If you want to design the system in order to have phase maring=45º, then yes. Design in order to have arg(A(j·ω)*B(j·ω))=-135º (yes, MINUS 135º)Thank you...But for arg(A(j·ω)*B(j·ω))=+ or - 180º... Should I keep it equal to 180 or 135, because we want min 45deg phase margin.
tisheebird - you are speaking of a circuit, correct?
Now - when you have some questions, don`t you think it could be helpful to show a diagram of the circuit under question?
I must admit - I do not feel much motivated to guess what you mean with a set of (incorrectly written) formulas like
A= G sCdRf/ 1+sRf( Cd+Cf)
B= 1+sRfCf/sRfCd...
G= gm1r03/ 1+sC2ro3..
"G" in your circuit diagram, seems like a MOSFET amplifier ? If so, the calculation is wrong since you have mixed somehow the OP Amp(has infinite gain at DC) with MOSFET amplifier. Your amplifier has finite gain at DC and is gm1·ro3.
If not, show please what is inside "G" amplifier.
If one has a finite gain amplifier with feedback, the open loop gain "A" is calculated by breaking the loop of the feedback.
The assumed first order gain characteristic isn't bad as an estimation. There's an additional cascode stage pole, but you can probably ignore it.
The other expressions are however only right if C2 >> Cf, otherwise the amplifier output impedance matters.
As a short cut, it's rather unlikely to drive an amplifier with a single gm stage into instability.
Your circuit is a shunt shunt topology. In order to calculate open loop gain "A" , you have to actually break the feedback loop and calculate it like it is shown in the book of Sedra-Microelectronic circuits.
In this case, as you have calculated the transfer function of the cascode amplifier and it has infinite input impedance, then you can write equations like in an op amp and solve them. See attached picture.
- - - Updated - - -
Vx in the attached picture is show here:
Unfortunately the amplifier is rather an OTA than an OP and the calculated G is only correct for Co >> Cf, as mentioned. The exact transfer function is even more complex than the complete loop gain.In this case, as you have calculated the transfer function of the cascode amplifier and it has infinite input impedance, then you can write equations like in an op amp and solve them.
Unfortunately the amplifier is rather an OTA than an OP and the calculated G is only correct for Co >> Cf, as mentioned. The exact transfer function is even more complex than the complete loop gain.
According to SapWin4
G = -gm*Ro*(1+(Cd+Cf)Rf s)/(1+(CdRo+CdRf+CfRf+CoRo)s + (CfCd+CoCd+CoCf)RfRo s²)
View attachment 136609
loop gain = -gm*Ro*(1+(CfRf s)/(1+(CdRo+CdRf+CfRf+CoRo)s + (CfCd+CoCd+CoCf)RfRo s²)
Unfortunately the amplifier is rather an OTA than an OP and the calculated G is only correct for Co >> Cf, as mentioned. The exact transfer function is even more complex than the complete loop gain.
According to SapWin4
G = -gm*Ro*(1+(Cd+Cf)Rf s)/(1+(CdRo+CdRf+CfRf+CoRo)s + (CfCd+CoCd+CoCf)RfRo s²)
View attachment 136609
loop gain = -gm*Ro*(1+(CfRf s)/(1+(CdRo+CdRf+CfRf+CoRo)s + (CfCd+CoCd+CoCf)RfRo s²)
v1:
+ ( - Hgm Cd Rf Ro + Cd Ro ) j w
- ( Cf Cd Rf Ro ) w^2
------------------------------------------------------------------------------
+ ( Hgm Ro +1 )
+ ( Cd Ro + Cd Rf + Hgm Cf Rf Ro + Cf Rf + Co Ro ) j w
- ( Cf Cd Rf Ro + Co Cd Rf Ro + Co Cf Rf Ro ) w^2
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