Electronic – op amp DC gain drop

current measurementinstrumentationoperational-amplifier

I got a really simple circuit, it is a voltage follower in cascade with an inverting amplifier. The input of the voltage follower is a current to voltage converter, measuring nanoamperes with a 1 M restore resistor. The voltage swing of the follower is 0 – 1 V, while the amplifier has a gain of 5.1 using 1k and 5.1 k resistors. Both circuits are coupled via a direct link, no resistors to ground or caps are used. I'm feeding the circuit with +/- 15 V. Signals from the current to voltage converter is really stable.

The problem is that the inverting amplifier has a gain of 3.3 only. I checked the amplifier using a power source and works fine with a 5.1 gain. Why the gain is being reduced so much using the follower? I checked the circuit using a voltage source as input for the follower and the circuit works fine.enter image description here

Best Answer

Some problems:

  1. R2 makes no sense. A TL081 isn't going to do much of anything with a 100 Ω load on its output. Just get rid of R2.

  2. OA2 doesn't do much useful. As you say, it's a voltage follower. Ideally, its output voltage is therefore the same as its input voltage. Such constructs are used to lower the impedance of a signal and to provide increased current drive. Neither makes sense here. You already have a opamp driving the signal. There is no point buffering it to have a different opamp drive the same signal

  3. You apparently want a gain of -2 from the OA3 stage. The gain is -R4/R3. That nails down one degree of freedom. You can't just ignore the other or pick arbitrary values. 100 Ω is again way too low. 10 kΩ for each R3 and R4 would give the same gain while not being such a excessive load on the opamps.

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