# Output Impedance of an Inverting Op Amp

1. Jan 20, 2016

### gl0ck

1. The problem statement, all variables and given/known data
Hello,
I just came across a design problem based on an inverting op amp.
The requirement output values are -25G and not greater than 1kOhm output impedance.
With inputs 1V p-t-p and 75Ohms input impedance
Choosing values for the gain seem fairly easy, but how these values affect the output impedance? I've looked online to see that people say the output impedance should be >> than the input impedance. This is what confuses me. Even more some people talk about output impedance and then they end up talking about output resistance. Are these two the same?

This is the op amp circuit that I'd been given:

Thanks!

2. Jan 20, 2016

### LvW

Each voltage-opamp has a low output resistance r,out and a very large input resistance r,in.
Applying negative feedback (as in your circuit) the input resistance of the whole inverter circuit is (practically) equal to r,1=(R2+1/jwC1) and the output resitance will go down to r,2=r,out/(1-LG) which is a very small value due to the negative feedback effect.
Note that LG is the loop gain of the circuit (LG has a negative value due to negative feedback).

3. Jan 20, 2016

### gl0ck

LvW, thanks for the reply, but it seems still unclear. As I am unsure of if the two terms (output resistance and output impedance) could be found by the same way? Also how would you know the A of the op amp in order to find the LG? Also are you saying that R2 =Rout/(1-LG), which is -R1/Rin ? and if so where is the capacitor C1 in this and does it affect the equation.

4. Jan 20, 2016

### Staff: Mentor

The output impedance of a circuit with negative feedback will be less than the OP-AMP manufacturer's rated output impedance without feedback. Knowing that is usually sufficient in most practical designs.

5. Jan 21, 2016

### LvW

LG= - Ao*(R2+1/sC)/(R1+R2+1/sC)

6. Jan 23, 2016

### rude man

If the op amp gain is infinite then it doesn't matter what the op amp output resistance is; the output impedance will be zero. Of course, there may be voltage excursion and frequencyresponse limitations.

7. Jan 23, 2016

### Staff: Mentor

I'm not following this. The output impedance shouldn't be dependent on any feedback, at first glance. Certainly it isn't for large signals that saturate the output drive transistors. And I'm not seeing it yet for small signals...

8. Jan 23, 2016

### rude man

Saturated output transistors don't count!

9. Jan 23, 2016

### Staff: Mentor

The output impedance of the OP-AMP isn't changed by feedback. It's the circuit incorporating feedback where impedances are affected by the feedback, and the direction this works is in our favour!
Small signal analysis does assume reasonably linear operation.

The OP-AMP output impedance lies inside the feedback loop, so if your load changes and it draws more current, output voltage tends to drop and this produces an increase in the error signal and the OP-AMP obligingly tends to restore the output voltage back to what it was before the load changed. Hence, we experience a lower output impedance; negative feedback makes circuit operation closer to the ideal.

10. Jan 24, 2016

### LvW

It is one of the well-documented benefits of voltage-controlled negative feedback that the output impedance of the whole circuit will be drastically reduced if compared with the output impedance of the active unit alone. The reduction factor is roughly the loop gain.
In case of saturation (large signals) we do not speak about any (dynamic) output impedeance because this is a small-signal (linear) parameter.

11. Jan 24, 2016

### Staff: Mentor

Thanks for the explanations. I need to think about this a bit more, and try a SPICE simulation. Thanks again.