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Question

The Debye length is a characteristic length for semiconductor. Arrange the doping density given below in such a way that Debye length changes from maximum to minimum.

(A) N = 1015 cm–3
(B) N = 1017 cm–3
(C) N = 3 × 1015 cm–3
(D) N = 1016 cm–3
(E) N = 5 × 1016 cm–3

Choose the most appropriate answer from the options given below :

This question was previously asked in
UGC NET 2023 Electronic Science Question Paper (13-Dec-2023) (Shift 1)
The correct answer is

(A), (C), (D), (E), (B)

 Debye length falls as the doping rises, so ordering it from maximum to minimum means ordering the doping from minimum to maximum — option 3.

\(L_{D}=\sqrt{\dfrac{\varepsilon kT}{q^{2}N}}\quad\Rightarrow\quad L_{D}\propto\dfrac{1}{\sqrt{N}}\)

Step 1 — sort the doping densities. Writing them all with the same exponent makes the comparison immediate:

LabelN (cm–3)As × 1015Rank
(A)10151Smallest
(C)3 × 101532nd
(D)1016103rd
(E)5 × 1016504th
(B)1017100Largest

Step 2 — invert. Since \(L_{D}\propto N^{-1/2}\), the largest Debye length belongs to the smallest N. Ordering the Debye length from maximum to minimum therefore gives

\(A\rightarrow C\rightarrow D\rightarrow E\rightarrow B\)

— exactly option 3.

What the Debye length means. It is the distance over which mobile carriers screen out an electric field, or equivalently the width of the transition region at the edge of a depletion layer, where the charge density falls off gradually rather than abruptly. Heavier doping supplies more carriers to do the screening, so the field is neutralised in a shorter distance — which is the physical reason for the inverse square root.

The numbers are worth a feel. At room temperature in silicon, \(N=10^{15}\) cm–3 gives about 130 nm, and \(10^{17}\) gives about 13 nm — a factor of ten for a hundredfold change in doping, exactly as the square root demands.

Why it sets a limit in device scaling. The abrupt-depletion approximation is only valid while the depletion width greatly exceeds \(L_{D}\). In a heavily scaled device the two become comparable, so the depletion edge is no longer sharp and the simple textbook analysis breaks down — one of the reasons short-channel modelling is difficult.

Hence, the order is (A), (C), (D), (E), (B).

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Similar Questions

  1. Consider the following statements :

    If an electric field is applied to an n-type semiconductor bar, the electrons and holes move in opposite directions due to their opposite charges. The net current is

    1. due to both electrons and holes with electrons as majority carriers.
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    3. the difference between electron and hole current.

    Which of these statements is/are correct ?

  2. Consider the following :

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  3. Match the following :

    List – IList – II  
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  4. A conducting line on an IC chip is 2.8 mm long and has a rectangular cross-section of 1 μm x 4 μm.. A current of 5 mA produces a voltage drop of 100 mV across the line. If the electron mobility is 500 cm2/V-s, the electron concentration is

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Important Questions from Carriers in Semiconductors

  1. The velocity with which electrons are emitted in the photoemission process

  2. The process of adding impurities to a pure semiconductor is called

  3. Mobility and conductivity are related by which of the following equations?

  4. How many electrons are there in the valence shell of a pure semiconductor?

  5. In a pure silicon, what is the time for an electron to drift $1\mu m$ in an electric field of 100 V/cm? 

    Assume electron mobility of $1350 \text{ cm}^2/V-s$

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