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Design >> Analog Design >> Two Stage Opamp
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Message started by hande.vinayak on Jan 31^st, 2007, 1:23am

Title: Two Stage Opamp
Post by hande.vinayak on Jan 31^st, 2007, 1:23am

Hi to all,
If u go through simple two stage opamp design,in that case can see that gm6 = 10*gm6 (approximately):where
gm1 = trnscounductance of one of the i/p pair trnsistors &
gm6 = trnscounductance of 2nd stg i/p trnsistor.
My doubte is if we want 60 deg PM,why we are selecting only 10 as a multipying factor why not other one?
In Holberg,he has given explanation for cc > 0.22 * (cload).Buut npthing for transconductance related anything.
Please,can anybody guide me in this regards?
Thnx.

Title: Re: Two Stage Opamp
Post by avlsi on Feb 26^th, 2007, 1:15am

Hi,

I think Cc = 0.22 X CL is a kind of value from experience. Infact, gm6=10Xgm1 is also from experience based on specs. Your output stage gm is dependent on SR also.

Title: Re: Two Stage Opamp
Post by aamar on Feb 26^th, 2007, 8:20am

avlsi yes we can get everything through experience but you have to search for the scientific meaning too.

the answer to the question is:

the 2 stage opamp has a right hand side zero which can cause for us stability problem, so to overcome it you have either to bring it to the left hand side using a compensation resistance with the miller capacitance or to keep it in the right hand but move it far away from the W0db (i.e. 10*W0db)

as the Z (zero) = gm6/Cc , and the W0db=gm2/Cc = GainBW product (approx.)

so gm6 should be 10 times larger than gm2.

GBW= Gain * BW = Gain*P1 = (A1*A2) * (1/A2*Cc*R1) where A1 is the gain of the first stage = gm1*R1 and R1 is the output resistance of the first stage.

so the result will be = gm1*R1*A2 / A2*Cc*R1 = gm1/Cc = gm2/Cc where 1,2 are the input symetrical differential transistors.

aamar