Topic: Analog Circuits
Both transistor $T_{1}$ and $T_{2}$ in figure have a threshold voltage of 1 V. The device parameters $K_{1}$ and $K_{2}$ of $T_{1}$ and $T_{2}$ are 36 $\mu\text{A/V}^{2}$ and 9 $\mu\text{A/V}^{2}$ respectively. The output voltage $V_{0}$ is
- A. 1 V
- B. 2 V
- C. 3 V
- D. 4 V
Answer: C. 3 V
Explanation: Both transistors have their gates connected to their drains, ensuring they operate in the saturation region. Therefore, $I_{D1} = I_{D2}$.
$K_1 (V_{GS1} - V_T)^2 = K_2 (V_{GS2} - V_T)^2$.
From the circuit, $V_{GS1} = 5 - V_0$ and $V_{GS2} = V_0$.
$36(5 - V_0 - 1)^2 = 9(V_0 - 1)^2$.
$4(4 - V_0)^2 = (V_0 - 1)^2$.
Taking the square root: $2(4 - V_0) = V_0 - 1 \Rightarrow 8 - 2V_0 = V_0 - 1 \Rightarrow 3V_0 = 9 \Rightarrow V_0 = 3~\text{V}$.