If you’ve ever wondered why an op amp has the little plus and minus symbols on it, its because at the heart of it, the device is a differential amplifier. The problem is that — ideally, at least — it has infinite gain so it works like a comparator and that’s not what you usually want. So we put resistors around the thing to constrain it and get useful amplification out of it. [Stephen Mendes] does the analysis for you about how the standard configuration for a differential amplifier works. He assumes you know the stock formulae for the inverting and non-inverting amplifier configurations and uses superposition.
[Stephen] mentions that’s the easiest way to do it and then goes on to do it sort of how we would do it as a check. We think that’s the easier method, but maybe its a matter of preference. Either way, you get the right answer.
With superposition, you basically ignore one input at a time. This reduces the circuit into an inverting amplifier when you short one power source to ground and a non-inverting when you short the other one. Superposition tells you that the real answer is the sum of those two answers.
The other way — we hate to call it the easier way — is to remember two simplifying assumptions about op amps. First, assume the op amp will do whatever it can do to make the plus and minus terminals have the same voltage. Next, assume the inputs are totally open which, for today’s op amps, isn’t too far from the truth.
Based on that, it is easy to see that V2 is divided by the voltage divider of R3 and R4. So you know the plus voltage. By our assumptions, then, you know the minus voltage too, because it should be the same. If you know that voltage and R1 you can compute the current through R1. Because the op amp terminals are open circuit (or, at least, we are pretending they are) you know the current through R1 must also be the current through R2. So we know the voltage across R2 which leads to knowing the voltage at Vo.
Let’s try a numeric version of that. Say V1=5 and V2=8. R1 and R3=10K and R2 and R4 are 20K.
- Vplus is 8V times R4/(R3+R4) or 5.3V
- That means Vminus is also 5.3V
- The current through R1 must be (5.3-5)/10000 = 33.3 microamps (sign depending on which way you consider it flowing)
- The current through R2, then must be the same, so 33.3 microamps
- To get that current through R2 with 5.3 V on one side, you need to solve .0000333 = (Vo-5.3)/20000
- If you carry enough digits, that’s 6V; if you rounded like I did you get 5.966V which is close enough
If you write out the math like [Stephen] does you’ll see the gain is 2 and since the difference in voltage is 3V, a 6V output makes perfect sense. Of course, assumptions are only good to a point. An op amp with +/-15V on the rails can’t produce 50V to wrestle the input to a particular value. But for common sense applications, those assumptions will take you a long way.
Is it easier one way or the other? We can’t say. But superposition is an important concept in its own right and can often tame an otherwise intractable circuit so even if you think it is harder to do it that way in this particular case, it is still worthwhile to follow the math.