How does this operational amplifier behave in this configuration?

Tangeton

Joined Oct 28, 2017
3 I need help understanding why this op-amp's gain calculation behaves in a certain way.

Please assume the non inverting input is connect to the ground ( V+ = 0).

My issue is, that when assuming this is an ideal op-amp, I have been told that the 10k resistor going to ground does not affect Vout/Vin or Rin. So Rin is 10k and vout/vin is -10. I don't understand why this is the case.

But now, if we say this op-amp is no longer ideal, for some reason, I've been told that actually that extra 10k resistor going to ground does matter when determining vout/vin, and treating it as potential divider network, using superposition, we get V- as a sum of two potential divider circuits given that A = 1000.

I really can't understand why it didn't matter when the case was ideal but in non-ideal case it made a difference... could anyone please explain what actually is taking place in both cases..?

Thank you.

dl324

Joined Mar 30, 2015
13,815
Welcome to AAC!

Is this homework?

Tangeton

Joined Oct 28, 2017
3
Welcome to AAC!

Is this homework?
No I am trying to understand lecture notes.

dl324

Joined Mar 30, 2015
13,815
No I am trying to understand lecture notes.
You can use the Zero Differential Input theorem to calculate the gain.

We treat opamps as ideal to simplify calculations. Real opamps have input bias currents and offset voltages that we need to deal with for more precise behavior. If the input voltage was sufficiently small, error caused by non-ideal behavior could be significant.

crutschow

Joined Mar 14, 2008
28,474
An ideal op amp has infinite gain, so the voltage at the minus input is always equal to the plus input, (which here is ground) when the amp has negative feedback, no matter what the output voltage.
Since the minus voltage is always zero, the resistor to ground can have no effect on the circuit operation.

A real (non-ideal) amp has a high, but finite gain, which means the voltage at the minus input is now not zero, but equals the output voltage divided by this gain.
Thus the resistor at the minus input will now have an effect on the circuit operation since the small voltage at the minus input causes a current to ground through the resistor.
This current subtracts from the feedback current (since the sum of the currents into the minus summing junction is zero), slightly changing the output voltage from the ideal.

• Tangeton