A semiconductor has equal electron and hole concentration of $2\times10^{8}\ \mathrm{m^{-3}}$. On doping with a certain impurity, the hole concentration increases to $4\times10^{10}\ \mathrm{m^{-3}}$. (i) What type of semiconductor is obtained on doping? (ii) Calculate the new electron and hole concentration of the semiconductor. (iii) How does the energy gap vary with doping?

(i) p-type (holes are now the majority carriers). (ii) $n_h=4\times10^{10}\ \mathrm{m^{-3}}$ and $n_e=n_i^{2}/n_h=10^{6}\ \mathrm{m^{-3}}$. (iii) The effective energy gap decreases on doping.

CBSE Class 12 · 2001 · SemiconductorsShort Answer

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A semiconductor has equal electron and hole concentration of

2 \times 1 0^{8} m^{- 3}

. On doping with a certain impurity, the hole concentration increases to

4 \times 1 0^{10} m^{- 3}

.
(i) What type of semiconductor is obtained on doping?
(ii) Calculate the new electron and hole concentration of the semiconductor.
(iii) How does the energy gap vary with doping?

Solution

(i) Since the hole concentration has risen above the electron concentration, holes are the majority carriers, so the doped material is p-type.
(ii) The intrinsic concentration is

n_{i} = 2 \times 1 0^{8} m^{- 3}

. By the mass-action law

n_{e} n_{h} = n_{i}^{2}

n_{e} = \frac{n _{i}^{2}}{n _{h}} = \frac{( 2 \times 1 0 ^{8} ) ^{2}}{4 \times 1 0 ^{10}} = \frac{4 \times 1 0 ^{16}}{4 \times 1 0 ^{10}} = 1 0^{6} m^{- 3} .

The new hole concentration is

n_{h} = 4 \times 1 0^{10} m^{- 3}

.
(iii) Doping introduces impurity levels in the forbidden gap, so the effective energy gap decreases.

p-type; n_{e} = 1 0^{6} m^{- 3}, n_{h} = 4 \times 1 0^{10} m^{- 3}; E_{g} decreases.

CBSE Class 12 Board · 2001 · Subjective

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CBSE Class 12 · 2001 · SemiconductorsShort Answer

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