Why Common Mode Voltage is the Average of Inputs in Op-Amp CMRR

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SUMMARY

The common mode voltage in operational amplifiers (op-amps) is defined as the average of the input voltages, calculated as (10V + 7V) / 2 = 8.5V in the discussed scenario. This average is crucial because it reflects how non-ideal amplifiers respond to common mode signals, which can slightly affect output voltage. The difference between open loop and closed loop amplifiers is significant, as open loop amplifiers ideally amplify only differential signals, while closed loop configurations can manage single-ended signals. Understanding the concept of pseudo differential signals and the importance of reference levels in analog circuits is essential for accurate circuit design.

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  • Understanding of operational amplifier (op-amp) fundamentals
  • Knowledge of common mode rejection ratio (CMRR)
  • Familiarity with open loop and closed loop amplifier configurations
  • Basic concepts of differential and pseudo differential signals
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Electrical engineers, circuit designers, and students studying analog electronics who seek to deepen their understanding of operational amplifier behavior and common mode voltage implications.

unseensoul
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An ideal op-amp amplifies the differential input signal which means that its CMRR tends to infinity. However in real applications common mode signals are also slightly amplified.

Suppose you apply a 10V voltage to the non-inverting terminal of the op-amp and a 7V voltage to the inverting one... It is said that the common mode voltage is (10+7)/2 = 8.5V

My question is why do we take the average of the inputs when measuring the common mode voltage? In this situation shouldn't 7V be the common signal?
 
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Keep in mind the difference between open loop amplifier and a closed loop amplifier, ie with or without feedback.

Open loop amplifier amplifies differential signals, but by using feedback it can handle single ended signals. Since an opamp is fundamentally a differential circuit you need to consider the difference between the two inputs. As you point out, you consider the open loop gain which is ideally infinite for differential signals and 0 for common mode signals.

For fully differential input signal, each side of the input swings equally but in opposite directions. The average of the two signals is always the same. That is the common mode voltage.

For non ideal amplifiers the output voltage will have a small dependence on where the average level sits. So if you changed the common mode from 8.5 to 6.5 you might get say 1 mV of change in the output.

You maybe thinking along the lines of what would be called a psuedo differential signal. There one input is kept at a constant voltage (7 V) in your example and the other input varies with respect to it. With a psuedo differential signal, the common mode level constantly changes so its contribution to the output will also vary. That's one reason why fully differential signals offer better performance.

For analog circuits you need to avoid thinking of 0V or ground as some universal reference point. In some cases it makes much more sense to consider the average level of the signal as the reference level.

A voltage/potential is always between 2 things and what you call 0 V could be much different than what I call 0 V in another circuit due to a difference in our "grounds" or if the circuits where floating with respect to ground (like a battery powered circuit). With different 0 V levels you can still get the correct amplifier output because the differential signal is the same for you and me.
 
Most likely this can only be answered by an "old timer". I am making measurements on an uA709 op amp (metal can). I would like to calculate the frequency rolloff curves (I can measure them). I assume the compensation is via the miller effect. To do the calculations I would need to know the gain of the transistors and the effective resistance seen at the compensation terminals, not including the values I put there. Anyone know those values?

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