Multiple choice

The channel resistance of an N-channel JFET shown in the figure below is 600$\Omega$ when the full channel thickness (tch) of 10$\mu$m is available for conduction. The built-in voltage of the gate P+ N junction (Vbi) is - 1 V. When the gate to source voltage (VGS) is 0 V, the channel is depleted by 1$\mu$m on each side due to the built in voltage and hence the thickness available for conduction is only 8$\mu$m

The channel resistance when VGS = - 3 V is

  1. 480$\Omega$
  2. 600$\Omega$
  3. 750$\Omega$
  4. 1000$\Omega$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

$At \hspace{0.2cm} V_{GS} = 0V, b = 4 \mu m \\ since \hspace{0.2cm} 2b = 8 \mu m \\ \text{Thus} \hspace{0.2cm} r' = \frac{\rho L}{W a} \times \frac{a}{b} = 600 \times \frac{5}{4} = 750 \Omega$

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