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I am trying to simulate a voltage-controlled current source in LTspice using the feedback current of an OP-amp as the load current. My circuit is:

enter image description here

As an output, I get something around I_L = 620pA. As an amplifier, I tried LT6300, which can give at least 500mA output, but I never get the right current, which should be $$I_L=\frac{V^+}{R_1}=\frac{V_{cc}}{R_1}. $$ I tried to use several other resistor values and the generic OPamps, but never got the right current (which with the values in the image should be 800mA DC). What am I missing?

Thanks.

Solution:

Uff, I focused so much on the fact that the OPamp should be a current source that I forgot the most fundamental things about amplifiers (to make sure that the output is lower than the rail voltage). I changed the OPamp to a generic one just to make the simulation run, and it did. In this configuration, I get the right values:

enter image description here

Thank you everybody.

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  • \$\begingroup\$ Think a little. You have connected +Vcc to non-inverting input. And you demand that the op-amp output will drive his output beyond Vcc voltage (to 48V). How can this be possible? Also, +/- 24V is too much for LT6300. And it will be unstable if the gain is smaller than 10. \$\endgroup\$
    – G36
    Commented Oct 2, 2023 at 14:14
  • \$\begingroup\$ @G36 god I can't believe that I missed that. Thanks! \$\endgroup\$
    – Arad
    Commented Oct 2, 2023 at 14:50

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If you look at the data sheet for the op amp you're using, you'll find that the output swing using +/-12V rails only goes to 10.7V. The data sheet, on first glance, doesn't offer an input range, but the input range typically is smaller than the rail voltages (if it included the rails, my assumption is that the data sheet would be squawking about it, as it's a good feature).

The way your circuit works is that the output needs to go higher than Vcc to hold the negative input terminal at Vcc. That isn't going to happen. It is unclear whether Vcc is an allowed input for this opamp -- my assumption is that it is not.

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  • \$\begingroup\$ Thank you! I edited the main post \$\endgroup\$
    – Arad
    Commented Oct 2, 2023 at 14:50

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