# OP Amp as differential amplifier

I'm learning about op amps, and I have the following circuit where the amplifier is used as differential amp

Since both inputs are the same, I was expecting the output to be zero. But instead, I got this (around 9 mV peak)

I don't understand where this output comes from. I simulated the circuit using livewire

– jonk
Commented Jun 8, 2020 at 4:53
• I would advise simulating with an actual differential source. your two input signals are now not really differential (and area only made differential because of the feedback). Commented Jun 8, 2020 at 5:48

There are several effects buried in that 1 Hertz output of 0.009 volts, for 2 volts input.

As was stated, the resistor tolerances matter. But this is a simulation.

The opamp has FINITE GAIN, probably only 100,000X or 100dB.

Your circuit has 40dB gain (100X).

The opamp has 100 ohms (?) Rout.

The circuit achieves (before gain) 0.009/100 = 90 microVolts Referred To Input of Common Mode Rejection at 1Hertz.

The voltage you see is due to the tolerances of the resistors in addition to a parameter of the opamp itself called common mode rejection ratio.

You have the amplifier set up for a gain of 100 so if R4 was connected to ground you would expect 200V pk-pk at the output (yes I know that is well beyond the capabilities of the amplifier).

You are getting ~20mV pk-pk so you have a rejection of about 10,000 times (or 40dB).

You only need an error of 0.01% in the resistors to result in that error so I think you are doing very well. If you adjust one of the input resistors slightly you may be able to improve it.

An important parameter of a differential amplifier is the common-mode rejection and as you have seen it requires extremely good matching to achieve rejections of even 40dB.

Wikipedia - Common Mode Rejection

• All absolutely true, but the OP’s output looks like simulation, not a scope shot. So the results might be due to other non-idealities in the model as resistors in simulation (unless using Monte Carlo or other tolerance analysis) should be identical. Commented Jun 8, 2020 at 2:14
• That was really helpful. Thank you! Commented Jun 8, 2020 at 2:18
• Yes, it is the output of a simulator. For a moment I thought that maybe some parameter in the software could be causing that output, but I wasn't sure @JohnD Commented Jun 8, 2020 at 2:21

It is a powerful idea to explain the abstract electronics circuits by simple electric concepts. Thus, the common-mode operation of the 4-resistor op-amp differential amplifier can be easily explained by the basic electric circuits of voltage divider and balanced bridge.

## Equivalent electric circuits

1. Voltage divider. The pairs of resistors R1-R2 and R3-R4 form two identical voltage dividers. Both are grounded (the first - by the virtual ground at the op-amp output; the second - by the real ground) and their inputs are driven by the same input voltage source.

2. Balanced bridge. Actually, the two voltage dividers form a balanced Wheatstone bridge with varying supply. Its unique property is that the voltage drops across the opposite resistances (R1<->R4 and R2<->R3) are equal; so the difference between them is zero.

## Operation

1. VIN = 0. Initially, the input voltage is zero. So the op-amp output voltage is zero and the R1-R2 voltage divider is virtually grounded.

2. VIN = var. When the circuit input voltage changes (common-mode), both op-amp input voltages simultaneously change with the same rate so that their difference is always zero. The op-amp output voltage does not change since it depends only on the difference between the input voltages... and this difference is zero. So the R1-R2 voltage divider remains virtually grounded.