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Design >> Analog Design >> Miller Compensation and Low Impedance Nodes
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Message started by SoliS on Jan 31^st, 2012, 2:47pm

Title: Miller Compensation and Low Impedance Nodes
Post by SoliS on Jan 31^st, 2012, 2:47pm

I am using the feed-forward zero technique for op amp compensation, very similar to that used in Johns & Martin Ch. 5.
(resistor in series with a cap around the output gain stage)

However, assume that there is a LOW impedance node driving the output gain stage (such as a source follower). This is the situation I am in now; I am using a SF to level-shift the output of the diff amp and then drive the second gain stage.

I noticed the compensation does not work if connected to the gate of the output stage (also the source of the SF). Someone told me it is because the node is low impedance. When I connect the compensation network to the gate of the SF device, everything works great again.

Could somebody explain why a low impedance node affects the compensation network this way?

Title: Re: Miller Compensation and Low Impedance Nodes
Post by PW on Jan 31^st, 2012, 11:41pm

Hi,

Title: Re: Miller Compensation and Low Impedance Nodes
Post by raja.cedt on Feb 1^st, 2012, 8:56am

hello,
the concept of creating zero by feed forward technique is rely on how much extra gm you have in the parallel path and how much cap you are driving. So i am not seeing any node impedance dependency. Any how why don't you explain with simple fig.

Thanks,
raj.

Title: Re: Miller Compensation and Low Impedance Nodes
Post by SoliS on Feb 1^st, 2012, 10:37am

Okay, here is a schematic. This is a UNITY GAIN configuration; Rfb is only there to compensate for the base current of the OTA (has a bipolar input stage).

This image has the compensation network connected to node A, and not node B. The compensation does not work at all when connected to node B (looks very close to uncompensated response).

I am testing stability with STB analysis in cadence; the probe is placed between Rfb and the + input.

MN1 acts as a source follower (level shifter), MN2 is the output gain stage (approximate gain of 30), MP1 is active load. I forgot the body connection on MN3 in this figure; it goes to ground. The OTA is a folded cascode type, PNP input stage.

Basically, someone explained to me that when connected to the low impedance node (B), the capacitor only makes a pole at very high frequencies. This sort of makes sense since the pole location depends on R*C, if R is small, then pole is high frequency. What I don't understand is how the equivalent R seen by the capacitor is low; I thought it was the high output Ro of the 3rd gain stage and active load that provided the high R - which is the same regardless of where the other end of the FB network is connected. Obviously something is wrong with my understanding. Also the explanation given to me could be wrong as well.

Title: Re: Miller Compensation and Low Impedance Nodes
Post by SoliS on Feb 1^st, 2012, 2:26pm

I have a theory about this but really have no idea if its even on the right track:

Connecting to Node B basically means that the only compensating capacitance for the Folded Cascode OTA is the parasitic gate cap of MN1; so the output stage may be compensated perfectly but the OTA is still unstable, which causes the whole loop to go unstable.

If that's true, then it means getting optimal compensation for this sucker isn't going to be as easy as looking up the formula for lead compensation resistance of the typical 2 stage opamp (Rc = 1/1.2gm), and it wont follow the formulas for folded cascode amps either.

Title: Re: Miller Compensation and Low Impedance Nodes
Post by Vladislav D on Feb 2^nd, 2012, 1:47am

For simplicity suppose the follower does not alter the frequency response. Look at the transfer function (feedback path) from the output to the node B through the cap, connected between them. It is (1/gm)/ (1/gm+Cf). So, if gm is high, this cap does not change the transfer function.
The statement that this is due to low output impedance of the follower is absolutely correct.

Title: Re: Miller Compensation and Low Impedance Nodes
Post by Vladislav D on Feb 2^nd, 2012, 2:57am

Another way of looking at the problem. For an opamp without a follower, there are 2 poles gm1/Ca and gm2/Cload. By placing Cc across the output stage, Ca=Cc and so, the pole is shifted to low frequency, improving frequency response. If you have a follower with a capacitor across the output stage, Ca keeps the same value, since output of the gm stage does not "see" the Cc capacitor

Title: Re: Miller Compensation and Low Impedance Nodes
Post by SoliS on Feb 2^nd, 2012, 7:10am

Vladislav D wrote on Feb 2^nd, 2012, 2:57am:

Thanks for your replies everyone, I suppose I should clarify that the two options are to connect the compensation network are:

(1) between Node A and Output node
(2) between Node B and Output node

from your post it looks like you may have been referring to the compensation network connected between A and B, which I am not doing. I also do not understand how you derived that the Feedback transfer function was (1/gm) / (1/gm + Cf). What is Cf (C feedback?), which transistor is this "gm" referring to, and how does the equivalent impedance of some node affect gm?

Title: Re: Miller Compensation and Low Impedance Nodes
Post by Vladislav D on Feb 2^nd, 2012, 12:37pm

SoliS wrote on Feb 2^nd, 2012, 7:10am:

I also do not understand how you derived that the Feedback transfer function was (1/gm) / (1/gm + Cf). What is Cf (C feedback?), which transistor is this "gm" referring to, and how does the equivalent impedance of some node affect gm?

Yep, instead of Cf, of course should be 1/sCf
gm is a transconductance of the source follower. Cf is a compensation capacitor I derive this transfer function by putting a voltage source at the output and see what voltage I get at node 'B'.
I don't connect Cc between the nodes A and B. I am just saying that placing Cc between output and 'B' will not work. Ideally, you would like zero output impedance of the follower. Now, imagine that you have it. In this case, how does the Cc affects voltage at the gate of MN2? The answer is it does not. So, there is no feedback and so, there is no pole splitting.

Title: Re: Miller Compensation and Low Impedance Nodes
Post by SoliS on Feb 2^nd, 2012, 12:47pm

Vladislav D wrote on Feb 2^nd, 2012, 12:37pm:

Ahh, that makes perfect sense. I do know that the equivalent resistance looking into the source of an active device is 1/gm (neglecting body effect). This is very helpful. So the voltage at node B will be dependent on the feedback transfer function and also on the equivalent small-signal resistance to ground at Node B, which is 1/gm, or "low."

Putting the test source at the output node makes all of this make sense. Of course in my case it will not be 1/sCf but rather Rc + 1/sCf. Should still work out similarly as just about everything at that node is strongly affected by the low impedance.

I finally got my hands on a copy of the Razavi book and chapter 10 talks about compensating a telescopic cascode OTA cascaded with a common source second gain stage, like I have here. It's very helpful so far; Simulations show that my level shifter has no real effect on the frequency response except for some weird poles and zeros that are 3 decades higher than the unity gain freq, so Razavi's analysis seems to hold true for my circuit as well. (though I am not driving a load capacitance, only a moderate resistance, so that saves me the trouble of one additional pole.

Title: Re: Miller Compensation and Low Impedance Nodes
Post by Xiaolang on Feb 24^th, 2012, 5:35pm

this is easy to understand.
there are three poles in the circuit (assuming OTA is one-pole system), Pa, Pb and Pout.

if connect Cc at the SF output and the opamp output, this will move Pb to zero, and push Pout far away, then Pa now is dominant, your circuit stability is determined by the position of Pb and Pout after compensation; might be still unstable.

if use the one you modified, the dominant pole is Pa, even closer to zero, the Pb is very far away, safely say, we can ignore it; Pout is pushed away to high freq, so the system is more stable.

Title: Re: Miller Compensation and Low Impedance Nodes
Post by sushanth on Mar 6^th, 2012, 4:24am

Hi Vladislav D,
But why do you say connecting Cc between 'B' and ouput will not help in moving the pole? Due to the gain offered by the output stage, still there will be pole splitting. And actually it is better, since the feedback signal will not affect the gate of MN2, which means we eliminate the zero which will occur if gate of MN2 is not low impedance.

Title: Re: Miller Compensation and Low Impedance Nodes
Post by Vladislav D on Mar 6^th, 2012, 7:21am

sushanth wrote on Mar 6^th, 2012, 4:24am:

Hi,
First of all, the zero u r talking about appears due to feedforward path, and one of the way to eliminate it is to put a buffer in the feedback path. This is a bit different from what is shown in the schematic.

Normally, during the compensation, the second pole goes to the higher frequency since output impedance of the second stage decreases due to feedback, formed by Cc. However, in this circuit. impedance at point B is equal to 0 and so, there is no feedback

Title: Re: Miller Compensation and Low Impedance Nodes
Post by sushanth on Mar 6^th, 2012, 9:24pm

Hi,

At node B the impedance is low @ lower frequency but @ higher frequency it sees the o/p impedance of the OTA. So, still the feedbcak due to Cc will have its effect.