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This circuit essentially takes a control input and then doubles it when presenting it to the load. I would like a circuit that controls the current to the load.

How would I modify the circuit to do current control with a sense resistor in series with the load?

So the transfer function would look something like this:

$$ V_{IN} = R_{sense}I_{load}$$

schematic

simulate this circuit – Schematic created using CircuitLab

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  • \$\begingroup\$ Are you asking how to make a general transconductance amplifier? \$\endgroup\$
    – Nino
    Aug 27, 2020 at 21:05
  • \$\begingroup\$ No, a transconductance amplifier is only one amplifier and uses a differential voltage, this one has two which means you can double the voltage to a load (only one amplifier). So I would like to use two amplifiers to double the current to the load (one amplifier is inverted from the other, so one is sinking while the other is sourcing) \$\endgroup\$
    – Voltage Spike
    Aug 27, 2020 at 21:12
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    \$\begingroup\$ @VoltageSpike I don't know if you've read this, but if you are looking for precision current control over two quadrants, then that's what I've worried about years ago and wrote a little something there. Just in case it's of interest. \$\endgroup\$
    – jonk
    Aug 27, 2020 at 23:35
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    \$\begingroup\$ @TonyStewartSunnyskyguyEE75 No, but you can come up with a conceptual diagram, and that's all I'm looking for. Don't need absolute things \$\endgroup\$
    – Voltage Spike
    Aug 28, 2020 at 3:27
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    \$\begingroup\$ @VoltageSpike have not understood my consistent demand for design specs? It is a priori. \$\endgroup\$ Aug 28, 2020 at 5:54

3 Answers 3

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So if I understand your question correctly, you don't want to regulate the voltage at all, but you want to regulate the current. So just take your output current and turn it into a voltage and put feedback around the amplifier to make it match a reference voltage that you scale proportional to the current.

I used your load resistor in this example, but you could just as easily add a sense resistor.

Depending on your swings, you'd need to bias R12 to halfway voltage maybe.

enter image description here

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Maybe a Howland current pump with an inverter to drive the other side of the load (rather than grounding it). Eg. (conceptual only, you can take it from here if you want to- there may be some stability issues to be dealt with).

schematic

simulate this circuit – Schematic created using CircuitLab

enter image description here

The load current V2/R1 is independent of the value of R_LOAD, and it is shown with a 15Vp-p output from a +10V supply.

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We don't know what's the application or specifications it must attend to, but ok, we can do more generic.

Your circuit amplifies only voltage.

It's easier to just throw a voltage controlled current source which can have its "gain" adjusted with no changes other than resistor values (for example, with a Howland Current Pump)

From link: Circuit

So either you put a voltage controlled current source with an op. amp. circuit in cascode with your circuit (gain=2 and then the op. amp. circuit) or adjust the gain in the op. amp. circuit and use just it.

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  • \$\begingroup\$ A howland current pump does not present a bi-directional current to a load \$\endgroup\$
    – Voltage Spike
    Aug 27, 2020 at 21:03
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    \$\begingroup\$ OP wants AC voltage doubling. \$\endgroup\$
    – Nino
    Aug 27, 2020 at 21:22
  • \$\begingroup\$ Didn't know that. \$\endgroup\$ Aug 28, 2020 at 1:32

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