Metal Semiconductor Schottky barrier

9.31 (a) Consider a metal—semiconductor junction formed between a metal with a work function of $4.65 eV$ and Ge with an electron affinity of $4.13 eV$. The doping concentration in the Ge material is $Nd = 6*10^{13} (cm^{-3})$ and $N_a = 3*10^{13} (cm^{-3})$. Assume $T = 300 K$. Sketch the zero bias energy-band diagram and determine the Schottky barrier height. (b) Repeat part (a) if the metal work function is $4.35 eV$.


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Position of Fermi level with respect to the Valence Band (Ev) is

$E_f = E_g/2 +KT*ln(N_d/N_i)$

or equivalent

$E_f = E_g/2 -KT*ln(N_a/N_i)$

$KT =0.0256 eV$ at $T=300 K$


For Ge we have ($T=300 K$)

$Eg =0.66 eV$

$N_i =2.4*10^{13} cm^{-3}$

$Ef =0.66/2 +0.0256*ln(6*10^{13}/2.4*10^{13}) =0.3535 eV$

(the Fermi level is about in the center of the gap)

$E_c-E_f =0.66-0.3535 =0.307 eV$

(The Fermi level is $0.307 eV$ below conduction band)

The Schottky Barrier is the difference in energies between Fermi level in semiconductor and metal upper band (see the figure).

$Phi(B) =W(metal) – Hi(semiconductor) -(E_c-E_f) =4.65 -4.13 -0.307 = 0.213 eV$

The figure with the bands before and after contact is above.