Question

Conisder the ideal amplifier circuits shown in the figure below.




Assuming all the MOS transistor to be idential with $K_a=\frac{{\mu }_u{C}_{ox}W}{2L}$ and the thershold voltage $V_{Th}.$ If $V_{out1}=V_{X1}+V_{Th}$ and $V_{out2}=V_{X2}+V_{Th}$, then the value of $\frac{V_{out2}}{V_{out1}}$ is equal to ______.

• A. 0.707

Answer 1

The correct option is A 0.707

For the circuit in figure. 1 the value of output voltage. $V_{out}$ can be given as $V_{out}=V_{X1}+V_{Th}$ and $V_{X1}=-V_{GS}$
where $I_D=K_a(V_{GS}-V_{Th}{)}^2$
$\sqrt{\frac{I_D}{K_n}}=V_{GS}-V_{Th}$
$-V_{GS}=-\sqrt{\frac{I_D}{K_n}}-V_{Th}$
$-V_{GS}+V_{Th}=-\sqrt{\frac{I_D}{K_n}}$
$V_{out1}=V_{X1}+V_{Th}=-\sqrt{\frac{I_D}{K_n}}=-\sqrt{\frac{I_{in}}{K_n}}$
now, for figure 2, since the two MOS circuit are in parallel thus,



$V_{D2}=I_{D2}=\frac{I_{in}}{2}$ and $V_{out2}=-\sqrt{\frac{I_{D2}}{K_n}}=-\sqrt{\frac{I_{in}}{2K_n}}$
now for both transistors
$I_m=\frac{V_m}{R_1}$
$\therefore \frac{V_{out2}}{V_{out1}}=\frac{1}{\sqrt{2}}=0.707$