Question

Consider the circuit shown in the figure below.

$\text{The MOS transistor serves as an electronic switch. If}{V}_{\text{in}}=0,V_G=1.8\text{and}{R}_L=100\Omega \text{, if the circuit is designed to attenuate an incoming small signal by 5\%, then the value of}\frac{W}{L}\text{is equal to\_\_\_\_\_\_\_\_\_.}$
$\text{[Assume}{V}_T=0.4V\text{and}{\mu }_n{C}_{ox}=200\mu A/V^2]$

• A. 673.85

Answer 1

The correct option is A 673.85

$V_{in}=0,\text{then we can assume the MOS is working as a voltage controlled register with controlling voltage}{V}_G=1.8V.$
$\text{now, Attenuation}=5\%$
$\therefore V_{out}=0.95V_{in}$
$\text{now,}\frac{V_{out}}{V_{in}}=\frac{R_L}{R_L+R_{on}}=0.95$
$\therefore R_{on}=5.26\Omega$
$\text{now,}{\mu }_n{C}_{ox}(V_{GS}-V_t).\frac{W}{L}=\frac{1}{R_{on}}$
$\therefore \frac{W}{L}=\frac{1}{{\mu }_n(V_{GS}-V_t).R_{on}}$
$=\frac{1}{200\times {10}^{-6}\times (1.8-0.4)\left(5.26\right)}$
$=673.85$