The AC response

Hi Ricky,

1.) I suppose the shown ac magnitude respose is the loop gain, correct?
Why didn't you show the phase response?
2.) Are you sure that the loop gain was simulated correctly? According to the slope of the magnitude response at unity gain the phase margin should me much lower (perhaps 45-50 deg).
3.) Does the step response belong to the closed loop circuit?

Ricky Chen wrote on Apr 29^th, 2012, 11:22pm:



Sorry the LOOP_GAIN plot has not same axis for both PHASE and Gain so I have attached the same with same axis in GAIN and PHASE

@ricky: Sorry, I didn't see the phase (forgot to scroll). Please, can you confirm that you have used the "double-injection technique" ? (Because the term "Middlebrook" is somewhat ambigious).

@raja: In most cases (in particular for opamps without transfer function zeros and with a continuous gain decrease) the phase response is directly related to the gain response and, thus, must not be shown explicitely. In these cases, the phase margin can be estimated using the SLOPE of the gain function only. In particular, if the 2nd pole is approx. at the unity gain frequency (slope changes from -20 to -40 dB/dec) the PM will be approx. 40...50 deg.

raja.cedt wrote on Apr 30^th, 2012, 1:53am:


1:- Why you imagine RHZ from transient response ? Can you tell the reason behind it ?
2:- Do you expect a peaking in Magnitude plot to reflect RHZ ?

[quote author=raja.cedt link=1335730561/0#6 date=1335775995]@buddy: hello buddy..i couldn't understand this

@ ricky: in theory you get small overshoot (around 3%), but when you implement with transistors you could expect slight change in the gm or some small signal parameter, and may be your overdrive of the input pair lower and with 30mv swing it may change some %ge. So try to apply very small step like micro volts or increase the transistor overdrive.

I actually did with 1uV input. No peaking is SHOWN.
I wonder why you tell that transistor is overdriven.
Pleas NOTE:- The input is being applied at the input-R of the OTA not at the OTA input. ( please see the schematic). Also the OTA's o/p stage is a CLASS-AB so slew is NOT that a problem  and feed-back makes the input of the OTA stand-still(virtual Ground).

1:- Middle-Brook theorem is NOT ambiguous at all.

I wrote that the TERM "Middlebrook" is not sufficient because his theorem (The general feedback theorem, GFT) consists of several parts (some with simplifications).

2:- Yes it is 'double injection technique' i.e. I found RRi(injecting current) and RRv( injecting voltage) at the break point.
This is clear from the schematic I have shown.

But it was not clear from the drawing - more than that, I must apologize for not beeing familiar with your simulation program (incl. schematic entry).


3:- From gain plot it is very difficult to figure the PHASE AMrgin.
As it may have pole zero pairs.

May answer to raja was a general one and not specifically related to your drawing.