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First post so sorry for any bad formatting.

I'm trying to figure out how to bias my signal going through a non-inverting op amp with a gain of 1. Because I want a gain one just 1, it would just be a unity gain buffer with the signal as the input.

1) Lets say the incoming signal is biased at 0V with signal going from -2.5 to 2.5V. How do I bias at 2.5 with signal going from 0 to 5V.

2) How about re-biasing from 2.5V to 1.65V?

Thanks for any input!

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If it's AC, the easiest way is to pass it through a capacitor, then add a high value resistor pulling the signal towards a virtual ground which you would set at 2.5v.

I searched a bit, and it's pretty easy to do with DC as well. You can use this circuit: enter image description here

Source/explaination here

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  • \$\begingroup\$ This is essentially a summing amplifier. Note that this configuration inherently has a substantially lower input impedance (30k in this case) than a plain unity-gain buffer where the input impedance of the circuit is simply that of the op-amp, which is generally extremely high. \$\endgroup\$ – ajb Nov 3 '18 at 6:26
  • \$\begingroup\$ However, the resistor values in the circuit do not give the results shown. At +2.5V input, the output of the circuit will only be 3.3V. The bias node (shown at 5V here) must be above 5V to pull the input of the op amp to 5V because of the divider. Alternatively, since the input is being divided, some gain can be applied to compensate. \$\endgroup\$ – ajb Nov 3 '18 at 6:31
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    \$\begingroup\$ Thanks for the response! This is for AC. So I would want my signal to go into the capacitor and then have a pull up resistor going to my wanted bias? And the output would be after the capacitor which goes into the unity gain buffer? How do I go about choosing the capacitor and resistor value? \$\endgroup\$ – John Le Nov 3 '18 at 6:51
  • \$\begingroup\$ The capacitor essentially forms a low pass with the resistor, so usually just something large like 10uF, and a high resistor value like 100k. \$\endgroup\$ – Drew Nov 3 '18 at 21:11
  • \$\begingroup\$ Sorry meant to say high-pass. The standard filter equations apply. \$\endgroup\$ – Drew Nov 9 '18 at 9:00

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