6.012 Recitation 14: I-V Characteristics of P-N Junction Diode (I)

Recitation 14 I-V characteristics of p-n junction diode (I) 6.012 Spring 2009 Recitation 14: I-V characteristics of p-n junction diode (I) What happens to the (minority) carrier concentrations when a bias is applied? Location-Dependence In order to see what happens to the minority carrier concentrations at different locations inside a diode when a bias V is applied, let us look at some diagrams: 1. QNR & SCR (or depletion region) V = 0, SCR region edge: xno ∀xpo 2. Under thermal equilibrium: Hole concentration in p-QNR:ppo = Na ni2 Minority electron concentration in p-QNR:npo = Na n2i Minority hole concentration in n-QNR:pno = Nd Electron concentration in n-QNR:nno = Nd The hole concentration and minority electron concentrations are constant throughout QNR. 1 Recitation 14 I-V characteristics of p-n junction diode (I) 6.012 Spring 2009 3. Built-in barrier (potential) φB = φno − φpo kT = ln q  Na · Nd ni2  4. Is there any relationship between the hole concentration on the p-side (ppo ) with the hole concentration on the n-side? It does not seem like at first look, but   Nd 1 Na · Nd , Na = ppo , 2 = φB = Vth ln 2 pno ni ni   ppo =⇒ φB = Vth ln pno =⇒ pno = ppo e−φB /Vth = Na · e−φB /Vth and similarly, npo = nno · e−φB /Vth = Nd e−φB /Vth They are related to each other by φB 2 Recitation 14 I-V characteristics of p-n junction diode (I) 3 6.012 Spring 2009 Recitation 14 I-V characteristics of p-n junction diode (I) 6.012 Spring 2009 What happens when V =  0? V >0 xn , xp shrinks, electric field reduces, barrier reduces φB = φB − V What can we say about the minority carrier concentration at the SCR edges? It turns out the relationship pno = ppo · e φ − VB th ; npo = nno · e−fracφB Vth still holds, but φB → φB . ∴ pn (xn )  pp  = important ·e φ − VB th = Na · e φ − VB th will talk about this later = Na · e − (φB −V) Vth V = Na e−φB /Vth ·e Vth    pno =⇒ pn (xn ) = pno · e V Vth Law of the junction pn (xn ) = (Na · e φ − VB np (−xp ) = (Nd · e th φ − VB th V V V V ) · e Vth = pno e Vth ) · e Vth = npo e Vth The “law of the junction” relates the minority carrier concentration at the SCR edge with    important the equilibrium minority carrier concentration and the applied bias (60 mV/dec) Some notes: 1. This is so only for the concentration at the SCR edge 2. For V > 0, this is usually called “minority carrier injection” because from the “law of the junction” relationship, by lowering the barrier, more majority carriers from one side move across the SCR and add on to the minority carriers on the other side 3. This can also be understood from the fact that, when V > 0, E ↓, Jdrift < Jdiff , more holes can diffuse from p → n, and more e− can diffuse from n → p 4. Low Level Injection: pn < nno and np < ppo (minority < majority) in order to have valid law of the junction 4 Recitation 14 I-V characteristics of p-n junction diode (I) 6.012 Spring 2009 5. What about the minority carrier concentration at other locations inside the QNR? • At the contact (wn , −wp ): since recombination/generation occurs very fast, carrier concentration in equilibrium • Inside QNR: we will talk about this on Thursday So now we have two data-points In Scale Ratio remains the same: np (−xp ) npo = pn (−xn ) pno V <0 Applying law of the junction: pn (xn ) = Na e−(φB −V)/Vth = Na e φ − VB th V · e Vth V = pno e Vth V np (−xp ) = npo e Vth Equations are the same, but since V < 0 this time, instead of having “minority carrier injection”, we have “minority carrier extraction”. 5 Recitation 14 I-V characteristics of p-n junction diode (I) 6.012 Spring 2009 The picture regarding minority carrier concentration: Exercise Na = 1018 cm−3 , Nd = 1016 cm−3 . Calculate minority carrier concentration at the SCR edge for V = 0, V = −0.12 V, V = 0.6 V First, pno = npo = n2i = 104 cm−3 Nd ni2 = 102 cm−3 Na Law of the junction as power of 10: np (−xp ) = npo · eV/Vth = npo · 10V/60 mV V = 0, npo , pno V = −0.12 V, =⇒ np (−xp ) = npo · 10−0.12/+0.06 = 1 cm−3 , pn (xn ) = 102 cm−3 V = 0.6 V, =⇒ np (−xp ) = npo · 100.6/0.06 = 1010 · 102 = 1012 cm−3  Na = 1018 cm−3 pn (xn ) = pno · 1010 = 1014 cm−3  Nd = 1016 cm−3 At what point is Low Level Injection valid? Let pn (xn ) = Nd = 1016 , eV/Vth = 1012 , =⇒ V = 12 × 60 = 720 mV 6