Hello I am planning to do some tests with a 18650. In these tests my goal is to measure required time to discharge them with different discharge current rates (1A, 2A, 5A and so on...)

The op amp work as a comparator between logical level set by the potentiometer and voltage supplied to the load. If load voltage is lower than logical, mosfet turns on, if lower turns off.

18650 battery nominal voltage is 3.7 V (if not mistaken) so that means maximum current is 3.7 A right? Which will be when mosfet is on all the times? If I want to increase discharge current I need to change the load resistor to a smaller one? Like putting two 1 ohm resistor in parallel?

Constant Current Load

OP amp will be connected to a bench power supply (+5V and 0 V)

  • \$\begingroup\$ Please provide a link to the MOSFET datasheet and the op amp datasheet, and tell us about the supply voltages for the op amp. \$\endgroup\$ Oct 4, 2018 at 0:42
  • \$\begingroup\$ The op-amp works as an op-amp. Given the gate capacitance of the MOSFET, it'll oscillate like mad. You want a resistor (probably about 470 ohms) from the op-amp to the gate, and another one on the '-' input. Then you want a capacitor, probably about 1nF, from the op-amp output to the '-' input. That should kill the oscillation. Please note that I'm designing this in my head, so the values may be off. \$\endgroup\$
    – TimWescott
    Oct 4, 2018 at 1:20
  • \$\begingroup\$ Someone asked a question recently with a similar setup. They were blowing up MOSFET's and trying to figure out why. It would be a good idea to go find that question and read through it to make sure you understand the power dissipation issues associated with some switch-type MOSFETS's. \$\endgroup\$
    – user57037
    Oct 4, 2018 at 3:21
  • \$\begingroup\$ electronics.stackexchange.com/questions/397467/… \$\endgroup\$
    – user57037
    Oct 4, 2018 at 3:31

1 Answer 1


I built a slightly more complex version of that circuit, and yes it will work, with some changes. There's some risk in using a MOSFET as a voltage-controlled resistor because of the negative temperature coefficient of Vgs(th). A safe current when pulsed may not be a safe current when run constantly. You have a fair chance of frying the MOSFET if you're using it to dissipate power rather than switching. A safer bet would be a darlington power transistor, like the TIP10* family (with a heat sink). But when you use that transistor, the LM358 won't be able to fully turn it off because the output can't go low enough--you can drop the output voltage by adding a couple diodes or LEDs after the op amp. I also suggest adding a resistor after the op amp output or a diode after the battery to protect the battery from being charged to a high voltage by the op amp output whenever you disconnect the sense resistor. A resistor will also protect the op amp from high output. (The exact problems that will occur depend on where you put the "off" switch.) But if you keep the MOSFET, you need a resistor on the op amp output since it's not nice to use a puny op amp to drive a capacitor. (You can think of a MOSFET gate as a capacitor.)

Next, your inputs will be outside the common mode input range of LM358 if you want to test a low current. This is easiest to solve by switching to a rail to rail op amp. You don't need a super fast one. The cheapest I could find is by Microchip. I think the model was MCP6001/MCP6002 or MCP6022. If you have a dual voltage supply, you can solve both problems (input and output voltage ranges) by the LM358 negative input with a negative voltage rail.

And is the max current you can drive 3.7 A? Yes, but it will be a bit higher when the battery is full, and lower (perhaps 3.3 A) when the battery is near empty. Don't drive it further than that. But a 5 W resistor won't work. Since power is I*V or I²*R, the resistor power will be around 15 W.

Something else to think about: is the positive voltage rail stable? You wouldn't want variation on the voltage rail to change your reference voltage. If it's not stable, you can use a 3.3 V LDO regulator on top of the potentiometer where you wrote "+5V".


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