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Additional info for Electrical principles - second edition (1958)
835 eV (c) Check the result in part (b) using the formulas for the built-in voltage developed in the text. Since the doping concentrations are given let us use the formula that involves these quantities. 0259 ln 1018 1016 (1010 )2 which is precisely the same result as found in (b). 1 Particle ﬂux and current ﬂow direction for holes and electrons in a p–n junction. 2 p–n homojunctions under bias In a p–n junction both drift and diffusion occur. It is best to start our discussion of the nonequilibrium behavior of a p–n junction by examining in which direction the currents ﬂow due to diffusion and drift.
Since the donor doping concentration is very much larger than the intrinsic concentration, using the approximation that the electron concentration is equal to the donor concentration is valid. 417 eV By evaluating Ei , the position of the intrinsic level relative to the valence band can be determined. 10 Energy band diagram as a function of position. L is the total length of the sample. Ef corresponds to the ﬂat dashed line in the ﬁgure. 987 eV above the valence band edge. 1 Show that the probability that a state E above the Fermi level, Ef , is ﬁlled is equal to the probability that a state E below the Fermi level, Ef , is empty.
Assume that the electron and hole mobilities are equal and that n i = 1010 cm−3 . 3 Determine an expression for the steady-state minority hole concentration as a function of the one-dimensional position x in a semiconductor sample of length, L. Assume that no generation–recombination events occur and that the applied electric ﬁeld is zero. Take the boundaries of the sample at x = 0 and x = L. Assume that the excess hole concentration at x = 0 can be written as δp(x = 0) = pn . Also assume that the excess carrier concentration at x = L is zero.