Topic: Electronic Devices
Determine the channel half-width for an n-channel silicon FET having Gate-to-Source voltage, $V_{GS}=V_P/4$, where $V_P$ is the Pinch-off voltage and drain current, $I_d=0$. (Consider Donor Concentration $N_D=10^{15}\text{ electrons}/cm^3$ and Channel half-width for $V_{GS}=0\text{ V is } 3\mu m$).
- A. $2.25\mu m$
- B. $3 \mu m$
- C. $1.5\mu m$
- D. $0.75 \mu m$
Answer: C. $1.5\mu m$
Explanation: The available channel half-width $b$ is given by $b = a(1 - \sqrt{V_{GS}/V_P})$, where $a$ is the metallurgical half-width. Given $a = 3 \mu m$ and $V_{GS} = V_P / 4$, we substitute to get $b = 3 (1 - \sqrt{1/4}) = 3 (1 - 0.5) = 1.5 \mu m$.