Note: This is a followup to a previous question that was left unanswered. The circuit has been remade on a breadboard and the symptoms are different

I have been trying to make a voltage limiter circuit work for ages now, there must be something I don't get about transistors. In the following circuit that takes a low frequency high voltage (50V max, 20V typical) sinewave to produce a stable 5V later on, the voltage limiter protects the big capacitor (16V breakdown).


simulate this circuit – Schematic created using CircuitLab

As simple as it is on breadboard (power supply DC in, no load), no matter the setting on the potentiometer the capacitor does not charge (Gate of Q2 is always the input). Is there anything I am doing wrong? I am tempted to make a PCB to rule out wiring as 2 breadboards have 2 sets of symptoms, but I'd like to be sure it's supposed to work.

Wiring: wiring

On a front view, the MOSFET's pinout is gate-drain-source.

  • \$\begingroup\$ If this is an implementation of the circuit by @Andy-aka electronics.stackexchange.com/questions/111983/… then I think R4 is on the wrong side of the pass transistor; it's very unconventional to leave no resistor path from Gate to Source. Andy used R2/R3 to determine gate drive of M2 (your M1), it looks like you replaced R2/R3 with R4 but connected to M1(M2) drain instead of source. \$\endgroup\$ – MarkU Sep 6 '14 at 0:37
  • \$\begingroup\$ Indeed that's what I did. Thanks for pointing out this is wrong; could you elaborate on leaving no resistor path from gate to source? \$\endgroup\$ – Mister Mystère Sep 6 '14 at 11:31

It looks to me like M1 will always prevent M2 from conducting, because M1's gate initially has a path to ground and it won't get a chance to rise to a higher voltage.

I'm no expert on FETs, but it seems to me that your transistor arrangement isn't robust; it tries to pull itself in two different directions and has no clear stable point to settle to. Additionally, the exact gate threshold voltage depends on temperature and that could upset any balance it might attain.

How about this modification: replace R4 with a simple comparator circuit (e.g. LM393) that compares the voltage of C1 (divided down using a voltage divider) with a voltage reference (e.g. a red LED would work, though there are much better references too). If the voltage goes above the reference, the comparator's open-collector output would pull M1's gate down, cutting off the voltage supply.


simulate this circuit – Schematic created using CircuitLab

  • \$\begingroup\$ Thank you for your answer. Then I must have been lucky with the other circuit which is behaving differently. I can see how your circuit is much better (how about a 10V Zener instead of D1 and no voltage divider?), but I don't get how it behaves much differently: previously, the gate voltage threshold was kind of comparing C1's voltage to a voltage reference even if it was a more progressive transition wasn't it? \$\endgroup\$ – Mister Mystère Sep 6 '14 at 11:15
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    \$\begingroup\$ The voltage divider helps because it brings the input to the comparator clearly below its VCC (and good voltage references tend to be e.g. 1.25V anyway), but a suitable comparator might work well enough rail-to-rail. You might actually get away with using just a third, N-channel FET instead of a comparator, relying on its VGS threshold to compare the voltage. But I don't really see a way to make it work with just two P-channel FETs. \$\endgroup\$ – Rennex Sep 6 '14 at 11:46
  • \$\begingroup\$ Right, so you're suggesting using the 5V generated downstream for the comparator's supply. But what happens at the beginning until C1 is charged beyond 5.5V? \$\endgroup\$ – Mister Mystère Sep 6 '14 at 11:53
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    \$\begingroup\$ When the capacitor's voltage is low enough, the comparator's output is floating, so M1's gate is pulled up by R2 and it doesn't conduct, so M2 conducts. \$\endgroup\$ – Rennex Sep 6 '14 at 12:03
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    \$\begingroup\$ Sorry, seems I can't edit the schematic at all. LM211 does look suitable, but it might be best to supply it direct from C1 instead of the downstream 5V (I missed that part of your previous comment). Otherwise, the 5V supply might be just turning on (or even shorted by a load) when the input voltage already goes too high, and the circuit couldn't cut off M2. \$\endgroup\$ – Rennex Sep 6 '14 at 13:08

No, the circuit can't work as drawn. If the capacitor is initially discharged, the gate of M1 is held at ground, regardless of the setting of R4. Therefore, as soon as the input voltage rises, one of the MOSFETs will turn on, depending on its exact threshold voltage.

If M1's threshold is a little lower than M2's, M1 will turn on and M2 will be held off. C1 won't charge at all.

If M2's is lower, then it will allow some current through to start charging C1, and even though M1 might also start conducting, it will have a lower Vgs across it than M2 does. Whether this succeeds in shutting off M2 is anybody's guess.

I really don't understand how you expect to get a specific cutoff voltage from this setup.

  • \$\begingroup\$ Thanks for the explanation. Obviously I was wrong about the only modification I've made to Andy Akka's overvoltage protection circuit he's suggested (putting the voltage divider on C1's side). What do you think about Rennex's circuit? Are there any ICs that would do all of that already? That may be a bit of a hack, but how about a 12V voltage regulator before C1 (if the input voltage is less than 12V, isn't the output following the input? Is the current nonzero?)? \$\endgroup\$ – Mister Mystère Sep 6 '14 at 11:29
  • \$\begingroup\$ Actually a 12V regulator might work well :-) I think it should conduct at lower input voltages too. But if the input voltage stays high and there is a load current on the capacitor side, the regulator will heat up as it's turning the excess voltage into heat, so that might be an issue. \$\endgroup\$ – Rennex Sep 6 '14 at 12:13
  • \$\begingroup\$ It's not obvious that the output will follow the input, and above all if it can source current; the datasheets I've seen were saying otherwise and I just don't know if it is ever possible. \$\endgroup\$ – Mister Mystère Sep 6 '14 at 21:28
  • \$\begingroup\$ I've just tested the circuit on a PCB, it works fine! The capacitor is discharging quite quickly with no input and no load though (2-3s to reach 0.5V), any suggestions to reduce that? \$\endgroup\$ – Mister Mystère Sep 20 '14 at 12:06

In the original circuit (by @Andy-aka Open a circuit when voltage exceeds a certain value? ) , R2/R3/M2 are the first stage: when Vin*R2/(R2+R3) > M2.VThreshold, then M2 turns on. The second stage is M1/R1 where R1 keeps M1 gate on unless M2 pulls it high. Notice that Andy drew the original circuit with R2/R3/M2 together on the left, then M1/R1 on the right. In Andy's circuit, M2's gate drive does not depend on whether M1 is on.

Your circuit is similar to Andy's (with different reference designators, M1 and M2 swapped), and you replaced resistor divider R2/R3 with an adjustable trimpot R4. This is a great idea for prototyping, since it gives you some flexibility to adjust the circuit. I wouldn't want to use a trimpot on a production circuit unless there was no other choice, but for a prototype it's a good idea.

However, you connected R4 to the C1 side (the pass mosfet drain terminal) instead of the BR1+ power input side (the pass mosfet source terminal). So when the overvoltage detection mosfet M1 detects overvoltage condition and turns on, the pass mosfet M2 turns off, and that in turn turns off M1. You want M1 to stay on during overvoltage conditions. But since M2 gate depends on M1 and M1 gate depends on M2, it's unclear whether this circuit will reliably work.

A schematic is a 2-dimensional language, and there are a handful of ways that parts are used together. When reading a schematic, we look for commonly used expressions: op-amp with negative feedback, BJT in common-emitter configuration, a diode across the coil of a relay... and for MOSFETs, something must predictably drive the gate. Mosfets are often used with external resistor between Gate and Source, to ensure that V(gs) is zero if nothing is driving the circuit. Engineers try to help make their circuit schematics easier to understand by organizing them into these commonly used blocks.

I think what happened here was a minor transcription error turned a sensible circuit into something that doesn't work. R4 should connect to M1 source, so that it is powered by BR1+ regardless of the state of M2.

  • \$\begingroup\$ Very good explanation thanks, I must admit this wasn't my brightest moment. \$\endgroup\$ – Mister Mystère Sep 7 '14 at 10:33
  • \$\begingroup\$ I've just tested the circuit on a PCB, it works fine! The capacitor is discharging quite quickly with no input and no load though (2-3s to reach 0.5V), any suggestions to reduce that? \$\endgroup\$ – Mister Mystère Sep 20 '14 at 12:05

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