Student Project Physics of Semiconductor Devices WS 2011/12 A Silicon p-n junction Markus Polanz 0630160 Introduction: In this student project the following exam question about a silicon p-n junction will be solved and discussed: a) Band diagram (no voltage applied) The pn-junction consists of an acceptor doped area (p-zone), which borders on a donator doped area (n-zone). In the p-zone the charge carriers are holes and in the n-zone they are electrons. At the crossing, the two sorts can recombine. Only immobile charge carriers are left in this area. These are the negative and positive space charges of the impurity atoms respectively in the so-called depletion zone. The electric field lines in this zone start at the donator atoms and end at the acceptor atoms. Because of the electric field there is a potential difference between the two space charge areas. There is a thermal equilibrium between the field current and the diffusion current. Thats why the fermi energy is on a constant level. The total potential difference between the left and the right zone is given by the built in voltage Vbi and results out of the diffusion of the charge carriers. b) Band diagram in forward and reverse bias If we apply a forward bias then charge carriers are pushed into the depletion zone and the depletion zone gets smaller. If we apply a reverse bias electrons are picked up on the right side and holes are picked up on the left side and the depletion zone gets bigger. In both cases there is a energy difference between the p-zone and the n-zone. The sign of the bias tells us if the energy is lifted or lowered. c) Concentration of holes For the following calculations some physical constants and informations about silicon are needed: kB = 1.38E-23 J/K …Boltzmann constant ε0 = 8.854E-12 F/m …dielectric constant e = 1.602E-19 C …elementary charge εr = 11.9…dielectric constant for silicon ni = 1.5E10 cm-3…intrinsic carrier concentration for silicon Nd = 5E15 cm-3…donator concentration at T = 300 K for silicon Na = 1E17 cm-3… acceptor concentration at T = 300 K for silicon On the p-side the concentration of holes np is equal to the concentration of acceptors Na: np = Na On the n-side we can calculate the concentration of holes np with the law of mass action: n2i np = Nd All in all we get for the hole concentrations np on both sides: p-side: np = 1E17 cm-3 n-side: np = 45000 cm-3 d) Calculation of the depletion width To calculate the depletion width W we need to determine the built-in potential Vbi: k B T Nd Na Vbi = ln( 2 ) e ni Now we are able to calculate the depletion width W: 2 ε0 εr (Nd + Na )Vbi W=√ e Nd Na Built-in potential: Vbi = 0.73 V Depletion Width: W = 0.45 μm