# Ideal operational amplifiers

#### MrAl

Joined Jun 17, 2014
8,353
Hi,

It would be good if you started this so we can see what is required.

It looks like the input resistor is 200k/20k=10k and that starts the ball rolling.
The op amps have zero output impedance, but because they spec 50k that probably means that the output has 50k in series with it, and that in turn means that there will be a change in input impedance with load and with input voltage in the closed loop mode.

It would be better if you would state how you would start it, that way we would have better idea what is required. Sometimes the previous course work dictates how we go about solving some of these problems. Ypu are the only one that would have that info on hand.

#### ErnieM

Joined Apr 24, 2011
8,167
Your suspicion the source impedance affects the signal is correct. Essentially it forms a voltage divider with the first stage input impeadance.

It is that reduced signal that is amplified.

Another way to look at this is the source impeadance is in series with the amp input resistance, so the effective R1 is greater, hence the gain is lower.

That is two ways to analyze the circuit that should give the same resultant gain of the first stage.

For part II I have no idea what loop is being closed.

Part iii leaves me perplexed as I would first determine the gain of the system to determine the equivalent.

#### MrAl

Joined Jun 17, 2014
8,353
Your suspicion the source impedance affects the signal is correct. Essentially it forms a voltage divider with the first stage input impeadance.

It is that reduced signal that is amplified.

Another way to look at this is the source impeadance is in series with the amp input resistance, so the effective R1 is greater, hence the gain is lower.

That is two ways to analyze the circuit that should give the same resultant gain of the first stage.

For part II I have no idea what loop is being closed.

Part iii leaves me perplexed as I would first determine the gain of the system to determine the equivalent.
Hi,

The only thing that would make sense is if the output was connected to the input to form the loop. The output resistance would be connected to the input forming a closed loop. That would be a typical connection scheme too.

#### The Electrician

Joined Oct 9, 2007
2,846
Hi,

The only thing that would make sense is if the output was connected to the input to form the loop. The output resistance would be connected to the input forming a closed loop. That would be a typical connection scheme too.
As I read the problem, only the signal source has 50k output impedance; the opamps have zero output impedance.

#### MrAl

Joined Jun 17, 2014
8,353
As I read the problem, only the signal source has 50k output impedance; the opamps have zero output impedance.
Hello,

Yes i agree, it looks like the 50k is the output impedance of the signal source at the input. I was looking for a way to make this question make sense, which now i am thinking that it is just very poorly posed. For example, with zero output impedance (of the whole amp itself) when the loop is closed the output will short out the input, unless there is some theoretical license we dont really know about here, which we could only know from previous work in the course.

#### WBahn

Joined Mar 31, 2012
26,398
About the only way that I can see that it makes sense is for the reader to assume the topology of each amplifier (i.e., the classic inverting and non-inverting topologies) and then use the values for the gain and the feedback resistor to determine the value of the other resistor. That may make sense within the context of the material that has been covered thus far.