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p-n junction photodetectors (a.k.a. photodiodes) Asymmetrical p+ - n or n+ - p junction can be used as a fast photodetector • Reverse bias applied to the p-n junction creates a depletion region with high electric field. • Photons absorbed in the depletion regions create electron-hole pairs, which are separated by the electric field and contribute to the photocurrent. Equilibrium conditions (in the dark, zero bias applied) – the diode current is zero The potential barrier BLOCKS the electrons in n-type material from diffusing into p-region The potential barrier BLOCKS the holes in p-type material from diffusing into n-region Junction under illumination Light is generating the e-h pairs -- + The light creates the carriers that move in the same directions as the minority carriers in the reverse biased junction (in the darkness). Therefore, under illumination, there is a photocurrent, which direction corresponds to the reverse current of the junction. The origin of this photo-current is the DRIFT of photo-generated carriers. Junction photocurrent Pinc φ0 = (1 − Θ R A hν Incident photon flux (the number of photons per unit area, per second): ΘR is the reflection coefficient of the top surface, A is the device area, Pinc is the incident optical power. The drift current density, Jdr assuming that all the carriers are swept out by the electric field in the depletion region: ( J dr = q φ0 1 − e−αW ) In case the depletion region is thick enough, i.e. αW >> 1, Jdr max = q ϕ0 The total drift current: ( I dr = Aq φ0 1 − e−αW ) ) The external quantum efficiency of the photodiode: The responsivity of the photodiode: R= η q η λ ( µm) = ext = ext ( A /W ) Pinc hν 1.24 I ph Photodiode response time The diode response time has two components: 1) transit time 2) RC-limited time constant 1) Photo-carrier transit time If the electric field in the depletion region is strong enough, both electrons and holes move with the saturation velocity, vS ≈ 107 cm/s. Transit time ttr ≅ W/vS Note that W depends on the applied voltage: For N a >> N d , (V0 − V ) = 1q Nd W 2 2ε 2) RC-time constant P-n junction capacitance E(x) V=0 V<0 -xp - xn xn V=0 xn V<0 E0 W VV0bi-- V V Depletion region capacitance Since the p-n junction has a space charge region there has to be a capacitance associated with it. The static and differential capacitances. First, consider a simple plain capacitor: The electric field between the plates is related to the charge per unit area: Q1 σ E= = ε ε0 ε ε0 ++++++ d ------ Q1 is the charge per unit area. The voltage drop across the capacitor, V = E × d; V= Q1 ε ε0 ×d Q= ε ε0 V d ×A , A is the cross section area of the junction The static capacitance C0 = Q/V For the plain capacitor Q= ε ε0 V d C0 = ×A ε ε0 A d The differential capacitance which is the most important for the speed of response Cd = ∂Q ∂V For the plain capacitor, Cd = ε ε0 A d ∂ Q εε 0 = A ∂V d = C0 Static capacitance of the p-n junction For the depletion region of a p-n junction, the exact separation between positive and negative charges is not obvious: the charges are distributed within the depletion width W. We use the definition C0 = Q/V to find the capacitance. The total (positive) charge, (n-side of the junction) QD = q×ND×Xn × A The total negative charge |-QA| = q×NA×Xp × A (p-side of the junction), Static capacitance of the p-n junction (cont.) The voltage across the junction 1 q N AN D (Vbi −V ) = W2 2 εε 0 N A + N D where W = xn + xp; 1 q For N A >> N D , (Vbi −V ) = ND W 2 2 εε 0 and W ≈ xn; The static capacitance (for |-V| >> Vbi, NA >> ND) Q = q×ND×W × A C0 = Q/V = q ND W A q ND W 2 2ε ε0 = 2ε ε0 A ε ε0 A = W (W / 2) This capacitance is equivalent to that of plain capacitor with d = W/2; Differential capacitance of the p-n junction Cd = ∂ Q1 ∂V Consider asymmetric p-n junction, NA >> ND (Vbi −V ) = 1 q ND W 2 2 εε 0 Q = q×ND×W × A ∂Q ∂Q ∂W = × ∂V ∂W ∂V ∂Q = q×N D× A ∂W ∂W 1 = ∂V ∂ V ∂W ∂V q ND W = ∂W ε ε 0 ⎛ qN DW ⎞ εε 0 A ∂Q Cd = = q×N D×A ⎜ ⎟= ∂V ⎝ εε 0 ⎠ W (omitting the voltage negative sign) Cd = εε 0 A W P-n junction as a variable capacitor (Vbi −V ) = p+ region W1 @ |V| =V1 W2 @ |V2| > |V1| 1 q ND W 2 2 εε 0 Cd = εε 0 A W W(V) C(V) Vbi - V The RC component of the photodiode speed of response: An intrinsic RC – time constant of the photoresponse: C= εε 0 A p+ region W1 @ |V| =V1 d W2 @ |V2| > |V1 | n- region RS = ρ W (d − W ) A τ RCi = C×Rs (A is the diode area) When the photodetector is connected to the external load (i.e. the amplifier), τRC = C (RS + RL), where RL is the load resistance Photodiode overall response time RC – time constant of the photoresponse: Transit time Total response time: τ RCi = C×Rs ttr ≅ W/vS τT ≅ ttr + τRCi p+ region W(V), ttr(V) W1 @ |V| =V1 d W2 @ |V2| > |V1 | RS(V) n- region C(V) Vbi - V As W increases, ttr increases but τRCi decreases because C decreases; hence there is an optimal W value for the fastest photoresponse.