Constant current load for LiPo capacity testing

[reyntjensm]

I'm trying to measure the capacity of some LiPo cells (no specs available for these cells). My idea is to make a constant current load and measure the time for the cell to discharge from 4.2V to 3.2V. I made a constant current load ( https://www.instructables.com/DIY-Constant-current-load/ ) and this is working fine. But only for a fixed voltage. So for instance the current is constant at 4.2V, as soon as i lower the voltage. The current stays constant but not all the way down to 3.2V. The current is constant only from 4.2V to around 3.6V. What am i doing wrong? How can i change the schematic so the current draw stays constant with a battery voltage between 4.2V and 3.2V? Attached you can find my schematic.

[reyntjensm]

Thank you for your feedback.
I don't really understand the relation between the supply voltage for the opamp and the current sense resistor? I have already tried different mosfet's and resistor configuration. The result doesn't improve

EDIT:

I removed everything from the breadboard and started over again. Now it's working fine. A bad breadboard jumper... Another reason to throw those breadboards in the dumpster. I can draw a constant current for a large range of voltages.
What is the limiting factor in this circuit? The voltage across the sense resistor can't be higher than half the opamp supply voltage? 

[Vovk_Z]

Current shunt resistance is too high. Use up to order less value. E.g. something between 0.1 .. 0.5 R. (It depends on a needed current and minimal battery voltage).

[Hiemal]

As others stated, the shunt voltage is too high; think of it in terms of ohms law.

You want most of your power to be dissipated in your pass element, aka, your FET.

The lower the shunt voltage, the less power is dissipated in the resistor. But, it comes with a tradeoff, you don't want it to be TOO small otherwise it'll be too small of a voltage for your op amp to handle properly.

For a small constant current load, a 1 ohm resistor would probably work alright. If you're trying to measure capacity, I would keep the discharge current to around 500 mA, to 1 amp, depending on the rough size of the cell (cells will have markedly lower capacity the higher the current draw is)

So, to get 500 mA, or 0.5 A, you'd use ohms law. You know your shunt resistor is 1 ohm, and you want 0.5 amps. V = I * R.

You'd need a reference voltage of 0.5v. You'll also dissipate about 250 mW in your resistor.

To make it even more accurate, I'd also use a voltage reference instead of just dividing off of your supply voltage. It'll be more reliable that way.