Unformatted text preview: 5.3 EEE 352 – Fall 2009 Homework 7 (a) A silicon semiconductor is in the shape of a rectangular bar with a cross sectional area of 100 µm2, a length of 0.1 cm, and is doped with 5 X 1016 cm
3 arsenic atoms. The temperature is T = 300 K. Determine the current if 5 V is applied across the length. (b) Repeat part (a) if the length is reduced to 0.01 cm. (c) Calculate the average drift velocity of electrons in parts (a) and (b). (a) From Fig. 5.3, we find that µe ~ 1300 cm2/Vs and from Fig. 5.4, we find that the resistivity is about 0.8 ohm
cm. Using the latter, we find that the current is 5.4 V VA 5 ⋅ 10−6 I= = = = 62.5µA . R ρL 0.8 ⋅ 0.1 € (b) If we reduce the length, we increase the electric field, which increases the current, as I= V VA 5 ⋅ 10−6 = = = 625µA . R ρL 0.8 ⋅ 0.01 (c) For the velocity, we need the mobility found above, and €
V 1300 ⋅ 5 = = 6.5 × 10 4 cm / s L 0.1 . V 1300 ⋅ 5 5 v( b ) = µ = = 6.5 × 10 cm / s L 0.01 v( a ) = µ (a) A GaAs semiconductor resistor is doped with acceptor impurities at a concentration of 1017 cm
3. The cross
sectional area is 85 µm2. The current in € the resistor is to be I = 20 mA with 10 V applied. Determine the length of the device. (b) Repeat part (a) for silicon. (a) From Fig. 5.4, we find the resistivity of the p
type GaAs to be 0.23 ohm
cm. Since we need a resistance of 500 ohms, we need ρL AR 85 × 10−8 ⋅ 500 R= →L= = = 18.5µm A ρ 0.23 (b) For Si, we find that the resistivity is 0.2 ohm
cm, and € R= ρL AR 85 × 10−8 ⋅ 500 →L= = = 21.3µm . A ρ 0.2 € 5.26 The hole concentration is given by p = 1016exp(
x/Lp) for x > 0 and electron concentration is given by n = 5X1014exp(x/Ln) for x < 0. The values for the two diffusion lengths are 5 µm and 10 µm, respectively. The hole and electron diffusion constants are 10 and 25 cm2/s, respectively. The total current density is defined as the sum of the hole and electron current density at x = 0. Calculate the total current density. By the above definitions, we have that ȹ D p(0) D n (0) ȹ J = eȹ h +e ȹ ȹ L Ln ȹ ȹ p Ⱥ ȹ 10 ⋅ 1015 25 ⋅ 5 × 1014 ȹ = 1.6 × 10−19 ȹ + ȹ −4 10−3 ȹ 5 × 10 Ⱥ = 5.2 A / cm 2 5.40 Germanium is doped with 5X1015 donor atoms per cm3 at 300 K. The € dimensions of the Hall device are d= 5 X 10
3 cm, W = 0.02 cm, and L = 0.1 cm. The current si 250 µA, the applied voltage is Vx = 100 mV,and the magnetic flux density is B = 0.05 T. Calculate (a) the Hall voltage, (b) the Hall field, and (c) the carrier mobility. From the parameters given, we find that 0.1 = 100V / m 10−3 2.5 × 10−4 Jx = = 2.5 × 10 4 A / m 2 −5 −4 5 × 10 ⋅ 2 × 10 1 RH = − = −1.25 × 10−3 m 3 / Coul. 21 −19 5 × 10 ⋅ 1.6 × 10 Ex = (b) The Hall field is € E y = R H J x Bz = −1.25 × 10−3 ⋅ 2.5 × 10 4 ⋅ 0.05 = −1.56V / m (a) and € Vy = E yW = −1.56 ⋅ 2 × 10−4 = 0.312 mV (c) The mobility is found from € ρ=
Ex 100 = = 4 × 10−3 ohm − m J x 2.5 × 10 4 R 1.25 × 10−3 µ=− H = = 0.313m 2 /Vs = 3130cm 2 /Vs ρ 4 × 10−3 5.42 Consider silicon at 300 K. A Hall effect device is fabricated with the following geometry: d = 5X10
3 cm, W = 0.05 cm, and L = 0.5 cm. The electrical € parameters measured are: I = 0.5 mA, V = 1.25 V, and B = 0.065 T. The Hall x x z field is – 16.5 mV/cm. Determine (a) the Hall voltage, (b) the conductivity type, (c) the majority carrier concentration, and (d) the majority carrier mobility. We first determine the internal paramters: 1.25 E x= = 250V / m 5 × 10−3 5 × 10−4 4 2 Jx = = 2 × 10 A / m 5 × 10−5 ⋅ 5 × 10−4 (a) The Hall voltage is € Vy = E yW = −1.65 ⋅ 0.05 = −83mV (b) From the above data, € E −1.65 RH = y = = −1.27 × 10−3 4 J x Bz 2 × 10 ⋅ 0.065 Since the Hall constant is negative, the carriers are electrons. The conductivity is thus n
type. € (c) The majority carrier concentration is 1 1 n=− = −19 eRH 1.6 × 10 ⋅ 1.27 × 10−3 = 5 × 10 21 m−3 = 5 × 1016 cm −3 (d) Finally, the mobility is given by € R H 1.27 × 10−3 ⋅ 2 × 10 4 µe = − = = 0.1m 2Vs = 1000cm 2Vs ρ 250 € ...
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 Fall '08
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