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小信号模型与理想偏差
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1 . Lecture #13 OUTLINE • pn Junctions – reverse bias current – deviations from ideal behavior – small-signal model Reading: Chapters 6.2 & 7 Spring 2003 EE130 Lecture 13, Slide 1 Diode Current due to Generation • If an electron-hole pair is generated (e.g. by light) in the depletion region of a Schottky diode or pn diode, the built-in electric field will sweep the generated carriers out, resulting in an additional component of current. Spring 2003 EE130 Lecture 13, Slide 2 1
2 .Review: Current Flow in Long-Base Diode • Under forward bias (VA > 0): – Holes are supplied (from the external circuit) through the p-side ohmic contact • Some of these recombine with injected electrons; the rest are injected into the n-side – Electrons are supplied (from the external circuit) through the n-side ohmic contact • Some of these recombine with injected holes; the rest are injected into the p-side Spring 2003 EE130 Lecture 13, Slide 3 Review: Current Flow in Short-Base Diode • Under forward bias (VA > 0): – Holes are supplied (from the external circuit) through the p-side ohmic contact • If the p-side is short (WP’ << Ln), ~all of the holes are injected into the n-side, and recombine with electrons at the n-side ohmic contact – Electrons are supplied (from the external circuit) through the n-side ohmic contact • If the n-side is short (WN’ << Lp), ~all of the electrons are injected into the p-side, and recombine with holes at the p-side ohmic contact Spring 2003 EE130 Lecture 13, Slide 4 2
3 . Reverse Bias Current • Consider a reverse-biased (VA<0) pn junction: minority carrier concentration x -xp xn – Depletion of minority carriers at edges of depletion region – The only current which flows is due to drift of minority carriers across the junction. This current is fed by diffusion of minority carriers toward junction (supplied by thermal generation). Spring 2003 EE130 Lecture 13, Slide 5 Alternative Derivation of Formula for I0 “Depletion approximation”: excess minority carrier concentration x -xp xn • I0 represents the rate at which carriers are generated within a diffusion length of the depletion region ∂n ∆n 2 n / NA =− p = i -LN -x p ≤ x ≤ -x p ∂t τn τn 2 ∂p ∆p n / ND =− n = i xn ≤ x ≤ xn + LP ∂t τp τp n 2 / NA n 2 / ND I 0 = qALN i + qALP i τ τn p Spring 2003 EE130 Lecture 13, Slide 6 3
4 .Deviations from the Ideal I-V Behavior Spring 2003 EE130 Lecture 13, Slide 7 Effect of R-G in Depletion Region • The net generation rate is given by 2 ∂p ∂n ni − np = = ∂t ∂t τ p (n + n1 ) + τ n ( p + p1 ) where n1 ≡ ni e ( ET − Ei ) / kT and p1 ≡ ni e ( Ei − ET ) / kT ET = trap - state energy level • R-G in the depletion region contributes an additional component of diode current IR-G xn ∂p I R −G = − qA∫ dx − x p ∂t R −G Spring 2003 EE130 Lecture 13, Slide 8 4
5 .• For reverse bias greater than several kT/q, qAniW 1 n p I R −G = − where τ 0 ≡ τ p 1 + τ n 1 2τ 0 2 ni ni I n Ip Spring 2003 EE130 Lecture 13, Slide 9 • For forward biases, I R −G ∝ qAniWe qV A / 2 kT In Ip Spring 2003 EE130 Lecture 13, Slide 10 5
6 . Effect of Series Resistance Spring 2003 EE130 Lecture 13, Slide 11 High-Level Injection Effect • As VA increases, the side of the junction which is more lightly doped will eventually reach HLI: nn > nno (p+n junction) or p p > p po (n+p junction) ⇒ significant gradient in majority-carrier profile Majority-carrier diffusion current reduces the diode current from the ideal Spring 2003 EE130 Lecture 13, Slide 12 6
7 . Review: Charge Storage in pn-Diode Spring 2003 EE130 Lecture 13, Slide 13 Small-Signal Model of the Diode Small signal equivalent circuit: i dv v R=1/G C i = Gv + C dt Small-signal conductance : 1 dI d d G≡ = = I 0 (e qVA / kT − 1) ≈ I 0e qVA / kT R dVA dVA dVA q kT G= I 0 e qVA / kT ≅ I DC / kT q Spring 2003 EE130 Lecture 13, Slide 14 7
8 .2 types of capacitance associated with a pn junction: 1. Cdep depletion capacitance 2. CD diffusion capacitance (due to variation of stored minority charge in the quasi-neutral regions) For a one-sided p+n junction, QP >> QN so Q = QP + QN ≅ QP: dQ dI τ p I DC CD = = τp = τ pG = dV A dV A kT / q Spring 2003 EE130 Lecture 13, Slide 15 Depletion Capacitance p n W conductor “insulator” conductor dQdep εs Cdep ≡ =A dVA W What are three ways to reduce Cdep? Spring 2003 EE130 Lecture 13, Slide 16 8
9 . Total pn-Junction Capacitance Spring 2003 EE130 Lecture 13, Slide 17 Cdep-vs.-VA (Reverse Bias) 1 W2 2(V − V ) = 2 2 ≅ 2bi A Cdep 2 A εs A qε S N Spring 2003 EE130 Lecture 13, Slide 18 9
10 . Example If the slope of the (1/Cdep)2 vs. VA characteristic is 2x1023 F-2 V-1, the intercept is 0.84V, and A is 1 µm2, find the lighter and heavier doping concentrations Nl and Nh. Solution: N l = 2 /( slope × qε s A2 ) = 2 /( 2 × 10 23 × 1.6 × 10 −19 × 12 × 8.85 × 10 −14 ×10 −8 cm 2 ) = 6 × 1015 cm −3 2 qV 0.84 kT N h N l ni kTbi 10 20 0.026 Vbi = ln 2 ⇒ N h = e = e = 1.8 × 1018 cm −3 q ni Nl 6 ×1015 Spring 2003 EE130 Lecture 13, Slide 19 10