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Low Voltage current source Analysis

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ccw27

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This is what I got using Blackmans formula:

Rout(a=0)=ro2
Rout(short)=-A*gm1*(ro1||RoB)
Rout(open)=-A(gm1*(ro1||RoB)-gm2*ro2)

Rout=Rout(a=0) [1+Rout(short)]/[1+Rout(open)]
Rout=ro2*[1-A*gm1*(ro1||RoB)]/[1-A(gm1*(ro1||RoB)-gm2*ro2)]

Actually I believe the expression is somewhat the same if you simplify it, maybe there is a sign error somewhere.
 

elbadry

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I made an approx. analysis and found it to be almost:

A*Rob

I am not quite sure this is correct though.
 

venkateshr

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I think it is -Rob.

a negative resistance ckt. i didnt use any blackman's theorm or other thing. strgtfwd analysis.anyway thanks for the good ckt.

Added after 34 minutes:

there is small correction . resistance is Ro1 || -Rob . i forgot the o/p imp of the mos.

Added after 53 seconds:

Ro2 instead of Ro1
 

pixel

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Very useful circuit, whoose idea is to make equal Vds voltages of two pmos, which results in ideal current matching. Amplifier is just used to set DC voltage gate voltage of current current mirror gate. So I see this circuit like ideally matched current mirror.
If you want to calculate resistance you can to put at node X Itest current generator and find Vx (R=Vx/Ix). If A is high, negative feedback makes that Vtest=V+=V-. As a result of matching, both pmos will have same current, and output resistances, and through RoB will also flow current Ix, which makes V-=-RoB*Ix.
Then Rout=-RoB.
In the case of finite amplifier gain, I have used symbolic circuit solver Syrup for Maple.
Rout=-ro*(ro+RoB+A*gm*ro*RoB)/(gm*ro^2*A-ro-RoB)
 

pixel

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Just to add:
You have two feedback loops (one positive and one negative), and I think that Blackman theorem can not help you.
 

    ccw27

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venkateshr

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can you please tell me waht that blackman's theorem is ? I am very curious to know that.
 

ccw27

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Just check Gray Meyer book
 

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