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I need to calculate the gain of this op amp circuit. It look like a differential op amp circuit, but i only have one resistor in the feedback loop, R3.

This circuit functions as a Voltage level shifter by adding 2.5V DC offset to the AC input. enter image description here

UPDATE: After making changes, (removing the 1K across the inputs and replacing the 47K by 100k all over), this is the result. Corrected Schematic and Simulation result

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  • \$\begingroup\$ With R4 there, my calculation shows 26mVpp. \$\endgroup\$ – Jason Han Aug 7 '17 at 9:20
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In your circuit there should not be 1k ohm between the + and - input.

If the input resistances are chosen such that, R2 = R1 and R4 = R3, then

  • \$V_{O} = \frac{R_{3}}{R_{1}}(V_{2} – V_{1})\$

enter image description here

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  • \$\begingroup\$ Thanks, you're right, the 1K does not come across the inputs. \$\endgroup\$ – Prithvi Raj Prakash Aug 7 '17 at 12:31
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How to determine the gain of this OP AMP circuit?

The formula is quite simple: -

enter image description here

And, if you have R1 = R2 then the gain magnitude is unity.

Because the +IN signal and -IN signal have to be (within reason) at the same potential, the 1 kohm resistor in your circuit has no effect. Are you aware of this or, do you have a valid reason to incorporate it?

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  • \$\begingroup\$ The 1k does not come across the inputs. Thanks for noting that. I initially thought that the resistor network was for dropping the voltage, but i was wrong. But the output is not inverted even after making the correction. It's in phase with the input. Shouldn't it be inverted? \$\endgroup\$ – Prithvi Raj Prakash Aug 7 '17 at 12:34
  • \$\begingroup\$ @PrithviRajPrakash look how you have connected your input signal up (V1) - the negative output is connected to the inverting chain hence two negatives make a positive and you have, what seems like, a non inversion. \$\endgroup\$ – Andy aka Aug 7 '17 at 14:01

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