Question

A bar of cylindrical Germanium has a diameter of 0.4 cm and a length of 5 cm. 5V of voltage is applied between the ends of the bar. You have the following information regarding the bar: Mobility of the electrons, le = 0.13 m2/V s Hole mobility, Mn = 0.05 m2/V s Number of free electrons per unit volume, Ne = 1.5 x 1016 electrons/m3 Number of holes per unit volume, Nn = Ne Calculate the following quantities: a) the conductivity o of the germanium bar b) the current I flowing in the bar c) the electron drift velocity ue and the hole drift velocity un d) the resistance R of the bar between the two ends e) the power dissipated in the silicon bar, P

Answer 1

radius (8) = 0.2cm Criven: - d = 0.4cm I 5cm V - SV E = V 5 s le = 0.13 m/vs. Ne = 1.5x 1% elma 1 lm = 1000/m len = 0.05 ml vos Nen Ne. Q Condue uctivity (ment Plep) q ( Nelen't Ny elep) q Ne (en + lep).q 1.5XL0 (0.43 +0.05)XIC 4.32 x 10 Semners

I = OEA = 4.32x10 Wy 1 area = 48 UT(0.2)² = 0.5cm? x 0.5X104 10-2 I = 2.17 A 9 drift velocity vd = ltr for electron = Ue = het for hole 12 = 0,12X10 13 m un line 0.0 5x16 5 m/s d Resistence (R) = pl

R = 4.32x109 S 0,5 23.148K2 R = 23.148 KR Power = 1² XR (P) - (2.17 X1 & 2 x 23.148X10 3 R = P = 0.100 Watt