Question

9.1. Show that the 'built-in' voltage in a p-n junction is given by $\frac{kT}{e}\log_e\frac{N_{cn}}{N_{ep}} = \frac{kT}{e}\log_e\frac{N_{hp}}{N_{hn}}$, where $N_{cn}$, $N_{hn}$ and $N_{ep}$, and $N_{hp}$ are the carrier densities beyond the transition region in the n and p-type materials respectively. [Hint: Use eqns (8.17) and (8.20) and the condition that the Frmi levels must agree.] Hence the total current $I_e = I_e(right \ to \ left) - I_e(left \ to \ right) = I_0[exp(eU_1/k_BT) - 1]$, (9.17) The total charge of the donor atoms is $Q = eN_Dx_n = \left\{2\epsilon e(U_0 + U_1)\frac{N_AN_D}{N_A + N_D}\right\}^{1/2}$. (9.20)

          9.1. Show that the 'built-in' voltage in a p-n junction is
given by
$\frac{kT}{e}\log_e\frac{N_{cn}}{N_{ep}} = \frac{kT}{e}\log_e\frac{N_{hp}}{N_{hn}}$,
where $N_{cn}$, $N_{hn}$ and $N_{ep}$, and $N_{hp}$ are the carrier densities
beyond the transition region in the n and p-type materials
respectively.
[Hint: Use eqns (8.17) and (8.20) and the condition that the
Frmi levels must agree.]
Hence the total current
$I_e = I_e(right \ to \ left) - I_e(left \ to \ right) = I_0[exp(eU_1/k_BT) - 1]$,
(9.17)
The total charge of the donor atoms is
$Q = eN_Dx_n = \left\{2\epsilon e(U_0 + U_1)\frac{N_AN_D}{N_A + N_D}\right\}^{1/2}$.
(9.20)

$9.1. Show that the 'built-in' voltage in a p-n junction is given by (kT)/(e)(Ncn)/(Nep) = (kT)/(e)(Nhp)/(Nhn), where Ncn, Nhn and Nep, and Nhp are the carrier densities beyond the transition region in the n and p-type materials respectively. [Hint: Use eqns (8.17) and (8.20) and the condition that the Frmi levels must agree.] Hence the total current Ie = Ie(right to left) - Ie(left to right) = I0[exp(eU1/kBT) - 1], (9.17) The total charge of the donor atoms is Q = eNDxn = {2ϵ e(U0 + U1)(NAND)/(NA + ND)}^1/2. (9.20)$

Added by Jaime G.

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