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AC response of uncompensated LDO

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nguyenvanthien

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Hi all,
I'm considering AC response of the uncompensated LDO. You can see this picture.
Can you show me that how many poles, zeros are in circuit? And formula to calculate them?
Thanks
 

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basically there are 2 poles and one zero(esr zero need to place resistance(10m) in series with output cap )


pole positions
1/(ctotal *rtotal) at every node

poles are
1)power fet gate pole
2)output pole location is 1/(cout*{ron*[rf1||rf2]}) ron if power fet in triode or else it is rds

for powerfet gate pole cgd will be multiplied by 2nd stage gain times A2
pole = 1/(R1*(C1+CGS+CGD*A2)

Depending on your error amplifier structure some others poles may come
 

Thank araya.jagadeesh!
I use the single ended 2 stages opamp in my LDO. After I run DC simulation, I have parameters:
Gm1=0.0269; R1=1329K;C1=18.2(fF); Mosfet MP has: Gm=384.1u; Rds=685.7K; Cgs=131.6(fF); Cgd=48.7(fF). So, applying your formula, pole at P1 is 9.23 Khz. It's very different from simulation result.
What's wrong?
I hope to receive your answer!acresponse.jpg
 

Hello arya.jagadeesh,
I think out pole is calculated: 1/[Cout*((Rf1+Rf2)//Rds)]. Cout=15pF; Rf1=23.5K; Rf2=49.3K; Rds=943K. So, output pole is 156.7Mhz.
Is this true?
 

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