
Physics of Semiconductor Devices 

The depletion approximation for a pn junction (1)This is a relatively long problem where you will derive some important results of semiconductor physics. The answers you submit are formulas not numbers. Since these are important results, you can find the answers you need in books on semiconductor physics. Try to resist the temptation of just looking up the answers because you should understand this derivation. Consult books or other students if necessary. Consider a silicon pn junction. Far from the junction on the pside, the Fermi energy is k_{B}Tln(N_{v}/N_{a}) above the valence band edge E_{v}. Far from the junction on the nside, the Fermi energy is E_{g}  k_{B}Tln(N_{c}/N_{d}) above the valence band edge. For zero bias voltage, the Fermi energy is the same on both sides of the junction so the valence band edge is higher on the pside than the nside. The difference in energy between the valence band edge on the nside and the pside is eV_{bi} where V_{bi} is called the builtin potential. Derive an expression for the builtin potential in terms of Na, Nd, Nc, Nv, kB, e, ni, and T. Use the expression: n_{i}² = N_{c}N_{v}exp(E_{g}/k_{B}T) to eliminate E_{g} from the equation. The functions sin, cos, tan, asin, acos, exp, ln, and sqrt can be used. For instance, 2*sin(e*Na)*exp(kB) is a valid (but wrong) answer. Be sure to include a * to indicate multiplication; write 2*e*Na not 2eNa to multiply 2 times e times Na. It is possible that you do not need all of the variables listed above. 