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For an n-channel silicon JFET with a = 2 × 10-4 cm and channel resistivity ρ = 5Ω –cm, μn = 1300 cm2/V-s and ε0 = 9 × 10-12 F/m, the pinch-off voltage, Vp, is nearly
1. 2.30 V
2. 2.85 V
3. 3.25 V
4. 3.90 V

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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 = Ndn 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

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