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What is Photo Diode and How does the photo diode work? What are the applications of Photo Diode?

Answer:

Photo Diode

A photo diode is a semiconductor device with a p-n junction and an intrinsic layer between p and n layers that converts light into current. Photo-diode produces photo-current by generating electron-hole pairs, due to the absorption of light in the intrinsic or depletion region. The photo-current thus generated is proportional to the absorbed light intensity. Photo-diodes usually have a slower response time as their surface area increases. Photo diodes can be used in two ways — in a photo-voltaic or photo conductive role. To use a photodiode in its photo-conductive mode, the photo-diode is reverse-biased; the photo diode will then allow a current to flow when it is illuminated.

Principle operation of Photo-diode

A photodiode is a PN junction, When a photon of sufficient energy (<1.1ev) strikes the diode, it creates an electron-hole pair. The intensity of photon absorption depends on the energy of photons – the lower the energy of photons, the deeper the absorption is. This mechanism is also known as the inner photoelectric effect. If the absorption occurs in the depletion region of the p-n junction, these hole pairs are swept from the junction – due to the built-in electric field of the depletion region. Thus holes move toward the anode, and electrons toward the cathode, and a photocurrent is produced. The total current through the photo-diode is the sum of the dark current (Dark current is a generated current in the absence of light) and the photo-current, so the dark current must be minimized to maximize the sensitivity of the device.

Operation modes of Photo-diode

Photo-diodes can be operated in different modes like Photovoltaic mode, Photoconductive mode, Avalanche diode mode.

1. Photo-conductive mode – In this mode device used in reverse biased. The depletion layer increased with the help of reverse voltage, which in turn reduces the response time and capacitance of the junction. The Photo-conductive mode is very fast, and exhibits electronic noise
2. Avalanche diode mode – This mode also used in a high reverse bias condition, which allow multiplication of an avalanche breakdown to each photo-generated electron-hole pair; which gradually increases the responsivity of the device.
3. Photo-voltaic mode – In this mode voltage is generated by the illuminated photodiode and it is also known as zero bias mode. It provides a small dynamic and non-linear dependence of the voltage produced.

Applications of Photo diode

• Surveying instrument
• Safety equipment
• Cameras
• Bar code scanners
• Medical devices
• Optical communication devices
• Position sensors
• Automotive devices

Two important characteristics of a photodetector are its quantum efficiency and its responsivity. These parameters depend on the material band gap, the operating wavelength, and the doping and thickness of P-N regions of the device.

QUNTUM EFFICIENCY-

It is defined as the ratio of the number of electron-hole pairs (ehp) generated to the total number of incident photons and is given by

η= Number of ehp’s generated/total no. of incident photons

η= Ip/q/Pop/h?          or

η= I_p x h?/q x P_op

where I_p is the photo-generated current, P_op is the incident optical power, h? is  the photon energy and q is the free carrier charge.

RESPONSIVITY

It is defined as the ratio of photo generated current to the incident optical power.it is given by the following formula:

Ʀ=I_p/P_op

Responsivity is related to the quantum efficiency by

Ʀ=ηq/h?

This parameter is quite useful, since it specifies the photocurrent generated per unit optical power.

 

Problem 1: The responsivity of a photodiode is 0.85 A/W and the i/p saturation is 3.5 mW. What is the photocurrent if the incident light power is (a) 1mW?

Solution: Given that photodiode responsivity  Ʀ = 0.85 A/W

P_op = 1 mW

Ʀ = I_P / P_op

I_P =   Ʀ =P_op

I_p = 0.85 x 1

I_P = 0.85 mA

Problem 2: What is the responsivity of an InGaAs photodiode if its quantum efficiency is equal to 70%? The energy gap of InGaAs is 0.75 eV.

Solution: Given that energy gap of InGaAs is 0.75 eV; quantum efficiency = 70%.

According to Einstein energy E_g = hc/l;

l = 6.6 x 10^-34 (J-s) x 3x 10⁸ (m/s) / 0.75 (ev) x 1.6 x 10^-19

l = 1664X 10^-9 m

So, the reponsivity in terms of efficiency is defined by

Ʀ = (h/1248) l

Ʀ = (0.70/1248) 1664 x 10^-9

Ʀ =0.933 A/W