I'm trying to find and plot the CMRR of the following circuit:
I'm getting zero for common mode gain, that gives an infinite CMRR for the circuit. When I followed normal feedback laws, I get (2/SRC)+1 as the differential gain.
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1 Answer
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Welcome to Electrical Engineering.
With such a circuit, you have a differential input (\$V_1 - V_2\$) and a differential output (\$V_3 - V_4\$). But you also have a common-mode input (\$\frac{V_1 + V_2}{2}\$) and a common-mode output (\$\frac{V_3 + V_4}{2}\$).
Then you can define the differential gain as \$\frac{V_3 - V_4}{V_1 - V_2}\$ and the common mode gain as \$\frac{V_3 + V_4}{V_1 + V_2}\$.
Then you get the CMRR of this circuit.
Hint: the common-mode gain is not 0.
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\$\begingroup\$ Thanks for the answer. The way we were told to do it is finding the output for V1 and V2 and then rearrange the Vout into G(common)*V(common) + G(dif)*V(dif). When I solved for Vout of this circuit. I got a pure differential gain. Am I doing something wrong? \$\endgroup\$ Commented May 16, 2021 at 12:08
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\$\begingroup\$ However, with your method I'm getting G(common) = 1 I'm not sure if I'm right \$\endgroup\$ Commented May 16, 2021 at 12:10
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\$\begingroup\$ I do agree with G(common) = 1, but I fail to understand your “When I solved for Vout of this circuit.”… What \$V_{\textrm{out}}\$? \$\endgroup\$ Commented May 16, 2021 at 12:22
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\$\begingroup\$ my bad, I'm using Vout = V4-V3 \$\endgroup\$ Commented May 16, 2021 at 12:37
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\$\begingroup\$ @AkilaUyanwatta I don’t think you should consider \$V_{\textrm{out}} = V_4 - V_3\$ and use that to derivate the CMRR. Or you might as well consider \$V_{\textrm{out}} = V_2 - V_1\$ and you get perfect CMRR with no component at all. \$\endgroup\$ Commented May 16, 2021 at 13:27
Q1andQ2are identical andCandCare identical, did you expect to get a non zero common mode gain ? How is the output defined ?V3-V4? or justV3? \$\endgroup\$