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, lle = 0.13 m2/Vs 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

Length of rod =1= 5 cm = 0.05 m

Diameter of rod = d = 0.4 cm 0.004 m

πd2 Area of crosssection of rod = A= -= 7 * 4 x 10-6 m 4

Magnitude of charge of electron charge of hole = q = 1.6 x 10-19

Mobility of electron = Me 0.13 m²/Vs

$Hole\,\,mobility=\mu_h=0.05\,m^2/Vs$

$No\,\,of\,\,free\,electrons\,per\,unit\,volume=N_e=1.5\times 10^{16}\,electrons/m^3$

No of free holes per unit volume = Nh = Ne = 1.5 x 1016 holes/m

a)

Conductivity = 0 = 9(eNe + Minh) = N(He + uh

= 1.6 x 10-19 x 1.5 x 1016 (0.13 +0.05) Siemens/m

$=4.32\times 10^{-4}\,Siemens/m$

b)

$Resistivity=\rho =\frac{1}{\sigma}=\frac{1}{4.32\times 10^{-4}}=2314.8\,\Omega m$

0.05 Resistance = R= R= p- = 2314.8 A 7T X 4 x 10-6 9210.3k2

$Voltage=V=5V$

$Current=I=\frac{V}{R}=\frac{5}{9210.3\times 10^3}=0.54\,\mu A$

c)

$Electric\, field\, in\, the\, rod=E= \frac{V}{l}=\frac{5}{0.05}=100\,V/m$

$Electron\,drift\,velocity=u_e=\mu_eE=0.13\times 100= 13\,m/s$

$Hole\,drift\,velocity=u_h=\mu_hE=0.05\times 100= 5\,m/s$

d)

$Resistance=R=9210.3\,k\Omega$

e)

$Power\,\, dissipated=P=\frac{V^2}{R}=\frac{25}{9210.3\times 10^3}=2.7\,\mu W$