Banha University

Recommendations

Info

Also EF – EFi = kT ln(Nd/ni) (iii) The position of Fermi level in p-type semiconductor: For p-type po = Na, then Also EFi – EF = kT ln(Na/ni) ___________________________________________________________________ Solution to Question 3 (18 points) a) Three samples of silicon at T = 300 K are the following: Sample 1 is n-type doped with arsenic atoms to Nd = 5x10 16 cm -3 . Sample 2 is p-type doped with boron to Na = 2x10 16 cm -3 .
Sample 3 is compensated silicon doped with both donors and acceptors described in the n-type and p-type samples. Find (i) the electron and hole concentrations in each sample, (ii) the conductivity of each sample (iii) The current density passing in each sample if an electric field of 25 V/cm is applied. Solutions (i) sample 1: no = Nd = 5x10 16 cm -3 Po = ni 2 /Nd = (1.5x10 10 ) 2 /5x10 16 = 4.5x10 3 cm -3 sample 2: po = Na = 2x10 16 cm -3 no = ni 2 /Na = (1.5x10 10 ) 2 /2x10 16 = 1.125x10 4 cm -3 sample 3: no = (Nd - Na)/2 + {[(Nd - Na)/2] 2 + ni 2 } 1/2 no = (Nd - Na)/2 = (5x10 16 – 2x10 16 )/2 = 1.5 x10 16 cm -3 Po = ni 2 /no = (1.5x10 10 ) 2 /1.5x10 16 = 1.5x10 4 cm -3 (ii) sample 1: σn = e no μn = (1.6x10 -19 )(5x10 16 )(1350) = 10.8 S/cm sample 2: σp = e po μp = (1.6x10 -19 )(2x10 16 )(480) = 1.536 S/cm sample 3: σn = e no μn = (1.6x10 -19 )(1.5x1016)(1350) = 3.24 S/cm (iii) sample 1: Jn = σn ε = 10.8 x 25 = 270 A/cm 2 sample 2: Jp = σp ε = 1.536 x 25 = 38.4 A/cm 2 sample 3: Jn = σn ε = 3.24 x 25 = 81 A/cm 2 __________________________________________________________________ b) (i) Describe how the space charge region is formed in pn junction diode. (ii) Derive an expression for the electric field and the electric potential in the pn- junction diode. (iii) Sketch the energy band diagram of zero-biased, reverse-biased and forward- Biased pn-junction. Solutions (i) Formation of space charge region in pn-junction diode: A single-crystal semiconductor material in which one side is doped with acceptor impurity atoms to form the p region and the the other side is doped with donor atoms to form the n region. The interface separating the n and p regions is referred to as the junction. Initially, at the junction, there is a very large density gradient in both the electron and hole concentrations. Majority carrier electrons in the n region will begin diffusing into the p region leaving behind positively charged donor ions and majority carrier holes in the p region will begin diffusing into the n region leaving behind
Page 1 and 2: Banha University First Term 2012/20
Page 3 and 4: where The emitted wavelength of lig
Page 5: ) Derive expressions for the positi
Page 9 and 10: The electric field is assumed to be
Page 11 and 12: Solution to Question 4 (18 points)
Page 13 and 14: possible to reach the collector wit
Page 15 and 16: (6) The figure shows the E - k rela

Banha University

You also want an ePaper? Increase the reach of your titles

Delete template?

Save as template?