CBSE Class 12 · 2005 · SemiconductorsLong Answer

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(a) Distinguish between metals, insulators and semiconductors on the basis of their energy bands.
(b) Why are photodiodes used preferably in reverse bias condition? A photodiode is fabricated from a semiconductor with band gap of

2.8 eV

. Can it detect a wavelength of

6000 nm

? Justify.

Solution

(a) Metals have overlapping conduction and valence bands (or a partly filled conduction band), so there are always free carriers.
Insulators have a large forbidden gap (

E_{g} > 3 eV

) so electrons cannot reach the conduction band.
Semiconductors have a small gap (

E_{g} \approx 1 eV

) so some carriers are thermally excited at room temperature.
(b) Photodiodes are used in reverse bias because, on illumination, the fractional change in the small reverse (minority-carrier) current is large and easily measured.
For detection the photon energy must satisfy

h ν \geq E_{g}

, i.e.

λ \leq h c / E_{g} = 1240/2.8 \approx 443 nm

.
The energy of a

6000 nm

photon is

E = \frac{1240 eV nm}{6000 nm} \approx 0.21 eV .

Since

0.21 eV < 2.8 eV = E_{g}

, the photon cannot excite a carrier across the gap.

No; 0.21 eV < E_{g} = 2.8 eV .

CBSE Class 12 Board · 2005 · Subjective

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CBSE Class 12 · 2005 · SemiconductorsLong Answer

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