The deadline is over. Thank you both SO much for your help and time!

 

Everything is looking good except for Ro_f. I get a much different answer with a full nodal analysis than what the classical feedback analysis, (RoA / (1+ β * Aa)), gives.

Let me ask MrAl what he gets for Ro_f.

Hi there,

I get a similar result for Rout as his original calculation when i used Rin=6175k not Rin=6175 ohms.
With Rin=6175k, I get Rout=0.01775
With Rin=6175 ohms, i get Rout=0.02046 ohms.

That, however, is assuming we use a zero ohm (or close to it) internal impedance generator. It would be slightly different if we used a generator with an internal impedance that matched the amplifier's input impedance which is sometimes the real case because the previous stage output impedance may be matched to this stage's input impedance.

If i use a generator with output impedance approximately equal to the calculations i saw here for the input impedance which was about 87k, then i get (with Rin=6175 ohms again):
Rout=0.1454 ohms.

So the output impedance is different depending on what is driving the input. Maybe they want us to drive it with that 131 ohm generator for the correct calculation. In that case the numbers shown previously are close to correct as they use a zero ohm generator. The actual Rout with a generator with 131 ohms series resistance comes out to 0.02083 ohms.

 

With no generator on the input (Rs = ∞, in effect), I get Rout = .2702 ohms. For the other conditions you mention, I get the same result.

Classical feedback theory says the open loop input impedance should be increased by the return difference (1+ β * Aa), and the open loop output impedance should be decreased by the same factor. If we break the feedback by removing R3 (retain the same input and output circuit impedances by connecting 17400 ohm resistors in parallel with R2 and R4), then the open loop Rout doesn't depend on whether there is a source connected or not. So its value when divided by the feedback factor doesn't depend on the input condition according to classical feedback theory. Yet ordinary circuit analysis says Rout does depend on the input condition.

Can we use the same feedback factor to calculate closed loop Rin and Rout? And how do we deal with R1 in calculating the feedback factor?

 

Hello again,

If i understand you correctly, i think this amplifier is different than a classic amplifier. For one, it has two feedback loops, one negative and one positive, so there would at least have to be two loops drawn to show it as a flow diagram. Since there is a direct connection between the input and output (through resistances) there must be at least some interaction.
In other words, it's more like a transformer where the output impedance depends on the input impedance and vice versa. The equivalent 'winding' resistance would be higher though so there would be less effect, but still at least some effect.
We proved beyond a reasonable doubt that the two impedances interact, so the theory you refer to must be either an approximation or is only applicable to circuits that adhere to a certain topological form. I could guess that it might be a regular negative feedback amplifier with no positive feedback at all or with positive feedback that does not actually alter the input current that much.

If you could provide a sample circuit where the theory does in fact apply, we could look at that and compare. That should tell us what it is that is different that makes the theory invalid for this particular circuit. Hopefully a simpler circuit that is quick to analyze

R1 has to enter into any of the equations through the input resistance of the op amp because that connects R1 to the feedback resistances and so to the output.

BTW interesting idea about doing the analysis with infinite input impedance. I'll have to try that too at some point.

LATER:
I tried the infinite input impedance generator, and got 0.270222 ohms output resistance, so our results agree again.
For the input impedance i got 86933.79878 ohms.

 

By classical feedback amplifier analysis, I mean this: https://en.wikipedia.org/wiki/Negative_feedback_amplifier

In the circuit of this thread, the feedback signal passes through resistors and R3, R2, Rid, R1||Rs form a voltage divider, guaranteeing that the signal applied to the + input will always be less that the signal applied to the - input. I think that can be treated as just a single negative feedback loop.

The problem I see with the TS's problem is that it isn't easy to partition the circuit into independent forward gain block and feedback block. It may be difficult to disable the feedback and still maintain the proper impedances. The feedback block (resistors in this problem) presents various impedances to input and output parts of the circuit, and for the classical formulas to work, those impedances must be the same whether the feedback is disabled or not.

I think the TS got good approximate results, but did he get the correct results for Rin and Rout without feedback? The no feedback case would seem to require that R3 be removed and appropriate additional resistances be added where needed to maintain the loading on input and output the same for the no feedback and feedback cases.

R1 is problematic. The schematic in post #1 shows Ri_f being calculated to the left of R1, so apparently R1 must be included in a calculation of Ri_f. A blind application of the classical feedback formulas would suggest that Ri_f is in the hundreds of megohms, but clearly it can't be.

So, what must one do to get the classical formulas to give exactly the same results as nodal analysis?

Anyway, the TS is gone, so a new thread should be started for further exploration of all this. It would be good to find a nice example circuit where the discrepancy (if any) between nodal analysis and classical feedback theory is apparent.

 

Hello again,

Well a quick inspection of that link shows that there are two basic topologies. One is series and one is shunt. What strikes me about the shunt type is it is set up similar to a transformer, which as you know, can transform impedances as well as amplify voltages. The shunt topology shows a current source at the input, and that is what helps reflect the load impedance to the 'primary' or in this case the input. Likewise it can reflect the input impedance to the output, which changes the total output impedance.
Probably looking into that would lead to more understanding of how the behavior we are seeing comes about and what the necessary factors are.
One of the other things that seemed interesting is that for the transformer it seems hard to come up with a flow graph because of that current source in the primary (in the model). That current source injects current into the input based on what is happening at the output. So there is an exchange of energy in both directions, unlike a regular flow graph would show. There must be a way to deal with this in simple terms through.

I'll look at this more today too and see what i can find on the web. I guess it couldnt hurt to start with a simpler circuit too.

Are you saying that the OP used a feedback factor and got the right values for input and output impedances? Where is the calculation of that, i'll take a look...

LATER:
I started with a simpler circuit and found that once i calculate the transfer function i can calculate the input resistance from:
Rin=Aol*Ri/Vout

and that excludes what we are calling R1 in the original circuit, and Ri is the input resistance of the op amp.
To include R1, all we have to do is put R1 in parallel with Rin above, giving us the total Rin for the circuit.
This boils down to two constants with Vin=1v:
Vx=K1*Aol/(1+K2*Aol)

then the input resistance is:
Rin=Ri*Aol/Vx

which rewritten comes out to:
Rin=Ri*(1+K2*Aol)/K1

which i think can be worked into the form:
Rin=(Ri/K1)*(1+K2*Aol)

where K1 is determined by the input impedance with the output short circuited.
This would put it in compliance with the Riol*(1+B*Aol) form where Riol is the open loop input resistance.

Aol is the internal open loop gain of the op amp.

The simpler circuit is almost the same but has only the usual two feedback resistors, and Ri of the op amp is the same, and R1 is the input parallel resistance as before, and again that has to be put in parallel with the above Rin calculation to get the total input resistance. Unfortunately the simpler circuit has zero output impedance so i'd have to include that next to see how that alters the calculation, which as it stands the input resistance does not change with output load. With some output impedance in the op amp, it should then show a change, and show how that works into the equation also.

A quick test shows that the output resistance of the op amp does affect the input impedance, even without load. As we add load, the impedance changes again. With this quick look, it looks like the output impedance changes the overall gain of the op amp in effect, so it would probably end up just changing one or both of the constants, K1 and K2. With that kind of change we would have the result we were after i think.
'Probably' the output resistance of the op amp and the load resistance act as a voltage divider where we effectively loose some gain in the value of Aol, making Aol look smaller than it really is. As we change the load, this effective Aol would change too, so the load must be included in the calculation of Rin of the circuit (less feedback with lower Rload values must change the input resistance, how could it not).