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1. Diffusion: the Flow of Charge Due to Density Gradients

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1. Diffusion: the Flow of Charge Due to Density Gradients 1. Diffusion: the flow of charge due to density gradients. a. Diffusion Current The diffusion currents are expressed in terms of the electron or hole diffusion coefficient Dn or Dp, as shown below: dn dn Jn,diff = (q)Dn = qDn dx dx dn dp J = (+q)D = qD p,diff p dx p dx b. Einstein Relation: kT D = μ n q n kT D = μ p q p 2. Total Current: J = Jdiff + Jdrift EXAMPLE1: We have a piece of uniformly doped silicon with a acceptor concentration of Na=2E16cm-3. By shedding light on one end (x=0) of the semiconductor, we creat an electron concentration of n(x)=1E13*exp(-x/2um). Calculate the electron diffusion current as a function of x. SOLUTION1: 2 Na = 2E16 μn = 1050cm / Vs kT D = μ = 0.026 1050 = 27.3cm2 / s n q n dn d x /2104 Jn,diff = qDn = 1.6 1019 27.3 1013 e () dx dx () = 0.22e x /2μm A / cm2 3. Integrated Passives a. Resistors -Diffusion Resistor: ion implantation/diffusion 100ohm/sq(unsilicided),10ohm/sq(silicided) -Poly Film Resistor: deposit a thin film of heavily doped poly-Si 10-100ohm/sq(unsilicided),1ohm/sq(silicided) b. Capacitors -IC MIM Capacitor: by forming a Metal plate-Insulator(thin oxide)-Metal plate structure. -Parallel Plate Capacitor: A C = d -Nonlinear capacitor: Q-V curve is not a straight line Small signal capacitance: differential capacitance df (V) Q = Q + q f (Vs) + v 0 dV s V =Vs df (V) C dV V =Vs 4. PN Junction a. Structure b. Distribution of charge, electric field, and potential -Depletion Approximation: The transition region is completely depleted of free carriers, the only charges exist are immobile dopants. This region is called “Depletion Region” qNa (x po < x < 0) 0(x) +qNd (0 < x < xno ) -Electric Field in Depletion Region: qN a x + x (x < x < 0) ()p 0 po E (x) = s 0 qN d x x (0 < x < x ) ()n 0 no s -Potential in Depletion Region 2 qNa p (x) = p + ()x + x p0 2s qNd 2 n (x) = n + ()x xn0 2s -Boundary Conditions: Potential continuous: 2 qNa qNd 2 p (0) = p + ()x p0 = n (0) = n + ()xn 0 2s 2s Electrical Field continuous: qNa x p0 = qNd xn 0 -Depletion Width & Built-in Potential 2sbi 1 1 X d 0 = xn 0 + x p0 = + q Na Nd 2sbi Na xn0 = qNd Na + Nd 2sbi Nd x p0 = qN N N a a + d Nd Na bi n p = Vth ln + ln ni ni Na Nd = Vth ln > 0 ni c. PN Junction Capacitor -PN Junction under Bias 2s()bi VD Na VD xn (VD ) = = xn 0 1 qNd Na + Nd bi 2s()bi VD Nd VD x p (VD ) = = x p 0 1 qNa Na + Nd bi C j 0 C = = s j V X 1 D d 0 bi qs Na Nd s C j 0 = = 2bi Na + Nd X d s C j (VD ) = X d (VD ) -EXAMPLE2: Consider the following silicon P-N junction: (a) If Na = 4E16cm-3 in the P-region and Nd = 1E17cm-3 in the N-region, under increasing reverse bias, which region (N or P) will become completely depleted first? (b) What is the reverse bias at this condition? (c) What is the small-signal capacitance (F/cm2) at the bias condition? SOLUTION2: (a) Under reverse bias, the depletion region will expand. The side which has fewer dopants will fully deplete first. Na*Wp=4E16*1.2>Nd*Wn=1E17*0.4 So N side will deplete first Na Nd (b) bi = Vth ln = 0.81V ni 2sbi Na xn0 = = 0.055um qNd Na + Nd VD 0.4 = xn (VD ) = xn 0 1 bi So VD=-42V (c) The small signal capacitance is determined by the depletion region width. Xn=Wn=0.4um, Xp=(Nd/Na)*Xn=(1E17/4E16)*0.4=1um Xd=Xn+Xp=0.4+1=1.4um 2 Cd=s/Xd=8.85E-14*12/1.4E-4=7.6E-9F/cm 5. PN Junction Diode -Thermal Equilibrium: small diffusion current: high barrier small drift current: few minority carriers and wide depletion region net current: none -Reverse Bias: drift current: remains small diffusion current: reduced exponentially with bias net current: small reverse current -Forward Bias: drift current: remains small diffusion current: increase exponentially with bias net current: large forward current -IV curve: -Minority Carrier Concentration with recombination, long base diode without recombination or short base diode: -Diode Current Densities qVA Dp D diff diff 2 n kT Jdiff = Jn + J p = qni + e 1 NdWn NaW p -Small Signal Model qV v ()d + d qVd qvd kT kT kT ID + iD = IS e 1 ISe e qv i d = g v D kT d d -Charge Storage 1 qI C = d d 2 kT -EXAMPLE3 Consider an ideal, long-base, silicon PN junction diode with uniform cross section and constant doping on either side of the junction. P side is heavily doped and N side has an doping concentration of 1E16cm-3 . For the n-side of the junction: (a) Calculate the density of the minority carriers as a function of x (distance from the junction) when the applied voltage is 0.589V (which is 23kT/q). (b) Find the minority carrier currents as functions of x (distance from the junction) SOLUTION3 2 n x / L x / L (a) p(x) = i ()e qVa / kT 1 e p = 1014 e p cm 3 . N d (b) Minority current: 2 dp(x) ni qVa / kT x / L p x / L p 2 J p (x) = qDp = q Dp ()e 1 e = 0.5e A/ cm . dx Nd Lp .
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