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miller op amp compensation (Read 15064 times)
aaron_do
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Re: miller op amp compensation
Reply #15 - Oct 14th, 2008, 11:36pm
 
Hi raja.cedit,

your equation gm*cc/(cc+cL1) is missing the output resistance of the first stage , r01. At low frequency, cc is open circuit, and because of r01, there is no feedback. So at DC, output resistance is r02, not 1/gm. Does that sound correct?

It seems like some kind of black magic that you can get a pole at gm2CL without having the output resistance as 1/gm2...

cheers,
Aaron

Aaron
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vivkr
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Re: miller op amp compensation
Reply #16 - Oct 15th, 2008, 11:00pm
 
Hi Aaron,

At high frequencies (around the unity gain and higher), the Miller compensation cap is practically a short. Since this is connected between the gate and drain of the second stage driving transistor, you have a gate-drain short at high frequencies, and this "diode" offers resistance of gm2 to the output load cap CL for a second pole at gm2/CL.

Of course, the real situation is slightly different since the Cgs of the second stage will change the action a little bit, but that's the basic idea.

No black magic here, or anywhere else...

Vivek
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raja.cedt
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Re: miller op amp compensation
Reply #17 - Oct 15th, 2008, 11:44pm
 
hi,
   i just want to know always miller compensation will give pole splitting or not(means this splitting depends on wether 1st stage , second stage has output dominant pole?)
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vivkr
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Re: miller op amp compensation
Reply #18 - Oct 16th, 2008, 1:22am
 
Hi,

Pole splitting will occur only when the dominant pole is set by the compensation capacitance Cc. If you increase the value of this cap from zero to a certain optimum value, the dominant pole will rapidly move lower and lower, while the nondominant pole will move outwards. Beyond a certain point, increasing Cc will not cause the nondominant pole to move further or not much further. The second pole is set at Gm2/Cl. After this point, you need to increase Gm2 to move the second pole out.

Note in particular that it is highly undesirable practice in general to go on increasing Cc, and Cc >= Cl will give diminishing returns. In a good design, Cc < Cl and preferably by a factor of 2-3. Otherwise, you are wasting power. In some exceptional cases (very low-bandwidth circuits), you may want to increase Cc a lot to reduce total inband noise. That's it.

Vivek
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aaron_do
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Re: miller op amp compensation
Reply #19 - Oct 16th, 2008, 6:02am
 
Hi Vivek,

I understand the part about the diode connection at high frequencies. What i don't get is this...

At low frequency gain of second stage is gm2rout. Pole frequency is gm2/C because the resistance is 1/gm2. Since transconductance is gm2, gain at pole frequency should be approximately gm2/gm2. This is clearly not 3dB less than gm2rout.

So where is the error?

cheers,
Aaron

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vivkr
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Re: miller op amp compensation
Reply #20 - Oct 16th, 2008, 11:07pm
 
aaron_do wrote on Oct 16th, 2008, 6:02am:
Hi Vivek,

Since transconductance is gm2, gain at pole frequency should be approximately gm2/gm2. This is clearly not 3dB less than gm2rout.

So where is the error?

cheers,
Aaron



I don't understand your question? Why should the gain be 3 dB lower than the DC value? Are you saying that you expect the gain of the 2nd stage to be 3 dB lower than its DC value at the frequency where the 2nd pole is to be found? In that case, your analysis is slightly off track. The gain would have been 3 dB lower for the 2nd stage at the frequency corresponding the its pole were there no feedback (in form of the Miller cap) across this stage. That changes the dynamics. So, gm2/Cl is the 2nd pole of a Miller-compensated opamp, not the open-loop pole of the 2nd stage. That is something different, and in certain compensation schemes like the Ahuja scheme where there is no direct feedback across the 2nd stage, that situation is seen more clearly.

I hope this answers your question.

Regards,
Vivek
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aaron_do
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Re: miller op amp compensation
Reply #21 - Oct 17th, 2008, 5:04am
 
Thanks. That basically answers my question. I like to simplify things as much as possible, and I was basically trying to compare the feedback case to an open loop case. Thanks for pointing out my error. I guess I know what i will be reading more about tomorrow...

cheers,
Aaron
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