6.012 Recitation 16: Small Signal Model of P-N Diode

Recitation 16 Small Signal Model of p-n Diode 6.012 Spring 2009 Recitation 16: Small Signal Model of p-n Diode Review Last week we learned about the IV characteristic of p-n diode: I = Io (eqVo /kT − 1)   Dp 1 1 Dn 2 + where Io = qAni Na wp − xp Nd wn − xn If we plot, Yesterday, we discussed the small signal model for p-n diode. Under forward bias, the small signal (ss) model of a p-n diode is: 1 Recitation 16 Small Signal Model of p-n Diode 6.012 Spring 2009 Cj is the junction capacitance as we talked about before, Cd is called “diffusion capacitance”, this is new Small signal circuit model of p-n diode: (two terminal device) In a small signal model, we have linearized conductance (resistance) and capacitances. 2 Recitation 16 Small Signal Model of p-n Diode 6.012 Spring 2009 Linearized Conductance (Resistance) 1 , gd γd = =⇒ γd gd = δIo · (eqVD /kT − 1)  δiD   =  δVD VD δVD VD q qVD /kT = Io · e kT q = · (ID + Io ) kT kT q = Vth = ID , q kT Vth = ID Vth constant, the larger operating current ID , the smaller is γd Depletion Capacitance (due to p-n junction)  Cjo = s = xdo  Cj (VD ) = qs Na Nd 2(Na + Nd )φB Cjo 1−   For forward bias limit VD to VD φB φB 2 capacitance/unit area! Diffusion Capacitance For this, we need to look at the majority carrier concentration as well. To keep quasi-neutral, nn (x) = Nd + pn (x) pp (x) = Na + np (x) 3 Recitation 16 Small Signal Model of p-n Diode 6.012 Spring 2009 It is a capacitor without two parallel plates! And the “+” and “-” charges are just mixed with each other! Isn’t that amazing! charge stored on the n-side: q pn Cdn 1. Cdn = 1 = −qNn = qA (wn − xn ) · 2 dqpn  wn − xn = ·  = qA dVD VD 2 n2i qVD /kT (e − 1) Nd ni2 q qVD /kT e Nd kT qA qA (wn − xn ) · pno eqVD /kT = (wn − xn ) · pn (xn ) 2Vth 2Vth 2. Write in terms of ID Cdn = Cdn = Define : T Tp = Cdn  Cdn = Similarly, Cdp  Dp n2i wn − xn wn − xn q · · eqVD /kT · · · Nd wn − xn 2 Dp kT 2 q (wn − xn ) · · IDp kT 2Dp transit time of holes through n-QNR: wn − xn length (wn − xn )2 = = 2Dp 2Dp /(wn − xn ) velocity q · TTp · IDp kT IDp · TTp Vth q · TTn · IDn kT qA Charges on both side are added together. The two capacitors are in parallel. q (TTn · IDn + TTp · IDp ) kT qA = ((wp − xp )npo + (wn − xn )pno )eqVD /kT 2Vth Cd = Cdn + Cdp = Cd = Cdn + Cdp Discussions: 1. Where do the extra majority carriers come from? 2. Majority current: -diffusion + drift 3. Cd ∝ eqVD /kT : for reverse bias, depletion capacitance dominate. For forward bias, diffusion capacitance dominate 4 Recitation 16 Small Signal Model of p-n Diode 6.012 Spring 2009 Exercise: p-n Diode wp = 0.5 μm Na = 2.5 × 1017 cm−3 Dn = 14 cm2 /s wn = 1.0 μm Nd = 4.0 × 1016 cm−3 Dp = 10 cm2 /s Vd = 720 mV ID = 50 μA Find Vd , Cj , and Cd . For this calculation, ignore xn , xp .  Jo = qni2 Dp Dn + Na wp Nd wn  = 5.79 × 10−11 A/cm2 ID = Io (eqVo /kT − 1)  Io · eqVo /kT = Io · 10720/60 = 50 μA =⇒ Io = 5 × 10−17 A =⇒ Jo = Io /A =⇒ A = 8.64 × 10−7 cm2 Vth 0.025 V kT Na Nd γd = = = 500 ω φB = ln = 840 mV ID 50 μA q n2i  qs Nd Na Cjo = A · 2(Na + Nd )φB  1.6 × 10−19 C × 8.85 × 10−14 F/cm · 11.9 × 4 × 1016 × 2.5 × 1017 cm−6 = A 2(4 × 1016 + 2.5 × 1017 )0.84 V · cm−3 = 8.64 × 10−7 cm2 × 5.88 × 10−8 F/cm2 Cj = 50.8 fF = 50.8 × 10−15 F Cjo 50.8 fF = = = 71.8 fF √1 1 − VφDB 2 Cd = = = = qA · (wp · np (−xp ) + wn · pn (xn )) 2Vth 1.6 × 10−19 C × 8.64 × 10−7 cm2 1020 1020 −4 (0.5 × 10−4 × + 1 × 10 × ) eqVo /kT 4 × 1016   2 × 0.025 2.5 × 1017 1.6 × 10−19 C 10−7 cm2 1.6 × 10−19 C 10−7 × 8.64 × 2 × 0.025 V × 8.64 × 2 × 0.025 V 1012 −4 14 15 (0.5 10−4 × 2.5  ×10 × 4 ×10 + 1 ×  × 10 ) cm cm−3 × 2.7 × 1011 = 746 fF 5 cm cm−3