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I am trying to design a voltage-controlled current source circuit. After some research, I decided to use an op-amp.

In the circuit shown below, when V1 is 0.3 V, then Vr is 0.3 V and the current is 0.3 A. These values are OK but when I use V1 is 2 V, Vr is 0.5 V.

What is wrong in this circuit? I also noticed that if I use a 12 V supply for the op-amp then there is no problem.

First: enter image description here

Second: enter image description here

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2 Answers 2

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Your input voltage is 2 volts and your current sense resistor (R3) is 1 Ω. This means you are trying to control the MOSFET at 2 amps. Then, if you look at the MOSFET's I-V characteristic from the data sheet you will see this (red-line by me at 2 amps): -

enter image description here

You should be able to see that you need a gate-source drive voltage of about 4.8 volts typically. Given that the source would need to be at 2 volts to make the circuit regulate, the gate voltage would need to be about 6.8 volts to barely satisfy regulation.

But, you have a 5 volt power rail and the op-amp cannot drive 6.8 volts. You need a higher voltage power supply or a MOSFET with a significantly lower \$V_{GS(THRESHOLD)}\$.

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    \$\begingroup\$ Does OP expect to deliver 2A from the MOSfet's drain? If so, then there's another problem...a 10-ohm load requires V3 to be 2V +4.5V + 20V, if drain current is 2A. For V3, a +12V DC supply is insufficient. \$\endgroup\$
    – glen_geek
    Sep 11 at 16:08
  • \$\begingroup\$ Good extra spot @glen_geek \$\endgroup\$
    – Andy aka
    Sep 11 at 16:15
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Look at the output of the opamp, better yet, look at the gate-source voltage. The IRF530 has a gate turn-on voltage around 5 volts. With the opamp power supply voltage set to 5 V, there isn't enough voltage to turn on the FET. There are modern MOSFETs with lower turn-on voltage.

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