I wanted to know if I'm correctly thinking.

The idea for current limiting is to force the opamp to output the correct voltage to control the MOSFET. So the maximum voltage possible is somewhere around 12v.

The idea is to have a negative feedback to a certain voltage and have a positive feedback before a sense resistor.

For example in the following example, I wanted to current limit the load at 3A with a resistor of 0.1 Ohm.

At 3A, the voltage dropped by the sense resistor should be 0.3v. With that in mind, we can then compute the voltage divider for the positive feedback.

If R1 is 100K and we want to get 0.3v at the positive input then we have to drop 11.7V through R1. Which will let 0.000117A through R1. With that in mind we can compute R2 with:

V = IR
0.3V = 0.000117A * R
2564 Ohm

With that, no matter which load will be present, the sensed voltage in the positive feedback will never be higher than the one at the negative feedback. Which in return limit the current.

Also, R2 could be a potentiometer to regulate the current limit or the whole voltage divider could be replaced by PWM input.

One last thing, In my simulation it kind of work and I guess I could probably build it to try it out but there is one thing I'm still unsure. For some reasons, I tried to reverse the feedback +/- and it still work. For that reason, I'm not sure if polarity of the opamp actually matter or it is really dependent on the op-amp spec.

From what I understand, op-amp are trying to make both input equal by changing its output. In this case, only the positive input is driven by the opamp while the negative input is fixed. But I guess my assumption here is wrong and the order actually matters.


simulate this circuit – Schematic created using CircuitLab


1 Answer 1


Your circuit isn't drawn very clearly so allow me to clean that up a bit so that we can better see what's going on:


simulate this circuit – Schematic created using CircuitLab

To learn how to draw more clear schematics, read this question

If the circuit works in the mode where the feedback works then indeed it will control the PMOS such that there will be 300 mV across R4 and then the circuit will limit the current to 3 A. This happens if the Load resistance is small enough (like 3.6 ohms or lower).

If the load has a higher impedance the feedback is not positive as you mention in your question. There is no working feedback in that situation.

What happens instead is that parts of the circuit reach their limit value. For example if the load impedance is 100 ohms as drawn, there can be no more than 120 mA flowing. So across R4 there will be less than 12 mV. The output of the opamp will then be at the lowest value that the opamp can manage and that is the negative supply rail so 0 V (zero Volt). That means that the PMOS has a very large Vgs and will be fully opened so there will be close to 12 V across the load.

There are some traps with this circuit like the inputs of the opamp need to be able to operate at a very low voltage, not all opamps can do that.

Another issue might be stability, when the circuit is in current regulation the loop is not that straightforward as the PMOS, load and R4 are part of the loop.

I tried to reverse the feedback +/- and it still work

Define "still work" because the current regulation cannot work like that also the PMOS will be switched off. What might be the issue is that your setup never really worked properly due the low input voltages which the opamp cannot handle properly.

  • \$\begingroup\$ I meant that it still worked as with low enough current it would still limit the current no matter if I used negative/positive feedback. I was planning to implement it with LM358 op-amp. I was planning to use that to limit the current of a DC motor (brushed or stepper). \$\endgroup\$ Commented Nov 23, 2018 at 9:25
  • 2
    \$\begingroup\$ I meant that it still worked as with low enough current it would still limit the current That does not make sense. When the load has a high impedance of for example 100 ohms then it is the load that limits the current, not the circuit. In that situation the circuit actually does nothing. \$\endgroup\$ Commented Nov 23, 2018 at 9:51
  • \$\begingroup\$ Damn right, I just retried the simulation and I must have made a mistake because as you say it doesn't work. I have an other question thought. I remember reading that using FET as a variable resistor will cause more heat than if it was used as on/off switch. This won't switch the mosfet on and off really fast no? \$\endgroup\$ Commented Nov 23, 2018 at 12:34
  • \$\begingroup\$ I have an other question thought Then write a new question. But before that, think about: Power = Voltage x Current. For a closed switch (so also for a FET used as a switch) the Voltage will be quite small. For a open switch the current will be quite small. What does that mean for the Power? A MOSFET will only switch on/off very fast if you switch it on/off very fast. Why would it do so by itself? \$\endgroup\$ Commented Nov 23, 2018 at 14:29
  • \$\begingroup\$ One thing that is not a problem here but might be in the future with another circuit - there is no positive feedback in this circuit. The MOSFET acts as an inverter in series with the opamp output. This effectively swaps the two inputs - the one marked - on the datasheet now is the non-inverting input. \$\endgroup\$
    – AnalogKid
    Commented Mar 21, 2019 at 1:34

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