problem in determining transfer function and ZVT

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akbarza

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Hi

I found the schematic(schematic1.pdf) in a file that I downloaded of net. but unfortunately, I deleted it and now I don't have access to it. The schematic is a capacitor feedback model of a circuit. I drew the schematic with ltspice software.

The problems are related to obtaining transfer function of it:

First method (node analysis): if I write node analysis in Vout then (I use Cl instead of Cload ):

Vout/vin = (Rout.cf.s -A.Rout)/(1+(cf+cl).Rout.s)



If I write node analysis in vin ( I assume input current to opamp is neglectable) then:

Vout/Vin = (1+(cf+cin).Rin.s) / (Rin.cf.s)

Obviously, two above TF's are not compatible.

In other side, we have three capacitor that one of them is not independent, so the denominator must be of order 2 namely we must have in denominator something like 1+b0.s +b1.s^2 .

But in above I had denominator of order 1.

What is my mistake?

Second method (zero value time constants or ZVT):

I uploaded file 7042.pdf (in this file the method ZVT has been described to obtain transfer function (TF)).

Problems:

  • First in page 7/47, the word independent has been written. Can you explain about its meaning?
  • I wanted to obtain Tf with ZVT method.
  • With a TF of the form H = (a0+a1.s+a2.s^2) / (1+b0.s+b1.s^2).
  • To obtain a1 according page 23/47 a1 equals with sum of some Ti.Hi terms( I am writing this text in word and then will copy/paste to site, so I changed notation)
  • Where Ti is characteristic time of i'th capacitor and it's value is ci.Ri. where Ri is total resistance seen from ci when for j<>i then cj=0. ( I used <> for is not the same).
  • For Hi, in page 22/47 is written Hi=H (when ci tends to infinite and for j<>i cj=0).
  • For cin capacitor in node Vin: Hi= H (when cin tends to infinite and cf , cl tend to zero)
  • As cin-->infinite then 1/(cin.s) =0, so vin=0.
  • As cf,cl -->0 then 1/(cf.s)-->inf , 1/(cl.s)-->inf. so capacitor feedback branch is broken and circuit is as vin-opamp(-A)-Rout-vout, so vin=0 leads to vout=0.
  • Now Hi,cin= vout /vin = 0/0, what can I do now?
  • Is true what I wrote in above lines? Guide me plz.

  • Excuse me for lengthy text.
  • Thanks
 

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In the circuit shown the input voltage Vin is directly imposed at the input of the amplifier, then regardless to the feedback circuitry, at the output of "-A" block there will be -A*Vin.
Possibly Rin is the series resistor between Vin and the amplifier input.
 
hi
as i mentioned, i do not access to the first source, but surly the schematic was as i drew.
thanks for mention
--- Updated ---

hi
you are right
in ZVT method, for obtaining characterestic time for a special capacitor or inductor, 1) nullify other energy storing elements( other capacitors and other inductors) and 2) nullify independent sources(short circuit voltage source and op)n circuit current source.
in short circuit vin, then Rin is connected to ground.
thanks
 
Last edited:

hi
I obtained H with two methods node analysis and ZVT .
with H=(a0+a1.S)/(1+b0.s+b1.s^2) , in determining of a1, two method give different values
with node analysis: a1=Rout.cf
with ZVT: a1= Rout.cf +(1+A).Rin
I checked them again for detection difference cause, but I can not find it.
do you have any opinion about it? which a1 is correct?
thanks
 

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