here's a datasheet
That will give you the important life curves for various loads, and other data that's not obvious.
It's a simple SPDT power relay with 240VAC coil, nothing special, but well-made and reliable within it's stated lifetime for proper loads. |
H: AC converter, step up or step down, which is more efficient, which loses more energy? why can't support large power and being small
I want to buy a AC voltage converter for frequent international travel with some small appliances, and should support step up and down between 110V and 220V; found this one https://amzn.to/2voExKD is portable but it supports max 100 Watts only, then found this one https://amzn.to/30UKV8b supports up to 2000 watts but it weighs 30 lbs very bulky not convenient to carry on to flights;
then found this one https://amzn.to/2U7FDF1 supports up to 2300 watt and lightweight to carry, but it supports step down from 220V to 110V only, not the step up;
so I wonder is it possible to make a product that
support both step up/down AC voltage converter, between 110V and
220V;
support large power consumption up to 2000 watts ideally;
also being lightweight and portable, easy to carry to flight
check-in luggage? (less than 3lbs, or as small as possible)
Update: I edited a little bit to be clear I wish to see a product has all the 3 features, especially large power appliances (like 1500 w induction top cooker), not only laptop/phone charger which I know that takes 100-240V variant AC input;
Thanks;
If not, could someone tell the rationales why can't make a product with all above 1 & 2 & 3 features?
If can only choose 2 features out of the 3, I would possible choose this https://amzn.to/2U7FDF1 supports up to 2300 watt and lightweight to carry, and one-way only step down from 220V to 110V; why I am not seeing a step-up only product? Is step-up harder to make than step-down only? which one of the step up/down is more energy efficient, and less energy loss inner heat dissipation?
I like to hear some in-depth answer about rationales behind step-up and step-down; that can explain why
the first product https://amzn.to/2voExKD is lightweight/portable but supports max 100 Watts only
the second one https://amzn.to/30UKV8b supports up to 2000 watts for both step up and down between 220V <=> 110V, but why it weighs 30 lbs why it needs to be made so bulky, is it related to heat dissipation? how much energy loss if this is really working at 2000 watts;
the third one https://amzn.to/2U7FDF1 supports up to 2300 watt step down from 220V to 110V only, it's lightweight to carry, Weight:1.3lb, is it because step-down is easier to make than step-up AC transformers? Does step-down only converters have better energy efficiency? Nearly no heat dissipation, Or 100% energy conversion? Is that the reason why it can be made so small (Weight:1.3lb)?
Thanks again;
AI: I will try to answer your questions the simplest way possible:
A step up converter is almost always an iron-core based transformer.
Simply because it's the simplest and most cost effective. A transformer easily converts 110V into 220V, or the reverse, at 60hz without any circuitry. The only problem is that transformers are big and bulky. A 500W transformer will weigh in excess of 5 pounds. A 2kW+ over 30lb. This is indeed because of heat dissipation and thermal constraints due to efficiency. The more powerful, the more power lost, so the bigger the transformer needs to be.
A step down converter is actually a very simple product. It has very little weight because there is no transformer.
Source: https://www.homemade-circuits.com/how-to-make-220v-to-110v-dc-converter/
This simple circuit can easily handle 2000W because it simply needs to lower the voltage. There is basically no loss at 15A maximum.
The issue with dual step up/down converters right now is that there is no cost effective way of converting 110VAC to 220VAC without using a transformer. There exists such products, but they are highly complicated and cost a fortune and isn't much smaller than a transformer of the same power capabilities. Someone correct me if they find one that is less than a few thousand dollars...
I did see actual papers on making a product like you want:
https://www.researchgate.net/publication/277673392_Single-phase_direct_AC-AC_boost_converter
Not sure if it is consumer available yet. |
H: Choosing Resistors for Brightness in Parallel LED Array
These LED’s I have are incredibly bright. So when using just one, I would typically use a 10-15k resistor in series to dim them down. Now I need to use those same LED’s in a parallel array. I used an LED array calculator to compile a circuit. The values I put through were taken from the data sheets. 20mA is the forward current, but that would be way too bright. The calculator chose 100 Ohm
In order to maintain roughly the same brightness per LED as in a series circuit with a 10-15k resistor, how do I calculate the new resistors? Should I simply use the same resistor values?
!
AI: Your assumption is correct, just use the same resistor values (10k-15k) in your parallel circuit.
In this simple circuit, from the point of view of the individual LED and current limiting resistor, they still see the 5V supply. Now that you have 14 in parallel, just copy your series circuit 14 times, and make sure the power supply can handle the load. |
H: Confirmation on Diode in parallel with a Resistor
In a circuit with a resistor parallel to a diode/resistor, will the voltage across both resistors be the supply minus the 2v drop of the LED?
Meaning you would calculate it still by using resistors in parallel:
1 ÷( 1/r1 + 1/r2). And then use supply minus 2v ÷ R to get the total current? Is this correct?
I am trying to teach myself electronics and find it very hard to find definitive answers on the things I am trying to learn. Therefore I am wanting to just try get confirmation on a question if I can.
I get confused as in a parallel split the voltage is unchanged on each wire, only current is halved. But since one lane has a voltage drop I initially assumed they would rejoin with different voltages which wouldn't make sense. And I have learnt that if there is 2 diodes it would only activate the one with the lowest forward biasing voltage and the other would stay off. .
I have done a few tests and found that seem to imply this is the case but none seem to definitively say it. If someone would be and to assist in confirming that this is the case that would be appreciated.
I have been learning about diodes and LEDs but keep running into problems that tutorials haven't discussed so it's tricky to find all the information. Apologies if I'm not posting correctly this is my first post. Also I know there is a circuit editor but I can't seem to see it in mobile. Thank you
EDIT: Schematic for a comment.
simulate this circuit – Schematic created using CircuitLab
AI: will the voltage across both resistors be the supply minus the .7v
drop of the LED?
No. You have drawn a nice schematics. The R2 resistor is direklty in parallel to the source voltage. So the voltage across the resistor will be same as the source voltage.
Proceed in this way. R2 is directly in parallel with the source voltage. Hence the voltage has to be the same as source Voltage
For R1, there are two components in series. The diode drop plus the drop across the resistor R1 has to be same as the supply voltage.
The voltage across the diode depends on many factors one of which will be the supply voltage itself and the other will be the forward Voltage of the LED. (About 2.4 V upto 4 V depending on the type of the LED, but definitely not 0.7 V as in normal diodes).
If there are two sides in parallel, the diode with the lower forward Voltage drop turns on first there by keeping he voltage across the second diode too lower than the forward Voltage drop of the second diode. Hence the second diode will not turn on (completely).
Start with the loop equations. Except I1 and I2 other parameters are constant and can be solved. I would approach the problem in this way. |
H: BQ297xy Overcurrent protection
Can someone explaine to me how dose this overcurrent protection works and how to calculate it?
BQ297xy Schematics
AI: From the datasheet:
The discharge overcurrent detection voltage (VOCD) is measured between
V–and VSS pins and triggered when the V2 voltage is increased above
VOCD threshold with respect to VSS. This delta voltage once satisfied
will trigger an internal timer tOCDD before the DOUT output drive
transitions from high to low.
There is no user setting where he/she can individually change the settings. The thresholds can be either set at the factory or use parts with pre fixed factory thresholds. |
H: Looking to generate 25 Hz square wave at about 90 V AC
I am looking to build a square signal generator tuned to ~25 Hz at a voltage of ~90 V AC that can output ~400 mA. I've searched around the internet and tried a few things but I've had no luck. My original idea was to have an Arduino and a LM741CN output a square AC wave and then attach that to a transformer to get it to 90 V AC. I achieved the square AC wave, but it only outputs about 2.5 V AC. Getting that far I was looking into transformers but low frequency transformers are not as easy to come by as I thought, I tried building one but the core I had on hand was not big enough to handle the amount of wounds needed.
My overall question here is: Was I heading in the right direction or is there a better way that I have not found yet?
AI: Sounds similar to a telephone ring tone generator, for example like this one, based on the LT1684:
The LT1684 is a more complex chip that actually generates a PWM sinusoidal wave.
Generating a square wave is easier, you can just use four MOSFETs or IGBTs and bootstrapped MOSFET gate drivers (eg. from IR). A full H-bridge with ~ +95VDC supply can be used.
Depending on the MOSFETs you can drive 100mA or 20A with a similar circuit. |
H: Making a Remote Control for My TV from Scratch
This might seem silly to many of you, but I lost my TV remote when I moved awhile back and instead of buying a universal remote for cheap, I want to make a remote for my TV from scratch myself. I'm talking about designing the pcb, soldering parts together, the whole shebang. I know nothing about electronics and thought it would be a great way to learn.
I just need some guidance on how to start or advice on whether this is even possible without having the original remote to decode the IR signals? Am I able to look up the IR codes for the receiver by looking up the model number of the TV? What components do I need and how do I organize them on the circuit board?
I know basic parts that I would need are an IR LED, transistors, push buttons, and a microcontroller. I'm not sure which specific transistors or microcontroller to use though or how to design the parts on the pcb. Pointing me in the direction of useful resources would definitely help. I've already read a few instructables and watched a few videos on making a remote, but none of them told me how to design one from scratch for a specific receiver without the original remote. Thanks in advance!
AI: Creating a remote control with an IR led and a IC for registering/sending codes is actually pretty easy using a PIC micro-controller or the likes. And programs like PicBasic are very easy to learn and use for programming.
The issue lies with a very important problem:
You need an actual remote that works with the TV to get the codes.
There is absolutely no source online for IR codes. You need a functional remote for this, which makes it useless creating a new remote from scratch in the first place....
I guess I'm getting old or don't research often enough but if you really want to make a remote there is a few options:
You can make an arduino remote control. This uses very low cost parts and a bit of learning:
https://importgeek.wordpress.com/2013/02/19/send-tv-remote-signal-using-arduino/
A more compact way of doing this is using a PIC micro-controller. (ex. PIC12F675 from Microchip) There is very little parts needed and has a smaller footprint. Programming is a bit harder, but many programs are already done for you. Only editing the HEX codes is needed for the most part.
As for the HEX codes needed for the TV to recognize, they can be found here:
remotecentral.com/cgi-bin/codes – Thanks to Mattman944 for pointing that out!
If you have a Samsung tablet, some have IR transmitters. If so, you can use this as an alternative remote using remote control apps. Or simply buy a new remote off ebay/amazon. |
H: Resistance of wire
I am currently studying Practical Electronics for Inventors, Fourth Edition, by Scherz and Monk. Chapter 2.5.1 How the Shape of a Conductor Affects Resistance says the following:
The resistance of a conducting wire of a given material varies with its shape. Doubling the length of a wire doubles the resistance, allowing half the current to flow, assuming similar applied voltages. Conversely, doubling the cross-sectional area \$A\$ has the opposite effect -- the resistance is cut in half, and twice as much current will flow, again assuming similar applied voltages.
Increasing resistance with length can be explained by the fact that down the wire, there are more lattice ions and imperfections present for which an applied field (electric field instigated by added electrons pumped in by the source) must shove against. This field is less effective at moving electrons because as you go down the line, there are more electrons pushing back -- there are more collisions occurring on average.
Decreasing resistance with cross-sectional area can be explained by the fact that a larger-volume conductor (greater cross-sectional area) can support a larger current flow. If you have a thin wire passing \$ 0.100 \ \text{A} \$ and a thick wire passing \$ 0.100 \ \text{A} \$, the thinner wire must concentrate the \$ 0.100 \ \text{A} \$ through a small volume, while the thick wire can distribute this current over a greater volume. Electrons confined to a smaller volume tend to undergo a greater number of collisions with other electrons, lattice ions, and imperfections than a wire with a larger volume.
I found this interesting because I have never heard of people considering the resistance of the wire in their calculations (as they would a resistor, or some other component) when doing electronics projects. I have often heard people discussing the appropriate wire gauge to use for a project, but this doesn't seem to be a matter of resistance calculation, and is more-so a matter of physical (not physics) considerations.
How common is it to consider the resistance of a wire? Is the resistance of a wire an important consideration when doing electronics work? Does this resistance need to be factored into calculations, as would other components (such as resistors)?
I would greatly appreciate it if people would please take the time to clarify this.
AI: Sometimes, a wire is negligible in terms of its resistance. Other times, impacts of the resistance of a wire can become significant. I'll first show the resistance of a wire, and how you can ignore it in most cases, and then show examples when its impact is significant, and finally a few applications.
The Resistance of a Wire
Ideally, the formula of the resistance of a conductor is...
$$ R = \rho \frac{L}{A}$$
Given the cross-section area (A), length (L), and resistivity (\$\rho\$) of the material. For copper, \$ \rho = 1.68 \times 10^{−8} \Omega \cdot \text{m} \$ at 20 °C.
For cylindrical conductors (like a wire),
$$ R = \rho \frac{L}{\pi r^2} $$
Example: What is the resistance of 5 cm of AWG-30 (0.255 mm in diameter) copper wire?
Answer: First, the radius of a AWG-30 wire is \$ 1.275 \times 10^{-4} \text{m} \$, find the resistivity of copper from a textbook, which is \$ 1.68 \times 10^{−8} \Omega \cdot \text{m} \$ at 20 °C. The formula yields \$ R \approx 0.0164 \Omega \$.
Example: What is the resistance of 5 cm of AWG-24 (0.511 mm in diameter) copper wire?
Answer: \$ R \approx 0.004 \Omega \$.
Remark 1: As we see, resistance of a wire is lower when the wire gauge is thicker. Specifically, when the diameter of a cylindrical wire doubles, its resistance decreases to one-fourth of the original wire. Thus, wire gauge is not only an indication of its shape. It is indeed a metric of its electrical property when its material (almost always copper) and length are given.
Remark 2: A quantitative calculation of wire resistance is not always performed. Sometimes rules of thumb are used. Often the consideration is only "whether the wire is thick enough", not "how much resistance/voltage drop/temperature rise does this wire have". On the other hand, to analyze a wire quantitatively, knowing its gauge is the first step. Not to mention that wires are sold by gauge, so people talk about "wire gauge" (or "trace width" in circuit board design) more often than wire resistance.
On a Printed Circuit Board, you can calculate the resistance of traces in a similar way from the thickness of copper and the length of a trace. The only difference: Wires are cylindrical, while traces are rectangular.
Example: What is the resistance of a 10-mil, 10-cm trace on a 1-oz circuit board?
Answer: 1 mil is a thousandth of an inch (0.0254 mm). A "1-oz circuit board" is a circuit board with 1 oz of copper per one square foot area, or a thickness of 1.37 mils. 10 mils is 0.254 mm, 1.37 mils is 0.0348 mm. Cross-section area \$ A = 2.54 \times 10^{-4} \text{m} \times 0.348 \times 10^{-4} \text{m} = 8.84 \times 10^{-9} \text{m}^2\$.
Thus, the resistance \$ R = \rho \times \frac{0.1 \text{m}}{8.84 \times 10^{-9} \text{m}^2} = 0.19 \Omega \$
When Resistance Can be Ignored
Most of the time, resistance of a wire is too low when you compare it to the resistance of other components and loads, so it's negligible and often safe to ignore. Moreover, \$ V = IR \$, the lower the current a load needs to take, the higher its equivalent resistance, so you also ignore the wire resistance if the current delivered by the wire is low, because it's equivalent to connecting a small resistor (a wire) to a large resistor (a device that takes current) - almost no effect.
For example, connect two 1,000 Ω resistors with a 5 cm, AWG-30 copper wire (a thin wire, 0.255 mm in diameter). If we measure the actual resistance between two resistors using an ideal ohmmeter with ideal probes, what would it be?
To calculate its effect, using the formula above for cylindrical wire resistance is often a waste of time, alternatively, we can look up the AWG-30 wire's resistance per unit length from an engineering table on Wikipedia, it says the resistance is "338.6 mΩ/m". In other words, the additional resistance contributed by the wire is \$ 0.3386 \Omega \times 0.05 \text{m} = 0.01693 \Omega \$. Ideally, the resistance should be 2000 Ω, but due to the existence of a wire, the measured resistance is 2000.01693 Ω, it's less than 10 parts per million higher, nearly undetectable.
Remark 3: In non-precision applications, a commonly used type of through-hole resistor is metal film resistor, 5% tolerance, with a temperature coefficient around 50-100 ppm for every 1 °C increase in temperature - the error introduced by the slightest change in temperature is still higher than your wire in this example.
Remark 4: For even the best general-purpose multimeter, like a Fluke 87, the maximum resolution of resistance measurement is 0.1 Ω, so even measuring the 0.01693 Ω wire resistance is difficult.
Another example is a microcontroller development board, which may require a 5 V DC supply and 50 mA current on average to operate. If you use five meters of AWG-30 to hook the power (positive electrode) and ground (negative electrode), the total resistance is \$ 0.3386 \Omega \times 5 \text{m} \times 2 = 3.386 \Omega \$. Total voltage drop across the 5-meter power wire and the 5-meter ground wire, is \$ 3.386 \Omega \times 0.05 \text{A} = 0.1693 \text{V} \$. Actual voltage supplied to the microcontroller board is \$ 5 \text{V} - 0.1693 \text{V} = 4.8307 \text{V} \$, or 96.6% of the original voltage.
Remark 5: A common voltage tolerance for digital electronics is +/- 5%.
If the power source itself is error-free, the drop caused by the wire still is well within the limit. Don't forget I used an extreme example here: 10 meters of extremely long and thin wires, which is not really a realistic scenario in most electronics experiments.
As you see, when using wires for interconnection, you can often ignore the wire resistance, and it's likely that you'll never see a mention about wire resistance in schematics. A similar situation occurs when you connect a cable through a socket, a connector, or a clamp - You'll also introduce additional contact resistance, but it's usually insignificant.
Remark 6: In the industry, the allowed contact resistance introduced by a connector is often 1 Ω. For a high quality connector, sometimes a 0.1 Ω contact resistance is specified.
When Wire Resistance Should Be Considered
But as the current delivered across a wire goes up, up to a point, you can no longer ignore the additional resistance from the wire. Again, due to Ohm's Law, it also happens when the absolute current is still small, but the resistance of other electrical components around the wire have decreased - it's just two sides of the same coin.
A high wire resistance has three harmful consequences:
The voltage drop \$ V = IR \$ across the wire becomes excessive and unacceptable, which moves the power supply voltage outside the range of specification. The device may stop working.
When the resistance of other electrical components are fairly low, the additional resistance of wire itself is simply too high to ignore.
The wire heats up by the current due to its resistance, and the "heater power" is \$ P = I^{2} R \$. This represents wasted power. If the wire resistance per unit length is too high, the wire cannot dissipate the heat quickly enough. Temperature will raise to a point when the wire becomes too hot and melt, creating a fire hazard.
Low voltage DC-power distribution
A common example is power delivered by a USB port. The nominal voltage of USB is 5 V, regulated to +/- 5% as usual. USB 2.0 allows a "low power" device to consume 100 mA, while a "high power" device can receive 500 mA of current. If one use USB as a power source for a charger, the current requirement is even higher, 2000 mA is typical nowadays.
Let's say we have a 1-meter USB cable of questionable quality, which uses two AWG-28 wires (0.361 mm in diameter) for power and ground. Its resistance is 0.42 Ω, when carrying 500 mA of current, we lose 0.21 V due to the cable. To complicate the situation, because the USB power is regulated to +/- 5%, the lowest permissible voltage is, in fact, 4.75 V, the received voltage at the other end of the cable can be as low as 4.54 V - error is much greater than 5% already.
To overcome this problem, the USB 2.0 standard has an additional voltage drop budget for cables.
The maximum voltage drop (for detachable cables) between the A-series plug and B-series plug on VBUS is 125 mV (VBUSD).
The maximum voltage drop for all cables between upstream and downstream on GND is 125 mV (VGNDD).
Functions drawing more than one unit load must operate with a 4.75 V minimum input voltage at the connector end of their upstream cables.
-- Universal Serial Bus Specification Revision 2.0
In other words, for any standard-compliant USB 2.0 high-power device, the manufacturer of this USB device either has to ship the product with a better cable with lower voltage drop, or has to design the device to work down to 4.5 V by any means necessary.
In this case, our device worked. A few days later, someone will find this USB cable and plugg it into a USB wall adapter to charge the smartphone at 2000 mA. Now the voltage drop across the cable is going to be 0.84 V, with only 4.16 V maximum available to the smartphone. The cable either won't work at all, or will charge the smartphone extremely slowly.
Remark 7: Often in practice, some USB chargers will intentionally regulates the USB to 5.25 V to allow more voltage drop on the cables, even it's strictly a violation of the USB standard (Update: This is no longer the case since 2014, the USB 2.0 VBUS Max Limit Engineering Change Notice bumped the maximal voltage to 5.5 V for all USB devices, the motivation is to allow the new USB-C cables to carry higher current. See, voltage drop issues can even lead to a change in the USB spec).
Remote Sensing
Cable drop is also a trouble in voltage regulator design. While it's easy to use an adjustable regulator chip to make a power supply and regulate it to +/- 2% or even lower. Unfortunately, just like previous USB example, your regulation only occurs at the output pin of the regulator, not the load.
Source: Remote Sensing is Important for Your Power Supply, by Keysight, fair use.
Additional wire resistance degrades accuracy of a voltage regulator, especially when the load is far away from it, or when the current is high. Typically, one should take special care when laying out the output traces for the regulator: Keep it as short as possible on a PCB.
But the error can never be fully eliminated, especially when the designer has no control over if there's a long cable in between. When it's critical to accurate regulate voltage at the load, one can employ a technique called "remote sensing" to solve the problem. The basic idea is adding two additional wires to "monitor" the "real" voltage at the other side. If the regulator sees a voltage lower than expected, it'll increase its voltage further to overcome the drop.
Source: Remote Sensing is Important for Your Power Supply, by Keysight, fair use.
The remote sensing wires at +s and -s can have the same resistance like the power wires (same thickness), but they are not affected by the voltage drop. It's true even if they have a much higher resistance (thin wires).
One way to think about it, is considering the fact that high current is running through the power wires, producing a \$ 10 A \times 0.015 \times 2 = 0.3 V\$ drop, but the sensing wires are only here to transmit a small signal - there's little current running across the sensing wire, so it produces almost no voltage drop across the cable.
Another way is thinking the equivalent input resistance of +s and -s of the sensing input. Ideally, its input resistance should be infinite (i.e. no current goes in, an ideal voltmeter, as if nothing is connected). In practice, a resistance of 1 megaohm (1 MΩ, 1 million ohms) is a realistic expectation. So the equivalent circuit is a small resistor (the wires) connected in series with a huge resistor (the regulator sensing input).
For example, in this schematic, although the sensing wires have total resistance of 200 Ω, but the sensing input resistance is 1 MΩ, many order-of-magnitude higher. The voltage seen by the sensing input is,
$$ V_\text{sensed} = 5 \text{V} \times \frac{1,000,000}{1,000,000 + 200} $$
The voltage drop exists, but it's only 0.02%, meanwhile, 99.98% of the voltage from the remote side is measured by the sensing input of the regulator.
Four-Wire Resistance Measurements
Sometimes it's necessary to measure the resistance of an extremely small resistor (lower than 1 Ω) using an ohmmeter. Resistance of the wires connecting between test probes and your ohmmeter become significant. One solution is to short-circuit the test probes before making a measurement - zeroing out the error. But this requires an additional step, it also introduces an additional source of possible error: the pressure applied between the probes can affect the resistance used for calibration.
A common technique for solving the problem is Four-Wire Resistance Measurement, or Kelvin Measurement.
We can think the output pins of a ohmmeter as a current source and a voltmeter - the current source keeps its output voltage at whatever value it needs for a specific current. Then the output voltage of the current source is measured by the voltmeter. Both the current and voltage are known, so the resistance is determined.
Due to the fact that we are measuring voltage directly across the output terminals of the meter, it cannot distinguish resistance from the resistor-under-test and resistance from the test probes.
Adding two additional wires fixes the problem, we can now measure the voltage at the far end across the resistor-under-test, not the output of our ohmmeter at the near-end. Unaffected by the probe wires, we can make an accurate measurement. It's similar to remote sensing design in voltage regulators.
Safety Considerations
This is the main consideration that dictates the wire size in utility power installation at homes. When a current passes through a resistor, not only a voltage drop is produced, but this voltage drop heats up the resistor as well. No matter whether the resistor is a resistor component or a wire, we must ensure the dissipated power \$ P = I^{2} R \$ does not exceed a maximum limit, otherwise the resistor will overheat.
If it's a wire, the wire can become dangerously hot and melt, creating a fire hazard. To find out the maximum current allowed to carry by a wire, first, the dissipated power in the wire is calculated, next, the flow of heat is identified - what is the ambient temperature of the environment, different materials have different thermal conductivity, etc. Finally, one determine a maximum operating temperature and use it to calculate the maximum permissible current, and finally a safety factor is included.
The actual calculation is fairly complex, and it also needs to follow the Electric Code with approval from regulatory agencies. Rather than calculating it from scratch, an engineering table is used. Again, the table on Wikipedia is a reference.
For example, at 20 °C ambient, a single, unbounded AWG-30 wire in a chassis of an appliance cannot carry more than 0.52 A of current in order to keep its operating temperature under 60 °C.
Remark 8: If you're designing a product, you must use a reliable handbook with engineering tables calculated according to your local regulatory agency's standards.
The current-handling capacity of traces on a PCB can be found by referring to an engineering table or a calculation program as well.
Application: Wire-Wound Resistor
Resistance of a wire is not always a nuisance, it has useful applications. Wire-wound resistor is a type of resistor made by winding a metal wire, usually nichrome for its resistivity on a core.
Source: Wirewound resistor, by ResistorGuide, fair use.
It has some advantages.
It's easy to produce highly accurate resistors, as its resistance is proportional to the length of a wire.
One can make high power resistors easily from a large wire.
It should be noted that a wirewound resistor has the same shape like an inductor, thus it has the highest inductance in all types of resistors. It should be only used in DC only, and audio-frequency circuit perhaps, but it's unsuitable for any AC circuits at a higher frequency.
Application: Shunt Resistor
Voltage drop due to resistance of a wire is sometimes helpful as well. An easiest way to obtain current measurement is connecting a low-value shunt resistor in series and measure the voltage drop across it, since \$ I = \frac{V}{R} \$.
Using a high-value resistor stops sufficient current from being delivered to a circuit-under-test, it's desirable to make the shunt resistance as low as practical. There will still be a voltage drop, called burden voltage in a multimeter, but low enough to be acceptable.
If you open a multimeter, you'll find a shunt resistor similar to this picture. As you see, it's just a glorified piece of wire.
Source: Open Air Resistor - Metal Element Current Sense, by TT Electronics, fair use.
If high accuracy is not needed, you can make a free shunt resistor by drawing a trace on a circuit board - the wire (trace) itself is your shunt resistor.
Source: Low ohmic shunt resistor direct on PCB copper layer, fair use |
H: Understanding how PCB CAD designs affect the final product
I'm looking to design my first printed circuit board and having trouble understanding PCB CAD systems conceptually. I've started on schematics with EasyEDA, KiCAD, and Fritzing without finishing, as there is a lot of detail and it's rather unclear what's important. In software terms, I don't know which fields are code with an actual effect on the product, and what stuff is comments that's just there for human communication, or to make a drawing look pretty?
There are large libraries of user-contributed part definitions and I'm reluctant to use them because I don't know how to evaluate them. What does quality look like and what's going to screw me up if it's wrong?
I have hand-wired a couple of projects by cutting FR4 and drilling a lot of holes in a couple of boards with less precision than I'd like. I have a simple model of how an unsophisticated through-hole based board design should work, where if the holes are in the right places and there is connectivity between the right holes, this is what actually matters? But there seems to be a lot of other stuff involved in drawing schematics and maybe simulation, design rules and so on, and I don't know how much of it is important.
I'd appreciate any explanations or references to conceptual introductions you might know about.
AI: To make a pcb you need to have the information required for every layer the board house needs (copper, soldermask, silk and the board outline). These layers are typically sent via so called gerber files.
The PCB Layout software uses footprints that define the interface with every component (Pads in a footprint define the places where the leads are soldered to, graphics on technical layers help for documentation like silk screen)
To define which pads need to be connected with traces (or other means) you use the connectivity information that you get from the schematic.
The schematic is build up from symbols that are the abstract view of the function of every component (The schematic is then the abstract view of the systems intended function)
Symbols define pins that are mapped to footprint pads (via the pin/pad numbers and the footprint field of the symbol). You define the connectivity information by using wires that are used to connect pins together. (or you can also use labels or power symbols)
I have made a KiCad version 5.1 specific tutorial that shows the minimum requirement for getting a finished board designed. https://forum.kicad.info/t/tutorial-introduction-to-pcb-design-with-kicad-version-5-1-getting-started/20600/
There are other similar tutorials out there. Some also as videos if you prefer that style. But make sure they really use the correct version of KiCad as the interface is quite different between different major versions. |
H: what is the Lt1086-5 Vin min?
I have LT1086 5V regulator
what is the Vin min to provide 5V output?( assume load current 500mA -800mA)
I tested it worked for 6V Vin,
but datasheet does not mention anything about Vin min
https://www.analog.com/media/en/technical-documentation/data-sheets/1086ffs.pdf
AI: You can consider the drop out voltage parameter to be a design constraint. For a output voltage of 5 V, 6.5 V should be the minimum input voltage.
Please note that, the 1.5 V is the drop considering load current of 1.5A. for a lower load current, the drop will be lesser. |
H: Can I re-use pins that are labeled as the same physical STM32G4 Nucleo pin?
Poorly written question. Let me elaborate.
The Nucleo-32 board I'm developing (STM32G431KB) in STM32CubeIDE shows on the CubeMX "pinout view" that there are 4 distinct pins PA6, PA15, PA5, and PB7, each with their own unique set of alternate functions. I assigned a function to each of those pins (although you don't see PB7 or PA15 populated in this image for reasons discussed below):
So after configuring my peripherals, I then went about wiring the board according to the ST User Manual. That's when I realized PA6/PA15 and PA5/PB7 are both assigned to a single pin for each pair (header CN3, pins 7 and 8):
My question is thus: am I prohibited from using a pin if I am simultaneously using a completely separate pin on that same physical header pin? For example, if I'm using alternate function SPI1 SCK on pin PA5, is there no possible way to use pin PB7 for anything at all? Or is there some sort of fancy GPIO muxing feature that I'm not aware of?
AI: For example, if I'm using alternate function SPI1 SCK on pin PA5, is
there no possible way to use pin PB7 for anything at all?
Of course there is. Just remove solder bridges SB2 and SB3 and you'll have separate PA5/PA6 on pins 8/7 of CN3 and PB7/PB15 on pins 7/8 of CN4 |
H: Selection criteria for microcontroller to read data from a car's CAN bus (from OBD2 to microcontroller)
I want to read data from a car's CAN bus using a micro-controller. The micro-controller is used outside the car (via OBD to micro-controller).
A few micro-controllers i found with CAN interfaces are Teensy 4.0, ArduinoDue with dual CAN interface shield,LPC1768 etc.
Im not sure as to:
On basis of what parameters to choose a microcontroller?
What are the components need to make a connection from the OBD port to microcontroller, do we need a CAN transciever and a controller with a microcontroller? which i found being used in
threadInterpreting CAN bus data from OBD port
Im new to embedded systems.Any suggestion would be really helpful.
Thanks :)
AI: If you choose a microcontroller with an integrated can controller then you need to buy only the CAN transceiver. If you want to use Arduino, then a CAN controller and a CAN transceiver or a single module with both CAN transceiver and CAN controller built in can be used.
Microcontroller decision is left to you depending on other product requirements such as memory, interfaces, performance and power.
Please take care of ESD, filtering and transient protection too.
Random example of a CAN controller |
H: How to rewind microwave transformer to isolation transformer
I'm thinking of getting an auto transformer and rectify it to make mains isolated 240V 8A DC power supply. I would like to know if its possible to rewind an old microwave transformer to 1:1 isolating transformer with maximum output of 2kW. I know that the waveform wouldn't be smooth so i'll be using beefy 450V 2000uF capacitor for rectification.
AI: The following is largely complementary to Dave Tweed's answer
The welds are generally along the surface and can be removed with an angle grinder "fairly easily". The laminations can be rebuilt using a clamp to hold the core together. Epoxy helps prevent lamination chatter but DO NOT introduce any additional air gap (or "epoxy gap") between laminations.
The magnetic shunt must be removed if the transformer is to have anything like normal regulation. Otherwise the transformer has an intentionally designd "droop" in its load characteristic.
Microwave ovens tend to run the iron rather hard - well up the saturation curve.
Adding a relatively small number of extra turns greatly reduces the magnetising current.
It may not be true in all cases but in at least some transformers the mains primary on its bobbin comes off easily as a complete unit. If the secondary is then removed - destructively or otherwise the available space left from the secondary & shunt would allow two identical primaries to be added.
To get 2 kW you'll need two typically sized transformers - or primaries from 4 identical ones installed on two cores. These can then be connected with primaries in parallel and secondaries in parallel.
If you add extra turns to the primaries and not the secondaries you'll get a degree of stepdown which will reduce the DC level that youd otherwise get. A sinewave will peak rectify to 1.414 the RMS ac value or in this case 240 x 1.414 =~ 340 V. Under load this will be somewhat lower and mean DC depends on the degree of filtering.
With 2000 uF filtering at 8A you'd probably get 10-20V of ripple.
V drop across a hald cycle at 8A =
Vdrop = t x i / c = 0.01s x 8A / 0.002F
= 40V.
That assumes instantaneous peak charging of the capacitor at the start of each half cycle - which would not be the case. But it gives you a feel for the order of ripple voltage involved. |
H: Canbus, Opencan and ISO BUS
Quick question,
I know about CanBus protocol. I'm interesting about ISOBUS and CanOpen.
I know that those are exentsion of the CanBus protocol. So my question is:
Do ISOBUS and CanOpen need different hardware?
Do ISOBUS and CanOpen can operate on an existing CanBus network?
Do the changes for those protocol is on the software layer or is it on the chip/hardware layer?
Thanks!
AI: No, both are ISO 11898.
No. ISOBUS is a derivative of J1939. But I do not own the documents of ISOBUS.
CANopen is unique above the data link layer, it basically eliminates the option to have multiple protocols on one bus. Most of them don't allow this actually. But ISOBUS and J1939 are compatible because they are designed to be.
Since both use CAN bus physical and data link layer, you only need software and maybe a different connector type.
If you are going to work with ISOBUS you should buy the standard! |
H: 4G LTE: Is it possible to make a high gain antenna that resonates well for two frequencies, where one isn't a close harmonic of the other?
I want a high gain antenna for Verizon LTE data to my JetPack 7730L but don't really see one as high as say this MIMO parabolic antenna that's probably design for 2.4 GHz WiFi. From all I know it's only possible to make an antenna that's turned well to one frequency but could also work at harmonics of that frequency like 1/4, 1/2, 5/8 (not sure of the gains for these).
For instance a 1/2 wavelength antenna for 2 GHz could also double as a 1/4 wavelength antenna for 1 GHz based on the equation...
\$\lambda =\frac{c}{f}\$
But is it possible to create an antenna that is tuned to a different frequency outside of the near natural harmonics? Like say
These frequencies I am interested... all the Verizon Wireless bands and I'm not sure if it's helpful at all the have the fraction of wavelength... but that's just the thought that first came to mind.
Frequency Fraction of Wavelength
-----------|------------------------
2100 MHz 1.00
1900 MHz 0.90
1700 MHz 0.81
850 MHz 0.40
700 MHz 0.33
PS: I'm not sure if all these frequencies are used for Verizon Wireless 4G LTE data, but at least these are frequencies they use in general.
AI: Yes, that's possible, and you'll find directive multi-band antennas commercially-off-the-shelf.
These internally mostly resemble Yagi-Uda antennas, but don't have the same director sizing as actual Yagi-Udas. Probably, they were designed starting as Yagi-Uda, and then stochastically optimized in simulation until they worked well enough on multiple frequencies.
By the way, this 90% already was covered in the answer about feeds to your last question.
So,
is it possible to make...
yes; whether or not it's possible for you depends on your willingness to learn how to simulate complicated antennas and your access to tools to build and to measurement chambers and costly measurement equipment to measure an antenna.
Really, buy a multi-band antenna. These exist. |
H: How to handle disconnected resistive sensor?
I'm trying to connect a PT100 thermo probe to an Arduino project to measure water temperature. After some research I have found and adopted a design using a Wheatstone bridge and an instrumentation amp, attached below. Using this arrangement I get 0V at the output @ -15°C and 3.3V (Vcc for the Arduino) @ 125°C, which is perfect for my application.
However, the temperature probe is designed to be disconnectable, which would create floating inputs. How could I alter the schematics to make sure that when the sensor is disconnected the differential inputs are at the same voltage, so the output also reads 0V?
AI: You need to protect the Arduino input by clamping the output voltage of the amplifier at an acceptable level. You may be able to do this with a couple diodes and resistor, or it might take more, it depends on a number of factors. As it sits, your amplifier can output from -12 to +12 or a bit less, depending on the op-amps.
Once you have it in digital form (it will be overrange for open-- +10V or +12V) you can have the input interpreted as you like. Usually we would prefer to have it either interpreted as an error (shut down the controller) or as overrange in a heating-type thermal system so the heater shuts down, but there are always exceptions. |
H: Hows does this 12v relay works with transistor voltage drop?
I'm working my way through "Make: Electronics" by Charles Platt and came across the following schematic:
As someone who just recently got started with electronics, what I can't wrap my head around is how R1 relay works after Q1's voltage drop. I have built the same circuit only with different power source and relay, both 5V instead of 12V as suggested by the book and what seems to be happening is that the emitter voltage is bellow the relay trigger voltage (12V in the schematics and 5V in my case since I'm using 5V relay and source). From what I have been reading around the internet and communities like this, the emitter voltage in this configuration is supposed to be limited to the base voltage minus ~0.4V or so. Due to the 10K resistor, once the transistor is triggered, the base voltage will be under 12V, thus the emitter voltage will also be under 12V and the relay will not close. That's basically how my thoughts are at the moment and I need help spotting my mistakes.
Edit: I believe this question is being misunderstood. I'm not asking how a self-locking relay works. I'm asking how a 12V relay is being triggered when the voltage on its coil (from Q1 emitter) is bellow 12V (I'm checking it with a multimeter and theres a voltage drop across the transistor).
AI: It seems the problem can be solved by lowering the value of the resistor between the transistor's base and +12V source. The voltage drop still happens, although it's not so big to the point of not triggering the relay as before. I'll wait for a few days since getting a more detailed explanation on the issue would be much appreciated and mark my own answer as accepted in case nothing better is posted. |
H: Can North American split-phase 220VAC supply power to a device designed for single phase power?
In North America I know we use split-phase 220V power. I am wanting to purchase an industrial power supply that has an input rating 100-240VAC, and has 3 input terminals labeled ⏚, L, N. I know the N is for neutral, which would apply to most of the world which uses regular single-phase 220V. But can I plug split-phase power (with its two hots, 180 out of phase) into those two L, N terminals safely?
AI: First of all: 220V is very dangerous! Do not mess around without proper knowledge and safety procedures and equipment. Be sure to never wire things without being unplugged or with the breakers at off! Measure voltages at all times before wiring or touching power cables. If you aren't sure, call a professional electrician.
As for the question: the answer is yes, you can safely plug split-phase 220V into the L and N terminals safely. The ⏚ label indicates Ground, or Equipment Ground. This is usually a bare wire in 220V wiring.
By using this diagram as an example you can see why we call 110V with the labels Line and Neutral. There is 110V between any Line and the Neutral. There is 220V between any two Lines.
:
(oempanels.com)
To wire your power supply, simply connect the two Lines (Hot) to the L and N respectively. ex. Line 1 (usually red) goes to the L label of the power supply. Line 2 (usually black) goes to the N label of the power supply. The ⏚ label indicates earth ground. This is usually the bare/green wire in the middle of a 220V cable, or the wire connected to the metal casing of a 220V electrical box.
ex:(copyright MS Paint) |
H: Question about the MOSFET saturation condition
I'm solving a problem in my textbook and I got stuck, and I have the solution manual of the textbook, so I went to see through the solution and I found a thing that I suspect to be wrong, due to my understanding, but It obviously could be wrong, so I wanted to get help, as I'm self-studying electronics.
At first, the book says that it will use this model for the MOSFET:
To my knowledge, and due to calculations that is previously in the book using the (switch-current source) model the boundaries of the saturation region are:
$$ V_T\le v_{in} \le V_T + \frac{-1 + \sqrt{1 + 2V_sRK}}{RK}$$
in this problem, it's about two cascaded MOSFETs:
now comes what I don't understand. I will quote the solution manual.
this condition on the saturation region is what I don't get. It's essentially the same as the upper bound of the condition that I provided above, but it has the boundaries on \$ V_{s} \$
How are both of them related as I can't see what I'm having wrong?
Also, I think that the details I provided are sufficient for my question, I can provide the whole question and answer from the textbook if needed.
AI: The first sentence you quoted from the solution manual has a typo. It should say,
First of all, if \$V_{in}\le V_T\$ then \$V_{MID}=V_S\$, so the second FET ...
That is, in this condition, the "\$V_{in}\$" for the second FET is \$V_S\$. |
H: Howland source, MAX current output?
How would one determine the maximum current output you could achieve from an improved Howland source given a specific op amp? In other words, what do I need to look at in the op amp's datasheet? (I am using a TL074)
AI: The op-amp has to supply all the output current plus a bit for the feedback. So you want to look at the current that the op-amp can supply. In the case of the TL074 it can supply 10V into 2K with a +/-15V supply, so that's 5mA, guaranteed.
It can probably supply quite a bit more current but there is nothing guaranteed, not even a typical number.
If you need high output current, you should probably pick a different op-amp with guaranteed specifications in that department. Op-amps such as the LT1014 are often specified with 600 ohm loads, so more like 15mA. There are also power op-amps that can handle amperes on the output stage. Or high voltage. |
H: Different between Inductor types?
first of all, I searched enough, but did not get the simple answer...
What's the different between color ring & power Inductor...
Is it all about wattage?
Color rings are mainly 0.25W, Don't know the other one's watt...
I'm trying to remake a "DC boost" circuit with MT3608(B6289)...
Can I use a color ring Inductor Instead of this...?
Schematic:
AI: The most important property of an inductor used in an inverter circuit is the current allowed to run through it. A tiny axial inductor may have the same inductance as a radial one but most likely it has a much smaller maximum current.
The inductance needed depends on the voltage spread you want to achieve.
I recommend to stick to the datasheet of the regulator chip closely. They usually give figures which inductors you should use for which voltage and current. |
H: How to solve a LED-resistor parallel with a resistor?
Hi there I am curious as to how you would solve this circuit to know voltage drops and currents. It came about from a previous question I had asked which showed my understanding was not quite what I thought it was.
I have looked at other questions of the exact same circuit but again I could not find an answer for myself. I will reference the sources I have found in the bottom. reference 2 seemed to answer my question but I don't understand how to solve the equation, reference 3 had a great explanation but subsequently didn't answer anything for me and reference 1 (and other sources) implied to view it as a voltage divider which would make sense looking at the circuit but when doing my own simulations it didn't seem to add up.
A lot of people talk about using 'thevenin' but I have never came across this before in all my online videos/guides/tutorials/textbooks/highschool physics classes. I am still a beginner and am trying to self teach so maybe I am just not at that level yet.
The LED is a red 2v.
simulate this circuit – Schematic created using CircuitLab
From reference 2 the circuit is the same with different values.'Eugene Sh' from that question states I would find the currents by having the voltage at the node after R1 as V1. Then finding the currents as
I1=(5−V1)/50
I2=V1/210
I3=(V1−2vLED)/200
Leaving: I1=I2+I3: (5−V1)/210 = V1/210+(V1−0.6)/200 and they solved it for a V1= X.
My First Question: Is that how you would solve every equation like this? And just use algebra to rearrange that sequence to make it: V1 = XXXXX? If so I dont know how to do that. as I get stuck moving the 2 divisions across. But if that is the way, I can look up algebra lessons or if someone wouldn't mind trying to show me how that would be appreciated.
Otherwise...
Reference 1 stated to look at it as a voltage divider
V1= Vin x (R3/R1+R3) Which for my circuit: `V1 = 5 x (21 ÷ (50÷210)) = 0.403V
But that seems way too low (unless I made a mistake). as i think my simulation says it should be 3.68V With a drop of 1.32V across R1.
My Second Question: Do you use this situation as a voltage divider? and if so what did I do wrong.
My Simulation:
My Third QuestionDoes it matter if R1 is before or after the ResistorLED-Resistor section? As I thought the total resistance would be the same regardless of R1's position before/after.
EXTRA THOUGHTS
My initial understanding was to drop the supply voltage by the LED's forward voltage and use ohms law. However From my previous question asked I learnt that both resistors ARE NOT in parallel. Therefore it is confusing for me with my understanding as I thought that the voltage would split at the node. But now I don't understand how to find the total current with the split not acting as parallel resistors. Originally R1 was after the split creating a 5v divide across each lane. But I changed it to have it uniform with my references. I understand All voltage in must equal negative voltage out. and that current prefers the least resistive path.
I was getting help on my old question but I thought it is more appropriate to create a new question. please let me know if I am posting questions incorrectly as I am still new here.
Thank you for everyone's time.
REFERENCES:
Reference 1
Reference 2
Reference 3
AI: I'll try and analyze your circuit in a way that requires you to learn just a few ideas.
Thevenin Equivalent of a Resistor Voltage Divider
A resistor voltage divider looks like this:
simulate this circuit – Schematic created using CircuitLab
On the left side we have two resistors in series between a power supply. What I'd like to know is what is the voltage at \$+V_\text{TH}\$. But when I ask that question, I have to say, "Relative to what other location in the circuit?" So I've labeled another spot (node) called \$-V_\text{TH}\$, which identifies the location I've picked as the "relative to" answer. I'm asking, "What is the voltage at \$+V_\text{TH}\$ with respect to the voltage at \$-V_\text{TH}\$?"
On the right side, I'm showing you a "cheat" that you will often find in circuits. That is that one of the nodes is named "GND." This becomes the "default reference point" whenever anyone is talking about the voltage at some other place in the circuit. We just "assume" that's the "relative to" location. So now, I can just ask "What the voltage at \$+V_\text{TH}\$?" and you are then supposed to insert in your head "with respect to GND" in your own head. It's just a "common" that is always inferred whenever anyone talks about voltages at any one point. (Voltages are always a "voltage here with respect to a voltage there," as they are always relative measurements and have no absolute meaning.)
With that in mind, we can work out the voltage at \$+V_\text{TH}\$. We know the current through the series circuit is \$\frac{5\:\text{V}}{1\:\text{k}\Omega+4\:\text{k}\Omega}=1\:\text{mA}\$. But \$1\:\text{mA}\$ through \$R_2\$ will cause a voltage difference from one end to the other end of the resistor of \$1\:\text{mA}\cdot 4\:\text{k}\Omega=4\:\text{V}\$. So it must be the case that \$+V_\text{TH}=4\:\text{V}\$ with respect to GND (or \$-V_\text{TH}\$.)
This is the often called the Thevenin voltage at \$+V_\text{TH}\$ (with that implied reference to GND, of course.)
There is also a Thevenin resistance. This is a little trickier to gather up, at first. But in this case it will, in effect, just be the resistance of the two resistors take "in parallel" to each other. There's a reason for this. But for now, just believe me. So the Thevenin resistance is \$1\:\text{k}\Omega\mid\mid 4\:\text{k}\Omega=800\:\Omega\$. This allows us to write:
simulate this circuit
Note that the voltage has changed. But we've also simplified the schematic so that there is only one resistor instead of two. This usually makes further analysis easier to do. But before proceeding, let's test this.
Let's consider two different resistor values that we'll place between the \$+V_\text{TH}\$ output wire and the GND wire. Suppose we use \$R_\text{LOAD}=800\:\Omega\$ and \$R_\text{LOAD}=1200\:\Omega\$. We'll analyze the first circuit and then we'll analyze the "Thevenin equivalent" circuit for both cases. So we'll have four results and we'll compare them.
simulate this circuit
On the upper-left, we have \$800\:\Omega\mid\mid 4\:\text{k}\Omega=\frac23\:\text{k}\Omega\$ that is in-series with \$1\:\text{k}\Omega\$. So the total current from the power supply will be \$\frac{5\:\text{V}}{1\:\text{k}\Omega+\frac23\:\text{k}\Omega}=3\:\text{mA}\$. This means that \$R_1\$ will drop \$1\:\text{k}\Omega\cdot 3\:\text{mA}=3\:\text{V}\$, leaving \$+V_\text{TH}=5\:\text{V}-3\:\text{V}=2\:\text{V}\$. From this, we find that \$I_\text{LOAD}=\frac{2\:\text{V}}{800\:\Omega}=2.5\:\text{mA}\$.
On the upper-right, we have a total current of \$\frac{4\:\text{V}}{800\:\Omega+800\:\Omega}=2.5\:\text{mA}\$. Note that all of the total current is flowing through \$R_\text{LOAD}\$. So this matches up with what we just calculated for the upper-left circuit.
On the lower-right, we have \$1.2\:\text{k}\Omega\mid\mid 4\:\text{k}\Omega=923 \frac1{13}\:\Omega\$ that is in-series with \$1\:\text{k}\Omega\$. So the total current from the power supply will be \$\frac{5\:\text{V}}{1\:\text{k}\Omega+923 \frac1{13}\:\Omega}=2.6\:\text{mA}\$. This means that \$R_1\$ will drop \$1\:\text{k}\Omega\cdot 2.6\:\text{mA}=2.6\:\text{V}\$, leaving \$+V_\text{TH}=5\:\text{V}-2.6\:\text{V}=2.4\:\text{V}\$. From this, we find that \$I_\text{LOAD}=\frac{2.4\:\text{V}}{1.2\:\text{k}\Omega}=2\:\text{mA}\$.
On the lower-right, we have a total current of \$\frac{4\:\text{V}}{800\:\Omega+1.2\:\text{k}\Omega}=2\:\text{mA}\$. Note that all of the total current is flowing through \$R_\text{LOAD}\$. So this matches up with what we just calculated for the lower-left circuit.
I think you can see, at least from these examples anyway, that it appears this "trick" works. There exists a Thevenin equivalent for a resistor divider and you can work it out using the above rules.
Hopefully, the above convinces you. But to be sure, you should repeat the above process any number of ways and verify that it "just works right." It will, I assure you. But you should repeat this process many times with varying values in order to be more fully convinced. It's worth the effort.
Applying the Thevenin Equivalent to your Circuit
If you think closely about your circuit, you'll see the following as true:
simulate this circuit
If we assume that \$V_\text{LED}\approx 2\:\text{V}\$ here, then we would find that \$I_\text{LED}=\frac{4\frac1{26}\:\text{V}-2\:\text{V}}{40\frac5{13}\:\Omega+200\:\Omega}=8.48\:\text{mA}\$.
This is pretty close to your simulation value. However, take note that your simulation yielded \$V_\text{LED}=1.91\:\text{V}\$. If we use that value instead, we find that \$I_\text{LED}=\frac{4\frac1{26}\:\text{V}-1.91\:\text{V}}{40\frac5{13}\:\Omega+200\:\Omega}=8.85\:\text{mA}\$. And that's almost an exact match with the simulation you tried.
Summary
If you work with this simple resistor divider's Thevenin equivalent for a while, you'll find it works well in every situation. So spend some time with it until you are comfortable with how it works for you.
There is much more to Thevenin equivalent circuits. But the voltage divider is a good way to start learning about Thevenin equivalents. I hope you will practice this, over and over, for a while to make sure you accept how well it works for you. Eventually, you can learn more about why it works like this. But for now, it's enough to see that it does work. |
H: Capacitor ID? Polar or Nonpolar?
I have been trying to repair an old 14 transistor radio from the mid-60's and have no sign of any image of the radio or schematics for it online.
One step I'm taking is to replace old caps (mostly the electrolytics) and I found a capacitor that appears to take the common form factor of an electrolytic capacitor, but has no clear marking for any kind of polarity.
The picture I have included shows the only marking on the capacitor and I would like to know if anyone can identify what types of capacitor it could be (electrolytic, tantalum, etc). In addition, an ID of the polarity could be useful as well. (One leg is longer than the other, but it was mounted with one hole further from the base) (Red coloring is my doing as well)
Another possibility is a test to determine polarity?
I know this is a tough ask, but I would appreciate if anyone could offer suggestions based off experience.
Thanks so much!
AI: The leg closer to the screen has a clear + on it, also longer leg is +ve by convention. It's an electrolytic cap. 10mfd is 10uF at 6V. |
H: Cost to run an Arduino device 24/7, simple maths question
I have built an Arduino window automatic closer/opener. When the device is on stand by mode, it draws 12V DC 0.025A.
My quesiton is, how much yearly I need to pay to have this on standby mode?
Would the equation below right?
12V, 0.025 A = 7.2Whour per day. It is then = 365*7.2 = 2628 W hr per year = 2.68kWhour per year.
If electrical cost in NSW Australia 27.56¢/kWh, so it will cost my 2.68*27.56¢ per yer = 72¢ per year.
Above calculation is based on the fact that the 240VAC to 12VDC is at 100% efficiency. I know it is never going to be 100%. Would I assume 90% ok?
So it will cost me 72¢/0.9 = 80.4¢
what do you think?
thanks all.
FYI, the reason did measure the AC current was because i was a bit dangerous and complicated to hoook up. but i did it at the end.
i found out at the end that it was drawing 220vac @ 3mAmp, so the cost would be about $1.7 a year.
however, when i open the electonic box. the 5 volt linear regulator was abit warm. and i wanna totoally minimise the heat. sill working on it
AI: It looks right.
\$ P = VI = 12 \times 0.025 = 0.3 \ \text W \$.
\$ E_{daily} = Pt = 0.3 \times 24 = 7.2 \ \text {Wh} \$.
\$ E_{annual} = 7.2 \times 365 = 2,628 \ \text {Wh} = 2.6 \ \text {kWh}\$
\$ Cost_{Annual} = 2.6 \ \text {kW} \times 27.56\ \text{c/kWh} = \text {AUD }0.72 \$. |
H: How can multiple ICs access shared RAM?
I'm trying to build a game console from scratch (as an exercise, not necessarily for practicality).
What I want to do is to have multiple "CPUs", in this case one being the Main CPU and the other one controlling the controller ports. My idea is to use shared RAM, where the "Interface Controller" CPU is polling the controller ports and then writing the results to RAM, where the "Main" CPU can read it.
What I'm trying to figure out is how I would connect the RAM to the system - I doubt I can just connect the address/data pins of both CPUs to the same pin on the RAM chip directly, but I'm not familiar with what arbitration methods exist.
What is commonly done here? Is there some kind of "intermediary" chip that serves as a "switch" to select which CPU is connected to the RAM at any given time? Or maybe some kind of "buffer" that the Interface CPU writes into, and which is then somehow flushed into the RAM? Or is there a simple "Bus" that allows multiple components to access it? (I'm thinking of SPI here, with it's chip select/SS pin)
Assume a system that's using simple CPUs (e.g., Z80, 6502) and SRAM (e.g., Cypress CY62256), so nothing that supports sophisticated protocols out of the box. And if it helps, one CPU would only ever write and one only ever read.
AI: In general: There are many ways and it's not trivial. You will need to design something yourself.
If the amount of data is very small (like controller inputs) you could use a register which stores the latest data. The input side of the register connects (somehow - you'll need to design some logic) to the interface CPU's memory bus, and the output side connects (somehow) to the main CPU's memory bus. Apart from this register, the two CPUs have entirely separate memory. Done. (It seems like a waste of a CPU though)
Most CPUs don't access the memory every single cycle, so you can make them take turns. When the main CPU wants to access memory, let it access the memory bus. When the interface CPU wants to access memory and the main CPU doesn't, let it access the memory bus. When the interface CPU wants to access memory and the main CPU does, pause the interface CPU. You'll still need to design some logic.
You could give each CPU its own memory bus, but design some logic (see the pattern?) to allow the main CPU to access some of the interface CPU's memory (or vice versa). You could pause the interface CPU whenever the main CPU accesses its memory (easier) or only when the main CPU tries to access the shared memory at the same time as the interface CPU (harder, but still doable).
In any case, the specific logic circuit can't be designed without knowing all the details of how the RAM and CPUs need to connect. "Assume we have a Z80 or 6502" is not enough information to actually design the circuit. We can only make approximations and draw block diagrams.
Here are some general tips:
You can pause a "retro" CPU by stopping its clock signal. (This is not always true for modern CPUs)
In order to connect/disconnect part of a bus, you want to search for a "bidirectional bus transceiver" chip, such as a 74xx245.
Feel free to do the above things as often as you want. If you had shared memory, you could disconnect it and reconnect it every single cycle, and the RAM won't care. (connecting it at the correct time is where the complexity is)
CPUs only access a specific address at a time. They don't care what other addresses are doing. Don't feel bad about disconnecting RAM from a CPU - as long as the RAM is there when the CPU actually accesses it, it won't care.
Dual port RAM, which has two busses, exists. I don't know how common dual port RAM chips are. It's pretty standard for FPGAs to have some dual port RAM inside them, so if you're building your console on an FPGA, this is a non-problem.
(thanks to Richard the Spacecat) Many CPUs are designed to share their bus already. With a Z80 you can use the /BUSREQ signal to ask the CPU to pause itself and electrically disconnect itself from the bus (so you don't need a bus transceiver). It runs for a few cycles first, and it tells you when it's disconnected by using the /BUSACK signal. |
H: Simulation of inverting Op-Amp
I am trying to build a circuit which is controlled from an analog input signal Vin to produce a -24V signal with >50mA output. My analog card can only produce ~10mA, therefore I generate the signal using an Op-Amp. I used the analog simulation tool from Microchip (MINDI) and the simulation result differs from my hand calculation.
The circuit represents the standard layout of a inverting amplifier. I removed most of filter capacitors and other resistors from the design guideline.
Building a non-inverting amplifier with the same principle gives the correct results.
Is there a bug in the calculation of the simulation or does my circuit has a significant error?
AI: I am trying to build a circuit which is controlled from an analog
input signal Vin to produce a -24V signal with >50mA output.
To produce -24 volts from an op-amp means that it has to have a negative power rail because an op-amp cannot conjure up any voltage that is outside the range of the power rails and, your negative rail is in fact ground (0 volts): -
The MCP6V51 is a rail-to-rail output device so, if the negative rail were (say) -25 volts you could expect -24.9 volts at best.
To produce circa 50 mA requires a more specialized op-amp because the MCP6V51 cannot normally be expected to deliver more than a few mA - it has a short circuit output current of -36 mA and, under normal load conditions, the data sheet implies about 5 mA in the table on page 4. |
H: Backfeed gate output to input with OR gate made out of BJT transistors is not working
I am very new into electronic.
Can you please explain why looping OR gate output back to input does not work when using BJT transistors for OR gate? but it works with OR gate ICs
Here is the schematic: http://everycircuit.com/circuit/6264984905187328
I am trying to achieve this setup, but wondering why it is not working when OR gate is made out of BJT transistors.
Simulation Screenshot when input is on:
Simulation screenshot when input is off:
AI: It is working exactly as it should from looking at your screenshots. When the input is high, the upper BJT is turned on, and current can flow to the LED.
When the input is low, the lower BJT is pulled down via the 5k resistor so both BJTs are off, and the LED will not illuminate.
So it looks to me like it is working as it should with this configuration. If you want to design an 'ideal' OR gate, you will need more than just 2 BJT transistors and a few resistors.
For example, just have a look at a TI OR gate (DATASHEET LINK): |
H: Using a small heatsink with a high power LED (50W)
Description
I want to use a 50 Watt LED like this one (42mm*40mm)
The space to integrate this component is limited. I have only 55mm*55mm*30mm in total.
So the LED + Heatsink size must not be higher than that.
I found one that is just with these dimensions: 55mm*55mm*13mm
However the description says Fit High Power Led 5 watt 10 watt.
The second option is to use a passive 40mm*40mm*20mm heatsink like this
The LED will be put in a tight spot: plastic under and metal (don't know which one) on the sides.
2cm away from the back (from the heatsink) would be some wires.
I can leave the top cover of my device open in order for the heat to go away, no problem for that.
Questions
Is one of these two heatsink appropriate in my case? Which one would be better? I have no idea how much heat produces an LED like this
What is the risk of using such heatsink with a 50 Watt LED? Could the LED blow up? Burn?
Is it possible the sheath of the wires on the back melt?
AI: If ambient temperature is 30°C and you want to keep the LED under 80°C, ignoring thermal resistance from junction to heat sink, you'll need a bit less than 1°C/W heat sink. Passive heat sinks with this thermal resistance exist but they're going to be huge and heavy (about half a kilo, and 200x70x50mm).
Using Mouser search engine, passive heat sinks with dimensions similar to your 40x40x20mm example score about 8-12°C/W so that won't work.
So you have to use a fan, like a CPU cooler.
The example heat sink with fan you link from banggood has no specs but it is way too small anyway.
This one could work but it is too big.
I have no idea how much heat produces an LED like this
About 50W, and the LED should be kept reasonably not too hot (like 85°C maximum).
What is the risk of using such heatsink with a 50 Watt LED? Could the LED blow up? Burn?
With the heat sinks given as examples it'll fry in less than a minute.
Is it possible the sheath of the wires on the back melt?
The LED will probably fry first.
You have to do the thermal design first because that will set quite a lot of constraints on your whole design, like the shape of the enclosure, air flow, etc.
If you don't do the thermal design first then you end up painting yourself in a corner.
Your options are either to pick a cheap cooling solution and then design around that, or use something a lot more expensive that will fit in your current design, probably something like a 1U server cpu cooler with heat pipes and a flat turbine fan that will be very loud.
If you need flat, but can tolerate length, then a GPU cooler from an obsolete GPU could work too.
You could put the fan somewhere else and duct the air through the fins of a passive cooler similar to the one you posted, but you'll have to check its thermal resistance versus airspeed.
It's always a good idea to add a temperature control, or at least a failsafe which turns the LED off if it is too hot, in case dust or cat hair clogs the vents.
Remember the air has to flow, so the enclosure needs enough holes to let the cool air in and the hot air out... and it needs to prevent the hot exhaust air from being sucked back into the fan and recycled... I'm not sure how you intend to build this but if the LED has its back against a wall, then you're going to have to blow the air from one side of the heat sink for example. |
H: Top VS bottom layer conventions
Are there any conventions regarding which side of board should be top and which should be bottom? The only convention I can think of is placing the user-facing side on top if the bare board is going to be visible (say, in a shield/hat).
To be clear, I am not asking what to put on top or bottom layer. My question is more akin to, "given a complete board, which side would you call top and which bottom, and why?".
AI: This is a frequent topic in our team and mechanical team. There is no guideline as such.
More or less it is truly opinion based
The PCB side which will finally face the user (examples are the PCBs which have buttons or displays facing the user) we call it top layer.
If there is no such way to identify, then the side where we have components assembled (one sided PCB assembly) is called top layer.
The side where we keep test points for production testing (for the ICT needles) is called bottom layer..but this is just what we follow. It is no standard.
If it is a Arduino shield, the PCB facing the top may be called top layer ( my convention)
It is truly user dependent.
Also, for example, the PCB with no components side can be called bottom layer as in lilypad PCB
If it is a bare PCB product, then the side at which it wil be kept visible most of the time can be called top layer
The side facing the sky or the user can be called top layer |
H: Limit the current draw from Op-Amp
Following up on this question I redesigned my circuit with the suggestions from the comments (see comments from @Andy_aka). I want to drive the output of the opamp between 0-24V with an Vin = 0-10V. I also need to draw currents arround 50mA. At low input voltages the power consumption of the opamp would probably damage the device. With a voltage follower I want to draw the current directly from my power source. I found some circuits on the internet and tried implementing them into my design
The voltages seem to be correct for the base and the load (R3). The current from the base is still higher than the current on the load. Should the main draw not come directly from the power source and not the output from the opamp?
UPDATE: I switched the direction of the transistor per @James suggestion and the current in the base is greatly reduced, where the Emitter (load) side draws the current from the collector. I also redraw the result window
AI: This is the sort of circuit you need to be considering: -
Of importance is that the op-amp's input common mode range MUST include the positive ground rail. This makes the choice of op-amp more difficult but there will be devices that fit the bill. |
H: Choosing a digital oscilloscope for a self learner
I'm about to start a course about electronic interfaces. My intention is to learn to debug hardware by doing measurements etc. with oscilloscopes and multimeters.
According to the course, the recommended device to buy (digital oscilloscope / multimeter) is the following:
National Instruments myDAQ
The main problem with the above is that it needs LabView to work (which is not for free). I want to buy a tool that I can use without paying exorbitant software licenses.
Looking on the internet, I have found this one:
PicoScope 2000 series.
I see that it is an oscilloscope, but I'm not sure if it will work as a digital multimeter.
Is the PicoScope a valid option to replace the NI myDAQ? If not, do you know anyone I can buy that doesn't need non-free software to run?
AI: National Instruments makes great stuff, I used NI products for 20+ years at work, but is a poor choice for a hobbyist.
IMO, a serious hobbyist should have a stand-alone scope and DMM, not something that attaches to a computer. I have always used Tektronix at work, so I wanted the same for home. I got the TBS1052B for $100 off from Amazon during their holiday sale last year.
If you aren't tied to one brand, there are many other choices that will give you more features for the same price. |
H: Class AB Amplifier - Q point
In a class AB amplifier, why is the Q point much closer to the cutoff voltage as compared to a Class A amplifier which is at the center. Wouldn't this clip the AC signal if its too close to Vce.
Also, notice -Vcc, would the transistor (TR2) still operate if -Vcc is grounded?
AI: Class A biasing usually refers to a single transistor. Yes, Class A requires \$I_Q\$ to be about half-way. Class AB usually refers to two-transistor (push-pull), where one transistor takes care of upper half, while the other transistor takes care of lower half. Class B would set \$I_Q\$ at zero...Class AB offsets \$I_Q\$ with extra \$V_{BIAS}\$.
Your source diagram shows \$I_Q\$ for only one transistor (perhaps the NPN). A more complete load-line diagram might include biasing for both NPN & PNP transistor overlaid, as below. Note that with no Input signal the NPN+PNP transistors both are biased with \$I_Q\$ current flowing from +VCC through NPN collector-to-emitter, through PNP emitter-to-collector, to -VCC. This current establishes a linear load line through the zero signal point:
The NPN is supplying current to Output during the signal swings that are above the average. The PNP supplies current to Output during the signal swings that are below the average.
For the example circuit, a single polarity DC supply would work. But be aware that this is a somewhat conceptual, simple schematic, and wouldn't likely be seen in the real world. |
H: What's the difference between these diode symbols?
What's the difference between these three different ways of symbolising a generic rectifier diode: solid, outline and outline with connection between anode and cathode?
AI: The filled symbol adheres to IEEE norms, the one with the line through it adheres to DIN/IEC norms, and the hollow one does not adhere to any norm. |
H: Calculating voltage through unknown resistor
I'm asked to get the voltage Ix and Vx, as you can see in the image, the voltage is on an unknown resistor, through loop analysis I got that Ix should be equal to 1. by KVL( Kirchhoff's Voltage Law) I had
$$5(Ix-6)+1(Ix)=0$$
although, I'm not quite sure if Ix and the 10A current source interact in this equation. so that is why I'm asking this.
For node analysis, please use this image (I'm not sure about N5 being a node btw)
AI: Well, we have the following circuit:
simulate this circuit – Schematic created using CircuitLab
Using KCL, we can write:
$$
\begin{cases}
\text{I}_1=\text{I}_3+\text{I}_4\\
\\
\text{I}_4=\text{I}_2+\text{I}_5\\
\\
\text{I}_6=\text{I}_2+\text{I}_3\\
\\
\text{I}_1=\text{I}_5+\text{I}_6
\end{cases}\tag1
$$
Using KVL, we can write:
$$
\begin{cases}
\text{I}_3=\frac{\text{V}_1-\text{V}_3}{\text{R}_3}\\
\\
\text{I}_4=\frac{\text{V}_1-\text{V}_2}{\text{R}_1}\\
\\
\text{I}_5=\frac{\text{V}_2}{\text{R}_2}\\
\\
\text{I}_6=\frac{\text{V}_3-\text{V}_4}{\text{R}_4}\\
\\
\text{I}_6=\frac{\text{V}_4}{\text{R}_5}
\end{cases}\tag2
$$
Now, we can solve for the unknowns.
Using your values we get:
FullSimplify[
Solve[{6 == I3 + I4, I4 == 10 + I5, I6 == 10 + I3, 6 == I5 + I6,
I3 == (V1 - V3)/1, I4 == (V1 - V2)/5, I5 == V2/2,
I6 == (V3 - V4)/R4, I6 == V4/3}, {I3, I4, I5, I6, V1, V2, V3, V4}]]
Which gives:
{{I3 -> -((2 (4 + 5 R4))/(11 + R4)), I4 -> (74 + 16 R4)/(11 + R4),
I5 -> (6 (-6 + R4))/(11 + R4), I6 -> 102/(11 + R4),
V1 -> 92 - 714/(11 + R4), V2 -> (12 (-6 + R4))/(11 + R4),
V3 -> (102 (3 + R4))/(11 + R4), V4 -> 306/(11 + R4)}}
Now, when we know that \$\text{I}_5=4\space\text{A}\$, we get:
$$\text{I}_5=\frac{6\left(\text{R}_4-6\right)}{11+\text{R}_4}=4\space\Longleftrightarrow\space\text{R}_4=40\space\Omega\tag3$$
So:
$$\text{V}_3-\text{V}_4=\frac{102\left(3+40\right)}{11+40}-\frac{306}{11+40}=80\space\text{V}\tag4$$ |
H: Uninterruptible Power Supply - Reset the load when power is restored
My system is powered by a redundant supply (UPS).
When the input power supply is removed, the system detects this event and it performs a safe shutdown (actually it is a microcontroller that enters a power-down mode after performing few operations).
I would like a recommendation for an IC that resets the MCU when the input power is restored to restart the system.
I found some MCU supervisor circuits (like CAT825, TLV803) that reset the MCU at power-on, but they also keep the MCU under reset when the input voltage falls below a threshold, which is not desirable on my situation.
I know it can be done with a "one-shot" circuit built with a 555 IC, but I am looking for a small dedicated IC.
Thank you.
AI: I've used these in various similar situations, and perhaps one of the would suit your application:
Edge-triggered interrupt pin on the MCU
ATTiny25/45/85 as a reset controller feeding an interrupt or reset pin on the MCU (use Tiny's internal oscillator to keep part count down; use its power-on reset and brown-out detector to send whatever signal desired.)
Although an ATTiny seems like overkill, it's sometimes convenient: they are pretty small (6-pin, 8-pin) and $0.25/unit!
It's worth noting that many MCUs (including ATTiny) have brown-out detectors, which are specifically for this kind of application.
At least ATmega328 and ATTiny25 (and I'm sure many others) work like this:
ATmega328P has an on-chip brown-out detection (BOD) circuit for monitoring the VCC level during operation by comparing it to a fixed trigger level. The trigger level for the BOD can be selected by the BODLEVEL fuses. The trigger level has a hysteresis to ensure spike free brown-out detection. The hysteresis on the detection level should be interpreted as VBOT+ = VBOT + VHYST/2 and VBOT– = VBOT – VHYST/2.
...
The BOD circuit will only detect a drop in VCC if the voltage stays below the trigger level for longer than tBOD
VHYST is 50 mV (Tiny) 80 mV (Mega)
VBOT is 2.7 ± 0.2 V or 4.3 ± 0.3 V, chosen in configuration fuses (also 1.8 V on ATTiny).
tBOD is a few microseconds, 2μs on ATTiny.
(image and quote from Atmel datasheets)
Many other MPUs' brown-out detectors will have very similar properties, though of course the details will vary. |
H: MOSFET and BJT transistors
What is the main difference between a mosfet transistor and a BJT transistor.
I know both are used for amplification and can be used as switches.
But where would you use one over the other and what are the benefits of each one ?
AI: This question is really broad.
MOSFETs switch faster as switches (they transition between zero and full conduction faster). BJTs are faster (have higher bandwidth) as amplifiers devices (BJTs can move around within the "partially on" region where linear amplification occurs faster.
BJTs need continuous signal current to control. MOSFETs do not (only voltage and transient currents to charge the
gate-source capacitor). This is a strike against BJTs in high power circuits where your control current is proportional to your load current, or in amplifiers where the source impedance is very high and can provide very little drive current.
MOSFETs behave as a resistance when fully conducting. BJTs behave as a diode. This is a benefit for lower voltage MOSFETs (<200~500V) since the voltage drop for resistance is less than the voltage drop for the BJT which means it runs cooler. But as you construct the transistors to tolerate higher voltages, the MOSFET resistance increases to a point where the voltage drop is larger than that of BJTs and the BJT becomes more efficient.
MOSFET minimun gate voltage to get it to conduct is higher than the minimum base-emitter voltage required to drive a base current into a BJT to get it to conduct (at least for now) . Important if you are using very low voltages to control a transistor switch.
MOSFETs have a parasitic anti-parallel body diode (i.e. they can only block
current in one direction). BJTs do not. This can be good and bad. In some circuits with inductive loads, it means you might be able to get away without an external flyback diode and instead rely on the body diode MOSFET, even though you really should have an external one optimized for the task.
MOSFETs can current share (put in parallel to share a load current) more easily than BJTs without thermal runaway.
BJTs have higher transconductance so tend to work better as amplifiers.
BJTs are lower noise. Better for sensitive amplifiers.
For integrated circuits, BJTs are easier to match (provides numerous advantages for analog circuits), but MOSFETs can
be fabricated are smaller (required for high density digital circuits). |
H: Garbage on RS232 and no the baudrate IS correct
I have a device like this https://www.vaisala.com/sites/default/files/documents/WXT530-Users-Guide-M211840EN.pdf.
Some facts:
I have an FTDI to USB board from Sparkfun which is tested to work with other serial devices, so it's not broken.
The device mentioned above is correctly wired for RS232 and is tested with an USB -> DB9 on 19200,8,N,1 to work correct both on Mac and Windows.
The serial device prints ASCII characters one line at the time which is ended with <crlf>
My problem:
When I wire the serial device to the FTDI to USB and I use the correct baudrate, I get garbage out. I have spent numerous hours to fix this and I can't figure out the problem.
If someone have ideas, that would be appreciated.
Things I have tried :
I tried 2 different USB to DB9 that work fine both on Windows and Mac
I tried the mentioned FTDI from Sparkfun both on Mac and Windows, same problem -> garbage.
On the Vaisala device, I have connected SGND (Serial Ground) with the Vin - no difference.
My ultimate goal is to connect the Vaisala serial device to the serial of an ESP8266, but until I understand this problem I can't continue. Actually I did try to connect it and I also get garbage here too.
AI: You have a USB to serial converter provides a serial interface that uses 5V or 3V TTL/CMOS voltage levels. The Vaisala device provides a serial interface that uses RS-232 voltage levels. These are incompatible. You need a USB to serial converter that also provides a serial interface with RS-232 voltage levels. |
H: How to measure DC voltage without knowing which pin is ground
I have a connector from the motor controller to the display on my e-bike that has five pins in it. The system is powered by a 52V battery.
I would like to find out which pin carries the battery voltage, and which one is the ground. The three other pins are much lower voltage, sending 52V to them would cause damage.
My issue is that I don't know how to safely make the measurements with a voltmeter without risking shorting the 52V pin with one that wasn't the ground. Can I just hook up my voltmeter's black probe to the (metal) frame of the bicycle and safely measure the pins in the connector one by one?
AI: I would like to find out which pin carries the battery voltage, and which one is the ground. The three other pins are much lower voltage, sending 52V to them would cause damage.
Modern digital multimeters typically have an input impedance of 10 MΩ. They are so sensitive that they can for many practical applications - such as your battery measurement - be considered ideal in that they pass no current while measuring the voltage. This means that the points being measured stay at their own voltage and are not shorted or pulled to the higher voltage while making that measurement.
Switch to DC volts and plug the leads into the correct sockets.
Connect the black lead to pin 1 and probe the others with the red lead. Record your measurements.
The pin with the lowest reading is most likely to be the battery negative.
The pin with the highest reading is likely to be the battery positive.
Subtract the lowest from the highest (do it properly if the low reading is negative) and the difference should be the battery voltage. |
H: Why does this circuit cut out when an extra load is added?
I have a circuit that consists of an Arduino controlling a small PAM8403 based amplifier which is powered via an L7805CV voltage regulator. When I press a button the arduino sends an audio signal to the amp and switches on a SPST relay. The whole thing is powered by four AA Energizer Lithium batteries supplying about 6.65V.
Without anything connected to the relay (except the signal and GND pins) this circuit draws about 0.22A. However, I want the relay to switch on a small motor and heating coil that I connect to the same battery pack as all the other stuff, so that the motor and coil get powered on when the button is pressed.
However, when I press the button, the amplifier cuts out and the circuit behaves erratically. The motor and coil run fine, however.
If I connect the motor and coil to the same battery pack, I measure a current of roughly 1.5A. This seems well within the range of what the batteries can supply according to the datasheet.
What am I missing here? I don't see why the rest of the circuit is affected by a load that should be well within the power supply's limits.
AI: Your motor is a lot less sensitive to voltage than your circuit is. You are probably looking at this:
But front-page "advertising" numbers can't be trusted, especially for numbers with more complicated behaviour. You should be looking deeper in the datasheet at this:
Your battery's internal resistance is 0.12Ohms, at best. At 1.5A you lose 0.18V per cell. If your nominal voltage is 1.5V per cell, then you are 5.28V.
If your battery is brand new, the internal resistance is 0.2Ohms and you have 4.8V when drawing 1.5A if using the nominal voltage. But your brand new battery probably starts at a high enough voltage to counteract this and more so your initial voltage is higher than nominal rather than lower.
Your LM7805 has needs a headroom (has a dropout voltage) of 2V on top of its output to operate when supplying 1A. You are drawing less current than that so the dropout voltage is less but you are still probably operating the regulator in a region where it's just struggling to pass all the input voltage straight to the output but it's a regulator, not a switch so even more voltage is dropped in that process as well.
According to the datasheet of the LM7805CV manufactured by ST at 220mA, the dropout is ~1.6V. |
H: Controlling brushed motor speed with variable load with a PWM
I am considering driving a brushed motor driver with an H-bridge and PWM. I'd like to roughly control the motor speed regardless of the motor's load (envision a small robot climbing/descending hills).
Since I know that if you power a brushed motor with a constant-voltage power supply, it will always rotate at about the same speed, with the current drawn reflecting the load (conversely, drive it with a constant-current power supply, and the motor speed and voltage will vary according to load).
Knowing the above, it seems simple to get feedback via an A/D to read the voltage being applied to the motor and have software tweak the PWM, to keep the voltage on the motor set according to the speed demanded.
I would need an LC in the circuit, because I would not be able to read the voltage that's got PWM noise on it.
Does the above make sense?
AI: Your idea is wrong and won't work. But there is a very similar idea that will work.
"a brushed motor with a constant-voltage power supply, it will always rotate at about the same speed" No. If you load it a lot, it will slow down a lot.
Your idea is to basically monitor the average voltage across the motor terminals and adjust the PWM so that a constant average voltage is applied to the motor in order to maintain constant speed with torque (this would also require monitoring the battery voltage as well). But this won't maintain constant speed with torque. It will just maintain an ratio between battery voltage and average voltage applied to the motor as the battery voltage decreases as it is drained. Also, if this is your goal it's easier to just to measure the battery voltage directly and adjust the PWM accordingly. No need to measure the messy voltage across motor terminals. But motor speed will still change as torque changes.
What you really want is to measure the motor's back-EMF (BEMF) and adjust the PWM to keep the BEMF constant. The BEMF is the voltage a motor generates when it is coasting. In other words, this is the voltage the motor generates when it is being back driven and acting as a generator. But now the question is, how do you measure the voltage across the motor while it is coasting coasting if you are driving it with a PWM waveform so it is not coasting?
Well, first, you do not want to continuously being sampling the motor terminal voltage in an uncontrolled manner while PWM is happening because the drive voltage (when the switches are closed and sending current through the motor) are mixed in with the BEMF voltage (when the switches are open and current is not going through the motor)
If you want to sample the BEMF while PWMing the motor you need to do the sampling when you know both HI side switches are open so that the motor isn't being driven, but you want one LO side switch to be closed so that you are measuring with respect to GND.
Your MCU may allow your PWM timer to trigger ADC samples so you can sample it directly and ignoring it when the motor current is not being forced through the motor and ignore it at all other times. Might need some RC filtering but not something like an LC.
If you can't do this, there are less elegant work arounds like having the PWM signal drive interrupts that cause the ADC to take a sample. At worst, you could directly feed the PWM signal from an output pin directly back into an interrupt pin. |
H: Microcontroller working voltage + brownout circuit
I'm working with LPC1769, but this applies to any MCU.
The datasheet states that the power supply range is from 2.4V to 3.6V.
The included brownout detection circuit triggers a warning interrupt below 2.2V and asserts the Reset below 1.85V.
What happens when the supply voltage is between 1.85V and 2.4V ?
Why the BO circuit was not designed to reset the MCU below 2.4V ?
I know that the operation is guaranteed down to 2.4V and that the MCU might work very well even at 1.85V, but this is a grey/risky area.
Thank you.
AI: The datasheet states that the power supply range is from 2.4V to 3.6V.
The included brownout detection circuit triggers a warning interrupt
below 2.2V and asserts the Reset below 1.85V.
At first glance this appears to be silly. What is the point of having a brownout voltage set below the minimum operating voltage of the chip?
However the datasheet also says "below 1.85 V... operation of the various elements of the chip would otherwise become unreliable due to low voltage.", which implies that it is reliable (though perhaps not meeting all specs) above 1.85 V.
What happens when the supply voltage is between 1.85V and 2.4V ? Why
the BO circuit was not designed to reset the MCU below 2.4V ?
We don't know exactly what might happen, but clearly it is expected to work well enough to respond to interrupts, and presumably take appropriate action in preparation for possible loss of supply voltage.
The BOD could have been designed to reset the MCU below 2.4V, but that would make it impossible to operate at 2.4V without risking brownout reset. That voltage would then have to be an absolute minimum rating for operation with BOD enabled, not a normal operating voltage. Which would be a problem, because the feature set includes "Single 3.3 V power supply (2.4 V to 3.6 V)". With no safety margin, how could they guarantee operation at 2.4V?
I'm working with LPC1769, but this applies to any MCU.
Other MCUs may be specified differently. For example the ATmega328p is specified for operation from 1.8 V to 5.5 V, and has nominal BOD voltages of 1.8 V, 2.7 V and 4.3 V. To reliably operate this chip at 1.8V you would to disable the BOD. |
H: Class A amplifier
So i have to make a circuit for my university, and it has to be made in electronics workbench. I tried to make a class A amp after this schematic:
And made this in EWB:
My problem is that my circuit weakens the signal rather than amplifies it (here is a picture of the oscilloscope):
The first one is the input (channel B) the second one is the output (channel A)
AI: It appears that your signal frequency is 20 Hz. This is too low for the sizes of the input and output capacitors. The input capacitor and the two base bias resistors form a high-pass filter of approximately 480 Hz. The transistor's input impedance changes the actual corner frequency, but it still is way above 20 hZ.
A similar situation is happening at the output, with the circuit's output impedance and the output coupling capacitor forming another high-pass filter.
As a simple test to see if this is contributing to what you are seeing, raise the test frequency to 1000 Hz. |
H: Resistor without terminals?
I have two terminals which I will measure resistivity between. I will be manually touching the terminal pair to different value resistors.
The terminal pair might look something like a 9v battery plug:
I don't want to be precise about where I touch the resistor (lining up the terminals with the resistor wires).
Is there some kind of "pad resistor" (the size of a small coin) in which I can just touch two terminals anywhere on the surface and have the same effect as if i were using set resistors?
Note: the reason I am using resistors is so I can have a series of pads with unique signature based on resistance over the terminals. If there is some other solution in which I can have unique "pads" be recognised by a microcontroller as different without using resistors, this will work also.
Thanks!
AI: different conductive materials have a different bulk resistivity. so you could use them in this. eg get thimbles mage from or coated with conductive rubber in 4 different formulations (or the same formulation in different thicknesses) but the readings you get will to some extent be pressure dependent.
a better solution is probably to instead make a pad with many terminals
that are all fed to a single resistor via a diode network, that way even if each of the contacts bridges multipler terminals you can still read the single resistor
simulate this circuit – Schematic created using CircuitLab |
H: What happens when two different voltages are connected to a car battery
I have purchase a battery charger solution that has a two part system. I am having a few problems with it which has led me to realise that the unit supplies two different voltages to the battery at any point in time.
The system has two units, they are both connected to the alternator and the battery in parallel.
The first unit is effectively a solenoid, it delivers up to 120Amps directly to the battery. It does not change the voltage, whatever voltage the alternator is outputting gets delivered to the battery.
The second unit is a "smart charger" DC/DC converter, it takes the alternator input and converts at the output to a particular charge profile. This can be up to 14.7 volts depending where in the charge profile it's at. It will pass a maximum of 20Amps to the battery.
So together the two units can deliver 140Amps. Of course this is limited by the amount the battery bank can take and the size of the alternator. MY alt is 115Amp, the most I have ever seen going into the battery is 63Amps which teters off quite quickly.
Apart from the fact that the two devices are connected in parallel they have no way of communicating with one another.
My question comes about because I always understood that you couldn't connect DC sources of different voltages in parallel.
So if my alternator is outputting 13.8 volts (which it seems to do all the time) and the DC/DC converter is outputting 14.7 volts, which it will do at some point in it's charge profile, what voltage does the battery see and how can this even work?
The product in question is a CTEK250 and CTEK120 for further reference.
Thanks
Block Diagram:
The earth cable from the SmartPass is tiny (0.05mm2??), while the positive cables need to be 35mm2. The cables to/from the DC/DC smart charger are all 4mm2.
AI: When the SmartPass conducts to allow current from the alternator to flow directly into the house battery bank, it shorts out the Smart Charger disabling it. This is similar to placing a copper wire right between the input and output of the Smart Charger. This forces the Smart charger to basically turn off. Note this is different from a short-circuit where you connect the output of a power source straight to ground. This is a short across a device which electrically removes the device from the circuit.
Put another way, electrons take the easiest possible path. If you provide an alternate path that takes practically zero effort to traverse, practically all electrons will take that path and practically no electrons will take the other paths leaving the devices in those paths without power to operate.
So in your scenario at least, you are not actually connecting different two voltage sources together (unless you consider the output of a functioning battery charger or alternator being connected to a battery to be that which it is. But you already know what happens there. The larger voltage wins and shoves current into the devices with lower voltage) |
H: How can different voltages produce the same short circuit current?
I hope I’m not just thinking past something trivial but I can’t make sense of this. In this video, https://m.youtube.com/watch?v=XDf2nhfxVzg , variable voltages from .1-20V are used and when he shorts out the wires, they all produce the same amperage. How does this work?
AI: In the video, the power supply can provide a maximum of 150 amperes.
If the resistance of the short is low enough, then more than 150 amperes would flow.
The cables used in the video are very thick and so have a very low resistance, and would allow more than 150 amperes of current to flow.
But, the power supply can only provide 150 amperes so you only ever see it putting out 150 amperes.
Take a couple of examples:
Set the power supply to 1 volt. Assume a cable resistance of 0.005 ohms. That would be a current of 200 amperes. The power supply can only deliver 150 amperes, so that's all we see in the video.
Now set the power supply to 20 volts, but keep the cables with 0.005 ohms. Ohm's law says you should get a current of 4000 amperes. The power supply can still only deliver 150 amperes, so that's all you will get. |
H: Switching a 5V power relay with optocoupler
I have looked at the schematics of a few 5V power relays and noticed different designs. Some use just a transistor to draw the higher current to switch the coil and some use optocoupler (surge protection) with a phototransistor.
Regarding this schematic:
Why do I need an extra transistor to switch the relay? Can I not just switch with the relay with the transistor of the optocoupler.
AI: Why do I need an extra transistor to switch the relay? Can I not just
switch with the relay with the transistor of the optocoupler.
The 4N25 has a minimum current transfer ratio of 20% and that means that if you push the absolute limit of continuous current into the photodiode (60 mA), the photo-transistor might pull 12 mA through the relay coil.
And, for most 5 volt relays that won't be enough current.
You might have noticed that the absolute maximum figure for continuous collector current is 50 mA and, you might wonder how that can that be achieved given the minimum CTR and maximum diode current. Answer: the typical CTR may be around 50 % but it might be higher in some cases so, it's important to design around minimum and maximum figures. |
H: Delay between steps, given speed in mm/min [stepper motor]
I'm building a device that have to move in orders of thousandths of mm per minute. I've got suitable stepper motor and driver, it is working correctly but I'm unable to figure out the formula needed to calculate delay between steps to get to proper speed. I have Nextion display connected to an Arduino, that sends the speed (in mm/min), but I just can't figure out how to calculate the delays. I've measured following:
1 step = 0.000004767mm
Driver is set to 1/6400 resolution
I have two end switches that will trigger interrupt once it reached end/start, so no worries about length. I also need the delay to be in us (microseconds). So far I've tried (0.000004676 / speed)x60000000. It seems to be OK in 0.0x range but when I go lower, it just doesn't work properly.
Thanks for replies.
EDIT: So the formula seems correct, but I don't get why 0.01mm/min is pretty much perfectly accurate and 0.009mm/min is done in 28s.
AI: From Arduino Wiki:
Floats have only 6-7 decimal digits of precision. That means the total number of digits, not the number to the right of the decimal point. Unlike other platforms, where you can get more precision by using a double (e.g. up to 15 digits), on the Arduino, double is the same size as float.
So doing math using floats with numbers like 0.000004767 and 60000000 is not gonna work with the standard float implementation in Arduino.
The easiest solution within Arduino is probably switching to integers and longs and use nanometers and nanoseconds as integers or use floats with micrometers and microseconds (then check if the variables are large enough to store the whole length if you need it, etc). |
H: Current limit resistors for Optocoupler controlled Mosfet
Reading different design rules regarding optocouplers used as switch I have a question about the current limiting resistors R4 and R6
One aspect is the protection of the devices (e.g. ~200Ohm for a standard LED over a 3V voltage drop @ 15mA). How can I calculate the correct values?
For the 4N25 the forward voltage is 1.5V and my digital pin can produce up to 20mA => $$R_4 = (5V-1.5V)/20mA \approx 175\Omega $$
where for the Mosfet can switch logic input (5V) but I do not know which value describes the limiting factor for calculation of the resistor R6?
My attempt is with \$V_{GS} = 2V\$ @ \$I_D = 250\mu A\$ with following consideration
$$ R_6 = (5V-2V)/250\mu A) \approx 12k\Omega$$
Edit: the datasheets 4N25 IRLZ34NPBF
AI: You should start at the mosfet Q2.
I think there is no need to drive the mosfet on a low voltage (2V). Always drive the gate at a voltage that is higher than the \$V_{GS}\$ given in the datasheet. A good guidance are the conditions for \$V_{GS}\$ used to specify the \$R_{DS(on)}\$.
When the mosfet is turned on/off at (very) low frequency, just drive it with the highest voltage available. (When driving at higher frequency, you may consider a bit lower voltage to reduce switching losses).
In your case this implies setting R6 to 0 Ω.
UPDATE With updated question, the following does not apply anymore
Note that this also implies R5 becomes superfluous as it is now parallel to R7.
Regarding the optocoupler: don't short the base connection to the emitter, but leave it open, or use a high ohmic resistor to control the sensitivity of the optocoupler.
UPDATE With updated question, this issue is addressed.
In order to get a \$V_{GS}\$ of 5V, (assuming R6 is shorted and R5 is removed), you need a current through the optocouper of only 5V / 10 kΩ = 0.5 mA.
Next, check the datasheet to find the minimum CTR: for the 4N25 it is 20%.
This means (in worst case) the optocoupler needs a forward current of 0.5 mA / 20% = 2.5 mA.
According to the datasheet, Fig 3, the forward voltage is about 1.1V.
So, worst case R4 should be equal to:
(5V-1.1V) / 2.5 mA = 1.56 kΩ
It may still work with a (bit) higher value as the CTR is typically higher. |
H: Where am I wrong in this symmetric circuit analysis?
Regarding the following circuit, SPICE calculates the current I3 through R3 as 3.333A.
I try to solve that myself and can only reach that result if I first set V1 as zero(short) and solve for the current and then set V2 as zero(short) and solve for the current. Then the total sum through R3 becomes 3.333A.
But there are some parts which are not yet clear to me.
How can we set V1 or V2 as short? How can the current produced by V1 can pass through V2 at all? As far as I know they can do that in AC small signal analysis but here we have DC currents. How come a theoretical voltage source or a power supply in real pass DC?
If I was asked this question I would first remove V1 and solve for the current through R3 and then remove V2 and solve for the current through R3 and add these currents. But that would give 5A which is not the case.
So my thinking is somehow not matching the results, because I cannot imagine how current produced by V2 can pass trough V1 or vice versa. I would think the power supply would block such current. And even if not the case, here V1 = V2 how come current flows through the voltage sources?
AI: How can we set V1 or V2 as short?
That's the rule when using superposition theorem.
If I was asked this question I would first remove V1 and solve for the
current through R3 and then remove V2 and solve for the current
through R3 and add these currents. But that would give 5A which is not
the case.
No, you have misused the superposition theorem.
With V1 removed (and a short put in its place), the current from V2 flows into R2 + R1 || R3 and that is 5 volts / 1.5 ohms = 3.333 amps (split equally between R1 and R3).
If you do the same for V2 you get the same but for R2 and R3 and, this ultimately means that 3.333 amps flows through R3. But you would have probably seen that if you get rid of V2 and connected the right node of R2 directly to V1 - you can do this because the voltage sources (V1 and V2) are equal in value.
How come a theoretical voltage source or a power supply in real pass
DC?
That's pretty much what is normally expected of them. They do it with ease. |
H: How do I find value of gain that stablizes the system from root locus?
Consider the following root locus:
which is obtained by the following code:
s = tf('s');
G = [2/(s+1) 3/(s+2);1/(s+1) 1/(s+1)];
k1 = 1;
k2 = 1;
D4 = [k1*(s+1)/s k2; -k1*(s+1)/s k2];
F4 = G * D4;
figure;
rlocus(F4(1,1))
how can I find the value of \$k1\$ that makes the closed loop stable from the root locus?
I have tried looking at the windows that opens in the plot:
and considered these valuse of gain as a bound for the gain to use,so I tried to use values of gain between \$0\$ and \$1.90\$, but the system is still unstable.
How can i solve this?
If I try to use the Routh criterion to find the value of \$k1\$ I have that the closed loop is:
\$\frac{F4(1,1)}{1+F4(1,1)}=\frac{-k(s-1)}{s^{2}+s(2-k)+k}\$
and then I use the Routh locus as follows:
\$s^{2} :\$ \$1\$ | \$1\$
\$s^{1} :\$ \$2-k\$ | \$0\$
\$s^{0} :\$ \$1\$ | \$0\$
where I have used the bar \$|\$ to indicate that the numbers belong to different columns in the table.
And so I should have that the closed loop is stable for \$k1<2\$, but if I use for example \$k1=1\$ the system is still unstable.
I have also done the Routh criterion for \$k2\$, which resulted in saying that I nedd \$k2>-0.28\$, but the system is still unstable.
What am I doing wrong?
also If I look for the gain margin, I have :
gm = margin(F4(1,1))
which gives as result:
gm =
2.0000
this if I consider \$k1=1\$ in \$F4(1,1)\$, so it agrees with my calculation using the Routh Criterion above, but why is it still unstable?
But for \$F4(2,2)\$ I have infinite gain margin, ans since I have done a partial decoupling, I have also the term \$F4(1,2)\$ for which also the gain margin is infinite. Could it be that the problem is this?
AI: I believe rlocus() only gives the stability of gain in the feedback loop (as in the form G(s)/(K*(G(s)+1)).) You can't find the root locus for any arbitrary gain that way. |
H: I lost a device's power adapter, what should i seek for a new power adapter?
i bought an expensive desk lamp from the US called Lumiy Lightsblade 1500s, i lost the power adapter, so my questions are as you are electrical engineers:
1) can i use it in israel (The standard voltage is 230 V and the standard frequency is 50 Hz.) ? does it lose its peak because the countries difference?
2) what power adapter should i seek for when buying?
AI: Yes, you can use it in your country.
Specs allow you to use any power adapter which outputs 12 VDC and can supply 2 A or more current. You need to buy a correct plug that can work with your light. |
H: Isolated MOSFET gate driver VCC and VDD power suppl
I am planning to use UCC21520 for HV SiC MOSFET based full bridge design. My operation frequency would be around 100kHz. I selected UCC21520 in order to achieve the isolation between the power and signal side. However, during design I am stuck at deciding how to select power supply for the VCC and VDDA/B.
http://www.ti.com/lit/ds/sluscq2b/sluscq2b.pdf
My questions are following:
If I use same power supply for VCC and VDDA/VDDB ( say 15V) would that work? I would use a laboratory 15V power supply ( isolated) and connect the ground of the power supply to the ground pin of the IC.
If I indeed need to use two power supplies say 3,3V +GND1 for VCC and 15V +GND2 for both VDDs, how do I deal with two new ground points? and how would they connected to the GND pin of the IC.
Speaking of grounds, I am planning to carry over the PWM signal to H-Bridge board from another PCB that has its own auxiliary power supply. I will carry over the PWM signals and and respective ground ( SGND) to the H_bridge PCB. How should I deal with this SGND with respect to aforementioned GND1 (and /or GND2).
Since, aim of using the IC is to achieve isolation from SGND and PGND, I am really confused how to deal with other grounds ? Your kind support is solicited.
AI: No, it won't work. the supply of the high-side transistor (VDDA) needs to be floating. That is normally achieved by using a bootstrap circuit or a dc/dc convertor.
You need two supplies for the secondary (VDDs). The low-side can be a "normal" 15V but the high-side needs to be floating.
It is normally recommended to joing signal gorund and power ground on one point only (called sometimes star connection) |
H: How can I add (kHz to MHz) high voltage signal to a high voltage DC offset?
I am working with MEMS structures and need to move a device (switch) in a operating point with quite high offset. (150V DC offset and a voltage swing of 60V) To generate the high HF Voltage (up to 5MHz) we use a System WMA-300 from Falco Systems.
Do get this DC offset on this system I was inspired by this answer to use a transformer (e.g. WB1010) to offset the input voltage.
In the end I want to record a response function of the device depending on the frequency (between 1Hz ideally (but 5kHz is fine) and 5 MHz) of the driving signal (between 120V and 180V). The signal that needs to arrive at the MEMS device should look as follows (Sorry for the badly drawn chirped sine).
When thinking about the solution, I was mainly afraid that somebody would operate the system wrongly applying a low frequency signal to the input coil thereby burning the primary coil. (50 Ohm source impedance, and 0.3Ohm coil resistance)
Do I need to prevent this by adding a High-Pass filter in front of the primary coil?
Is there a better idea how to create this signal?
There is a better way to do it (see the accepted answer), seems the transformer is overkill.
AI: The solution depends upon how quickly you need to vary the DC bias (if at all).
Just add a resistor in series with the DC source to the MEMS then a capacitor from the AC source to the MEMS.
The value of the resistor depends upon the leakage current in the MEMS, since they typically have very low leakage this could be 100k ohm.
The capacitor needs to have a reactance that is small relative to the resistor at the lowest frequency of operation. A value of 0.01uF would be appropriate for the frequencies you mention. |
H: Is it worth it or safe to design your own power supply?
I am currently building some kind of game console based on a Raspberry Pi, with some features (one discussed here), right now I am using for this project a Meanwell power supply.
However, I was thinking to design soon or later a PCB instead of wiring a power board and an arduino uno, since it can be so cheap nowadays to design it with KiCAD and then print & send from China.
So make a board with some microcontroller for I/O, measurements and so, and a power supply board too, in order to make all the system embedded.
Is it worth the hassle ? As far as I know a transformer, bridge rectifier and a circuit with several LM2596 or other switching power supply IC to generate 3.3V, 5V and 12V doesn't seem hard to make, but are there any downside other than risks of shock ?
Edit : This is only for my personal use, I won't mass produce it or sell !
AI: If you're building a few for personal use, I'd say it's not worth it unless part of your goal is to learn power supply design techniques and principles. Even then, it's much safer to get a wall wart to convert your line to, say, 12VAC for safety.
If you're planning to go into production with it, you should consider your volumes first. The lion's share of the development cost will be the safety and conformance certifications needed to plug it into the wall, which can run into thousands of dollars. Even so, you'll be hard pressed to match the cost performance of a commodity OTS supply with your lower volume custom design. |
H: Learning process automation starting from general engineering background
I'd like to study process automation by myself, could you suggest a good entry-level book to start with?
I have engineering college math level and I remember a bit of Laplace transforms to model systems. I have successfully implemented many simple PID and fuzzy logic control with microcontrollers, like tank-level control, pH and more.
My goal are to model systems, test automation strategies in Simulink (instead of time-domain simulations like I did until now), and be able to solve multivariate/interactive situations through decoupling (RGA etc.)
Any suggestion would be very appreciated!
Best,
AI: As an entry-level book I can recommend Modern Control Engineering from Ogata. It teaches system modelling and control theory, just what you're looking for.
Also, take a look at Scilab and Xcos, which are open-source alternatives to Matlab and Simulink respectively. Since these are open source, there is a lot of free information for these platforms on process automation. |
H: Function generator in the nano or micro volt range
I'm looking to get a hold of a wave function generator that can produce signal in the micro or nano volt peak to peak range. The lowest I've been able to find is in the millivolt range. Do these exist?
AI: At relatively low frequencies you can simply divide down the voltage from an ordinary function generator. Very accurate, cheap and fairly low output impedance is easy.
For example a 100K resistor and a 10 ohm will give you 10uV from 100mV function generator output with 10 ohm source impedance.
You can use two stages to get even lower, but below 100nV or so it's hard to measure.
The Johnson-Nyquist noise of a 10 ohm resistor is 12nV (RMS) at 1kHz bandwidth (at room temperature) so there's a limit to how far you can go, but if you're asking this question I suspect there isn't an issue.
Another approach which works only over a range of frequencies is to use transformer. For example, you could have a 1000:1 divider feeding a pot core transformer with another 100:1 ratio. If the output winding is low resistance (maybe a turn or two of copper wire) it will have very low noise. Getting to work over the audio frequency range is not hard for such low voltages but you should do the design calculations. |
H: How much average power delivered to each antenna?
I am confused if the characteristic impedance of line 1 is simply 50 ohms. Also, I know that \$P_{av} = P_{av}^i + P_{av}^r\$, but I do not have the voltage amplitude \$|V_o^+| \$ so I am unsure how to proceed.
AI: In comments you said,
I know that the two loads simplify to 75/2. And that from there, the input impedance equals that of the new load.
So you know the effective impedance of the whole load network.
From that you can find the power delivered to the network. The only place in the network where this power can be absorbed is in the loads.
Now, looking at the symmetry of the problem, is there any reason to believe that antenna 1 receives more power than antenna 2, or vice versa? |
H: How to connect a photovoltaic to a heating element directly
I am trying to connect a photovoltaic panel directly to a heating element (coil) without using a battery or an inverter and switch it on or off by using a transistor or a thyristor.
I am well aware that the power won't be constant throughout the day and not have power at all in the night, so you don't have to warn me about that.
The panel supplies a max of 37V and a max current of 8.3A. First I was looking for a high current MOSFET and found quite a few. At home I had IRF3205 which are rated for currents up to 110A and 55V.
Just to be on the safe side, I connected two of them in parallel by following two schematics I found on stack overflow. You can see them in the pictures below:
And here's the link to the stack overflow post: Parallel MOSFETs
I placed the MOSFETS on a large heatsink and then I connected the photovoltaic panels. The MOSFETS have a diode between source and drain which got shorted in just a few seconds after I connected the power from the panels.
I also have a large diode connected to the heating element to prevent reverse current.
I never connected the +5V DC to the gate of the transistor which means it was always closed.
I have also tried using TIP35C which shorted out just like the IRF3502.
Can anyone suggest a solution to this problem with any kind of transistor, or maybe using thyristors and an optocoupler?
Just to clarify some things, the transistor or thyristor will be connected to a microcontroller which will control the switching of power from the photovoltaic panel to the heating element.
AI: As shown with load in source lead, Vgate must be driven to Vload + Vgson to turn FET on.
For > 20V supply FET will be destroyed.
SO Place load in Drain side of cct. PV+ - load - FET_drain ... .
Floating gate may easily be driven to Vin by leakage.
FET will be destroyed.
SO Connect gates to ground with a resistor at all times (say 10k but anything smaller also OK).
Consider connecting a say 12V zener gate to ground so gates can NEVER be driven above about 12V.
5V gate drive is marginally low for IRF3205.
Probably OK but not as fully enhanced (on) as you may like.
Fig 1 & 2 in the datasheet show Vdson vs Vgs and Id.
At 8A with Vgs = 5V expect about 0.1 Ohm Vds
FET dissipation = V x I = 0.1 x 8 = 0.8W - say about 1 Watt/FET.
You have not said what package you are using.
A modest amount of heatsinking will be advisable.
simulate this circuit – Schematic created using CircuitLab
http://www.irf.com/product-info/datasheets/data/irf3205.pdf |
H: Newbie Adhesive Question
Our GPS trackers on our dogs were broken over the weekend and upon analysis I found some pins that were removed from the chip. It appears that there was some type of adhesive glue holding them onto place, but don't know what kind/type to purchase to attempt to repair. I'm not even sure if these are the correct terminologies, so looking this up is a bit difficult. Does anyone have any pointers as to what the adhesive was holding these onto place? They weren't soldered. It was like it was glue?
AI: The solder broke. That is likely to happen if there's mechanical stress on the solder alone … as in this case.
While you could try to repair that yourself with a hot air pistol, I recommend to take that piece to a repair cafe and let someone do it who does this all the time. The difference between repaired and gone forever is about 10 seconds, or 10 millimeters.
I also think this design is pretty shitty, as there is always mechanical stress on a charging connector. This connector wasn't intended for that use but for contacting a battery instead. Proper connectors for mechanical stress have additional tabs at least, if not bolt holes.
These are just pogopins on a spacer, it seems. They are intended for lab use … |
H: STM32F411 - Timer interrupt not firing
I setup TIM3 on my STM32F411 using the following code:
void Enable_Timer_3()
{
__TIM3_CLK_ENABLE();
TIM3->PSC = 8399; // Prescaler value, 8400 - 1, 84 MhZ / 8400 = 10000 Hz
TIM3->EGR = TIM_EGR_UG; // Update prescaler
TIM3->ARR = 9999; // Period, 10000-1, 1 second period
TIM3->DIER = TIM_DIER_UIE; // Enable interrupts for this timer
TIM3->CR1 = TIM_CR1_CEN; // Enable this timer
}
The timer counts just fine - inspecting the TIM3->CNT register shows it correctly counting from 0 to 9999 in the span of a second. But my interrupt handler is not called.
My handler:
void TIM3_IRQHandler(void)
{
TIM3->SR = 0;
HAL_GPIO_TogglePin(LD2_GPIO_Port, LD2_Pin);
}
It is set as the TIM3 interrupt in the interrupt vector:
.word TIM3_IRQHandler /* TIM3 */
Did I do something wrong?
AI: I do not know much about coding.. but I cannot see the place where you are enabling the interrupts..
The interrupt for timer TIM3 can be enabled by adding the following line to the timer setup function:
NVIC_EnableIRQ(TIM3_IRQn); |
H: ADC protection best practice
Say you were wanting to measure a signal with a higher voltage than the ADC could handle. Is there some best practice or commonly used circuitry that goes beyond a simple voltage divider to protect the ADC?
Or is it common for professional designs to simply rely on a voltage divider?
AI: simulate this circuit – Schematic created using CircuitLab
If you have need a resistive divider, it can limit current instead of an explicit current limiting resistor.
The diode rail clamps can only clamp to the rails if there is voltage on the rails. They won't offer protection if the ADC is unpowered. But zeners or TVSs which breakdown in reverse will provide protection when not powered (as well as powered, but they might not clamp to as constant voltage as the diode rail clamps. |
H: L298N heating up when driving a Nema 17 with Arduino
I've just bought a Nema 17 (Motor NEMA 17 40mm 1,2A 2,6kg/cm 1,8º 42STH40 1204A CNC) to build a pet feeder, and to control it I'm using a L298N driver I've also bought. I've wired everything together as shown in the following photo. I attach a 12V, 1.5A power adapter to the female jack over to the right. The Arduino gets powered up from the 5V + GND from the L298N
I'm using Arduino's Stepper library and it works ok. However the L298N board quickly heats up and gets very hot, I cannot even touch the heat sink. I'm pretty sure I'm doing something wrong. Why does it get so hot?
Edit 1: Here are the specs of the motor.
4-wire bipolar operation
Compatible with NEMA17 standard
Dimensions: 42 x 42 x 40mm
Motor shaft diameter: 5mm
Step angle: 1.8 degrees
Step Count per revolution: 200
Standard Voltage: 2V
Phase current: 1.2A
Phase resistance: 1.7 ohm
Holding torque: 4N.m (minimum)
Weight: 290g
AI: The cheap L298 modules don't have a chopper or current setting so you have to add a big power resistor in series with each winding if you want relatively high performance. You could also lower the supply voltage to about 5V from 12V which would work, but the motor would not be able to accelerate as fast.
To get 2V with a 12V supply, you would need to throw away 10V in the resistors and chip which is really wasteful. If you want to try it, you could try 1A or 1.1A, with something like a 6.8\$\Omega\$ or 7.5\$\Omega\$resistor on each winding, rated for 10W or so.
Or buy a stepper driver that has a current setting and chopper circuit. You can probably source one of those for not much more than the two power resistors. Some of them will allow you to increase the supply voltage to 24 or 36V which will give you better performance. You need to set the current (see the instruction for how to do that) to something like 1A or 1.1A) and the driver will automatically PWM the power to the motor to keep the current at close to the correct level from idle to maximum speed. Some also have an idle current reduction feature that reduces the current through the coils to a lower level so the motors won't easily turn (as they would if you completely turned the current off), but they run cooler.
Edit:
The series resistor (or PWM but I'm not discussing that here) means that the time constant L/R is reduced significantly (L being the coil inductance). In the case of a 6.8 ohm resistor in series with a 1.7 ohm coil, it's reduced by a factor of 5. |
H: How to draw/include ESD susceptibility symbol on my designs?
I am using Allegro 16.6V.
I want to draw/include ESD susceptibility, non-recyclable and recyclable symbols on my design.
How can I do this?
AI: I assume it is on the PCB you want include the symbols.
You can make them as mechanical symbols and add the graphics there.
The symbols need to be prepared as BMP file in MS paint or something similar.
Create a new mechanical symbol.
Under the File->Import->Logo menu you can import a BMP file to the silkscreen layer.
Select the correct file and click Import.
Adjust location and scaling parameters. For smaller logo set scaling factor < 1
Click Modify to apply changes.
Click OK when done.
Save the symbol to your library.
Now you can add the symbol to your design by selecting Place->Mechanical symbols and choosing the correct symbol. |
H: Zener voltage clamping
I would like to understand how a zener diode clamps the voltage.
I1 = I2 + I3, in order to calculate I2 I need the zener's resistance which is not doable since it is not a resistor. However I don't need to make such a calculation because I can just assume that Vd of zener is 5.1V and therefore Vd across R2 is going to be 5.1V, so order to calculate current in this circuit O would just use Vt = 500I + 5.1 which is I = (Vt-5.1)/500
My question is: Why do I need to assume that voltage drop of a resistor in parallel to zener diode will be equal to a drop of zener itself?
AI: If you want to solve the question manually, you need to do iterative calculations:
You start with assuming the zener isn't there, calculate the voltage drop across R2.
If the voltage drop across R2 is smaller than the rated clamp voltage (5.1V in your case) of the zener diode, you can stop and neglect the current through the (leakage) zener diode
If the voltage drop across R2 is about or larger than the rated clamp voltage of the zener diode, check the datasheet of the zener diode and hope there is a graph that shows the relation of clamping voltage to current.
It is easiest to (first) neglect the current through R2 i.e. assume R2 to be open.
Start with the rated clamping voltage (5.1V), calculate the current through R3 (Vcc-5.1V)/R3, check the voltage-current graph, adjust the zener clamp voltage and iteratively find the correct clamping voltage and corresponding current.
Then, calculate the current through R2 using the clamp voltage found in previous step. If this current is negligible compared to the current through R3, you can stop. If not, you should recalculate the whole thing involving the current through R2, using the zener's voltage-current graph, but you're likely not way off the clamping voltage you found earlier.
If you want to solve the question easy, you just simulate it with LTSpice or a comparable tool and have it solved in no time. |
H: How to wire a single wire alternator to a bridge rectifier?
I'm trying to make a 25cc weed eater generator (for fun) and I'm thinking of using an alternator to take the rotary output from the engine and turn it into electricity. I need a way to take the 12-14V AC from the alternator's output and turn it into ~12V DC to charge a 12V battery and power an inverter. The alternator I have my eyes on is a 35A ~12V output, but it only has one wire for output. I'm thinking I need to use a bridge rectifier to take the AC into DC, but I'm not sure If I can hook the single wire AC to a bridge rectifier. Is there a way to wire it to a bridge rectifier properly, or is there a better way I can go about doing this?
AI: A car alternator produces a DC output so you don’t need a bridge rectifier. The one you have linked is for a Chevy. Single wire is usually positive and metal chassis of alternator negative. |
H: Choose Extension Lead
I am trying to understand better the characteristics I need to look at when choosing an extension lead (i.e. an extension cord that plugs into a socket, and gives me multiple sockets on the other end).
I know it is best to avoid using these, but I have to as I do not have enough sockets.
I have budgeted for expensive extension leads, as I want it to be as safe as possible, so the issue is how to get the best possible.
Is the only risk overheating?
As far as I understand, the thicker the extension lead, the safer it is because it is less likely to over heat. Is there any downside in getting the largest found cable?
Some have a fuse, is it useful?
Can I use extension lead bought in a country where mains is 220 V, in a country where mains is 230 V?
Any other tips?
AI: Is the only risk overheating ?
Over heating is number one risk even incase of branded extension cord. Please make sure that there is sufficient air cooling for the cable. Problem is that when the cord is fully loaded with high current the heat generated needs good passive ventilation atleast. When the cord is coilable, the coils in the inner rings gets over heated and will be a chain reaction with higher temperature and increased resistance and higher dissipation as well to an extent of fire.
As far as I understand, the thicker the extension lead, the safer it
is because it is less likely to over heat. Is there any downside in
getting the largest found cable?
Definitely a plus provided that the cable is really a very good conductive cable than a thin cable with a very thick plastic sleeve.
Some have a fuse, is it useful ?
Fuses save lives. Yes.
Can I use extension lead bought in a country where Volts are 220V, in
a country where Volts are 230V ?
Yes for sure. |
H: Data Slicer for Differential Manchester
I have a differential manchester data stream (channel1) in the form of packets. I'm showing a capture of one packet. All packets begin with a stream of 1's. This is key, as per Maxim application note 3671: "
This circuit functions well when the received data stream has enough
extra bits at the beginning of a packet or frame (in the form of a
preamble or synch pattern) that it can afford to lose while the R1-C4
circuit charges up to the correct slicing threshold.
It is then suggested a fundamental data slicer can decode manchester encoded signals.
I need to convert this to to something an arduino, etc. can work with.
So I built the data slicer and tried various R and C value. I get a waveform as shown in channel 0. I need to obviously change my circuit, but will the fundamental circuit even get the job done?
Other Information:
This is a follow up to this question. This user stated in a comment:
It is trivial to make a Manchester decoder from simple logic and 3/4 1shot.
I think he is referring to a "one-shot multivibrator op-amp circuit", with specific voltage cut-offs?
UPDATE
This FCC filing shows the schematic for reading and writing the DeLaval Alcom bus. I have reproduced my version below minus the protection portion.
AI: The articles talk about a comparator. LM358 is an op-amp, not a comparator. In general, op-amps work extremely poorly as comparators. |
H: diesel generator load calculation?
I notice in my father’s farm , there is a diesel generator that powers 30 hp water pump as well as water pressure control or ( water pusher something like this) that is also 20 hp.
This is the diesel generator and its specs
Does anyone has PDFs, websites, or any videos that explain load calculations for the D.G, please?
I am interested in learning more about generators and motors and how each load draws current and also make sure that the DG is big enough for the load.
AI: All engine powered generator usually rated in kVA
from your spec given on picture, the maximmum is 65kVA, and minimum 24kVA.
load calculation is actually can be calculated by Ohm law.
but there is common procedure the maximum rated capacity only can be achieved by increasing the engine's speed / rpm to the max and for short period of time (<30minutes of full load).
which is not even worth it to try, either your engine will be blown or the generator will overheat.
thumb rules of basic generator : for continous operation only 50% from Full Cap Load. which is on your generator is 32.5kVA
and remember kVA is apparent power generated by generator. say, if your pump (load) is rated at 20HP = 14914Watt is not loaded on your generator by 14.9kVA but more than that.
which is calculated by your pump power factor. for your consideration the old pump usually have 0.5~0.7PF respectively with deaging starting capacitor on your pump + loading in your pump (water inside the turbine).
so 14,9/0.5(put the worst number of your PF just in case) = 29.8kVA (it's almost reaching the margin of your reccomended load). but that will be reduce after the motor fully spinning (PF increasing linear with the load in your pump decreased).
If you have the good oldie pump with 0.9PF don't worry, your generator is big enough for 30hp+20hp = 37285W = 41.43kVA = 63.74% from Full load. |
H: LTSPICE how to use the TSTOP and TSTART in Subckt?
For years I am using an automated potentiometer in LTSPICE that changes the position of the wiper from one side to another proportionally with time, I specify that time in two variables used in Subckt of the potentiometer, Tstart and Tstop. Those variables appear at the potentiometer.asy, so right clicking the part before the simulation I can change them according to my need. That in fact work very well for observing as a circuit behave when changing something like that. For other uses where a simple changing resistor value, I use a simple resistor and define its value as R=0.001+Time*10k, so if running a 100ms simulation, that resistor changes from 1mΩ to 1000.001Ω, what is neat for some testing and adjusting purposes. This approach with the resistor works, but its value depends on time, and I need to pay attention, for example running a simulation of 1s and wanting the resistor to change from 100Ω to 1kΩ during this 1s, I need to enter the correct formula, such R=100+time*900. This is not the case on the potentiometer subckt, since its total value is already defined, the change of the wiper must happen within a specific period of time, not just "time". What I wish is for the wiper to swing from 0 to 100% for the simulation time specified at the .TRAN
So, it is possible to use "time" as a value into a subckt calculation, what else can we use?
Now, I wish to use the potentiometer TSTART and TSTOP variables at my lib (potentiometer subckt) to copy the values stated at .TRAN "tstop" "tstart"...
How can it use those values if I omit them at the ASY?
The potentiometer subckt used is:
.subckt potentiometerx 1 2 3
.param TFTI=Tstop-Tstart
R1 3 2 R={if(time>Tstop,Rtot,Rtot*(time-Tstart)/TFTI*int(0.9999+((time-Tstart)/TFTI))+1m)}
R0 1 3 R={if(time>Tstop,0.0001,Rtot-(Rtot*(time-Tstart)/TFTI*int(0.9999+((time-Tstart)/TFTI))+1m))}
.ENDS potentiometerx
AI: A subcircuit is just another netlist, so whatever is allowed in a schematic, is allowed in a subcircuit. What you can't do is pass time as a parameter, i.e. x=time, because .params are evaluated prior to simulation start.
What you have inside your subcircuit are some behavioural sources which can have time passed in their expressions, expressions which act as functions, and the parameters within can be passed on to the subcircuit (Tstart, Tstop). Passing those parameters can be done by first declaring them globally (in the schematic):
.param Tstart=<...> Tstop=<...>
and then passing these as values to the subcircuit
Tstart={Tstart} Tstop={Tstop}
and the simulation card as:
.tran 0 {Tstop} {Tstart} <optional_time_step>
If I misunderstood your question, please let me know. |
H: What is the general procedure for matching networks when there are multiple circuits are connected back to back?
If I have two circuits like the one shown above (Rx antenna -> block1 -> block2 -> Tx antenna), where all should I add matching networks. How should I look at the impedance while doing the matching network? Let's say, if I am adding a matching network in front of Block1, do I have to find the impedance of block1+block2+antenna looking toward the output side or do I have to find the impedance of just block1?
AI: You need to match impedance between each block and between the antennas and blocks (every connection in the diagram). As long as you do that, the impedance of block1+block2+antenna will be equal to the nominal terminal impedance of block1 itself. In this case, you can design matching networks for each connection based on the nominal impedances. However, if you omit one of the matching networks, the effective terminal impedance of block1 will no longer be equal to the nominal impedance because of reflections from block2 and antenna, and indeed it will be difficult to calculate. So matching at each connection is the universal practice of RF engineers. |
H: Logic-level transistor not switching
I am trying to use a transistor to drive a lamp from a Raspberry Pi (3B+) and I went after one that can be switched by the 3.3V logic.
I found the FQP50N06L to have a max Vgs threshold of 2.5V, so it should work (right?).
Here's the circuit I am using:
simulate this circuit – Schematic created using CircuitLab
The lamp is rated 0.75A @ 5V, so this setup should be working, but it's not. The lamp turns on if I use 5V at the gate and goes dim for 4.5V or so.
I would suspect I chose the FET poorly, but I'd appreciate someone else's input on the matter before going hunting yet again for another transistor and found out the circuit is wrong.
So what could be the problem here?
EDIT: Yes, I forgot the word "threshold" on "max Vgs threshold."
AI: If the lamp is really 0.75A @ 5V, then it's resistance is closer to 7 ohms (when hot).
The Vgs of 2.5V is the threshold, where it will just start to conduct, so you have not chosen a good MOSFET. However, it probably would work, if the load resistance was a constant 7 ohms. But, it is not a constant 7 ohms, a cold filament can have 10 times less resistance than a hot filament. So, you never get enough current to heat up the filament. |
H: Why do AC SSRs have lower AC Voltage limits and what does this mean?
I want to use something like a Crydom D2450 SSR to switch AC current supplied by a variable transformer. This device is spec'd at 24-280VAC on the output. Why is there a lower output voltage limit (24VAC)? My device seems to work as low as a couple volts out of the transformer.
AI: Why is there a lower output voltage limit (24 VAC)?
Because some voltage is required to operate the triac.
simulate this circuit – Schematic created using CircuitLab
Figure 1. An instant turn-on SSR features an opto-triac and a regular triac.
TR1 requires some headroom to turn on. The SSR may work at lower voltages but the turn-on point will be late in the cycle because the voltage has to rise high enough for TR1 and TR2 to turn on.
Figure 2. Resultant waveform if supply voltage is too low. Image mine.
The same effect is seen at higher voltages but because the voltage is rising at a much faster rate the turn-on happens much earlier in the cycle and is barely noticible in the overall waveform.
Figure 3. The zero-cross SSR features circuitry to suppress turn-on if the voltage has risen above a certain level. Source: How opto-triacs and zero cross works.
You can read more in my article linked above. |
H: UART signal is "rounded"
I am trying to debug a UART driver for the STM32F405. It does not work above a certain baud rate (about 50 kBd). I connected it to a logic analyzer with analog capability, and I saw that the signal is rounded; it takes time for the signal to rise. It seems that at higher baud rates the signal does not rise high enough for it to be read.
How can I allow the UART to function at higher speeds? I'm not sure if this is a hardware problem, or if it can be corrected in software.
Screenshot of logic analyzer, MCU transmitting at 38400 baud.
Screenshot of logic analyzer, MCU transmitting at 115200 baud.
More details:
I am reading the serial connection with a FTDI chip connected to my desktop.
I sniffed this connection with a Saleae Logic 8 connected to the same lines as the FTDI.
In the screenshots shown, the board is sending "testing" over the serial.
The MCU is part of an OpenPilot Revolution flight controller.
AI: Check that the GPIO pin for UART TX is configured for alternative output push pull mode. It looks like it is configured for alternative output open drain. |
H: Making a 'graph-like' circuit schematic
I would like to make a parallel circuit with two resistors to most-closely resemble how a 'graph' would look, for example:
However, when I try doing something like this in Everycircuit it fails (though it works if the - side of the battery terminal is also connected to ground). For example:
A couple questions here:
Is my drawing a valid circuit? Or is it not valid until the 'negative side of the battery is grounded' ?
If it's valid, why doesn't Everycircuit accept that? Are there any other circuit programs that would accept it? I have an easier time understanding circuits when drawn closer to the top drawing than in the 'box-like' schematics.
AI: Your circuit is open so no current can flow. You must connect the battery negative to the other end of the circuit.
The convention is that we use the ground symbol to signify points that are connected on the common reference. This can be handy in a complex schematic as it avoids many wires, connections and cross-overs.
simulate this circuit – Schematic created using CircuitLab
Figure 1. Your circuit drawn conventionally in three slightly different formats. Electrically they are all the same.
You have used the earth symbol which, in my part of the world (Europe) signifies that it is actually connected to an earth rod or, as the symbol suggests, metal plates buried in the ground. (Come to think of it, yours doesn't suggest that as you have it rotated 90°. They should always point down to the earth.)
In drawing schematics the convention is that the higher voltage is generally at the top of the page and negative at the bottom. That way current flows from top to bottom and voltage reduces from top to bottom. You can invent your own scheme but, similar to inventing your own language, you'll find it difficult to communicate with others.
simulate this circuit
Figure 2. Another way - prompted by @DKNguyan. |
H: Possible ways to represent a circuit schematic
I have the following circuit with a 9V battery and two resistors, R1, R2, in parallel. Here is how it looks:
Are the following three ways all valid ways to represent this circuit?
Explanation:
In the schematic, current flow is shown, which is always positive to negative.
In #2 I've basically removed the negative terminal of the battery and put it at the bottom of the schematic.
In #3, I've moved the negative side to the right of the diagram and smoothed out the lines, so it looks more like a graph/loop and less like a schematic.
If #3 is a valid way to depict circuits (which for me is the simplest to understand intuitively), what are some resources that show how to decompose a circuit into that?
AI: You would also not use giant circles with +/- inside them. You would use commonly understood symbols. A tiny circle with a +/- beside it on the outside is fine. It makes it clear it is just a pin, and not a component unto itself. The large circles with +/- makes it look like it is its own component.
For #2, you would normally put the + at very top instead of to the side. For readability and consistency.
#3 has the problem that the large circles with +/- inside makes it look like its own component. With multiple wires running straight out of the circle, all of which are unlabelled, it is not clear if these are different connections or multiple connections to the same point.
#1 only really has the problem that the battery symbol is not standard but it's easy enough to figure out what it is supposed to be when drawn like that: One component, with two clearly labelled terminals + and -.
Curved lines for the sake of being curved also does not work for more complex circuits multiple loops, or 3-way, 4-way, or n-way intersections. Or long lines that must drop off at different locations. In larger circuits, it's very important that your eyes easily follow where things are going which won't be the case with arbitrarily curved lines that sway back and forth everywhere in a non-uniform manner. How would you draw this circuit with curved lines?
simulate this circuit – Schematic created using CircuitLab |
H: Why do series resistors have a single current running through them?
In a circuit such as the following:
The current is the same through all the resistors and is:
V = IReq
9 = I (R1 + R2 + R3)
9 = I (100 + 300 + 50)
9 = I * 450
I = 9/450 or 1/50 Amps
Why is this the case though? For example, why wouldn't there be three or four or two current running through the circuit? Why is all the current equalized with the circuit? Note that I'm looking more into the why (perhaps physics?) behind why this happens and not just so much a link to Ohms law or something.
AI: Suppose some charge exits the right terminal of R1, where can it go besides into the left terminal of R2?
If the current exiting the right terminal of R1 isn't equal to the current entering the left terminal of R2, then charge must be accumulating in the wire between the two resistors.
To model that possibility, we could include a parasitic capacitor between that node and the ground node in our model. Then we'd see that when that happens (assuming the current through R1 is greater than the current through R2), the voltage on that node will increase. Which will cause the current through R2 to increase until things balance out again.
By not including the parasitic capacitor in our model, we're saying that this capacitance is so small as not to matter for the way we're using the circuit. And any capacitive charging needed to get the resistor currents balance happens so fast after connecting the 9 V source that we can ignore it. |
H: What is a three-pronged LED?
I've used two-pronged LEDs and Diodes, but I've come across a three-pronged one:
What is this component called, and how would it work?
AI: Two-colour LED that either share anode pins or cathode pins. That's why 3 pins instead of four. |
H: Equivalent circuit of a transformer
In equivalent circuit of a transformer, there are two inductances, \$L_p\$ and \$L_s\$. These are called Leakage inductances, as the name suggests due to the leakage flux from both the coils.
1) Are these self inductance?
Cause, when we calculate the voltage across an inductor we include two components, one is due to self induction, \$-L \frac{di}{dt}\$ and the other due to mutual \$M \frac{di}{dt}\$. Are \$L_p\$ and \$L_s\$ only due to leakage flux or they generally represent self inductances of both the coils? If yes, what about the rest of the flux, won't that induce emf?
2) In the book I'm following,
In an ideal transformer, assuming there's no leakage of flux, the
induced emf in the primary coil will be given by $$e_1 = \dfrac{d\lambda_1}{dt} = N_1\dfrac{d\phi}{dt}$$ and for an ideal
transformer $$v_1=e_1$$and thus \$e_1\$ and therefore \$\phi_1\$ must be
sinusoidal of frequency \$f\$ Hz, the same as that of the voltage
source. So, $$\phi = \phi_{max} \sin\omega t \implies
e_1=N_1\dfrac{d\phi}{dt}=N_1\omega\phi_{max} \cos\omega t$$ Therefore,
induced emf leads the flux by \$90^\circ\$
How can the induced emf lead the flux by \$90^\circ\$? An induced emf is created only when there's a change in the flux, which is produced only when there's a current flow, which means the flux has to be produced first to create an induced emf.
3) Why does resistor reduce the lag in a RL circuit? I understand why current lags by 90° with the voltage across the inductor. (After watching this) But I don't understand why that lag, would be diminished by the presence of a resistor,
the resistor would just decrease the amplitude of the current.
I thought that, there would be lesser voltage across the inductor in a RL circuit, as some would fall across the R, and hence the change would also be less which produces smaller opposition. But then if I apply that reduced voltage across L alone in a separate circuit, there's a solid 90° lag, it's not reduced. If there were a capacitor, it would push the current more, and therefore it would reduce the lag, by why does the resistor reduce the lag in a RL circuit?
AI: Are Lp and Ls only due to leakage flux or they generally represent
self inductances of both the coils? If yes, what about the rest of the
flux, won't that induce emf?
Lp and Ls are leakage inductances due to imperfect coupling between the primary and secondary. Lm is the prime mover when it comes to flux because it is the magnetization inductance: -
Above image from here.
How can the induced emf leads the flux by 90∘
The current in Lm is 90∘ lagging the primary voltage because
$$V_{primary} = L\dfrac{di}{dt}$$
Current and magnetic flux (\$\Phi\$) are in phase but, there is a further 90∘ shift in the induced secondary voltage due to: -
$$V_{secondary} = N\dfrac{d\Phi}{dt}$$
Why does resistor reduce the lag in a RL circuit?
For a very high frequency stimulus the inductor would be seen as open circuit compared to R hence, as frequency rises, the 90∘ lag gets smaller. |
H: Charging current/voltage for lipo batteries
I have a 500mah 1S lipo battery.
I wanted to charge it with my lab bench power supply.
Easy, I thought:
I want to charge at 2C (which is 1A) so I simply set my power supply to 4.20V and 1A limit and done.
What actually happened is that the battery charged about 0.9C and it was slowing down the closer it got to 4.2V.
This means that to charge at a constant 2C current the whole way I have to use voltage much higher than 4.2V, but a fully charged lipo cell must not be over 4.2V or it does the FLAME ON thing.
So my question is, how do normal chargers do it? Do they simply bump up the charging voltage until they reach the desired current (so they work in constant current mode) and they every couple of seconds the stop charging and measure the battery voltage to find out if it is 4.2V and if it is, then they stop charging, and if it is not they jump back into constant current mode?
Thanks
AI: To charge any lithium based battery you generally should not be using a bench power supply unless it has a special "Charge lithium based cells" function.
But in a pinch (there's no other option) I would consider slow charging an option, set the voltage to 4.2 V and limit the current to C/10. So for a 1000 Ah battery, charge with 100 mA up to 4.2 V. So on the supply: V = 4.2 V, Imax = 100 mA. That will slowly charge the cell and charging will become slower when the cell reaches its full charge. I would not do this on a regular basis though, only when there's no other option and you must have a charged cell.
To properly charge Lithium based cells, read this article.
Most chargers use either voltage regulation, current regulation or even both at the same time depending on the state of the battery (usually derived from its voltage and temperature). Fast charging gets even more complex as the full charge current should only be applied when the battery is between 30 % and 80 % full (those percentages can vary depending on who you ask).
This means that to charge at a constant 2C current the whole way I
have to use voltage much higher than 4.2V
Charging at 2C is fast charging, you should not be doing that "the whole way", see my remark about fast charging. Even when fast charging the current must go down when the battery is getting fully charged. If you don't, the battery will overheat and might start smoking but for sure you will stress it and limit its lifetime.
The "FLAME ON" thing doesn't happen that easily, you can "safely" charge a LiPo battery to slightly more than 4.2 V but that does stress the battery and that will limit its lifetime. The 4.2 V is a decent compromise between lifetime and battery capacitance. |
H: What does this circuit with an analog video signal to the emitter and a TX signal to the base of a an NPN do?
The CAM1_AHD is the signal coming from a camera and it goes to the TVI receiver as well as to the emitter of the transistor. The video signaling scheme is analog. The TVI receiver is TP2825 and in the datasheet the pin 42 which is what the PTZ_TXD is connected to, says "Upstream data signal output. Unconnected when not used".
The TVI receiver is modeled as shown in the screenshot below
AI: The video cable from camera is also used to transmit pan/tilt/zoom control data back to camera if it contains such a feature. The logic level TXD output data from the video receiver chip is buffered by the transistor which drives the video cable with AC coupled data signal. |
H: Impedance matching an amplifier near 1 dB compression point
I am using a RF amplifier (model: Minicircuits LZY-22+) to drive 6.5 MHz oscillation in a coil. I am using an antenna tuner (model: MFJ-969) to impedance-match the output of the amplifier to the coil. Up to some power (~40 dBm), I can get pretty good matching with almost no reflection (SWR < 1.1). But as I try to increase power beyond 40 dBm, SWR goes up and it's hard to keep it below the value of 2 while I try to optimize the matching again. I noticed that this transition seems to be close to the 1dB compression point of the amplifier.
Q1: Does the nonlinearity of the amplifier near 1 dB point actually mean that the impedance is changing, and hence I have to re-optimize?
Q2: Why would the SWR show an increasing lower bound near the 1 dB point, even after trying to re-optimize the tuner?
AI: Q1: Does the nonlinearity of the amplifier near 1 dB point actually
mean that the impedance is changing, and hence I have to re-optimize?
Yes, the impedance changes. Impedance matching is a concept based on the fact that the system behaves linearly. That means doubling the input signal will double all internal signals and the output signal. Also any reflections will double.
Up until the 1 dB compression point the amplifier behaves in a linear (enough) way so that the impedance matching works as expected.
However at the 1 dB compression point the output signal of the amplifier is 1 dB lower than what it should have been if it was an ideal amplifier. That output signal being lower than expected is non linear behavior and that harms the impedance matching.
You could try and tune for that specific situation (the 1 dB compression point) but even if you find such a point it will mean that at lower power signals, the impedance matching will be less optimal. You'll need to compromise in that respect, good small signal matching or best you can get large signal matching or a compromise between those two.
Q2: Why would the SWR show an increasing lower bound near the 1 dB
point, even after trying to re-optimize the tuner?
I am unsure why this is but I think it is a consequence of the non-linear behavior I explained above. |
H: Calculating power consumption of optocoupler
I am having trouble calculating the power consumption of this optocoupler (4n25). In the datasheet the input power dissipation stated as \$ P_{diss} = 100mW \$ and forward current of \$I_F = 60mA\$. In the figure below are two setups with different resistor values.
changing the resistor value of R3 results in:
\$R3 = 100\Omega\$: \$U_{in} = 1.24V, I_{in}=37.22mA \$ => \$ P_{R3} \approx 47mW\$
\$R3 = 10k\Omega\$: \$U_{in} = 1.01V, I_{in}=0.398uA \$ => \$ P_{R3} \approx 0.402\mu W\$
Looking at Fig3 and inserting the \$U_{in}@25°C\$ results in
\$R3 = 100\Omega\$: \$I_F\approx 50mA \rightarrow P_{opto} = 62mW\$
\$R3 = 10k\Omega\$: \$I_F\approx 1mA \rightarrow P_{opto} = 1.01mW\$
(The graph only goes to 1.2V but for my thought experiment I extrapolated the values).
The consumed power of the input circuit would be
$$ P_{tot} = P_{R3} + P_{opto}$$
Is this approach correct? If this is the case it would be potentially dangerous to remove the resistor.
Also: Is it best practice to choose a the highest resistor which still guarantees that the opto functions as intended?
EDIT: This post is based on this question
AI: You're quoting the ABSOLUTE MAXIMUM RATINGS for \$P_{diss}\$ at the input and \$I_F\$. You should not use these parameters as design parameters or actual values, but consider them as values which should not to be exceeded. Like the footnote says:
Stresses in excess of the absolute maximum ratings can cause permanent damage to the device. Functional operation of the device is not implied at these or any other conditions in excess of those given in the operational sections of this document. Exposure to absolute maximum ratings for extended periods of the time can adversely affect reliability.
Regarding your calculations: \$I_{in} = I_F\$ for all situations, so, you should use the same value for the current to calculate the \$P_{R3}\$ and \$P_{opto}\$.
Then, yes,
The consumed power of the input circuit would be $$ P_{tot} = P_{R3} + P_{opto}$$
Is it best practice to choose a the highest resistor which still guarantees that the opto functions as intended?
Yes, like shown in Current limit resistors for Optocoupler controlled Mosfet
When you use a resistor value that just works for the absolute worst situation, then it works for all situations.
For linked question, a minimum current of 2.5 mA was required. It cannot hurt to increase this to e.g. 5 mA as the maximum allowed forward current is 60 mA. But still, there is no need for as the worst situation is already covered. |
H: RS232 Communication Voltage Level
I understand that RS232 Communication voltage levels should be between 3 and 25 Volts for positive voltages and the equal negative for negative voltages, but is there a source that can explain the real-life wave forms?
It is pretty obvious seeing a 'perfect' signal, but in real life, I deal with signals that have oscillating and varying voltage levels.
and I do not know what would count as a noisy refutable signal and what doesn't.
AI: The Wikipedia article has an image that's fairly representative of what you should see in real life:
The rise and fall times are non-zero, the high and low levels are not necessarily all that stable or smooth (especially with charge-pump drivers) and there's a bit of overshoot/undershoot.
But if you look at the part of the waveforms between +3 and -3V (and a fair ways beyond that) they're very clean. +/-3V are the detection levels of the receiver, so it doesn't care what happens beyond those levels, +15 is the same as +3. The additional voltage gives you noise immunity. |
H: High side MOSFET driver with transformer
I am reading the application note " Fundamentals of MOSFET and IGBT Gate Driver Circuits" written by Texas Instrument.
In order to control High Side MOSFET, the application note suggests to use this kind of circuit :
Why I just cannot use this kind of circuit ?
It seems simpler from my point … But this is probably not the case ?
Thank you very much !
AI: If you are controlling the pulse width but not using a DC restore circuit, it will mean that at a high value of duty cycle, the peak voltage from the transformer will not be enough to properly turn on the MOSFET.
Fact: the output of a transformer must have an average value of 0 volts and if the original pulse shape is 10 volts for 90% of the time, it will become 1 volt for 90% of the time after passing through the transformer: -
This is a simple simulation that shows what I mean. The input to the transformer goes between 0 volts and +10 volts (blue trace). It is capacitively coupled (1 uF) to the transformer primary. The red trace is the output of the transformer into a 1 kohm load. |
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