OFCS OPTICAL DETECTORS 11/9/2014 LECTURES 1

OFCS OPTICAL DETECTORS 11/9/2014 LECTURES 1

1-Defintion & Mechanisms of photodetection It is a device that converts the incident light into electrical current External photoelectric effect: Electrons are freed from the surface of a metal by the energy absorbed from an incident stream of photons Examples : -Vacuum photodiode -Photomultiplier tube Internal photoelectric effect: In semiconductor junction devices free charge-carriers (electrons and holes) are generated by absorption of incoming photons Examples: -PN junction photodiode -PIN photodiode -Avalanche photodiode 11/9/2014 LECTURES 2

Input optical power Output electrical current 2-Photodetector properties (1) Responsivity (ρ) Is the ratio of the output current / voltage of the detector to its optic input power: ρ = i / P [A/W] or ρ = v / P [V/W] (2) Spectral response Is the detector responsivity as a function of wavelength ρ(λ). Responsivity changes very rapidly with wavelength and depends on the particular detector therefore different detectors must be used depending on the wavelength emitted by the source (3) Rise time t r Is the time for the detector output current to change from 10% to 90% of its final value assuming the optic incident power variation is a step function P i 90% 10% t t 11/9/2014 LECTURES 3 t r

3-Vaccum photodiode: -When there is no light the current and voltage drop across the resistor is zero -When incoming photons are absorbed they give up their energy to the electrons in the metal -Some of these electrons gain enough energy to escape from the cathode surface and reach the anode and so closing the circuit and a current flow as long as there is a flow of incident photons -The minimum amount of energy required to emit the electrons from the cathode should be greater than the work function Wg of the material of the cathode hf Wg hc/λ Wg λ 1.24 / Wg [um, ev] Cathode - Light of energy hf - + V - Electrons - i R L Anode + v 11/9/2014 LECTURES 4

Quantum efficiency of the detector: The ratio of the number of emitted electrons to the number of the incident photon number of emitted electrons number of incident photons η = (i/e) / (P/hf) i = ηepλ/hc, ρ = i/p ρ = ηeλ/hc [A/W ] Example: Calculate the responsivity of the detector with η=1% at =0.8um Solution: v = i R L = ρpr L = (ηeλp/hc) R L ρ = ηeλ / hc = 0.01x1.6x10-19 x0.8x10-6 /6.626x10-34 x3x10 8 = 6.4 ma /W 11/9/2014 LECTURES 5

4-Photomultiplier tube (PMT): -PMT is similar to the vacuum photodiode but they have greater responsivity because of an internal gain mechanism due to the multiple anodes called dynodes -The fast moving electrons hit the metal dynodes causing the release of additional (secondary) electrons -The gain is defined as the number of secondary electrons per incident electron - Consider : = gain per dynode, and The total gain will be : N = number of dynodes The current will be given by : M = N Cathode - i=ρ P= M (ηeλ/hc) P Light of energy hf V - - - 200V - + 100V - - - - Electrons Dynodes 400V - - - - 300V i R L Anode + v 11/9/2014 LECTURES 6

Example: A PMT has nine dynodes with gain of 5 each. It is used to detect an optical power of 1 w at 0.8um. The cathode is 1% efficient and the load is 50Ω. Calculate the responsivity, the current and the output voltage, assuming the responsivity with dynodes is 6.4 ma/w Solution: M = N = 5 9 =1.95x10 6 ρ = (6.4 ma/w) 1.95x10 6 =12.5 ka/w i = ρp = (12.5mA /W) 1W=12.5mA v =ir L = 12.5mAx50=0.725V 11/9/2014 LECTURES 7

5- PN photodiode: Semiconductor photodiode -The device is reversed biased, hence the electric field developed across the pn junction drifts the holes and electrons to the p and n sides and the depletion region is created on either side of the junction. This barrier prevent the majority carriers to cross the junction in opposite direction of the field ( a little minority carriers will move creating a reverse leakage current) Photon hf > E g Electron E g =W g Hole - p n +. -When a photon (hf > W g ) is incident in or near the depletion region, a photogeneration of electron hole pair is formed ( electron is excited from the valence to conduction band leaving a hole in the valence band) Electron energy Photon hf E.. -The carrier pairs generated near the junction are separated and drift under the effect of the electric field,hence it produces current in the external circuit in addition to the leakage current Depletion 11/9/2014 LECTURES 8 region W g =E g

6-PIN photodiode Inserting a wide intrinsic semiconductor layer between the p and n regions effectively widening the depletion layer. Most of the photons are absorbed in that layer because it is long Most of the voltage drop is across the intrinsic layer. This creates a high electric field in the intrinsic layer. Now there is much less effect of the delay caused by diffusion and the response time is much faster than that of a pn photodiode hf W g c hc W g 1.24 W Material Wavelength range ( m) Intrinsic layer - p n + Peak response λ ( m) Peak responsivity ρ (A/W) Silicon 0.3 1.1 0.8 0.5 Germanium 0.5 1.8 1.55 0.7 c g InGaAs 1.0 1.7 1.7 1.1 11/9/2014 LECTURES 9 E.

I-V characteristic of PIN photodiode and its circuit analysis -When there is no optic incident power there is a small reverse current called the dark current I D or reverse leakage current. It is temperature dependent. i.e., increases with an increase in temperature. It determines the lower limit of the detectable incident power Example: PIN diode with responsivity 0.5A/W and dark current 1nA, minimum detectable power assuming an equal current to the dark current is detectable will be : P= I D /ρ =1nA/0.5mA=2uW -20-10 0 0.5 0 P = 10 W 20 30 40 For a silicon diode with responsivity 0.5A/W I D Photoconductive region Diode Current i d ( A) -20 Photovoltaic region 11/9/2014 LECTURES 10-5 -10-15 Diode Voltage v d (volts)

+ - OPTICAL FIBER COMM. SYSTEM -Load line equation : V B + v d + i d R L = 0 -Assume that the incident optic power is 10uW, the operating point is at the intersection of the load line with the characteristic curve at 10 mw: v d = -15V, i d =5uA -Saturation occurs when v d = 0, that is when : i d =V B / R L and the saturation P = 10 W power will be given by :P max = i d / ρ = V B / ρ R L 20 30 40 v d - R L = 10 6 11/9/2014 LECTURES 11 V B - + + i d -5-10 -15-20 R L -20-10 0 0.5 0 I D V B = 20 V i d ( A) v d (volts)

The 3-dB bandwidth is : f 3-dB =0.35/t r = 0.35/2.19R L C d =1/2πR L C d The larger the load resistance, the smaller the bandwidth Hence in design we choose R L according to : Large R L to increase the output voltage Small R L to reduce the response time Small R L to increase the receiver s dynamic range R Lmax =V B / ρp max 11/9/2014 LECTURES 12

Is a semiconductor junction detector that has an internal gain which increases its responsivity over other detectors Principle of avalanche current multiplication The reverse bias is very strong and so when a photon is absorbed in the depletion region, a free electron and a free hole are created The large bias causes them to accelerate with huge speeds and gain kinetic energy. The energy is large enough so that only part of it is enough to raise other electrons across the energy band-gap, increasing the number of free carriers and so the generated current The new generated electrons themselves causes more electrons to be set free The avalanche gain M is given by : 7-Avalanche photodiode APD M= 1/[1-(v d /V BR ) n ] V BR is the diode breakdown voltage, v d the diode reverse bias voltage and n is an estimated parameter >1.Relation of current and responsivity is the same as that of the photomultiplier 11/9/2014 LECTURES 13

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