Problems are listed in approximate order of difficulty. A single dot (•) indicates straigh...

Problems are listed in approximate order of difficulty. A single dot (•) indicates straightforward problems involving just one main concept and sometimes requiring no more than substitution of numbers in the appropriate formula. Two dots (••) identify problems that are slightly more challenging and usually involve more than one concept. Three dots (•••) indicate problems that are distinctly more challenging, either because they are intrinsically difficult or involve lengthy calculations. Needless to say, these distinctions are hard to draw and are only approximate.

••• Electric field in the depletion region. Consider a silicon pn junction diode, with donor concentration Nd on the n side, and equal acceptor concentration Na on the p side (Na= Nd). In equilibrium, the uncompensated impurities on either side of the junction form two slabs of equal and opposite charge density as shown in Fig. 1. The potential difference between the two sides of the depletion region is the contact potential V0 = 0.7 V, and the total width of the depletion region is 2L. (a) Use Gauss’s law to derive an expression for the electric field E as a function of position within the depletion region. Sketch the magnitude of the electric field versus position. [Hint: Apply Gauss’s law to the cylinder shown in the figure.] (b) Derive a formula for the half-width L of the depletion region in terms of the contact potential V0. [Hint: Recall that ∆ V = − ∫ E dx.] (c) Assuming Nd = Na = 1022 m−3 (a typical impurity concentration), what is the width of the depletion region and what is the magnitude of the electric field at the junction? The dielectric constant of silicon is K = 11.7.

FIGURE 1

Depletion region in a pn junction.