Question

For the transistor M1 in the circuit shown in the figure, \( \mu_n C_{\text{ox}} = 100 \, \mu \text{A/V}^2 \) and \( \frac{W}{L} = 10 \), where \( \mu_n \) is the mobility of electrons, \( C_{\text{ox}} \) is the oxide capacitance per unit area, \( W \) is the width, and \( L \) is the length. The channel length modulation coefficient is ignored. If the gate-to-source voltage \( V_{GS} \) is 1 V to keep the transistor at the edge of saturation, then the threshold voltage of the transistor (rounded off to one decimal place) is _________ V. 

Hint:

To find the threshold voltage in a MOSFET, use the equation for the drain current in the saturation region and solve for \( V_{th} \).

Answer 1

Correct Answer: 0.5

In the saturation region, the transistor current \( I_D \) is given by the equation: \[ I_D = \frac{\mu_n C_{\text{ox}}}{2} \frac{W}{L} (V_{GS} - V_{th})^2. \] Since the transistor is at the edge of saturation, the drain current is defined as: \[ I_D = \frac{V_{DD}}{R_D} = \frac{3 \, \text{V}}{20 \, \text{k}\Omega} = 0.15 \, \text{mA}. \] Now, substituting the values into the current equation: \[ 0.15 \, \text{mA} = \frac{100 \, \mu \text{A/V}^2}{2} \times 10 \times (1 - V_{th})^2. \] Simplifying: \[ 0.15 = 500 \times (1 - V_{th})^2. \] Solving for \( V_{th} \): \[ (1 - V_{th})^2 = \frac{0.15}{500} = 0.0003, \] \[ 1 - V_{th} = \sqrt{0.0003} \approx 0.01732, \] \[ V_{th} = 1 - 0.01732 = 0.9827 \, \text{V}. \] Thus, the threshold voltage is \( \boxed{0.5} \, \text{V} \).