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Dv/dt = 5/100 =0.05V

V * 100 =0.05 * 100= 5V

This is what I did. I can't get it right. It's a measurement and instrumentation course, specifically an electronic engineering course.

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  • \$\begingroup\$ Why do you have dv/dt? What does time have to do with it? Your units are not matching at all. \$\endgroup\$
    – jusaca
    Commented Feb 19, 2021 at 8:51
  • \$\begingroup\$ Please help me out, I just tried it out with time... what do you have about it? \$\endgroup\$
    – Piko
    Commented Feb 19, 2021 at 8:53
  • \$\begingroup\$ @Piko: It is a ratio. Simply one voltage divided by another. The trick here is the units. Both voltages have to be in the same unit. You have volts and millivolts. Convert the millivolts to volts before going further. \$\endgroup\$
    – JRE
    Commented Feb 19, 2021 at 10:06
  • \$\begingroup\$ The units cancel. It's like with regular variables. \$\frac{20X}{4X} = 5\$ . When you divide volts by volts, the result has no unit. \$\endgroup\$
    – JRE
    Commented Feb 19, 2021 at 10:09
  • \$\begingroup\$ I appreciate your help \$\endgroup\$
    – Piko
    Commented Feb 21, 2021 at 19:24

1 Answer 1

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The amount of output voltage error (in your case 5mV) in proportion to the input common mode voltage (in your case 100V) is the common mode rejection ratio. This has nothing to do with time, so no need to specify some dV/dt.

\$ CMRR = \frac{V_{out}}{V_{com_i}} = \frac{5~mV}{100~V} = 50 \cdot 10^{-6}\$

As you can the, the CMRR (Common mode rejection ratio) has no unit, because it is a ratio.

Typically the value of the CMRR is not given as a linear number, but in the logarithmic measure of dB:

\$ CMRR_{dB} = 20 \cdot log(CMRR) = -86~dB \$

Notice that dB is not a unit! It only is scaling the linear value of the CMRR to a logarithmic scale.

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  • 1
    \$\begingroup\$ Thanks alot, I appreciate... \$\endgroup\$
    – Piko
    Commented Feb 21, 2021 at 19:22
  • \$\begingroup\$ Thanks alot, I appreciate... \$\endgroup\$
    – Piko
    Commented Feb 21, 2021 at 19:23

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