Why is a photo-diode invariably reverse biased, when it is used as a photo-detector?
A. The power consumption is much reduced compared to reverse biased conditions.
B. electron hole pairs can be produced by the incident photons only if the photodiode is reverse biased.
C. Light variations can be converted into current variations if the photodiode 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 photodiode will be spoiled if it is operated under forward biased conditions.
Answer
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Hint: If the photo-diode is forward biased and the light falls on the diode, the current produced by the voltage dominates the current produced due to the light. Whereas if the diode is in reverse biased, there is a comparable amount of change in the current when the light falls on the diode.
Complete answer:
Let us first understand what a photo detector is.
From the name itself we come to know that a photodetector is something related to photons. A photodetector is a device that is used to detect the presence of any electromagnetic wave or radiation.
A photodiode is used in a photodetector for detecting the photons.
A photodiode is a p-n junction diode which works on the photoelectric effect. When light of suitable wavelength falls on the diode, the photons collide with electrons and electron-hole pairs are created.
If the photodiode is connected to a battery, then due the creation of the electron-hole pairs by the light, the current in the circuit increases.
Now, when the diode is in the forward biased condition, there is already a good amount of current in the circuit before the light falls on the diode. When the light falls on the diode, the change (increment) in the current is very small compared to the initial current (I.e. before light falls on the diode). Hence, the change is not noticeable.
Whereas, when the diode is in reverse biased condition, the current in the circuit is very low. Hence, when the light falls on the diode, the change in current is noticeable and the light or the electromagnetic wave can be detected.
Therefore, a photodetector is more efficient when the photodiode is reverse biased. The reason is that 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.
So, the correct answer is “Option D”.
Note:
We have concluded that for the photodetector to be efficient, the photodiode must be reverse biased. However, the potential difference created across the diode must be less than the breakdown voltage of the diode.
Complete answer:
Let us first understand what a photo detector is.
From the name itself we come to know that a photodetector is something related to photons. A photodetector is a device that is used to detect the presence of any electromagnetic wave or radiation.
A photodiode is used in a photodetector for detecting the photons.
A photodiode is a p-n junction diode which works on the photoelectric effect. When light of suitable wavelength falls on the diode, the photons collide with electrons and electron-hole pairs are created.
If the photodiode is connected to a battery, then due the creation of the electron-hole pairs by the light, the current in the circuit increases.
Now, when the diode is in the forward biased condition, there is already a good amount of current in the circuit before the light falls on the diode. When the light falls on the diode, the change (increment) in the current is very small compared to the initial current (I.e. before light falls on the diode). Hence, the change is not noticeable.
Whereas, when the diode is in reverse biased condition, the current in the circuit is very low. Hence, when the light falls on the diode, the change in current is noticeable and the light or the electromagnetic wave can be detected.
Therefore, a photodetector is more efficient when the photodiode is reverse biased. The reason is that 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.
So, the correct answer is “Option D”.
Note:
We have concluded that for the photodetector to be efficient, the photodiode must be reverse biased. However, the potential difference created across the diode must be less than the breakdown voltage of the diode.
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