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Im trying to calculate the output of an opamp, the data is

  • Closed loop gain = \$20dB\$
  • CMRR= \$90dB\$
  • Common signal applyed = \$-60dBV\$
    so the idea is to use the next equation
    \$v_{o}=A(v_{id}+\frac{A_{cm}v_{ic}}{A})=A(v_{id}+\frac{v_{ic}}{CMRR})\$
    or

\$CMRR=20log\frac{A}{A_{cm}}\$
I undestand that \$A=10\$

but I dont know how to calculate the differential voltage and the common voltage.
Someone can clarify this, so the problem can be set right?

Update
if it is used
\$CMRR=20log\frac{A}{A_{cm}}\$
\$90=20log\frac{10}{A_{cm}}\$ It is supposed to find the common mode gain, but then what?

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    \$\begingroup\$ You need a circuit to do that \$\endgroup\$ May 3 at 3:54
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    \$\begingroup\$ The three values are the only data available, besides I dont know how to read the minus sign in the common signal of -60 dBV, whats supposed to mean? \$\endgroup\$
    – avelardo
    May 3 at 3:57
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The path gain or loss for a differential input amplifier are designed to be matched over the spectrum so that common mode input noise does not get converted to an equivalent differential mode input and appear at the output. This imbalance is called Common Mode Rejection Ratio.

Although it is referred to as a positive attenuation ratio it is a negative gain referenced to the input ratio, such that differential gain amplifies this imbalance to a single-ended output.

It is important to understand, the same CMRR imbalance exists on parasitic LC coupling on cables and board layouts so balance is important.

Given; \$~~~A_{dm} = 20 ~dB, ~~~CMRR = 90 ~dB,~~~ V_{cm-in} = -60~dBV ~~~~~~~_{(Ref: ~0~dBV~=~1~V)} \$

\$~~~~~A_{cm} ~[dB] = A_{dm}~ [dB] - CMRR~ [dB] \$

\$V_{cm-out}~[dB] = A_{cm}~ [dB] + V_{cm-in} ~[dB] \$

\$ ~~~\therefore V_{cm-out} = -130~dBV =( 20 - 90 ) + (-60) \$
\$_{~~~~~~~~~~~~(Ref:~~~1 \mu V~ = ~- 120 ~dBV ~~( = 20 ~log ~( 1~\mu V/1~V ) )}\$

  • This only applies to the ideal layout.

  • take note in future to determine all the factors that affect differential input impedance and balance error, as this limits your CMRR from component tolerances, parasitic coupling and differences in individual wire impedance to ground.

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    \$\begingroup\$ I don't understand what an output common-mode voltage, in a differential to single-ended opAmp refers to. The output as pointed out should be Av_{idm} + A_{cm} v_{ocm}. And for that, we require the differential mode input. \$\endgroup\$
    – HyperBean
    May 3 at 10:26
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    \$\begingroup\$ Good comment. I added more info. I hope it is clear. . CMRR is the ratio of unbalanced gains for each differential input that gets amplified by the closed loop differential gain and converted to a single-ended output. It also applies to cables and layouts, yet unspecified we can estimate it \$\endgroup\$ May 3 at 13:31
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    \$\begingroup\$ We consider the common mode to be either a virtual ground of a differential input or the output 0V reference to the output or the difference between grounds interconnected such as earth ground. One solution is to have a high common mode input impedance like Op Amps and a low differential impedance, or if not possible use a CM choke and shunt cap. I suppose what I ought to have said was it is the (DM) output due to a CM input attenuation and gain. \$\endgroup\$ May 3 at 14:18

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