Help me find the output impedance of a circuit

• Jul 17, 2007

• #1

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firsttimedesigning

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I am trying to find the output impedance of a circuit(file: gain-boosting). I drew a small signal model for it (file: ss gain-boosting). However, I dont think my small signal is correct because I kept having trouble finding the right output impedance. Please take a look at my small signal model and point out where I have made a mistake. thanks.

 

• Jul 17, 2007

• #2

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hardings

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gain-boosting

I guess it is because of the input, one terminal of VIN should be connected to RO1?

 

• Jul 17, 2007

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firsttimedesigning

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Re: gain-boosting

I modified the picture but I still could not find the right equation for Rout.

 

• Jul 17, 2007

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barath_87

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Re: gain-boosting

you need not struggle so much...... to find the o/p impedence ....
You can tell it intutively right........
Forget about the op-amp you know that the o/p impedence looking from the drain for a source degenerated circuit is gm2*ro2*ro1
This is boosted by the open loop gain of the op-amp(AV).........
So the o/p impedence is Av*gm2*ro2*ro1......
Hope this helps you.........

 

• Jul 17, 2007

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hardings

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gain-boosting

Ve=Vx, and then do KCL for node X,
For learning, you can use this way.
but for designing, use barath_87's way.

 

• Jul 18, 2007

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firsttimedesigning

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Re: gain-boosting

Thank you hardings for replying.
there is one thing I dont understand though. why does Ve = Vx?
I thought Vin - Vx =Ve...but then I think I am wrong...

 

• Jul 18, 2007

• #7

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A.Anand Srinivasan

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Re: gain-boosting

i think Vin=Vx+Ve is the right equation... because KVL would result in that equation only...

 

• Jul 18, 2007

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firsttimedesigning

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Re: gain-boosting

Thank you barath_87 for replying.
There is one thing I dont understand, I looked up for the equation that gives output resistance of a common-source amplifier with source degeneration,
Ro = Rs + ro[1+(gm+gmb)Rs]

Ro = o/p impedance
ro = o/p resistance of the transistor
Rs = the resistance of the resistor connected to the source. In the graph that I have drawn, Rs = ro1.
gm = top-gate transcoductnace
gmb = body effect transcoductance

you mentioned that the output impedance is gm2*ro1*ro2 for a circuit with source degeneration.
Ro = Rs + ro[1+(gm+gmb)Rs]
Ro = ro1 + ro2[1+(gm2+gmb2)ro1]
So I guess that gm2*ro1 is much bigger than 1 and gmb =0.
Ro = Rs + ro[(gm)Rs]
Ro = ro1 + ro2[(gm2)ro1]
Also gm2*ro1*ro2 is much bigger than ro1.
Ro = ro2(gm2)ro1
Is that right?

 

• Jul 18, 2007

• #9

barath_87

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Re: gain-boosting

Yes you are very correct.....
I just made a few approximations....

 

• Jul 18, 2007

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hardings

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gain-boosting

But you are analysing AC character, not DC.
For source degeneration, Vin is a dc level voltage, in other word, vin is ground for AC analyse.

 

• Jul 18, 2007

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firsttimedesigning

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Re: gain-boosting

Thanks for replying. Hardings, you are right. I forgot that Vin becomes ground when doing AC analysis. Thank you so much for pointing that out.

Added after 1 minutes:

Ok so I have modified the ss gain-boosting file. I still can not derive the equation for Rout.

the equation that i have is (KCL at node X)
Ve/ro1 + Ve/Rin + gm2Vgs + Ve/ro2 = 0

what i really want to know, though, is that what is the relationship between Vgs and A1Ve. Are they equa??

 

• Jul 18, 2007

• #12

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hardings

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gain-boosting

I think Vgs is not equal to A1Ve,

A1Ve is the AC voltage from Vgate to GROUND.
Vgs is the AC voltage from Vgate to Vx

Now you can finish your calculation.

 

• Jul 18, 2007

• #13

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Re: gain-boosting

ok, so i derived the equation but i still have one question

Vx = it * (ro1||Rin) = Ve

Vt = Vx + i2ro2 = Vx + ro2 [it-gm2(Vgs)]
Vt = Vx + ro2 [it - gm2 (AoVx - Vx)]

Vt = Vx + ro2 [it + gm2 Vx (Ao - 1)]
Vt = it * (ro1||Rin) + ro2[it + gm2 it (ro1||Rin)(Ao-1)]
Vt/it = (ro1||Rin) + ro2[1 + gm2 (ro1||Rin)(Ao-1)]
assume Ao >> 1
Vt/it = ro1 + ro2[ 1+gm2(ro1Ao)]
assume gm2ro1Ao >> 1
Vt/it = ro1 + ro2gm2ro1Ao
assume gm2*ro2*ro1*Ao >> ro1
Vt/it = gm2ro2ro1Ao

so the question that i want to ask is that when i go from
Vt = Vx + i2ro2 = Vx + ro2 [it-gm2(Vgs)] to
Vt = Vx + ro2 [it - gm2 (AoVx - Vx)]

i used Vgs = AoVx - Vx. However, if i used
Vgs = Vg-Vs = 0-Vs = -Vx
then I would not get the right equation since
there is no Ao. So does that mean Vg is not
zero. I would think Vg is zero since Vb is grounded
for AC analysis. Also, I updated the ss gain-boosting file

 

• Jul 19, 2007

• #14

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hardings

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gain-boosting

VG is not zero, why you think vg is 0?
Only when VB=VX, VG is zero, but here VB=0,
VX is not 0.

 

• Jul 20, 2007

• #15

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Re: gain-boosting

I am sorry I was confused.

 

• Jul 24, 2007

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Re: gain-boosting

You'll also find that the exact value is somethin like gm2(1+A)ro1ro2 but since A>>1, that's the reason for the approximation to just (gm2)A