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Design >> Mixed-Signal Design >> derive the transfer function from fully-diff cir.
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Message started by trashbox on Jan 7^th, 2008, 1:17am

Title: derive the transfer function from fully-diff cir.
Post by trashbox on Jan 7^th, 2008, 1:17am

Hi all,
I am studying a fully-diff circuit and wanna derive the transfer function from the attached circuit, would you please give me any hints or formulas? Thanks.

Title: Re: derive the transfer function from fully-diff c
Post by Berti on Jan 7^th, 2008, 11:55pm

hmmm, just use nodal analysis. You will get a system of equations which you can solve for
the output (voutp-voutm) and divide by the input (vinp-vinm) in order to obtain the transfer
function.

Title: Re: derive the transfer function from fully-diff c
Post by trashbox on Jan 10^th, 2008, 1:46am

Hi Berti
Thanks. I have solved this problem.

Title: Re: derive the transfer function from fully-diff c
Post by thechopper on Jan 14^th, 2008, 12:10pm

Hi

You can easily derive it by finding using thevenin equivalent circuits. You can use the concept of hemi-circuit, based on the fact you are analyzing a fully diff (purely symmetrical circuit).
If you split the input circuit (Z1/Z3/Z1), you'll get a simple resistor divider equal to Z3/2÷(Z3/2+Z1). That is the input impedance of your input voltage source.
Z2 does not play a role in the transfer function assuming the op amp's input impedance is >> Z2.

From this you get H=-(Z4*(Z3/2 + Z1)) / (Z3/2)

Hope this helps
Tosei