Chapter 11 gives the chemical potential's offset from each band edge in a doped…

PHYSICS · Subjective · Fermi University Physics · Condensed Matter, Materials & Soft Matter · Semiconductor physics and devices · Semiconductor physics and devices

Condensed Matter Physics, Ch The junction: diodes, transistors, lamps and solar cells

Problem

Chapter 11 gives the chemical potential's offset from each band edge in a doped crystal:

E_c-\mu=k_BT\ln(N_c/N_d)

on the n side, and by the mirror argument

\mu-E_v=k_BT\ln(N_v/N_a)

on the p side. Chapter 11 also gives

n_i^2=N_cN_ve^{-E_g/k_BT}

. Using only these three results, together with the fact that

\mu

becomes level across the junction, derive

V_{\rm bi}=(k_BT/e)\ln(N_aN_d/n_i^2)

by a second route, through the gap, rather than through the minority density as the chapter does. Then use your result in the form it takes before

E_g

is eliminated to show that

eV_{\rm bi}

is less than

E_g

whenever

N_d<N_c

and

N_a<N_v

, and explain in one sentence why that must be true.

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