Question

Consider the MOS transistor circuit shown in the figure below


$\text{The two N-MOS transistor are identical with threshold voltage}{V}_T=1V\text{and}\frac{{\mu }_n{C}_{ox}W}{L}=2mA/V^2.\text{Assuming the value of}\lambda =0,\text{the value of drain to source voltage for transistor}{T}_2(V_{DS2)}\text{is equal to \_\_\_\_\_V.}$

Answer 1

$V_{GS2}=-V_2$

$I_{D2}=\frac{V_2-(-5)}{1k\Omega }=\frac{1}{2}\times 2\times (-V_2-1{)}^2$

$\therefore V_2+5=V_2^2+2V_2+1$

$\Rightarrow V_2^2+V_2-4=0$

$V_2=1.55V,-2.56V$

$\because V_{GS2}\text{cannot be negative, thus}{V}_2=-2.56V$

$\text{Thus,}{V}_2=-2.56V$

$\text{Now,}{I}_{D1}=I_{D2}=\frac{V_2-(-5)}{1k\Omega }=\frac{-2.56+5}{1k\Omega }=2.44$

$\therefore 2.44\times {10}^{-3}=\frac{1}{2}\times 2\times {10}^{-3}(5-V_1-1{)}^2$

$V_1^2-8V_1+13.56=0$

$V_1=2.44V,5.56V$

$\therefore V_1=2.44V$

$\because V_{GS}\text{cannot be negative.}$

$\therefore V_{GS2}=V_1-V_2=2.44-(-2.55)=5V$