Field Effect Transistors Questions and Answers

121.

In a JFET the change in drain current is due to the applied

A.Electric field between S and D
B.Electric field between G and S
C.Magnetic field between S and D
D.Magnetic field between G and S

Exam Year(s)

UGC NET - 2012,

Answer: B

In a JFET the drain current is controlled by the electric field between the gate and the source: the reverse-biased gate junction sets the width of the depletion region, which narrows or widens the conducting channel, so the drain current changes with the gate-to-source voltage. A magnetic field plays no part in the operation of the device, which rules out options C and D. The electric field between source and drain drives the current along the channel, but it is not the controlling quantity that the question asks about, so option A is also wrong. The answer is option B.

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122.

Consider an n-channel depletion mode MOSFET having following parameters :

\(V_{TN} = -2.5\) V and \(K_{n} = 1.1\) mA/V\(^{2}\)

If \(V_{GS} = 0\)V and \(V_{DS} = 0.5\) V, then current \(I_{D}\), is

A.9.96 mA
B.6.43 mA
C.2.48 mA
D.5.56 mA

Exam Year(s)

UGC NET - 2012,

Answer: C

The drain current is 2.48 mA, option (C). With \(V_{GS} = 0\) and \(V_{TN} = -2.5\) V, the overdrive is \(V_{GS} - V_{TN} = 2.5\) V. As \(V_{DS} = 0.5\) V is less than 2.5 V, the device works in the triode region, where

\[ I_{D} = K_{n}\left[2(V_{GS}-V_{TN})V_{DS}\right.\]

\[ \left. -\,V_{DS}^{2}\right] \]

\[ I_{D} = 1.1\left[2(2.5)(0.5) - 0.25\right] \]

\[ = 1.1\times2.25 = 2.48\ \text{mA} \]

Options (A), (B) and (D) are not given by the triode equation. The saturation formula \(K_{n}(V_{GS}-V_{TN})^{2}\), which would give 6.9 mA, applies only when \(V_{DS}\) is at least 2.5 V, which is not the case here.

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