Correct Answer - Option 2 : 2.85 V
Concept:
The pinch-off voltage of a JFET is given by:
\(V_p=\frac{qN_da^2}{2\epsilon}\) ---(1)
Nd = Donor concentration
a = Channel width
Since σ = Ndqμn and \(\rho=\frac{1}{\sigma}\), we can write:
\(\frac{1}{\rho}=N_dq\mu_n\)
\(N_d=\frac{1}{\rho q\mu_n}\)
Substituting this in Equation (1), we get:
\(V_p=\frac{qa^2}{2\epsilon}\times \frac{1}{\rho q\mu _n}\)
\(V_p=\frac{a^2}{2\epsilon\rho \mu _n} \)
Calculation:
Putting on the respective given values, we get:
\(V_p=\frac{(2\times 10^{-4})^2}{2\times 9\times 10^{-14}\times 11.7\times 5 \times 1300} \)
Vp = 2.92 V
The nearest value from the given options is 2.85