what is the approximate conversion efficincy between current

what is the approximate conversion efficincy between current and photons in the LED ?(assume you can detect the LED is on when it emits 0.1lumens, lumen=W/683, and a red photons has an energy of 2.0eV).

Solution

A)In a LED, a part of the electrical energy supplied to it is used to excite electrons to higher energy levels. When such an excited electron in a higher energy level recombines and falls back to a lower energy level, a photon with energy Eph is emitted.

where h is the Planck’s constant,(=c/) is the frequency of the light emitted,c is the speed of light in vacuum and is the wavelength of the light emitted by the LED.The wavelength of the light depends on the band gap energy of the semiconductor material used for the LED.

A Photodiode (PD) converts light energy into electrical energy and hence can be used to measure the intensity of light. A photodiode is sensitive to the incident light for a certain range of wavelength. The current developed in the photodiode is linearly proportional to the intensity of light, up to a certain limit. When light falls on the sensitive area of a semiconductor photodiode, some of the incident photons free some of the electrons within the semiconductor material. The ratio of the number of free electrons generated per second (Ne) to the number of incident photons per second (Np) is termed as the quantum yield or efficiency of the PD and is denoted by pd.

Expression for Efficiency

Let us select a photodiode (PD) with a square shaped (with each sidea) sensitive area. A LED emits light in a cone with cylindrical symmetry as shown in the Figure 2.The intersection of the cone (in which the LED emits light) and a plane perpendicular to the axis of the cone is a circular disc. We can divide this circular disc into number of circular strips of small width equal to the side a of the square shaped sensitive area of the PD.The area of each such circular strip which is at a distance ri from the axis of the cone is given by (2ria+a2).Suppose the PD is placed at a distance ri from the axis of the cone which is the axis of symmetry.The current I(ri)in the PD (i.e.IPD) kept at a distance ri from the axis of the cone is given

I (ri) = Nee=Npqpe   ---------        (3)

where e is the charge of an electron.Let the radiant power received by the PD be (ri). Since Np is the number of photons (of energy h) received per unit time by the PD,(ri) is given by

(ri) =Np. h