I have an issue with the PGA207 instrumentation op-amp.

Basic connections of the PGA207

I connected it as a voltage buffer, as seen in the following image:

Voltage buffer in design

When I connect a 1 kHz, 2Vpp sine wave to the non-inverting input (pin 5) and measure the output on pin 11, the signal amplitude is precisely halved.

I measured the rails and each is within the minimum range (datasheet says minimum ±4.5V, and I measured ~4.9V and ~-4.9V on the rails). The input (pin 5) and output (pin 11) are disconnected from the rest of the circuit (I put 0 Ω links in to remove for testing).

Currently, I am not driving the two gain control pins (15 and 16), so I should be getting a gain of 1. I also tried pulling the gain control pins high one after the other and it does not seem to reliably change the gain according to the schematic. "01" lifts the amplitude slightly and "10" makes a big jump to outside the Pico range. By default, as mentioned before, the programmable gain of "00" gives me an actual gain of 0.5.

Any other debugging tips or tricks I can try would be appreciated.

  • 1
    \$\begingroup\$ an instrumentation amplifier has the feedback circuitry on the die. So by connecting V- to the output, you've altered the feedback network. \$\endgroup\$
    – Rahmany
    Aug 30, 2022 at 8:19
  • \$\begingroup\$ How would I be able to make a summing amplifier using this chip? \$\endgroup\$ Aug 30, 2022 at 17:47
  • 1
    \$\begingroup\$ Assuming the signals are referenced to the same potential, then it's the same way you would make a normal summing amplifier but without the feedback. en.wikipedia.org/wiki/… i.e. without Rf. Signals should be fed to the V+ pin and their reference to V- \$\endgroup\$
    – Rahmany
    Aug 30, 2022 at 22:36
  • \$\begingroup\$ Thanks. Will try that! \$\endgroup\$ Aug 31, 2022 at 6:28

2 Answers 2


You should not be feeding the output back to the Vin(-) input- this is an instrumentation amplifier- the feedback is all internal.

Connect the Vin(-) terminal to a quiet ground if you don't need it.

Now you have:

Vout = Vin(+) - Vin(-) (G=1)

But you've connected Vin(-) to Vout, so Vout = Vin(+) - Vout,

or Vout = Vin/2.

  • \$\begingroup\$ Thanks for the quick reply. I will try that fix tonight and confirm asap. \$\endgroup\$ Aug 30, 2022 at 8:02
  • 1
    \$\begingroup\$ This did the trick, thanks. \$\endgroup\$ Aug 30, 2022 at 17:46

I connected it as a voltage buffer

You made a circuit that would be totally suitable for an op-amp but unsuitable for an instrumentation amplifier given it's main use as a differential amplifier.


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