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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

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  • \$\begingroup\$ You must define at least 4 useful parameters, Gain, Vin, frequency, output current, then compare with current limits and slew rate limit, then you will see OA2 is redundant as well, poor design specs for circuit \$\endgroup\$ – Tony Stewart Sunnyskyguy EE75 Apr 13 '18 at 18:11
  • \$\begingroup\$ I wonder how you get a voltage swing of 1V. 100nA multiplied with 1M Ohm only gives 0.1V? How do you measure the input voltage or current to check the amplification? What is the internal resistance of your meter? Maybe the error is introduced by the meter? \$\endgroup\$ – Kitana Apr 13 '18 at 20:44
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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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  • \$\begingroup\$ Hello, and thank you very much for the answer. I got rid of R2 and the voltage follower. The circuit shown has R3 1 K and R4 4.7k to get 4.7 gain. Problem insist and im getting at the end a 3.3 gain. I'm trying using a non inverter amplifier and get a good response but using an inverting amplifier at the end of the of the current to voltage converter won't work fine. I'm not sure how the change in the input impedance in both designs affect its performance. \$\endgroup\$ – Leonardo Lameda Apr 13 '18 at 18:56
  • \$\begingroup\$ @Leo: Since you have considerably changed the circuit, this is now a new question. You should post a new question, with a schematic showing exactly what you have implemented, and explain exactly what results you are getting. You can refer to this question as background, but the new question should be able to stand on its own. \$\endgroup\$ – Olin Lathrop Apr 13 '18 at 20:44
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Looking at your circuit, the last 2 amplifiers may not be necessary and your problems may be caused by component selection. In general the feedback resistor should be in the K range not M.

OA2 has a gain of +1 OA3 has a gain of -1

By feeding the signal into the positive terminal you will have a non inverting amplifier with the benefit of a high impedance input. The leakage current is much lower than your current source. Convert your current source to a voltage by placing a resistor from the + terminal to ground.

The output of the op amp is low impedance so driving a 100 ohm resistor is not a problem but you are wasting a lot of power. Is R2 really necessary?

To adjust the non inverting amplifier (OA1) gain, just make it look like OA3 and ground R3.

Not sure where your 5.1 K resistor is on the diagram. Gain calculation for DC is straight forward but if the input is AC, you need to check the open loop gain which is frequency and temperature dependent. Error is introduced when the loop gain (open loop - close loop) is small and approach R1/R2 when it is large.

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