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# SEMICONDUCTOR DEVICES ECEN 3320

GPA 3.9

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This 12 page Class Notes was uploaded by Mrs. Lacy Schneider on Thursday October 29, 2015. The Class Notes belongs to ECEN 3320 at University of Colorado at Boulder taught by Bart Van Zeghbroeck in Fall. Since its upload, it has received 19 views. For similar materials see /class/231776/ecen-3320-university-of-colorado-at-boulder in ELECTRICAL AND COMPUTER ENGINEERING at University of Colorado at Boulder.

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Date Created: 10/29/15

Example 46 Solution A 1 cm2 silicon solar cell has a saturation current of 103912 A and is illuminated with sunlight yielding a shortcircuit photocurrent of 25 mA Calculate the solar cell efficiency and ll factor The maximum power is generated for V LP 0 IseV V 1 1 h 7mIXeVmV dVa P K where the voltage Vm is the voltage corresponding to the maximum power point This voltage is obtained by solving the following transcendental equation 1 IP11 I l Vm V Using iteration and a starting value of 05 V one obtains the following successive values for Vm Vm 05 0542 0540 V and the efficiency equals Vm1m Pin The current 1m corresponding to the voltage Vm was calculated using equation 461 and the power of the sun was assumed 100 mWcmz The fill factor equals ll factor M W V I 062gtlt0025 00 so VmVn n 054gtlt002413 83 where the open circuit voltage is calculated using equation 461 and I 0 The short circuit current equals the photocurrent Problem 215 Solution A piece of ntype silicon Nd 101 cm393 is uniformly illuminated with green light 7 550 nm so that the power density in the material equals 1 mWcmz a Calculate the generation rate of electronhole pairs using an absorption coef cient of 104 cm39l b Calculate the excess electron and hole density using the generation rate obtained in a and a minority canier lifetime due to Shockley Read Hall recombination of 01 ms c Calculate the electron and hole quasi Fermi energies relative to E based on the excess densities obtained in b a The generation rate is given by a P 7F 277 x109 cm393s391 ph 0p b The excess canier density then equals the product of the generation rate and the minority carrier lifetime 5 Since equal numbers of electrons and holes are generated the excess minority canier density equals the excess majority canier density 5n5pG 1p277x 1015 cm393 The quasi Fermi energies are then obtained from N d 5 n n 0p Fri El kT1nlkT1n 4l7meV 71 z and FF El kTn kTn Lap 324 meV n1 n1 where the total carrier density is obtained from the sum of the thermal equilibrium density and the excess carrier density The thermal equilibrium hole density is obtain from the electron density using the mass action law Problem 218 Solution Consider the problem of nding the doping density which results in the maximum possible resistivity of silicon at room temperature m 1010 cm393 un 1400 cmzVsec and up 450 cmzV sec Should the silicon be doped at all or do you expect the maximum resistivity when dopants are added Ifthe silicon should be doped should it be doped with acceptors or donors assume that all dopant are shallow Calculate the maximum resistivity the corresponding electron and hole density and the doping density Since the mobility of electrons is larger than that of holes one expects the resistivity to initially decrease as acceptors are added to intrinsic silicon The maximum resistivity or minimum conductivity is obtained om do dwnnwpp dwnnwann 0 z dn 1 dn dn which yields u n pnl 057 n 57 x109 cm393 n The corresponding hole density equals p 176 m 176 x 109 cm393 and the amount of acceptors one needs to add equals N a 120 m 120 x 109 cm393 The maxrrnurn resistivity equals 394 chm qunnupp arm1587 pmax Problem 225 Solution Electrons in silicon carbide have a mobility of 1000 cmzV sec At what value of the electric eld do the electrons reach a velocity of 3 X 107 cms Assume that the mobility is constant and independent of the electric eld What voltage is required to obtain this eld in a 5 micron thick region How much time do the electrons need to cross the 5 micron thick region The electric eld is obtained from the mobility and the velocity L007 30 kVcm gz z V 3 X 10 Combined with the length one nds the applied voltage V5L30000X5 X 10394 15 V The transit time equals the length divided by the velocity tLv 5 x103943 x107 167 ps Problem 226 Solution A piece of silicon has a resistivity which is speci ed by the manufacturer to be between 2 and 5 Ohm cm Assuming that the mobility of electrons is 1400 cmzVsec and that of holes is 450 cmzV sec what is the minimum possible canier density and what is the corresponding canier type Repeat for the maximum possible canier density The minimum canier density is obtained for the highest resistivity and the material with the highest canier mobility ie the ntype silicon The minimum canier density therefore equals n 892 x1014 cm393 gun pmax 16 gtlt1019 X1400 X 5 The maximum canier density is obtained for the lowest resistivity and the material with the lowest canier mobility ie the ptype silicon The maximum canier density therefore equals p 4 694 x1015 cm393 amp0min 16x1019 gtlt4SOgtlt 2 Problem 230 A 20 um thin piece of gallium arsenide consists of two regions with a different carrier life time namely 139 20 ns for 0 S x S 10 um and r 00 for 10 um S x S 20 pm The material is illuminated with light so that 1022 cm39zs391 electronhole pairs are created at x 20 um Calculate the steady state electron and hole density at x 0 10 and 20 um Assume LG Hp 1000 cmzVs and T 300 K Solution Since the material is undoped we can assume that the electron and hole densities as equal to each other and much larger than the intrinsic carrier densities The diffusion equation in the region with r 20 ns then becomes d 2 7 abc2 00SxS 10 um I with solution nn0coshLi 0 SxS 10 pm Where 710 is the electron density at x 0 pm and Ln is the diffusion length The diffusion equation in the region with 139 00 is d 2 7 abc2 0 with general solution nn0cosh1mAxilOLm for 10 S x S 20 um The electronhole pair generation results in an electron current dn J G D D A n q 0p q n dx q n So that the constantA equals GOPDn387 X 1021cm393s391 The slope of the carrier density at x 0 must be continuous so that E n0 sinh10um A dx x10m Ln so that 710 LnGOPDn sinh10umLn The carrier density at each point is nx 0 pm 110 148 x1018 cm393 71 x 10 um 710 cosh10umLn 318 x1018 cm393 nx 20 um 710 cosh10umLn A 10pm 702x1018 cm393 Problem 11 Solution Calculate the wavelength of a photon with a photon energy of 2 eV Also calculate the wavelength of an electron with a kinetic energy of 2 eV The wavelength of a 2 eV photon equals hc 6626 gtlt10734 Js X 3 X108ms Eph 1602x10 19Cx2ev where the photon energy 2 eV was rst converted to Joules by multiplying with the electronic charge The wavelength of an electron with a kinetic energy of 2 eV is obtained by calculating the deBroglie wavelength A h 6626 xro 34 Js p 762 X10725kg ms Where the momentum of the particle was calculated from the kinetic energy p 42m 2x911x10 31kgx16x10 19cX2 eV 764x10 25kg ms 062 pm 087nm Problem 13 Show that the spectral density um equation 124 peaks at E ph 282 kT Note that a numeric iteration is required Solution ha The spectral density um can be rewritten as a function of x E k3T 3 x3 2 h2n2c3 W x 1 The maximum of this function is obtained if its derivative is zero to or duw 3x2 x3 eXpx 0 dx eXpx1 eXpx 12 Therefore x must satisfy 3 3eXp x x This transcendental equation can be solved starting with an arbitrary positive value of x A repeated calculation of the le hand side using this value and the resulting new value for x quickly converges to 9cmax 282144 The maximum spectral density therefore occurs at E kT282144kT phmax xmax Problem 17 Solution Calculate the lowest three possible energies of an electron in a hydrogen atom in units of electron volt Identify all possible electron energies between the lowest energy and 2 eV The three lowest electron energies in a hydrogen atom can be calculated from 136 eV n n2 withnl2 and3 resulting in E1 7136 eV E2 34 eV andE3 l51eV The second lowest energy E 2 is the only one between the lowest energy E and 72 eV Problem 18 Derive the electric eld of a proton with charge q as a function of Solution the distance from the proton using Gauss s law Integrate the electric eld to nd the potential r m i 47 8 or Treat the proton as a point charge and assume the potential to be zero far away from the proton Using a sphere with radius r around the charged proton as a surface where the electric eld 5 is constant one can apply Gauss s law Er4727 r2 i 8 0 so that 50 47 r 8 0 The potential is obtained by integrating this electric eld from to Resulting in r m mm 4LTW q 00472780quot 47 or where the potential at in nity was set to zero Problem 19 Solution Prove that the probability of occupying an energy level below the Fermi energy equals the probability that an energy level above the Fermi energy and equally far away from the Fermi energy is not occupied The probability that an energy level with energy AE below the Fermi energy E F is occupied can be rewritten as exp 1 kT fEFAE W T lexp exp 1 kT kT 1 1 1 1 1 E AE e AE1 16 EFAE EF f F kar Xquot kT so that it also equals the probability that an energy level with energy AE above the Fermi energy E F is not occupied 092608 Example 14b Solution For the cylinder shown in Example 14 calculate the potential versus radius and the capacitance per unit length Assume the potential is zero at r Va The potential is the integral of the eld Assuming that the potential is zero at radius r p r 2 r r ln for r gt n 280 re and 2 2 r p r0 r for r lt n 480 The capacitance is the ratio of the charge to the potential difference Assuming a ground plane at radius r the capacitance equals QL 2723980 Cr 0 r 111VVo Using Q pn39rOZL B Van Zeghbroeck 1998

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