Question

Consider the circuit shown in the figure below.

$\text{The transistors}{M}_1\text{and}{M}_2\text{are in saturation have}{V}_T=1V,\text{and the process transconductors parameter}$

$k_a^{\prime }={\mu }_n{C}_{o}x=100\mu A/V^2\text{. If}{\left(\frac{W}{L}\right)}_1=1.5{\left(\frac{W}{L}\right)}_2=20\text{and}\lambda =0\text{, then the value of voltage V is equal to}$

• A. 0 V
• B. 0.2 V
• C. -1.8 V
• D. -1.6 V

Answer 1

The correct option is D -1.6 V

Ans : (b)

$\text{Assuming transistors}{M}_1\text{and}{M}_2\text{are in saturation region.}$

$\text{Thus,}{V}_{GS1}=V_{GS2}=-V_x$
$\text{and}V=V_1-V_2$
$\text{but}\frac{I_{D1}}{I_{D2}}=\frac{{\left(\frac{W}{L}\right)}_1}{{\left(\frac{W}{L}\right)}_2}=\mathrm{1.5...}\left(i\right)$
$\text{and}{I}_{D1}+I_{D2}=200\mu A...\left(ii\right)$
Thus from (i) and (ii), we get,
$\therefore I_{D1}=120\mu A,I_{D2}=80\mu A$
$\text{now,}{V}_1=5-40\times {10}^3\times 120\times {10}^{-6}=0.2V$
$V_2=5-40\times {10}^3\times 80\times {10}^{-6}=1.8V$
$\therefore V=0.2-1.8V=-1.6V$