Hi,

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.

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.

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?

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.

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.

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.