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Can I use the feedback capacitor as the compensation cap in this integrator?

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hebu

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The integrator is as attachment. The opamp is one stage folded cascode OPamp.
May I use the feedback capacitor to be the compenstation cap?
 

Re: One question

Yes!

The feedback response fives you +90 degree up to the frequency where the input cap of the diff inputs and the resistor set the pole. So the phase margin of the open loop increases. So look for a resistor which gives with the input cap a higher pole than the second/third parasitic pole within the opamp.
 

One question

The loop gain of your amplifier is:

Gm*Ro*Rs*jwC/(1+Ro*jwC);

so increase C will not compensate your circuit as miller cap; However, when RojwC is much greater 1, you cancel out the dominant pole in your circuit with the zero;
 

Re: One question

In practice you don't want to use the feedback cap as the compensation cap at all: the feedback cap and the input resistor forms a RHP zero which will be typically at higher frequency than your dominant pole frequency, yet lower than your UGBW. The non-dominant poles will come back after this zero and degrades your phase margin quickly and make your circuit not robust.
 

Re: One question

So, how should I do for this compenstation? make a loading cap inside OPamp?
My OPamp is only single stage OPamp.
 

Re: One question

willyboy19 said:
In practice you don't want to use the feedback cap as the compensation cap at all: the feedback cap and the input resistor forms a RHP zero which will be typically at higher frequency than your dominant pole frequency, yet lower than your UGBW. The non-dominant poles will come back after this zero and degrades your phase margin quickly and make your circuit not robust.

Where is the RHP zero?

Can you write out the transfer function?
 

Re: One question

neoflash Posted: 19 Jul 2005 21:59 Post subject: One question

--------------------------------------------------------------------------------

The loop gain of your amplifier is:

Gm*Ro*Rs*jwC/(1+Ro*jwC);

so increase C will not compensate your circuit as miller cap; However, when RojwC is much greater 1, you cancel out the dominant pole in your circuit with the zero;

--------

Gm*Ro*Rs*jwC/(1+Rs*jwC);

Replace the Ro with Rs in the denominator, is it right?
 

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