Question 10.SG.Q.24: In the circuit shown below, for the MOS transistor, μn C ox......
In the circuit shown below, for the MOS transistor, \mu_n C _{ ox }=100 mA / V ^2 and the threshold voltage V_{ Th }=1 V. The voltage V_x at the source of the upper transistor is
(a) 1 V (b) 2 V
(c) 3 V (d) 3.67 V
Given that for the upper transistor
\frac{W_1}{L_1}=4
So,
\begin{gathered}V_{DS 1}=6-V_x \text{and} V_{GS 1}=5-V_x \\V_{GS 1}-V_{Th}=5-V_x-1=4-V_x\end{gathered}
Therefore,
V_{ DS 1}>\left(V_{ GS 1}-V_{ Th }\right)
Hence, the transistor is in saturation region. Also, given that for the transistor at the bottom
\frac{W_2}{L_2}=1
Here, the drain is connected to the gate, hence, the transistor is in saturation.
The current flowing through both the transistors is the same. Hence,
\begin{aligned}& \mu_{N}C_{ox}\left(\frac{W_1}{L_1}\right)\left(\frac{V_{GS 1}-V_{Th}}{2}\right)^2 \\=& \mu_{N}C_{ox}\left(\frac{W_2}{L_2}\right) \cdot\left(\frac{V_{GS 2}-V_{Th}}{2}\right)^2\end{aligned}
Substituting the different values in the equation above, we get
4 \frac{\left(5-V_x-1\right)^2}{2}=1 \frac{\left(V_x-1\right)^2}{2}\left(\because V_{GS 2}=V_x-0\right)
On simplifying the above equation, we get
\begin{aligned}& 4\left(V_x^2-8 V_x+16\right)=V_x^2-2 V_x+1 \\& 3 V_x^2-30 V_x+63=0\end{aligned}
Therefore, V_x=3 V
Ans. (c)