The Howland current pump uses an op-amp in the configuration below with a resistive feedback network which gives me the gain show below: enter image description here

But if I decide to swap the feedback resistor for an instrumentation amplifier for less noise and better resolution, what will the new gain be? I've tried searching but can't seem to find anything on this.


simulate this circuit – Schematic created using CircuitLab

  • 1
    \$\begingroup\$ Your proposed circuit isn't going to work, because the input is shorted to the in-amp output. \$\endgroup\$
    – The Photon
    Commented Nov 5, 2018 at 6:06
  • \$\begingroup\$ Describe your theory behind this "concept", please. What do you imagine it does and how? What does the input source see here vs your earlier circuit? \$\endgroup\$
    – jonk
    Commented Nov 5, 2018 at 6:10

1 Answer 1


What you need to do is add a couple of resistors


simulate this circuit – Schematic created using CircuitLab

If the instrumentation amp has a gain G, then, since the current through R1 must equal the current through R2,

Vin/R1 = G iL Rs/R2,

where iL is the load current.

Rearranging the terms gives

iL = Vin(R2 /R1 G Rs)

Note that, strictly speaking, an instrumentation amp is not required, since Rs is grounded, and a simple non-inverting op amp would do the job. In practice, an instrumentation amp would be a good idea, since tiny differences in ground resistance will have a noticeable effect due to the large gain of the amp.

Also note that this configuration will almost certainly oscillate like crazy. The phase shift caused by the instrumentation amp will need careful compensation.


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