A given op-amp has an differential gain of 110 dB and a CMRR rating of 100 dB. What should be the open loop common mode gain of this op-amp ?

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LPSC ISRO Technical Assistant Electronics 07 Aug 2016 Official Paper
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  1. 1.10 dB
  2. 10 dB
  3. 210 dB
  4. 105 dB

Answer (Detailed Solution Below)

Option 2 : 10 dB
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Detailed Solution

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Concept:

CMRR (Common mode rejection ratio) is defined as the ratio of differential-mode voltage gain (Ad) and the common-mode voltage gain (AC).

Mathematically, in dB this is expressed as:

\(CMRR(dB)= 20\log \left| {\frac{{{A_d}}}{{{A_c}}}} \right| \)

Ad = Differential gain

Ac = Common mode gain

Using the property of log, we can write:

CMRR (dB) = 20log (Ad) - 20log (Ac)

CMRR (dB) = Ad(dB) - Ac(dB)

Calculation:

Given:

Ad(dB) = 110 dB

CMRR (dB) = 100 dB

We can write:

CMRR (dB) = Ad(dB) - Ac(dB)

100 dB = 110 dB - Ac(dB)

Ac(dB) = 110 dB - 100 dB

Ac(dB) = 10 dB

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