What is the output impedance??

[aryajur]

The circuit is the simple differential amplifier where the output is taken differentially as Vout1-Vout2 and the input is given differentially Vin1-Vin2. What is the output impedance across the output. Is it ro||Rd or is it 2*ro||RD ?? And why.

 

[flatulent]

You should draw the equivalent circuit and apply a voltage or current source across the output. You will end up with the 2x option. It is an important learning experience to do these hand calculations because the results will stick in your mind better.

 

[Circuit_seller]

Hi

Differential output impeadance is 2 * ro||Rd and single ended output impeadance is ro || Rd.

Regards

 

[aryajur]

Thanks a lot, I did that only and ended up with 2x option as suggested by both of u guys, when I had asked the same question to one of the TA's of this course I am taking he had told me that it is ro||Rd and not the 2x option so I got confused. Thanks a lot !

 

[terryssw]

Please correct if I am wrong, but it seems that different things can be seen by inspections: if we applied differential output voltage Vo, then we have + and - of single-ended voltage vo/2 applied to output (plus some CM voltage), and the single-ended small signal current should be + and - of (vo/2)/(ro//RL), thus the differential current is vo/(ro//RL) and thus the differential mode output impedance should be the same as ro//RL.

Another point of view is that it is well known (even in differential amplifier) that the gain is gm1(ro//RL). again assume that we have applied differential input signal vin and thus the singled-ended input of + and - of vin/2. Then it will produce singled output current, by the transconductor, with + and - of gm1*(vin/2). So the differential output current must be gm1*vin, and to conform with the gain the output impedence must be ro//RL.

I haven't done the calculation by hand, but it seems something wrong through the inspection?

 

[segabird]

There is something confused me.
Did Rout variable with frequency due to the A(gm*Rout) is variable with it?

 

[srivatsan]

Rout does not change with the frequency. But resistance and capacitance together forms RC circuit (obviously...) and it signifies "low-pass", high-pass.. so gain changes with frequency.

 

[hanerde]

I'm confused too!

 

[radiohead]

I agree that it depends on the output terminal definition.

if single ended: r0//Rd
if differential: 2xr0//Rd

Terry, by dividing the test voltage source, you only get half of the input impedance. The same input impedance is seen on the other node. Therefore, you will get the 2x input impedance.

 

[terryssw]

I think divided the test voltage source will not half the calculated output impedance, since the ro should independent of what your source is applying. Also if node A impedance is ro, node B impedance is ro, then the differential output impedance is ro*2? Can you give some proof on it?

 

[aryajur]

I think what terry is pointing out is if we put a differential voltage on both the outputs. If we put +vo/2 on one output and -vo/2 on the other output and from this arrangement calculate the output impedance as shown in the image attached, it comes out to be ro||Rd.
This is the same the single ended output impedance. But if we take the whole circuit and put a Vo across the two outputs we get the resistance as 2*(ro||Rd) because Vo sees 2 of these in series.
Its a little confusing this way because when we put the differential sources at the two outputs the voltage across the outputs is again Vo but as a differential resistance it sees just ro||Rd. This is a little strange.

 

[terryssw]

Yes, this is exactly what my idea is. Also, I think if you treat the two ro as in series, the output impedance is of course 2*ro, but can you really treat it as series? The ro on each output node is referred to AC ground.

 

[aryajur]

This is a good point we cannot put a single source and use the differential circuit which grounds the tail point at the virtual ground. In that case we will have to use the full circuit in that case.
Now what interests me is that what will be the output resistance in that case and what is the significance of that value? Any comments on that terry??

 

[terryssw]

I think one can referred to Razavi's book in Chapter 4 that the differential gain Av of the differential pair is Av = gm1*(ro//RL), so to match the equations the output resistance should be ro//RL.

 

[aryajur]

Hello Terri and everyone,
The output impedance of a diffamp is actually 2(ro||Rd) and I have included the proof in the image attached. The eguation explanation above is wrong since in that we consider the current twice in the equation.

 

[terryssw]

Thanks for your kindly information, and actually I don't make any hand calculation on this. Can you provide the source for the formula derivation?

By the way, I have a look on Razavi book p. 108 eq. (4.10) which state that the differential transconductance Gm is equal to sqrt(uCox W/L Iss)=sqrt(2ID u Cox W/L)=gm1 or gm2. So this seems have some contradition between the results, in which the prove here proves Gm = gm1/2. Can someone make some comment on this?

 

[aryajur]

Yes in Razavi book on Pg 108 he does that. Actually the conflict lies because of the way Gm is defined by Razavi in the equation 4.10
He says that

Gm = d (ΔId) / d (ΔVin)

ΔId = Id1 - Id2 ; ΔVin = Vin1 - Vin2

But Gm is actually according to 2 port networks:

Gm = Iout/Vin

Where Iout = short circuit output current. But ΔId definition of Razavi takes the current twice, so thats why Gm comes out to be 2 times.

 

[pappumaity]

The circuit is the simple differential amplifier where the output is taken differentially as Vout1-Vout2 and the input is given differentially Vin1-Vin2. What is the output impedance across the output. Is it ro||Rd or is it 2*ro||RD ?? And why.

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