When I search for opamp-based voltage to current converters, I find scads of circuits with a transistor driver on the output of an opamp or variations on the Howland circuit.

I'm beginning to doubt my sanity -- it seems like the following would work just fine, as long as the device to be driven doesn't need to reference ground:


simulate this circuit – Schematic created using CircuitLab

Aside from the lack of ground reference and the other usual opamp caveats (stability in the feedback loop, maximum drive current and voltage), are there any things to watch out for using this configuration?

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    \$\begingroup\$ The problem is that you don't specify what the DUT is. Such a configuration, as any other, has its limits, which depend on the nature of the DUT. For example, if the DUT is a capacitor, then you have built an integrator that will saturate if Vin has a nonzero offset (i.e. DC component) and this will force you to add components to discharge it. Please, state clearly the problem you are trying to solve. This would allow us to give you more focused and meaningful answers. \$\endgroup\$ Commented Apr 9, 2018 at 7:09
  • \$\begingroup\$ @LorenzoDonati: I think I covered that in that "usual opamp caveats" clause above. The issues you cite (i.e. a saturating integrator) would be true for any voltage to current converter. I'm asking if there are any issues with this particular configuration. \$\endgroup\$ Commented Apr 9, 2018 at 14:04
  • \$\begingroup\$ ...but having said as much, I'm planning on driving an inductor. :) \$\endgroup\$ Commented Apr 9, 2018 at 14:06
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    \$\begingroup\$ Vcm range is probably an overlooked requirement. Also driving an Inductor means Vio null is a requirement as it can saturate with a constant Vo offset. \$\endgroup\$ Commented Aug 3, 2021 at 21:09

2 Answers 2


Yes, there's a constant current going through your "DUT", proportional to the input voltage, provided the op-amp has enough output range at the desired current, provided the op-amp inputs are within the common mode range, and provided your input can supply all the current required to go through the "DUT".

Each of those aforementioned things (plus the lack of a ground reference) is a separate constraint on applications, limiting situations where it is useful.

It's still useful in some applications, for example, in making a log amp, but most often it's not very useful or would require excessive complexity and cost elsewhere to make up for the limitations (for example, bipolar supplies where a single supply would do, or an extra instrumentation amplifier measuring the DUT, the inamp with bipolar supplies and various internal saturation limits to restore a ground reference)


Looks good. Just remember that a negative Vin will result in a positive output to P2 in an attempt to keep the op-amp (-) pin the same as the grounded (+) pin.

So a positive voltage into Vin would do just the opposite. Op-amps always try to keep both inputs at the exact same potential, so make sure your 'DUT' can handle voltage in both directions, and its resistance match's 'R' or is not to far from it.

If the DUT has a high resistance the op-amp will have substantial gain-it may saturate to the + or - rail in an attempt to keep the inputs at the same potential.

I do not think you will use more than +/- 15 volts to power the op-amp, so as long as the DUT can tolerate that voltage you are safe to experiment. Keep 'R' above 2K ohms to limit the current into the DUT and within safe levels the op-amp can handle (+/- 5 mA max load).

  • \$\begingroup\$ "a positive voltage into Vin would do just the opposite." Heh -- notice I didn't drawn an arrow on the current label -- that was intentional. \$\endgroup\$ Commented Apr 9, 2018 at 14:07
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    \$\begingroup\$ What you have 'drawn' is an inverter with a DUT as a feedback element. DUT can have a linear or LOG response. \$\endgroup\$
    – user105652
    Commented Apr 9, 2018 at 20:03

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