# Discharging 18650 battery at constant rate

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?

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

• Please provide a link to the MOSFET datasheet and the op amp datasheet, and tell us about the supply voltages for the op amp. Oct 4, 2018 at 0:42
• 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. Oct 4, 2018 at 1:20
• 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. Oct 4, 2018 at 3:21
• electronics.stackexchange.com/questions/397467/… Oct 4, 2018 at 3:31

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.