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I have the following circuit that I am building, a constant current load):enter image description here

I am using a 9 volt battery and a 1.5 volt battery to power the circuit. My load under test is a AA battery. Based on this video (which is where this schematic is from), I want to know at what point I have to connect the leads of my amp meter to measure the constant current load (as the video does). I want to know where I have to connect those leads, so that when I turn my potentiometer, the constant current changes too (as seen in the video).

I tried to connect the positive lead of the amp meter @ point A as shown in the picture, but my meter reads out 0 amps.

Thank you for your help!

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  • \$\begingroup\$ You have the non-inverting input of the RH op-amp shorted to ground. You should put the amp meter in the collector circuit. And the 50K pot should go to some kind of reference voltage otherwise you'll only get Ib * 50K maximum or about 1mV maximum voltage (similar to the offset voltage). \$\endgroup\$ – Spehro Pefhany Feb 15 '16 at 23:20
  • \$\begingroup\$ Thank you. I made the changes to my schematic. What do you mean the you say put the op amp in the collector circuit and that the pot should go into a reference. I am very new to electronics, so thanks a lot for you help \$\endgroup\$ – user5139637 Feb 15 '16 at 23:25
  • \$\begingroup\$ You changed to to 7V from open circuit, so it does go to a reference (now). I said put the amp meter in the collector circuit but I meant 'drain' circuit (of the discrete MOSFET). \$\endgroup\$ – Spehro Pefhany Feb 15 '16 at 23:26
  • \$\begingroup\$ Please correct me if I am wrong, but I connected my amp meter the way I have shown in the picture above, but I am still getting 0.0 amps \$\endgroup\$ – user5139637 Feb 15 '16 at 23:49
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    \$\begingroup\$ Set the pot for 1V out, measure the voltage at non-inverting input of each op-amp, each should be 1V or very close, then check the output of the RH op-amp, it should be a few volts. There's no point having a 7V source on the pot, the resistor voltage can never exceed the battery voltage, so 0.7V would make more sense (and reduce the resistor). But that's not why you are getting 0A. \$\endgroup\$ – Spehro Pefhany Feb 15 '16 at 23:54
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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

schematic

simulate this circuit – 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.

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  • \$\begingroup\$ Thank you so much for your answer. I looked into the transistor and have found a logic level MTP3055VL. And I will make sure to use C1 & C2 in my circuit. Aside from these changes, what else do I have to make? I want to have a constant current of 20mA (to discharge my load at). I have updated my circuit to include a 51Ω resistor. The Vin to the op amp is ≈ 1V. So the constant current will be 1V/51Ω ≈ 20mA. Please correct me if my logic is wrong. With these changes, can I still use my original circuit with the 7V power supply? Thank you again \$\endgroup\$ – user5139637 Feb 16 '16 at 17:10
  • \$\begingroup\$ Oh. Well, 20 mA makes a big difference. With a 51 ohm sense resistor I'd guess you'll be OK, and you don't even need a different FET, although I'd still go for a smaller sense resistor, just to give you some margin on the FET. Try something like 27 ohms and 0.54 volts. \$\endgroup\$ – WhatRoughBeast Feb 16 '16 at 18:09
  • \$\begingroup\$ Thank you. I am in high school and am very new to electronics, so I do not understand what you mean when you mention "margin on the FET." If you could kindly explain the theory or give a link \$\endgroup\$ – user5139637 Feb 16 '16 at 18:19
  • \$\begingroup\$ Sorry. And I was wrong to suggest it. At 20 mA, the voltage drop across the FET will be very small, less than 0.1 volt. So the thing you have to worry about is the battery getting discharged to the point that it produces less than 1.1 volts or so, which for 20 mA will be pretty near full discharge. \$\endgroup\$ – WhatRoughBeast Feb 16 '16 at 18:26
  • \$\begingroup\$ When you say the "battery getting discharged" you mean the load, right? Or do you mean the battery powering the circuit (i.e. the op amps, pot, etc.) \$\endgroup\$ – user5139637 Feb 16 '16 at 18:29

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