Constant Current Load ampere measurement
Your circuit has a few problems.

1) You probably don't realize it, but the pot used in the video is a 5 or 10 turn pot. This allows fine adjustments which a 1 turn pot cannot make.

2) You have not specified the current you want to work at, nor have you specified the battery chemistry. I'll take the default position of an alkaline AA battery and 1 amp max.

3) An AA battery will discharge very quickly at 1 amp, and its voltage will start dropping fast. This is a problem for your circuit, since the battery MUST put out at least one volt in order to drive 1 amp through a 1 ohm resistor. Even worse, the MOSFET will also act as a resistor, further increasing the voltage the battery must put out. If you check the IRF510 data sheet, Typical Characteristics Figure 1 shows the (typical! not guaranteed!) voltage drop from drain to source for various currents and gate drives. At one amp, 1 volt will be produced at the resistor. The LM324 will probably swing to about 2 volts below its power supply, or something like 5 volts max for a 7 volt supply. I note from comments that you're getting 4.5. I suggest you measure the 2nd op amp inputs - if both are at the same voltage you might do better, but if the + input is measurably higher than the - input, that is all the IC will do. This will leave about 4 volts applied from gate to source. Referring to the figure, you'll see that the drain-source voltage will be in the neighborhood of 0.5 vols. This, in turn means that your circuit will only work with an absolutely fresh battery, and only for a short time. As soon as the battery starts to discharge, the battery voltage will drop below the value necessary to draw 1 amp.

You should have paid a bit more attention to about 7 minutes into the video, where he discusses the use of a logic MOSFET. The IRF510 is not remotely a logic MOSFET.

Is all lost? Nope. You can try







simulate this circuit &ndash; Schematic created using CircuitLab

To start, putting two batteries in series will give you a nominal 18 volts, which will allow much more gate drive for the MOSFET. The 100k in series with the pot gives a maximum voltage there of about 6 volts. The combination of R3 and R4 will drop that to about 0.33 volts. 

Coincidentally, reducing the sense resistor to 0.33 ohms (you can connect 3 1-ohm resistors in parallel to do this easily) will provide 0.33 volts when the current is 1 amp. 

The combination of better gate drive and lower resistor voltage should allow your circuit to operate over a much larger portion of the battery discharge cycle.

Also note the addition of a few capacitors. These, especially the one on the op amp, will allow much more stable operation, and you should always use one on each op amp package, as close to the package power and ground pins as you can.