# a method for simulating the output resistance of the operational amplifier

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#### Junus2012

Dear Friends, perhaps this is the most topic I ever posted regarding simulating the op-amp output impedance . but this time I have found a new method

from the A.C open loop characteristics test, I find the f-3d

as we now f-db= 1/2.pi.Ro.Co

..Ro = 1/2.f-3db.Co.pi

since we know Co which is Cl, then the only the known thing is Ro that what we need

I have test this method and I compared it with the method you told me (by putting an ac source on the output) and the result are exactly the same.

But here I would ask, this method is very easy, why non of the references are suggested it???

I would discuss with you this issue

thank you very much

Ro = 1/2.f-3db.Co.pi

But here I would ask, this method is very easy, why non of the references are suggested it???

... because it's only (exactly) valid for a circuit with just one single (and: first order) pole - aka the dominant pole. It might be a good approximation, if a second and any further (i.e. non-dominant) poles (and any zeroes) are far away from the dominant pole.

• Junus2012

### Junus2012

Points: 2
as we now f-db= 1/2.pi.Ro.Co
..Ro = 1/2.f-3db.Co.pi
since we know Co which is Cl,

Junus, may I kindly ask you to use words instead of symbols - just to avoid misunderstandings?
What is Co and what is Cl in your posting (in spite of your writing "since we know"...).
Otherwise, it´s hard to discuss with you.

Co: output capacitor load (or the compensation capacitor)

Ro: output resistance of the operational amplifier (which what we need)

yes I am sorry , Cl is what wanted to say Co, the same

Junus, may I kindly ask you to use words instead of symbols - just to avoid misunderstandings?
What is Co and what is Cl in your posting (in spite of your writing "since we know"...).
Otherwise, it´s hard to discuss with you.

- - - Updated - - -

... because it's only (exactly) valid for a circuit with just one single (and: first order) pole - aka the dominant pole. It might be a good approximation, if a second and any further (i.e. non-dominant) poles (and any zeroes) are far away from the dominant pole.

erikl, you are right . I did this test on a current mirror OTA where it has very far separated poles.

Any way , what about if the poles are not far? I think the equation of f-3db is the same and hence we can use the same principle. what you say

Junus, thank you for clarification.
Question: From your explanations I understand that you determine the 3-dB frequency of the open-loop gain and - assuming to know the C value ("load or compensation capacitor") - you derive a resistance which you set equal to the desired output resistor Ro. Is this correct? If yes, you are simply wrong.

Yes, LvW that is exactly what I mean, but why it is wrong???, i have tried it and i compared it with the conventional method and the result is the same

However i have also mentioned that i am using current mirror OTA with far separated dominant poles

Junus, thank you for clarification.
Question: From your explanations I understand that you determine the 3-dB frequency of the open-loop gain and - assuming to know the C value ("load or compensation capacitor") - you derive a resistance which you set equal to the desired output resistor Ro. Is this correct? If yes, you are simply wrong.

Any way , what about if the poles are not far? I think the equation of f-3db is the same and hence we can use the same principle. what you say
No: in case of more than 1 single pole these other poles also determine the f-3dB frequency. If these other (non-dominant) poles are far away, your measuring method may still achieve a good approximation. If they are not far away, however, they also affect the f-3dB considerably, so your a.m. relationship does no longer hold.

• Junus2012

### Junus2012

Points: 2
No: in case of more than 1 single pole these other poles also determine the f-3dB frequency. If these other (non-dominant) poles are far away, your measuring method may still achieve a good approximation. If they are not far away, however, they also affect the f-3dB considerably, so your a.m. relationship does no longer hold.

In this context, the most important question (in case of two poles) is if the first pole (that means: The measured 3-dB frequency) is caused
* by the internal amplifier circuitry (compensation), or
* by the external load capacitor (together with the Rout).

In this context, the most important question (in case of two poles) is if the first pole (that means: The measured 3-dB frequency) is caused
* by the internal amplifier circuitry (compensation), or
* by the external load capacitor (together with the Rout).

Dear LvW
The used OTA is a current mirror OTA as I mentioned before, I attached a picture of it. As you for sure know that this kind of OTA is a self compensated OTA, the output resistance is high enough to put the dominant pole at the lower frequency. because it is self compensated then the term Co is referred to the load capacitor . Junus, regarding determination of the open-loop output impedance it makes a fundamental difference whether you speak about an opamp (as in your first post) or an OTA.
In the first case (opamp) the output impedance is rather small - and it makes no sense (and does not work) to apply the method you describe.
However, in case of an OTA such a load-oriented measurement is - for my opinion - the classical method (C or R loading).
Therefore, I dont understand the last sentence in your first post (But here I would ask, this method is very easy, why non of the references are suggested it?
Did you check all the references?

OK LvW

I agree with you regarding the discreption of OTA or Op-amp, still many people use the word of Op-amp even when talking about the OTA. any way, I mean the OTA. now would you confirm me the method I posted to be right with only OTA ??? . if yes then why no reference like a book or any article described this as a method for simulating or measuring the output impedance of the OTA.

Please if still any thing not clear you can tell me
thank you

Junus, regarding determination of the open-loop output impedance it makes a fundamental difference whether you speak about an opamp (as in your first post) or an OTA.
In the first case (opamp) the output impedance is rather small - and it makes no sense (and does not work) to apply the method you describe.
However, in case of an OTA such a load-oriented measurement is - for my opinion - the classical method (C or R loading).
Therefore, I dont understand the last sentence in your first post (But here I would ask, this method is very easy, why non of the references are suggested it?
Did you check all the references?

why no reference like a book or any article described this as a method for simulating or measuring the output impedance of the OTA.

Hi Junus, I think the answer is quite simple:
The OTA is a (non-ideal) current source - and it is a normal and logical step to measure the output resistance of such a source using a certain load (resistor or capacitor).
Thus, most authors consider it as not necessary to describe such a basic procedure.

• Junus2012

### Junus2012

Points: 2
Thank you LvW,

The important thing for me is that you are confirming this method as true to simulate the output resistance of the OTA.

thank you all guys.
Have a nice weekend

Hi Junus, I think the answer is quite simple:
The OTA is a (non-ideal) current source - and it is a normal and logical step to measure the output resistance of such a source using a certain load (resistor or capacitor).
Thus, most authors consider it as not necessary to describe such a basic procedure.

In contrast to real measurements that might involve a number of practical problems, simulation of output resistance is idealizing circuit behaviour in several regards. It uses ideal sources and bias circuits, and it's performed as AC analysis linearizing the circuit under test in the given operation point.

Simulation circuits can be directly derived from definition of measured quantities. Measurement of output resistance with an ideal current source is a simple example. It follows directly the resistance definition R = V/I. Apply a current of 1 A and get the resistance in ohm. Achieving correct DC bias is the only issue involved with this measurement.

• Junus2012

### Junus2012

Points: 2
Thank you once again friends

I enjoyed a lot with your discussion.

erikl, LvW and FvM, you are my teachers , you always assisting me

many thanks for you all

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