Calculate impedance of a feedback circuit with Blackman's technique

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SUMMARY

The forum discussion centers on calculating the impedance of a feedback circuit using Blackman's theorem and the Vtest/Itest method. The primary equation discussed is Zin (with feedback) = Zin (with feedback circuit disabled) * (1 - RRshort) / (1 - RRopen), where RR represents the return ratio. Participants debate the implications of feedback on the circuit's impedance, particularly regarding the characteristics of NMOS transistors and the operational amplifier's gain. The consensus is that accurate calculations require considering the amplifier's gain and the circuit's biasing conditions.

PREREQUISITES
  • Understanding of Blackman's theorem for circuit analysis
  • Familiarity with feedback circuits and return ratios
  • Knowledge of NMOS transistor characteristics
  • Basic principles of operational amplifiers and their gain
NEXT STEPS
  • Study the implications of feedback on circuit impedance in operational amplifiers
  • Learn about the characteristics and behavior of NMOS transistors in feedback circuits
  • Research the detailed applications of Blackman's theorem in circuit analysis
  • Explore the Vtest/Itest method for impedance calculation in various circuit configurations
USEFUL FOR

Electrical engineers, circuit designers, and students studying feedback systems and operational amplifier circuits will benefit from this discussion.

  • #31
I must admit that - up to now - I didn`t need Blackman`s theorem. So, I took this opportunity to become familiar with it.
If somebody is still interested to see how it woks for the present task, here is the solution:

The impedance Z between two points of a feedback system is:
Z=ZD(1+Tsc)/(1+Toc)
with ZD=Impedance between both points without feedback , and
Tsc=Loop gain magnitude with a short across the selcted points; Toc=Loop gain magnitude with an open circuit across both points.

For the present case: We need the resistance between the pos. opamp input (point 1) and ground (point 2).

Therefore: ZD=R and
Tsc=0 because A=0 (both opamp input terminals at ground potential) and |Toc|=A*0.1*gm*R.

This gives

Z=R*(1+0)/(1+A*0.1*gm*R)=R/(1+A*0.1*gm*R)
(confirmed by symbolic analyzer, see my previous post)
 
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  • #32
LvW said:
I must admit that - up to now - I didn`t need Blackman`s theorem. So, I took this opportunity to become familiar with it.
If somebody is still interested to see how it woks for the present task, here is the solution:

The impedance Z between two points of a feedback system is:
Z=ZD(1+Tsc)/(1+Toc)
with ZD=Impedance between both points without feedback , and
Tsc=Loop gain magnitude with a short across the selcted points; Toc=Loop gain magnitude with an open circuit across both points.

For the present case: We need the resistance between the pos. opamp input (point 1) and ground (point 2).

Therefore: ZD=R and
Tsc=0 because A=0 (both opamp input terminals at ground potential) and |Toc|=A*0.1*gm*R.

This gives

Z=R*(1+0)/(1+A*0.1*gm*R)=R/(1+A*0.1*gm*R)
(confirmed by symbolic analyzer, see my previous post)
Thanks for your research! Only thing I would point out is that gm is a strong function of Vgs = 0.1AV where V is the quiescent + amp input voltage ("point 1"). So the operating point of the circuit still has to be solved for, then gm = 2k(0.1AV - VT) can be determined. Of course, you'd need k also.
 
  • #33
Yes, of course. My calculation (blindly) assumes an operating point in the linear FET region.
 
  • #34
LvW said:
Yes, of course. My calculation (blindly) assumes an operating point in the linear FET region.
OK. Now you'd be subject to Vds modulation but I'd say enough of this, thanks for enlightening us all on the (in?)famous Blackman method! :woot:
r m
 
  • #35
Waxterzz, are you going to try get a result with the Vtest/Itest method?
 
  • #36
This has been a fine, collaborative effort. Well done, all contributors!
thumbsup.gif
 
  • #37
Yes - and caused by an interesting question from the OP (forcing us - includiung me - to remember and apply Blackmans rule).
 
  • #38
LvW said:
Yes - and caused by an interesting question from the OP (forcing us - includiung me - to remember and apply Blackmans rule).
@ LvW - just as a final note - the fact that the upper FET can operate as a cascode connection (by suitably setting its gate bias) you can indeed operate the lower FET in the linear mode while keeping Vds essentially constant - meaning your assumption of operating in the linear mode while keeping gm constant, is fully validated.
Over and out!
rm
 
  • #39
rude man - thank you for this final remark.
 

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