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Differential and common mode input resistance of the fully diffferential amplifier

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Junus2012

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Dear friends,

I would like to ask you haw to simulate

1. Input differential impedance of the fully differential ampliifer.
2. Common mode input impedance of the fully differential ampliifer.

Thank you
 

Do you have a method please to simulate the differential input and common-mode input capacitance of the fully differential amplifier?
Use ideal_balun in AC analysis.

Yin_diff=Idiff/Vdiff
Yin_com=Icom/Vcom

Cin_diff=imag(Yin_diff)/(2*pi*freq)
Cin_com=imag(Yin_com)/(2*pi*freq)

This is an answer.
There is no easy answer than this.

Surely learn and understand linear circuit basic.
 
Last edited:

but you used skill for simulation,
I usually use the schematic
No.
It is based on schematic.

is the differentia output impedance of the fully differential amplifier
is twice the impedance taken from each output individually? (Rout_diff = 2 Rop=2Ron)
Yes, as far as you can understand diffential mode correctly.

Differential drive is different from floating drive.
 
No.
It is based on schematic.

Yes, as far as you can understand diffential mode correctly.

Differential drive is different from floating drive.

No.
It is based on schematic.

Yes, as far as you can understand diffential mode correctly.

Differential drive is different from floating drive.


Dear Pancho,

I have one issue,

I am simulating the differential output impedance of the fully differential amplifier by using voltage-controlled voltage source (VCVS) instead of using the balun, however, I am getting different result from connecting balun.

please see my attached both test benches, I presume both test benches are equivalent, unless if I am wrong

New Doc 66.jpg
 

I presume both test benches are equivalent
No, they are completely different.

Testbench using balun is correct.
However testbench using VCVS is wrong.

unless if I am wrong
Of course, you are wrong.

Can you understand VCVS ?
It is unidirectional.
You can not see output of Amplifier from VCVS input.
 
No, they are completely different.

Testbench using balun is correct.
However testbench using VCVS is wrong.

Of course, you are wrong.

Can you understand VCVS ?
It is unidirectional.
You can not see output of Amplifier from VCVS input.

Dear Pancho,

Thank you for correcting me,

As you said, the VCVS is unidirectional, in which I can use to plot my output differential signal, I tried it to compare both of them in transient simulation and both are compatable. However, my mistake in the simulating of the output impedance that I can not use the VCVS to load my ampliifer because it is unidirectional so can not actually load my amplifier
 

Testbench using VCVS does not drive output of amplifier differentially.
It does drive output of amplifier as floating.

On the other hand, testbench using balun does drive output of amplifier differentially.

Observe postive and negative output nodes of amplifier in Transient Analysis.
You can see difference between differential and floating.
 
Last edited:
Testbench using VCVS does not drive output of amplifier differentially.
It does drive output of amplifier as floating.

On the other hand, testbench using balun does drive output of amplifier differentially.

Observe postive and negative output nodes of amplifier in Transient Analysis.
You can see difference between differential and floating.

Dear Pancho,

I understand your point,
I believe with you kind explanation that VCVS can not load my amplifier and hence cant drive the output load differentially. I was avoiding this problem by driving the load at each output individually and I use the VCVS just to see the difference in the output signal after driving the load.

Kindly see my below images named method 1 and method 2. in Both cases I am getting an identical result and fit to the theoretical calculation. In this test benches, the feedback connection or the input circuitry is not showing

New Doc 66_2.jpg

That means in the method 2 I am not using the full functionality of the balun in the correct short way, I think it should look like the image below (method 3)
New Doc 66_3.jpg

is the last method 3 is correct and the one in your mind ?, if it is yes then how much I should the value of RL and CL as compared to method 1 or 2.

Please keep in your mind I am using a single supply operation (VDD = 3.3 V and so VCM=1.65 V)

Thank you very much once again
 

Attachments

  • New Doc 66_3.jpg
    New Doc 66_3.jpg
    149 KB · Views: 145

in Both cases I am getting an identical result and fit to the theoretical calculation.
I don't think so.
Method1 : Vo_diff = (Uo+-Uo-) + VCM
Method2 : Vo_diff = (Uo+-Uo-)

VCM is not required in method1. Surely consider direction of VCVS.

That means in the method 2 I am not using the full functionality of the balun in the correct short way, I think it should look like the image below (method 3)
Connect (2*RL)//(CL/2) to d node of balun.
 
Last edited:

Dear Pancho,

I am sencierely thankful to you kind help and your paticnce to answer my question, please forgive me is you find my questions are very week, but I am learning from you and your answeres are very useful to me.

I will later di a simulation for you to show how results from 1 and 2 are identical, nevertheless I am leaving using VCVS toward balun.

Concerning your answer regarding the load equivelant, I am now bit confused, therefore I have plot the three possible connection in the image below, so please you can tell me the value for each connection comparing to the method 1 that represent my reference circuit for comparesion

f.jpg

Thank you very much once again
 

Load of method1 is floating load not differential
Load. There is no common mode load.

So method1 is not equivalent to both method2 and method3.
If you ignore common mode,
RL=5kohm, CL=20pF in method2,
RL=10kohm, CL=10pF in method3.

See https://www.edaboard.com/showthread.php?352410

I will later di a simulation for you to show how results from 1 and 2 are identical,
nevertheless I am leaving using VCVS toward balun.
Not necessary.
Simply you are misunderstanding VCVS.
 
Last edited:
Load of method1 is floating load not differential
Load. There is no common mode load.

So method1 is not equivalent to both method2 and method3.
If you ignore common mode,
RL=5kohm, CL=20pF in method2,
RL=10kohm, CL=10pF in method3.

See https://www.edaboard.com/showthread.php?352410

Not necessary.
Simply you are misunderstanding VCVS.

Dear Pancho,

Again thank you very much for your fast response,

Now I learned from you about the load equivalent value.

One issue to discuss with you,
you mentioned that method 1 is not fully differential output rather it is floating load. However, see these two image below from Allen Holberg, he is referring to it as fully differential output

differential.PNG

differential_2.PNG

Thank you in advance
 

Dear Pancho,

Thank you for your explanation,

I have finished the design of signle ended amplifier, my new task is to design a fully differential amplifier wich have the same charastericitcs of the signle ended one with the same load condition,
kindly see from the attached image below please to determince the equivalent load (same effect) of the fully differential from the single ended

single-differential.jpg

Thank you very much
 

to determince the equivalent load (same effect) of the fully differential from the single ended
What do you want to mean by “same effect” ?

If you mean an equivalency as signal processing, RL=5kohm, CL=20pF.
 
What do you want to mean by “same effect” ?

If you mean an equivalency as signal processing, RL=5kohm, CL=20pF.

Yes I mean an equivelant to signal processing, and you already answered it

Many thanks to you
 

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