Thursday, January 06, 2011

Two Questions (MCQ) on Photo Diodes

“He who joyfully marches in rank and file has already earned my contempt. He has been given a large brain by mistake, since for him the spinal cord would suffice.”

– Albert Einstein

As you know, photo diodes are special purpose p-n junction diodes. They are very popular photo detectors used for detecting optical signals in communication systems.

Today we will discuss two multiple choice questions on photo diodes:

(1) In using a photo diode as a photo detector, it is invariably reverse biased. Why?

(a) The power consumption is much reduced compared to reverse biased condition

(b) Electron hole pairs can be produced by the incident photons only if the photo diode is reverse biased

(c) Light variations can be converted into current variations only if the photo diode is reverse biased

(d) When photons are incident on the diode, the fractional change in the reverse current is much greater than the fractional change in the forward current

(e) The photo diode will be spoilt if it is operated under forward biased condition

Whether the photo diode is reverse biased or forward biased, the number of electron hole pairs produced by the incident photons is the same. In other words, the change in the diode current is the same in both cases. But in the reverse biased condition the current drawn by the diode in the absence of the photons is extremely small, of the order of nanoamperes or microamperes where as in the forward biased condition this current is significant, of the order of tens of milliamperes. The fractional change in the current because of the incident photons is therefore large and easily measurable if the photo diode is reverse biased. The correct option is (d).

(2) The maximum wave length of photons that can be detected by a photo diode made of a semiconductor of band gap 2 eV is about

(a) 620 nm

(b) 700 nm

(c) 740 nm

(d) 860 nm

(e)1240 nm

The wave length λ (in Angstrom unit) of a photon of energy E (in electron volt) is given by

λE = 12400, very nearly.

Therefore, λ = 12400/E

[The above expression can be easily obtained by remembering that a photon of energy 1 eV has wave length 12400 Ǻ and the energy is inversely proportional to the wave length].

Since E = 2 eV we have λ = 12400/2 = 6200 Ǻ = 620 nanometre.

Photons with wave length greater than 640 nm will have energy less than 2 eV so that they will be unable to produce electron hole pairs in the semiconductor of band gap 2 eV. So the correct option is (a).

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