3 Minority carrier profiles (the hyperbolic functions) Consider a

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1 Microelectronic Devices and Circuits October 9, Homework #3 Due Nov 9, Te pn junction Consider an abrupt Si pn + junction tat as acceptors cm -3 on te p-side and donors on te n-side. Te minority carrier recombination times are e = 490 ns for electrons in te p-side and =.5 ns for oles in te n -side. Te cross-sectional area is 1 mm. Assuming a long diode, calculate te current,, troug te diode at room temperature wen te voltage, V, across it is 0.6 V. Wat are V/ and te incremental resistance (r d ) of te diode and wy are tey different? Junction capacitance of a pn junction Te capacitance (C) of a reverse-biased abrupt Si p + n junction as been measured as a function of te reverse bias voltage, V r, as listed in Table 6.4. Te pn junction cross-sectional area is 500 m 500 m. y plotting 1/C versus V r, obtain te built-in potential, V o, and te donor concentration, N d, in te n-region. Wat is N a? 3 Minority carrier profiles (te yperbolic functions) Consider a pnp JT under normal operating conditions in wic te E junction is forward biased and te C junction is reverse biased. Te field in te neutral base region outside te depletion layers can be assumed to be negligibly small. Te continuity equation for oles, p n (x), in te n-type base region is D d p dx n pn p no 0 [ were p n (x) is te ole concentration at x from just outside te depletion region and p no and are te equilibrium ole concentration and ole recombination lifetime in te base. a. Wat are te boundary conditions at x = 0 and x = W, just outside te collector region depletion layer? (Consider te law of te junction). b. Sow tat te following expression for p n (x) is a solution of te continuity equation

2 p x n p no W x x sin sin ev L L exp 1 pno 1 [ kt W W sin sin L L were V = V E and L = [D ]. c. Sow tat Equation 6.7 satisfies te boundary conditions. 4 Te pnp bipolar transistor Consider a pnp transistor in a common base configuration and under normal operating conditions. Te emitter-base junction is forward biased and te base-collector junction is reverse biased. Te emitter, base, and collector dopant concentrations are N a(e), N d(), and N a(c) respectively were N a(e) >> N d() N a(c). For simplicity, assume uniform doping in all te regions. Te base and emitter widts are W and W E, respectively, bot muc sorter tan te minority carrier diffusion lengts, L and L e. Te minority carrier lifetime in te base is te ole recombination time. Te minority carrier mobility in te base and emitter are denoted by and e respectively. Te minority carrier concentration profile in te base can be represented by Equation 6.7. a. Assuming tat te emitter injection efficiency is unity sow tat 1. E ead ni cot W L N d L ev exp kt E. C ead ni cosec W L N d L ev exp kt E W 3. sec L 4. were time. t W t is te base transit D b. Consider te total emitter current, E, troug te E junction, wic as diffusion and recomb-ination components as follows: E E eve eve so exp Ero exp kt kt

3 Only te ole component of te diffusion current (first term) can contribute to te collector current. Sow tat wen N a(e) >> N d(), te emitter injection efficiency,, is given by 1 E ro E so ev exp E kt 1 How does < 1 modify te expressions derived in part (a)? Wat is your conclusion (consider small and large emitter currents, or V E = 0.4 and 0.7 V)? 5. Potoconductivity and speed Consider two p-type Si samples bot doped wit atoms cm -3. ot ave identical dimensions of lengt L (1 mm), widt W (1 mm), and dept (tickness) D (0.1 mm). One sample, labeled A, as an electron lifetime of 1 s wereas te oter, labeled, as an electron lifetime of 5 s. a. At time t = 0, a laser ligt of wavelengt 750 nm is switced on to illuminate te surface (L W) of bot te samples. Te incident laser ligt intensity on bot samples is 10 mw cm -. At time t = 50 s, te laser is switced off. Sketc te time evolution of te minority carrier concentration for bot samples on te same axes. b. Wat is te potocurrent (current due to illumination alone) if eac sample is connected to a 1 V battery? 6. Compound semiconductor devices Silicon and germanium crystalline semiconductors are wat are called elemental group V semiconductors. t is possible to ave compound semiconductors from atoms in groups and V. For example, GaAs is a compound semiconductor tat as Ga from group and As from group V, so tat in te crystalline structure we ave an "effective" or "mean" valency of V per atom and te solid beaves like a semiconductor. Similarly GaSb (gallium antimonide) would be a -V type semiconductor. Provided we ave a stoiciometric compound, te semiconductor will be ideally intrinsic. f, owever, tere is an excess of Sb atoms in te solid GaSb, ten we will ave nonstoiciometry and te semiconductor will be extrinsic. n tis case, excess Sb atoms will act as donors in te GaSb structure. Tere are many useful compound semiconductors, te most important of wic is GaAs. Some can be doped bot n- and p-type, but many are one type only. For example, ZnO is a -V compound semiconductor wit a direct bandgap of 3. ev, but unfortunately, due to te presence of excess Zn, it is naturally n-type and cannot be doped to p-type.

4 a. GaSb (gallium antimonide) is an interesting direct bandgap semiconductor wit an energy bandgap E g = 0.67 ev, almost equal to tat of germanium. t can be used as an LED (ligt-emitting diode) or laser diode material. Wat would be te wavelengt of emission from a GaSb LED? Will tis be visible? b. Calculate te intrinsic conductivity of GaSb at 300 K taking N c = cm -3, N = cm -3, e = 5000 cm V -1 s -1, and = 1000 cm V -1 s -1. Compare wit te intrinsic conductivity of Ge. c. Excess Sb atoms will make gallium antimonide nonstoiciometric, tat is, GaSb 1+, wic will result in an extrinsic semiconductor. Given tat te density of GaSb is 5.4 g cm -3, calculate (excess Sb) tat will result in GaSb aving a conductivity of cm -1. Will tis be an n- or p-type semiconductor? You may assume tat te drift mobilities are relatively unaffected by te doping. 7. Scottky and omic contacts Consider an n-type Si sample doped wit donors cm -3. Te lengt L is 100 m; te cross-sectional area A is 10 m 10 m. Te two ends of te sample are labeled as and C. Te electron affinity () of Si is 4.01 ev and te work functions,, of four potential metals for contacts at and C are listed in Table 5.5. a. deally, wic metals will result in a Scottky contact? b. deally, wic metals will result in an Omic contact? c. Sketc te -V caracteristics wen bot and C are omic contacts. Wat is te relationsip between and V? d. Sketc te -V caracteristics wen is omic and C is a Scottky junction. Wat is te relations-ip between and V? e. Sketc te -V caracteristics wen bot and C are Scottky contacts. Wat is te relationsip between and V? 8. Peltier coolers and figure of merit (FOM) Consider te termoelectric effect sown in Figure below

5 in wic a semiconductor as two contacts at its ends and is conducting an electric current. We assume tat te cold junction is at a temperature T c and te ot junction is at T and tat tere is a temperature difference of T = T T c between te two ends of te semiconductor. Te current flowing troug te cold junction absorbs Peltier eat at a rate Q' P, given by Q P = [5.95] were is te Peltier coefficient for te junction between te metal and semiconductor. Te current flowing troug te semiconductor generates eat due to te Joule eating of te semiconductor. Te rate of Joule eat generated troug te bulk of te semiconductor is Q J L A [5.96] We assume tat alf of tis eat flows to te cold junction. n addition tere is eat flow from te ot to te cold junction troug te semiconductor, given by te termal conduction equation Q TC A T L [5.97] Te net rate of eat absorption (cooling rate) at te cold junction is ten Q net cool = Q P 1 / Q J Q TC [5.98] y substituting from Equations 5.95 to 5.97 into Equation 5.98, obtain te net cooling rate in terms of te current. Ten by differentiating Q net cool wit respect to current, sow tat maximum cooling is obtained wen te current is A m [5.99] L and te maximum cooling is Q max cool A 1 T L [5.100]

6 Under steady state operating conditions, te temperature difference, T, reaces a steady-state value and te net cooling rate at te junction is ten zero (T is constant). From Equation 6 sow tat te maximum temperature difference acievable is 1 T max [5.101] Te quantity /k is defined as te figure of merit (FOM) for te semiconductor as it determines te maximum T acievable. Te same expression also applies to metals, toug we will not derive it ere. Use Table 5.6 to determine te FOM for various materials listed terein and discuss te significance of your calculations. Would you recommend a termoelectric cooler based on a metal-to-metal junction? efore came ere was confused about tis subject. Having listened to your lecture am still confused. ut on a iger level. Enrico Fermi ( ; Nobel Laureate, 1938) Will you?