Question

Consider a differential amplifier as shown below,

Where the first set of signal is $V_1=50\mu V,V_2=-50\mu V$ and the second set of signals is $V_1=1050\mu V,V_2=950\mu V$. If the common mode rejection ratio is 100, then the percentage difference in output voltage obtained for the two sets of input signals is

• A. 25%
• B. 10%
• C. 15%
• D. 20%

Answer 1

The correct option is B 10%

The output voltage of differential amplifier is given as,
$V_0=A_d{V}_d+A_c{V}_c$
Where, $A_d$ = Differential gain
$A_c$ = Common mode gain
$V_d$ = Differential input voltage = $V_1-V_2$
$V_c$ = Common mode input voltage = $\frac{V_1+V_2}{2}$

$V_0=A_d{V}_d[1+\frac{A_c{V}_c}{A_d{V}_d}]=A_d{V}_d[1+\frac{1}{p}.\frac{V_c}{V_d}]$

Where, $p=\frac{A_d}{A_c}$ = common mode rejection ratio

Set of signal 1,
$V_d=50\mu V-(-50\mu V)=100\mu V$

$V_c=\frac{50\mu V-50\mu V}{2}=0$

$V_{01}=100\mu V.A_d[1+0]=100A_d\mu V$

$V_c=\frac{1050\mu V+950\mu V}{2}=1000\mu V$

$V_d=1050\mu V-950\mu V=100\mu V$

$\therefore V_{02}=A_{d}100\mu V[1+\frac{1}{100}\times \frac{1000\mu V}{100\mu V}]=110A_d\mu V$

% difference $=\frac{V_{02}-V_{01}}{V_{01}}\times 100=\frac{110-100}{100}\times 100=10\%$