# Ass Saa Aas

Topics: P-n junction, Semiconductor, Diode Pages: 3 (859 words) Published: April 24, 2013
Misr International University – Faculty of Engineering
Electronics and Communications Department Subject: Solid State Electronic Devices Code: GEE 235

Sheet 2

pn Junction

1 Calculate the built-in potential barrier, Vbi, for Si. pn junction if it has the following dopant concentration at T = 300 K, ND = l014 cm -3, NA = 1017 cm-3, ni=1.5x1010 cm-3.

2 An abrupt silicon pn junction at zero bias has dopant concentration of Na = 1017 cm-3 and Nd = 5 x 1015 cm-3. T = 300 K. a. Calculate the Fermi level on each side of the junction (w.r.t) the intrinsic Fermi level. b. Sketch the equilibrium energy band diagram for the junction and determine Vbi from the diagram and the results of part (a). c. Calculate Vbi, using dopant concentration, and compare the results to part (b). d. Determine, xn , xp and the peak electric field for this junction.(ni =1.5x1010 cm-3)

3 A silicon abrupt junction in thermal equilibrium at T = 300K is doped such that EC - EF = 0.21 eV in the n region and EF - EV = 0.18 eV in the p region. a. Draw the energy band diagram of the pn junction.

b. Determine the impurity doping concentrations in each region. c. Determine Vbi.
(NC=2.8x1019 cm-3, NV=1.04x1019 cm-3, ni=1.5x1010 cm-3.)

4 Consider the uniformly doped pn junction at T = 300 K. At zero bias, only 20 percent of the total space charge region is to be in the p region. The built-in potential barrier is Vbi, = 1.20 V. For zero bias, determine: a. NA

b. ND.

5 Consider a uniformly doped silicon pn junction with doping concentrations NA=5 x 1017 cm-3 and ND = 1017cm-3.
a. Calculate Vbi, at T = 300 K.
b. Determine the temperature at which Vbi, decreases by 1 percent. (ni=1.5x1010 cm-3, Nc=2.8x1019 cm-3 , Nv=1.04x1019 cm-3 Eg=1.12 eV)

6 An abrupt silicon pn junction has dopant concentrations of NA = 2 x 1016 cm-3 and ND = 2 x lo15...