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Design >> Analog Design >> output resistence of 2-order opamp ?
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Message started by zhonghan on Jan 28^th, 2006, 9:06pm

Title: output resistence of 2-order opamp ?
Post by zhonghan on Jan 28^th, 2006, 9:06pm

I am confused by such a question : after miler compensating ,the output resistence of 2-order opamp reduces to RII/(gm*RII)=1/gm , then if the gain of the second order remains original, does it mean the transconductance enlarge RII*gm time ?

Title: Re: output resistence of 2-order opamp ?
Post by vivkr on Jan 30^th, 2006, 5:13am

Not quite clear what you mean by R|| in your post, but I suppose you are saying that
the AC impedance at the output is reduced to 1/Gm due to the shorting action of the
compensating cap at high frequencies.

If so, then you are right. This does happen. But at low frequencies, this is not the case,
and the gain of the 2nd stage is gm2.R||. As the frequency goes up, this gain falls off
after a point Note that the second pole is not dependent on the Ro of the devices
in the 2nd stage, but on the Gm of the 2nd stage and the load cap.

Hope this helps.

Vivek

Title: Re: output resistence of 2-order opamp ?
Post by zhonghan on Jan 30^th, 2006, 7:54pm

thanks for you interpretation, Vivek , you have solved part of my confusion .

I thought the second pole Gm/Cload were deduced from small signal mode , so the total output resistence of the second stage could be gotten as 1/Gm no matter the frequency is high or low . could you tell me what caused my error?

Title: Re: output resistence of 2-order opamp ?
Post by vivkr on Feb 1^st, 2006, 7:36am

Hi Zhonghan,

It is true that the second pole is derived from small-signal analysis. However, one assumes a priori that
the second pole is atleast beyond the unity-gain frequency of the opamp, preferably sufficiently away
to give good stability.

As the frequency lies beyond the unity-gain bandwidth (Gm1/Cc), the cap Cc is now acting like a short,
shorting the gate and drain of the second stage driver transistor (neglecting the Cgs of this device).
So, the impedance seen by the load cap Cl is Gm2/Cl which gives you the second pole.

I think you forgot that you were deriving the pole under these assumptions, and hence made the error
in assuming the second stage impedance to be too low.

Incidentally, the effect which causes this second pole is undesirable in many ways (poor PSRR).
If you were to use another architecture (Ahuja, JSS, Dec. 1983), the second pole would lie much
higher and this shorting effect would be gone.

If you can tolerate the extra current needed in the intermediate level-shifter of this scheme, and
don't mind the possibility of a small systematic offset, then I would recommend using it. It makes
life much simpler, especially compensation, and boosts your performance quite a lot.

Anyway, all that is unrelated to your question. I hope the confusion is cleared up now.

Regards
Vivek

Title: Re: output resistence of 2-order opamp ?
Post by zhonghan on Feb 2^nd, 2006, 2:49am

Hi Vivek ,thank you for your detailed explanation .

According to my understanding, you mean that the second pole is different when signal frequency changes. At low frequency , it is 1/(RII*Cload), and when frequency becomes sufficiently high, it would be a higher value ( Gm2/Cload ). If this change takes place after the first pole(Gm1/(Av*Cc) ) ,then I think the low second pole may influence the frequency response.
Do you think so ?

Title: Re: output resistence of 2-order opamp ?
Post by vivkr on Feb 2^nd, 2006, 4:19am

Hi Zhonghan,

I think my reply was not clear enough. The second pole is fixed. The pole location cannot depend on the frequency.
It is a characteristic of the system, otherwise the whole idea of linear systems would become unworkable.

What I meant was the following:

While computing the second pole, you always make an assumption that the opamp is reasonably well-designed,
which means that the second pole is expected to be a bit beyond the unity-gain frequency. Then, you compute the
location of this pole with this assumption. This allows you to simplify the circuit by assuming that the cap Cc
is effectively a short, and creates a drain-gate short at high freequencies, giving you the simple expression of
Gm2/Cl for the second pole.

Note that you can go ahead and derive the second pole location without this assumption, and you will get a number
very close to Gm2/Cl (detailed analysis can probably be found in Gray&Meyer or Johns&Martin).

It is not the pole location that changes with frequency, but the output impedance of the 2nd stage that does. So, it
is relatively high at low frequencies, giving you DC gain, and at high frequencies, it starts to roll off, causing the
gain to fall. However, since the 2nd pole must be beyond the unity-gain bandwidth, the effect of this rolloff will
not dominate the opamp's gain vs. frequency curve.

This is why the 2nd pole is called the "non-dominant pole".

Hope this helps.
Vivek

Title: Re: output resistence of 2-order opamp ?
Post by zhonghan on Feb 2^nd, 2006, 4:57am

This time you make me completely understand the issue. You are right , at low frequencies , the second pole is still Gm2/Cload not 1/(RII*Cload), and at high frequencies the rolling off of AC gain derives from two aspects. One is the reducing of output resistence derived from shorting effect of Cap Cc , another is the shorting effect of Cap Cload .

Thank you again for your patience.