# Determining the phase margin of a negative series-shunt feedback circuit

I have been given a transfer function: $H(s)={A_0\over (1+s/w_{p_1})(1+s/w_{p_2})}$

With the condition that $w_{p_1} << w_{p_2}$

It's designed for this feedback circuit that has a feedback factor $K$ that is equal to 0.5 I calculated both of the open and closed-loop gains without considering any capacitor since I do not see any at the circuit? How should I deal with the given poles if there is no large capacitance in the circuit? Should I make calculations considering the amplifier-inside capacitors? How can I find the phase margin of such as circuit?

Open and closed-loop gains that I have found: $A_{open}=g_{m1}(ro_1//(R_1+R_2)$ $A_{closed}=A_{open}/(1+0.5A_{open})$

• without considering any capacitor since I do not see any at the circuit? Well, I see at least the Cgs of the NMOS. How can there be any poles/phase shift if you have no capacitors? And that's what you found: no poles. – Bimpelrekkie Nov 21 '17 at 15:25
• Sorry for my stupid mistake, I thought that I can just deal with the low-frequency area for this question. After I include the small capacitors into account, as a mid-band gain A0 in the transfer function should I use the open-loop for closed-loop gain? I think open-loop midband gain makes sense for this? – user8925869 Nov 21 '17 at 15:30
• For the transfer function of the closed loop circuit (as it will be used as part of other circuits): use the closed loop transfer. To evaluate open loop transfer for stability, phase margin: use the open loop transfer function. So it just depends on the condition for which you want that "mid-band gain" to be valid. Is that gain relevant for open loop or closed loop? – Bimpelrekkie Nov 21 '17 at 15:48
• because of the information of feedback factor. I think that the asked thing is about closed-loop transfer function. Do I have to calculate the poles wp1 and wp2 to find phase margin? – user8925869 Nov 21 '17 at 16:35
• Opening the loop with transistors-based designs is not that obvious depending on the configuration. Dr. Middlebrook in his general feedback theorem (GFT) showed that the return path can significantly load a circuit and opening it may disturb the original circuit, distorting the computed closed-loop analysis. More information can be found here web.archive.org/web/20160401041250/http://ardem.com/gft.asp. – Verbal Kint Nov 21 '17 at 19:30