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I have the following circuit and wanted to find out the transfer function of the whole circuit (which includes the non-ideal op-amp as well - modelled as a VCVS - H(S) is DC gain with some poles).

enter image description here

I tried the following:

1-To model it as a simple Op-Amp with feedback network with the transfer function as shown below: enter image description here But with this, I could only model an opamp with a constant dc gain and with poles. But the Input capacitance and output resistance of opamp (non-idealities) are not included.

2-To include the non-idealities of the op-amp (input capacitance and output resistance), I tried to analyze with a different approach with the feedback network modelled as a Two-port network - something like below: enter image description here

Somehow I got the h12,h11,h22 by solving the twin-T bridge (using Delta start transformation -> combining the parallel resistances -> then finding voltage gain, input and output resistances) using mathematica, but the relations got more complicated and it was of third order which I feel it was not correct.

I modelled the two-port network using h-parameters since it was a voltage-current feedback. But should i consider it as a different feedback structure (like a voltage-voltage feedback) and then try? or Is there any other method to get the whole transfer function?

Any help or guide in getting a simplified expression for the whole transfer function (with feedback and non-idealities) of the above circuit will be helpful.

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  • \$\begingroup\$ That's is not a great configuration to start analyzing. Also you show the inverting schematic instead of using positive feedback, while the block diagram uses the non-inverting inputwith negative feedback \$\endgroup\$ – Sunnyskyguy EE75 Jun 15 '18 at 23:26
  • \$\begingroup\$ Get a free sim tool. \$\endgroup\$ – Andy aka Jun 16 '18 at 9:09
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    \$\begingroup\$ @sundar Sapwin does symbolic transfer functions, but you can expect a complex, "high entropy" expression. I find that complex expressions are often dominated by a few terms, and resemble simpler expressions with only a few RC time constants. I guess at the simpler expression with hand analysis and SPICE, and then I check that it is correct by evaluating both expressions numerically. \$\endgroup\$ – DavidG25 Jun 18 '18 at 17:20
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    \$\begingroup\$ This is going to be an ugly transfer function with a 5th-order denominator (or 6th with the op-amp pole). I would use the FACTs to have the whole thing expressed in a well-ordered polynomial form. However, going for a 5th-order system without experience is tough. I would recommend going through cbasso.pagesperso-orange.fr/Downloads/PPTs/… and increase complexity. Exercise yourself to determine the transfer function with a perfect op-amp and the FACTs then plug the op-amp in again. I usually combine Mathcad and LTspice to check my calculations. \$\endgroup\$ – Verbal Kint Jun 20 '18 at 19:41
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    \$\begingroup\$ You can also have a look at this paper where the op-amp model is plugged in a type-2 compensator: how2power.com/newsletters/1702/articles/… \$\endgroup\$ – Verbal Kint Jun 20 '18 at 19:45
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The way I approach these is using a known opamp configuration and solve the system in sections using Z impedance's for components:

enter image description here
Source: http://sim.okawa-denshi.jp/en/opampkeisan.htm

The equation for this is:

\${\Large \frac{V_{out}}{V_{in}}=\frac{Z_{2}}{Z_{1}}}\$

Which I'll call Z7 and Z8

\${\Large \frac{V_{out}}{V_{in}}=\frac{Z_{8}}{Z_{7}}}\$

Now solving for a single T filter:

schematic

simulate this circuit – Schematic created using CircuitLab

you get:

\${\Large \frac{V_{a}}{V_{b}}=\frac{Z_{3}^2}{Z_{3}^2+Z_{1}Z_{3}-Z_{1}Z_{2}}}\$

Now if we duplicate the section with Z4,Z5 and Z6 (Z4 in the place of Z1 in the above pic, Z5 for Z2, Z6 for Z3), since Va and Vb would be the same for both top and bottom t filters we can treat them as parallel impedances

Here are the two parallel impedance's:

schematic

simulate this circuit

\${\Large Z_{top}=\frac{Z_{3}^2}{Z_{3}^2+Z_{1}Z_{3}-Z_{1}Z_{2}}}\$
\${\Large Z_{bot}=\frac{Z_{6}^2}{Z_{6}^2+Z_{4}Z_{6}-Z_{4}Z_{5}}}\$

and the parallel impedance's (with the negative terminal of the op amp being Va from above and Vb being Vout:

\${\Large \frac{V_{negterminal}}{V_{out}}=\frac{1}{\frac{1}{Z_{top}}+\frac{1}{Z_{bot}}}}=Z_{7}\$

\${\Large \frac{V_{negterminal}}{V_{out}}=\frac{1}{\frac{Z_{1}}{Z_{3}} +\frac{Z_{4}}{Z_{6}} - \frac{Z_{1}Z_{2}}{Z_{3}^2} - \frac{Z_{4}Z_{5}}{Z_{6}^2} + 2 }=Z_{7}}\$

(confusing yet?) back to the original transfer function: \${\Large \frac{V_{out}}{V_{in}}=\frac{Z_{8}}{Z_{7}}= \frac{Z_{filt}}{Z_{7}}}\$

where \${\Large Z_{7}= \frac{1}{\frac{Z_{1}}{Z_{3}} +\frac{Z_{4}}{Z_{6}} - \frac{Z_{1}Z_{2}}{Z_{3}^2} - \frac{Z_{4}Z_{5}}{Z_{6}^2} + 2 }}\$

and the whole enchilada:

\${\Large \frac{V_{out}}{V_{in}}=\frac{Z_{8}}{Z_{7}}= Z_{filt}*(\frac{Z_{1}}{Z_{3}} +\frac{Z_{4}}{Z_{6}} - \frac{Z_{1}Z_{2}}{Z_{3}^2} - \frac{Z_{4}Z_{5}}{Z_{6}^2} + 2 )}\$

Now substitute in all of the impedance's for their corresponding values and you get your final equation

\${\Large Z_{1}=\frac{1}{C_1 s} }\$
\${\Large Z_{2}=\frac{1}{C_2 s} }\$
\${\Large Z_{3}=R_3 }\$
\${\Large Z_{4}=R_1 }\$
\${\Large Z_{5}=R_2 }\$
\${\Large Z_{6}=\frac{1}{C_3 s} }\$

\${\Large Z_{8}=R_{in} }\$

Now for a few notes, I double checked this but have been liable to make mistakes. If you follow the process that is not wrong, whith so many variables it's hard to keep track of them all.

Second thing: I eliminated V3 because it doesn't make sense, if you set the negative terminal to V3 (by having a voltage source on an opamp terminal) then it sets the negative node to that voltage and turns the whole circuit into a comparator based off of Vin3 and ground. This effectively shuts down any feedback you would have.

Third thing: I neglected Cin because I didn't understand your diagram, it should be easily convertible for solving Z8 for a high pass filter or low pass filter depending on your configuration. I just used Rin instead.

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  • \$\begingroup\$ Thanks very much for your answer. Well, Vout/Vin = Z2/Z1 occurs when its ideal right? I wanted the transfer function (with the non-ideal characteristics of op-amp included - Gain of op-amp, CL, Cin), Anyways - sim.okawa-denshi.jp/en/opampkeisan.htm link is very helpful. \$\endgroup\$ – sundar Jun 18 '18 at 19:23
  • \$\begingroup\$ The gain always occurs and will still be very close to Z2/Z1 when the amplifier is non-ideal especially at DC. Additional loading capacitance will change the overall filter so minimize that. Find your transfer function gain at DC and check it with a spice package at DC, run an AC analysis and see if the frequency effects make sense for your application. Another thing that is not factored in is component parasitics, which usually don't matter until after 10Mhz \$\endgroup\$ – laptop2d Jun 18 '18 at 20:46
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    \$\begingroup\$ @sudar meta.stackexchange.com/questions/126180/… \$\endgroup\$ – laptop2d Jun 18 '18 at 21:03
  • \$\begingroup\$ I used SAPWIN finally and got the desired TF. But its just too long to post in comment. But finally ignoring low values and after substituting values for R,C and non-idealities in op-amp, I got - (-4.66595*10^17 - 1.09197*10^9 s - 0.211201 s^2)/(3.20047*10^15 + 6.8081 s + 1. s^2), which i verified using the link that you provided - sim.okawa-denshi.jp/en/dtool.php and got a notch filter characteristic. Thanks very much for the pointer. However, I will verify your result with my results and see if they match (just for validity). \$\endgroup\$ – sundar Jun 18 '18 at 23:25

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