ELG5103 Optical Communications Systems Assignment II 1 Due Tuesday 4 April 2017 Physical Constants 𝐠= 1.602 × 10−19 Magnitude of the electron charge ℎ = 6.625 × 10−34 𝐵 Planck’s constant 𝐠= 1.380 × 10−23 𝐵−1 Boltzmann’s constant. 𝐠= 3.00 × 108 𝐵−1 Velocity of light in a vacuum. Question 1 1) Derive a formula that describes the responsivity as a function of vacuum wavelength of an ideal quantum detector. (10 marks) 2) Briefly, describe the processes in a practical pin-photodiode that cause the quantum efficiency to be less than unity and how considerations of maximizing the quantum efficiency and speed of response influences its structure. (20 marks) 3) Calculate the ideal responsivity of a InGaAs (energy gap 𝐵 = 0.75 𝐵) photodiode when operating with incident light of vacuum wavelength: a)  = 1550 𝐵; b)  = 1700 𝐵 . (10 marks) Question 2 1) List and briefly describe the main sources of noise encountered when measuring an optical signal. Give relevant formulae and suggest remedies where applicable. (10 marks) 2) A high-speed pin-photodiode a responsivity of ℛ = 1 𝐯𝐠and capacitance 𝐠= 1𝐵 operates in the photoconductive mode into a resistive load 𝐵 = 50  . Calculate the bandwidth and sensitivity of the resulting photoreceiver circuit at a temperature 𝐠= 300𝐮 (10 marks) 3) The same photodiode is connected to a transimpedance amplifier with an open-loop gain of 𝐠= 99 and feedback resistor 𝐵 = 1 𝑗 find the new value of sensitivity assuming a noise-free wideband amplifier. (10 marks) 4) The feedback resistor is found to have a shunt capacitance of 200 fF. How does this affect your results? (10 marks) 5) Describe the options available and their practicality to improve receiver sensitivity towards the quantum limit. (10 marks) Question 3 Provide a physical justification of a simple equivalent circuit of an ideal pin photodiode specifying the I-V characteristics of its principal elements.ELG5103 Optical Communications Systems Assignment II 2 (10 marks) Use your equivalent circuit to derive an equation for the voltage 𝑤elivered by a photodiode operating in the photovoltaic mode to a load drawing a current 𝐠at an incident optical power 𝐠𝐮 Use your result to find expressions for short-circuit current 𝐵𝐠= 𝑼𝐽0; the open circuit voltage 𝐠𝐵 = 𝑼𝐽0; the product 𝐵 = 𝐍 𝐵𝐠𝐵; the electrical power delivered to the load 𝐵 = 𝐵; and the power ratio: 𝐠= 𝐵 𝐵 (10 marks) This ratio depends upon the incident optical power 𝐠𝐠through the photogenerated current 𝐵 = ℛ𝐠𝐬 where ℛ is the responsivity (found in Question 1) and on the current drawn by the load 𝐮 These currents are conveniently normalised by the dark reverse saturation current 𝐵 of the ideal photodiode: 𝐠= 𝐍 𝐵 ; 𝐠= 𝐵 𝐵 The parameter 𝐠is very large ≫ 106 for practical values of photocurrents and reverse saturation currents. Use Matlab to plot 𝐨𝐻 𝐩 versus 𝐠∈ [0,1] for 𝐠= 10𝐠with 𝐠= −1, 0, 1 … . .12 to illustrate that: lim 𝟶0 (𝑭ax ∈[0,1](𝐩) = 1⁄2 lim 𝟶∞ (𝑭ax ∈[0,1](𝐩) = 1 max 𝘈[0,1] (𝐩 is a monotone increasing function of 𝐍 The graph {𝐬 𝐨𝐻 𝐩|𝐠∈ [0,1] } for large 𝐠approaches a straight-line of unit slope. In the photovoltaics community, 𝐠̂ = max 𝘈[0,1] (𝐩 is known as the ‘fill-factor’. (10 marks) Piece your results together to give an expression for a critical optical power 𝐠𝐠for which the electrical power 𝐵 delivered to the load equals the optical power 𝐠𝐠incident on the photodiode. It follows that the model predicts 𝐵 > 𝐠𝐠for optical powers 𝐠𝐠> 𝐠𝐠in violation of the 1st law of thermodynamics. Given an explanation of the reason for the failure of the model (Hint: Evaluate the potential energy 𝐵 𝐵 for 𝐍 𝐠= 𝐠𝐠making use of the expression for the ideal responsivity derived in Question 1 and compare with the energy gap of the semiconductor). (10 marks) In addition to the ‘heat’ input from a hot reservoir (the incident optical power) and the work out (electrical power delivered to the load), the 2nd law of thermodynamics requires a ‘heat’ output to a cool reservoir (the ambient). Describe the physical processes responsible for converting the excess energy of the photon into this heat output. (10 marks) Question 4 This question concerns a fibre optic link operating at a bit rate 𝐠= 1 𝐵 𝘒1 using a standard fibre with a dispersion of 𝐠= 17 𝐵 𝐵−1 𝐵−1 and an attenuation 𝐠= 0.25 𝐵 𝐵−1 at a vacuum wavelengthELG5103 Optical Communications Systems Assignment II 3 𝐠= 1.55 𝐬 and a zero dispersion fibre (𝐠= 0) with an attenuation 𝐠= 0.5 𝐵 𝐵−1 at a vacuum wavelength 𝐠= 1.3 𝐍 1) Briefly describe the main sources of attenuation and dispersion in an optical fibre data link and how the limitations these place on performance may be alleviated. (20 marks) 2) You have a transmitter that operates at a vacuum wavelength 𝐠= 1.55 𝐬 has a spectral width Δ𝐠= 1 𝐵, and an output power of 10 mW. The receiver requires –20 dBm of input power to achieve the desired bit error rate. What is the length of the longest link you can build? (10 marks) 3) You have another transmitter that operates at a vacuum wavelength of 𝐠= 1.3 𝐬 has a spectral width Δ𝐠= 2 𝐵 and an output power of 1 mW. Assume the same receiver as before. What is the length of the longest link you can build? (10 marks) Question 5 Consider a standard double heterojunction laser diode. Well below threshold this device operates as a light-emitting diode, whereas well above threshold it behaves as a laser. Why is the overall power conversion efficiency (emitted light power leaving the device divided by electrical power) typically much better for laser operation than for LED-operation? (20 marks) Dr. T. J. Hall 21 March 2017