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I've used this circuit as an input stage for a digital oscilloscope. From left to right, the first op-amp provides high input impedance, the second op-amp is used to add DC offset to the signal (which should have no offset), the third op-amp is used to amplify/attenuate the signal.

The problem is when I put a voltage source of 1V on the bottom input of the summer op-amp (the second one) to add a DC offset of 1V to the signal, a DC offset of 2V is added instead. Why does this happen and how can I fix it?

You can view and edit the circuit simulated here.

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  • \$\begingroup\$ Your last opamp has a gain of -2. If that's where you're measuring then the result you're getting is expected. \$\endgroup\$ – brhans Jan 5 '16 at 15:31
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Your offset is added before the gain is applied- you can think of it as 1.0V referred to the input (input offset).

If you want to provide output offset (for example, to center the signal in the range of a unipolar ADC) then you should add the offset after the gain is added.

One way to do that is to leave the output amplifier gain constant, adjust the offset to compensate (halve it in your case, or use a 2K resistor in series with the offset rather than 1K and stay with 1V), and change the gain by adjusting the resistor shown below:

enter image description here

Note that if you actually build this circuit you'll need a bipolar supply for at least the first op-amp since it otherwise won't be able to follow the input below ground, and it will typically have to be a bit larger (as much as a few volts or as little as some tens of mV depending on the op-amp) than the largest signal expected at the input. So if you want to handle a +/-10V signal with some op-amps, you might choose +/-15 supplies.

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