6.012 Recitation 15: P-N Diode I-V Characteristics (II)

Recitation 15 p-n diode I-V characteristics (II) 6.012 Spring 2009 Recitation 15: p-n diode I-V characteristics (II) V Yesterday, we talked about the diode I-V relationship: I = Io (e Vth − 1). Today, we will look more closely how is this relationship and what is Io ? Brief Review of the “Law of the junction” When bias voltage V is applied (V can be either positive or negative): law of the junction −→ the minority carrier concentration at the SCR edge n2i · eV/Vth Na n2 = i eV/Vth Nd np (−xp ) = npo · eV/Vth = pn (xn ) = pno · eV/Vth 1 Recitation 15 p-n diode I-V characteristics (II) 6.012 Spring 2009 At the contact, recombination/generation occur very fast np (−wp ) = npo pn (+wn ) = pno We know two data points of the concentration, what is in between the QNRs? From yesterday’s discussion: Linear profile vs. position. This is because, approximating there is no recombination/generation inside QNR (and SCR). =⇒ current need to be constant. dp dn For minority carriers, only diffusion current =⇒ or constant. dx dx Note these are log scale, we try to plot linear dependence on log scale; it should not be linear on log scale 2 Recitation 15 p-n diode I-V characteristics (II) 6.012 Spring 2009 How to Calculate Current Jp (xn ) = −qDp dpn   dx x=xn pno    pn (xn ) − pn (wn ) pno eV/Vth − pno = qDp = −qDp xn − wn wn − xn 2 Dp n pno Jp (xn ) = qDp (eV/Vth − 1) = q i · (eV/Vth − 1) Nd wn − xn wn − xn =npo    np (−xp ) − np (−wp ) npo (eV/Vth − 1) = qDn = qDn −xp + wp wp − xp dnp   dx x=−xp n2 Dn Jn (−xp ) = q i · (eV/Vth − 1) Na wp − xp Jn (−xp ) = qDn The diode current is carried out by both electrons and holes. They need to be summed up Itotal = (Jn + Jp ) · A, A is cross-sectional area of diode   Dp 1 Dn 1 Itotal = qA · ni2 · + · · (eV/Vth − 1) Na wp − xp Nd wn − xn   Dp 1 Dn 1 2 Io = qAni · + · Na wp − xp Nd wn − xn Note: 1. Io is a pretty small value, 10−15 A. With a positive voltage, say 0.6 V, eV/Vth = 10V/60 mV = 1010 , we will get a fairly large current 2. Just by looking at the equation of Io , can we tell which part is Jn ? Which part is Jp ? 1 Dn Jn is electron diffusion in p-QNR =⇒ Na wp − xp Dp 1 Jp is hole diffusion in n-QNR =⇒ Nd wn − xn Be careful where to use Dn , Dp ! 3. For an asymmetrically doped diode, n+ p or p+ n, the Jn and Jp can differ a lot. 3 Recitation 15 p-n diode I-V characteristics (II) 6.012 Spring 2009 Overall Picture of Diode Current Jtot = Jn + Jp Through the diode, the electron current needs to be constant throughout, and the hole current needs to be constant throughout. This means: the majority hole current on p-QNR side = minority hole current on n-QNR side       drift + diffusion only diffusion the minority electron current on p-QNR side = majority hole current on n-QNR side       diffusion only drift + diffusion For students who are interested in the majority carriers, can discuss a little bit about it. Exercise Na = 5 × 106 cm−3 , Nd = 1017 cm−3 , wp = 0.3 μm, wn = 0.3 μm. If V = 0.75φB , find Jn , Jp . First Dn and Dp : • Dn is electron diffusion coefficient in p-region: doping 5 × 106 cm−3 , μn , μn = 900 cm2 /Vs, D kT = =⇒ Dn  22.5 cm2 /s μ q • Dp is hole diffusion coefficient in n-region: Nd = 1017 , μn = 350 cm2 /Vs, Dp = 8.75 cm2 /s 4 Recitation 15 φB V p-n diode I-V characteristics (II) kT Na Nd = ln = 0.025 ln q n2i = 0.75φB = 0.591 V xpo = 2s φB Nd = qNa (Na + Nd )  5 × 106 × 1017 1020 6.012 Spring 2009  = 0.789 V 2 × 1 × 10−12 F/cm × 0.79 V × 1017 cm−3 = 0.118 μm 1.6 × 10−19 C × 5 × 1016 × (5 × 1016 + 1017 ) cm−6 1 Na · xpo = Nd · xno , =⇒ xno = xpo = 0.059 μm 2 1 xn = xno 1 − V /φB = 0.059 μm × = 0.0295 μm 2 xp = xpo 1 − V /φB = 0.059 μm   Dn 2  forward bias V  Vth Jn = qni (eV/Vth −1) Na (wp − xp )   1.6 × 10−19 × 1020 × 22.5 · C × cm−6 × cm2 /s th = · eV/V   5 × 1016 cm3 (0.3 − 0.059) × 10−4 cm −10 Jp −10 109.85 9 = 2.98 × 10 A/cm · 10 = 2.98 × 10 × 7.07 × 10 A/cm2 = 2.10 A/cm2 Dp = qn2i (eV/Vth − 1) Nd (wn − xn ) 1.6 × 10−19 × 1020 × 8.75 = · eV/Vth = 5.17 × 10−11 A/cm2 × 109.85 = 0.36 A/cm2 1017 × (0.3 − 0.029) × 10−4 cm 2 9.85 5