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Question

List – I (Structure/Characteristics)List – II (Reasons)  
a. n-channel JFET is better than p-channel JFETi. Reverse bias increases along the channel
b. Channel is wedge shapedii. High electric field near the drain and directed towards source
c. Channel is not completely closed at pinch-offiii. Low leakage current at the gate terminal
d. Input impedance is highiv. Better frequency performance since μn >> μp

Codes :

This question was previously asked in
UGC NET 2014 Paper 2 History Question Paper (28-Dec-2014)
The correct answer is

a-iv, b-i, c-ii, d-iii

 Each structural fact about the JFET has one physical cause, and the four pair up uniquely.

(a) n-channel beats p-channel → (iv) because μn >> μp. Electrons are two to three times more mobile than holes in silicon, so for the same geometry an n-channel device has a higher transconductance and a shorter transit time, hence a better frequency response and a lower on-resistance. That single fact is why n-channel devices dominate the whole of solid-state electronics.

(b) The channel is wedge shaped → (i) because the reverse bias increases along the channel. Current flowing from drain to source produces a voltage drop along the channel, so the gate-to-channel reverse bias — and therefore the depletion width — is greatest at the drain end and least at the source end. The conducting path narrows progressively towards the drain, giving the characteristic wedge.

(c) The channel is not completely closed at pinch-off → (ii) because of the high field near the drain. This is the point most often misunderstood: "pinch-off" does not mean the channel is blocked. If it truly closed, no current could flow at all — yet the drain current is at its maximum there. What happens is that the channel narrows to a very thin constriction across which a strong longitudinal field exists, and carriers are swept through it at saturation velocity. Any extra drain voltage is dropped across that constricted region, which is why the current saturates instead of vanishing.

(d) High input impedance → (iii) because the gate leakage current is tiny. The gate-channel junction is reverse biased, so only the reverse saturation current flows — picoamps to nanoamps. The input resistance is \(V_{GS}/I_{G}\), which reaches \(10^{8}\) to \(10^{9}\ \Omega\). This is the JFET's headline advantage over the BJT, whose base junction is forward biased and draws real current.

CharacteristicCauseCode
n-channel superiorHigher electron mobility(iv)
Wedge-shaped channelBias varies along the channel(i)
Channel never fully closedHigh field at the drain end(ii)
High input impedanceReverse-biased gate leaks very little(iii)

The order (iv), (i), (ii), (iii) is option 1.

Hence, the correct match is a-iv, b-i, c-ii, d-iii.

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