Output resistance of equivalent op amp circuit

[george2525]

Hello

Can anyone tell me what the output resistance marked "RAo" in figure c) is please?

The answer states it as ro + RL||R22

This makes no sense to me

It may be a mistake but its realy bugging me so if anyone could give me a good answer i'd be very grateful

please give a basic method if possible

Thanks for any help!

 

[Harald Kapp]

The given answer is not what comes to mind as obvious and is wrong in my opinion, as you observed.

Since this is the homeworkforum, you will not be given complete answers.
Firstly: What is your proposed solution and how do you arrive at it? Then let's discuss the method.

 

[george2525]

well its not homework

Im preparing for an exam and they have provided some example problems with answers. In this case I think they must be wrong

The given answer seems to be ignoring the dependant voltage source uV1

If this is correct then i make it R22||ro||RL

But even if I am right I am very confused about how to deal with the dependant source here. In the past I have always used the Vx/Ix test method but that doesnt seem to work here.

If you can tell me what to do with the dependant voltage source that will probably be the best help

Thanks

 

[Harald Kapp]

If this is correct then i make it R22||ro||RL

Right, that's the same solution I come up using two different methods.


Method 1:
Replace the voltage source by a short circuit (an ideal voltage source has Ri = 0 Ω). ‚Look into‘ the output from the right side. What do you see (in terms of the network with the short circuited voltage source)?


Method 2:
Consider the unloaded ciruit. The open circuit voltage Vo is given by the voltage source and the resistive divider:
Consider the circuit when the output is short circuited. The short circuit current is given by Is.

The output resistance then is Rout= Vo/Is

If you can tell me what to do with the dependant voltage source that will probably be the best help

The fact that v0 is a dependant voltage source is imho irrelevant for the analysis of figure c).
On the contrary, the dependence of vo on the feedback by R1 and R2 if completely taken into account and given sufficiently open loop gain of the amplifier will eliminate the influence of ro on the output voltage (that's what the feedback is for) by increasing the vo if the voltage drop across ro due to load currrent increase. And vice versa.. Thus ro is reduced by the gain of the amplifier.

 

[george2525]

OK so are you saying that if the voltage source is dependant and you want to find output resistance you can still replace it with a short circuit in the same manner you would if it was a Normal voltage source?

I find it confusing because the A circuit is connected to the B circuit but figure c) implies that it is not

in the past I would have drawn the whole circuit out. shorted Vs , then apply Vx at Vo and included the value of Uv1 in the calculation. The value of uV1 would be a result of Vx and everything would cancel

 

[Harald Kapp]

OK so are you saying that if the voltage source is dependant and you want to find output resistance you can still replace it with a short circuit in the same manner you would if it was a Normal voltage source?

That's what figure c) implies.

I find it confusing because the A circuit is connected to the B circuit but figure c) implies that it is not

The b) circuit is the equivalent circuit for the feedback circuit marked by the blue box in figure a). It is therefore not 'connected' to a), but part of a).

The assumption I=0 in b) makes V1=0 and therefore µ*V1=0, too. This leads to the above observation for Rout.
In practice V1=f(vo) via the feedback mechanism. This leads to a rather more complex eqaution to be solved.

The value of uV1 would be a result of Vx and everything would cancel

Fo µ>>>1, that's how opamps are typically used.

Read this. I think it confirms my above statement..

 

[Ratch]

Hello

Can anyone tell me what the output resistance marked "RAo" in figure c) is please?

I can do that.

The answer states it as ro + RL||R22

This makes no sense to me

It may be a mistake but its realy bugging me so if anyone could give me a good answer i'd be very grateful

It is a simple typo on Sedra and Smith's part.

please give a basic method if possible

Thanks for any help!

View attachment 26791

There is a dependent voltage source in that provlem that is confusing and bemusing to you. Even so, the applied source method you are using is a clunky way to do it, especially when you have a dependent source.

You need to use a relatively unknown theorem called by different names, two of which are the General Immittance Theorem (GIT) or the Port Immittance Theorem (PIT). It basically states that in a linear circuit, if you know the transfer function between two ports, you can easily find the input and output immittance between those two ports. The theorem works because of Tellegen's theorem. I can prove it if anybody really gives a damn.

Anyway, the theorem says that the transfer function has to include the source immittance of the source and the load immittance of output. Then you find the -Rs solution of the denominator, and that gives you the input impedance. Similarly, find the -RL solution of the denominator, and that gives you the output immittance. Unless the circuit is complicated, it can be done by inspection. OK, let's see how it works. The first line of the attachment displays the input and output functions . Lines 2 and 3 show the solutions of Vo and Vi. Line 4 shows the transfer function. Examine the denominator of the transfer function. It has every parameter that affects the input and output immittance. Then we set the denominator to zero and find -RL, the output impedance. It is ro||R22 . Adding RL to it gives all three resistors in parallel. Since Rs shows up as a single term with no coefficients, you can just put your finger over the Rs term, and read the input immittance as the sum of the remaining terms.


This theorem works for any transfer function, V/V, V/I, I/I, I/V. It doesn't matter whether you represent the parameters in Laplace, Steinmetz, or any other representation you can imagine. It also does away with the angst of dependent sources. As long as you can write a transfer function for a linear circuit, you can find the input and output immittances. So GIT with with the problem and use the PIT. As always, ask if you have any questions about this method.

Ratch

 

[george2525]

That looks very interesting

does the transfer function have to have a special form for that to work though?

What is the form? it seems that you have to make sure there are no fractions anywhere on the top or bottom?

Also do you always solve for negative RL for output

and negative RS for input ?

Thanks!

 

[Ratch]

That looks very interesting

does the transfer function have to have a special form for that to work though?

What is the form? it seems that you have to make sure there are no fractions anywhere on the top or bottom?

Yes, just make sure the transfer function has no polynomial fraction terms in the numerator or denominator.

Also do you always solve for negative RL for output

and negative RS for input ?

Thanks!

Yes, the factors that make up the denominator are equal to zero, so naturally any term within those factors is going to be the negative of the remaining terms.

I like the way this method bypasses any hassle with dependent sources.

Ratch

 

[Herschel Peeler]

I can do that.



It is a simple typo on Sedra and Smith's part.



There is a dependent voltage source in that provlem that is confusing and bemusing to you. Even so, the applied source method you are using is a clunky way to do it, especially when you have a dependent source.

You need to use a relatively unknown theorem called by different names, two of which are the General Immittance Theorem (GIT) or the Port Immittance Theorem (PIT). It basically states that in a linear circuit, if you know the transfer function between two ports, you can easily find the input and output immittance between those two ports. The theorem works because of Tellegen's theorem. I can prove it if anybody really gives a damn.

Anyway, the theorem says that the transfer function has to include the source immittance of the source and the load immittance of output. Then you find the -Rs solution of the denominator, and that gives you the input impedance. Similarly, find the -RL solution of the denominator, and that gives you the output immittance. Unless the circuit is complicated, it can be done by inspection. OK, let's see how it works. The first line of the attachment displays the input and output functions . Lines 2 and 3 show the solutions of Vo and Vi. Line 4 shows the transfer function. Examine the denominator of the transfer function. It has every parameter that affects the input and output immittance. Then we set the denominator to zero and find -RL, the output impedance. It is ro||R22 . Adding RL to it gives all three resistors in parallel. Since Rs shows up as a single term with no coefficients, you can just put your finger over the Rs term, and read the input immittance as the sum of the remaining terms.

View attachment 26801
This theorem works for any transfer function, V/V, V/I, I/I, I/V. It doesn't matter whether you represent the parameters in Laplace, Steinmetz, or any other representation you can imagine. It also does away with the angst of dependent sources. As long as you can write a transfer function for a linear circuit, you can find the input and output immittances. So GIT with with the problem and use the PIT. As always, ask if you have any questions about this method.

Ratch

Interesting. If the output voltage source is the output driver then then it makes sense that the answer is RO + RL||R22. All the output current goes through R Out, then is split between RL and R22. Assuming V In = 0 V.

 

[Ratch]

Interesting. If the output voltage source is the output driver then then it makes sense that the answer is RO + RL||R22. All the output current goes through R Out, then is split between RL and R22. Assuming V In = 0 V.

If Vin = 0, then no current will be present in the output. Are you saying the book answer is correct, and the calculated answer of ro||R22||RL is wrong?

Ratch

 

[Herschel Peeler]

If Vin = 0, then no current will be present in the output. Are you saying the book answer is correct, and the calculated answer of ro||R22||RL is wrong?

Ratch

That does seem to be the question. No matter what the output voltage does the formula change? What is the actual difference in voltage at the output under either circumstances? How much does R22 change with, say, 1 V in?
Sorry if this is covered in the other posts. I haven't read them all yet.
Reading.

 

[Ratch]

That does seem to be the question. No matter what the output voltage does the formula change?

The denominator of the transfer function determines the input and output immittances. The denominator is affected by the circuit component values, not the voltages applied to the circuit.

What is the actual difference in voltage at the output under either circumstances?
How much does R22 change with, say, 1 V in?
Sorry if this is covered in the other posts. I haven't read them all yet.
Reading.

The questions you are asking appear to be related to determining the immittances by the applied source method. That is not the method I am advocating. The GIT method does not care what the voltage differences are. It is based on Tellegen's theorem.

Ratch

 

[Herschel Peeler]

That does seem to be the question. No matter what the output voltage does the formula change? What is the actual difference in voltage at the output under either circumstances? How much does R22 change with, say, 1 V in?
Sorry if this is covered in the other posts. I haven't read them all yet.
Reading.

Ro + RL || R22 makes sense if you just consider the output. Nut, as suggested, this is not the whole picture. Yes, with 0 V out the resistance is Ro || RL || R22. I can see that. But there is no output and no current. With no current out do we have a divide-by-zero problem in V/I?

With Ro || RL || R22 and no voltage in or out we have 0 divided by 0. Is this an irreconcilable problem?

So let's plug in 0.1 volts with a gain of 10 in a real circuit.

 

[Ratch]

Ro + RL || R22 makes sense if you just consider the output. Nut, as suggested, this is not the whole picture. Yes, with 0 V out the resistance is Ro || RL || R22. I can see that. But there is no output and no current.

If you drive another amp with the voltage at Vo, it will appear to the amp like the Vo has an impedance of Ro || RL || R22 no matter what the value of Vo is. Immittance does not change with output voltage.

Ratch

 

[Herschel Peeler]

If you drive another amp with the voltage at Vo, it will appear to the amp like the Vo has an impedance of Ro || RL || R22 no matter what the value of Vo is. Immittance does not change with output voltage.

Ratch

Well RL is the input resistance of the next op amp; meg ohms??? Next to meaningless in the equation.

Considering a real circuit ...

 

[george2525]

Ratch:

OK i'm liking this method

I want to be extra clear here as you are right that its not easy to find much info on it.

When you say "polynomial" what you mean is - make the transfer function a polynomial with "RL" as the variable right?



so this is effectively : (constant*RL) / (constant*RL + constant)

then you take the denominator = 0 and solve for -RL

Now for Rs you are saying that its now a polynomial in Rs

then you ignore (roR22+roRL+R22RL) because it cannot be 0

therefore Rs+R11+Rid = 0

Is this all ok?

Thanks this is useful stuff

 

[Ratch]

Well RL is the input resistance of the next op amp; meg ohms??? Next to meaningless in the equation.

What if the next amp is a lower impedance BJT instead of an op-amp? Anyway, the theorem gives the input/output impedance of the circuit. It does not tell you what to do with that information.

Considering a real circuit ...

View attachment 26804[/QUOTE]

Consider a real analysis. The voltage-shunt feedback of resistors R3 and R4, together with the tremendous gain of the op-amp, reduce the 75 ohms of the op-amp itself, into a circuit output impedance of almost zero. No matter what other impedance you hang onto the output of the op-amp, as long as the op-amp can supply the current, the output impedance of the circuit will be zero. See the analysis below using the PIT. This analysis was done using a finite amplification A. As a reality check, notice what the transfer function becomes when A approaches infinity. Look familiar?


Ratch

 

[Ratch]

Ratch:

OK i'm liking this method

I thought you would.

I want to be extra clear here as you are right that its not easy to find much info on it.

Yes, isn't it amazing that hardly anyone except yours truly has picked up on this method?

When you say "polynomial" what you mean is - make the transfer function a polynomial with "RL" as the variable right?

See the written definition below. Also look at post #18 of this thread.

View attachment 26807

so this is effectively : (constant*RL) / (constant*RL + constant)

then you take the denominator = 0 and solve for -RL

When starting out, just be sure both the numerator and denominator are free from fractions. When you become more familiar with the method, then you can take shortcuts.

Now for Rs you are saying that its now a polynomial in Rs

then you ignore (roR22+roRL+R22RL) because it cannot be 0

(roR22+roRL+R22RL) can be any value, but Rs in the other expression can make the denominator zero. That value is the negative of the input impedance.

therefore Rs+R11+Rid = 0

Is this all ok?

Thanks this is useful stuff

Whatever value of Rs makes the denominator equal to zero is the negative of the input impedance.

Here is a more formal definition the the PIT.

>> For a linear two-port of known transfer function that includes all network elements, the output immittance is the negative of the load immittance and the input immittance the negative of the driving generator immittance, as the transfer function denominator is set to zero. <<

The PIT is especially useful when pesky dependent sources are encountered.

Ratch

 

[The Electrician]

You need to use a relatively unknown theorem called by different names, two of which are the General Immittance Theorem (GIT) or the Port Immittance Theorem (PIT). It basically states that in a linear circuit, if you know the transfer function between two ports, you can easily find the input and output immittance between those two ports. The theorem works because of Tellegen's theorem. I can prove it if anybody really gives a damn.

This seems like a big "if" to me. Getting the transfer function is probably the most difficult part of the problem.

If a homework problem wants various gain functions as well as input/output immitances, then the transfer function will have to be calculated to get those gain functions anyway. Using the GIT is then a small additional effort. But if all that is wanted is the input/output immitance, the effort of obtaining the transfer function should be considered part of using the GIT; using the GIT is then no longer a small additional effort.

This theorem works for any transfer function, V/V, V/I, I/I, I/V. It doesn't matter whether you represent the parameters in Laplace, Steinmetz, or any other representation you can imagine. It also does away with the angst of dependent sources. As long as you can write a transfer function for a linear circuit, you can find the input and output immittances. So GIT with with the problem and use the PIT. As always, ask if you have any questions about this method.
Ratch

Won't one have to put up with the angst of dependent sources in obtaining the transfer function, which we have to do before we can use the GIT?

Do you have Dr. Stockman's "Theorem Book": http://ieeexplore.ieee.org/stamp/stamp.jsp?arnumber=6323775 (reference 1).