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Frequency Compensation of two stage of OPAMP (Read 4166 times)
bharat
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Frequency Compensation of two stage of OPAMP
Sep 01st, 2006, 6:29am
 
For conventional two stage opamp
First stage is standard active current mirror as load, diff pair and tail current source
Second stage common source amplifier. For the freq. compensation we connect the Cap between first stage and second stage. The idea is if we break this cap C through Miller, part of it would go to the o/p of first stage and make it dominant pole, while the other one will go to the o/p of second stage. Text book says that this pole will go towards infinity---
Now the question is: Cap at the output will reamin the same value C(1-1/A) is approx C and R is output impedance of CS (common source stage). Since R is not changing and C is increasing, how the second pole will go towards infinity.
Also, in small signal analysis, the output impedance of CS with negative feedback is same (am I missing something here????)

thanks
-Bharat
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caosl_zju
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Re: Frequency Compensation of two stage of OPAMP
Reply #1 - Sep 4th, 2006, 5:04am
 
hi, which book says the output pole will go to infinity? I have never readed that statements!
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bharat
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Re: Frequency Compensation of two stage of OPAMP
Reply #2 - Sep 4th, 2006, 11:19pm
 
Tell me, which book do you read? so that I can write you the page no. and line no. as well  :)
caosl_zju wrote on Sep 4th, 2006, 5:04am:
hi, which book says the output pole will go to infinity? I have never readed that statements!

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sheldon
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Re: Frequency Compensation of two stage of OPAMP
Reply #3 - Sep 5th, 2006, 12:00am
 
Bharat,

  The original explanation was a little simplified, there is a detailed discussion
in the following sources,

James E. Solomon, “The Monolithic Op Amp: A Tutorial Study,” IEEE Journal
of Solid-State Circuits, vol. sc-9, pp 314-332, Dec 1974

Grey, Hurst, Lewis, and Meyer, "Analysis and Design of Analog Integrated
Circuits", fourth edition, page 639-644

When analyzing ampliers, the Miller Approximation allows designers to estimate the
effect of an impedance connect across the amplifier as a capacitor, zA, at the input
node. As a result the pole due the input node of the second stage is reduced, moves
towards zero and the pole at the output of the second stage moves towards infinity.
Since the frequency difference of the two poles is increased, the poles are said to
be split.

For the reason that the second pole moves towards infinity, you should review the
derivations in the references above. Just remeber that C[zA] is not the only terms
in the expression for the second pole, rather C cancels some nasty terms in the second
pole and thus the pole moves towards infinity.

                                                                                 Best Regards,

                                                                                     Sheldon
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packiaraj
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Re: Frequency Compensation of two stage of OPAMP
Reply #4 - Sep 14th, 2006, 10:15pm
 
Hi Bharat,
The reason why the output pole move towards infinity is due to the resistance seen by output load capacitance CL at high frequencies reduces to 1/gm (At high frequencies the compensation capacitor shorts the input and output of second stage, leaving a diode-connected device)...... (You may refer Pg:364 in Razavi)

-Packiaraj.V.
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