2.2 DEPLETION REGION
2.2.1 Abrupt Junction
Build-in Potential and Depeletion-Layer Width. When the impurity concentration in a semiconductor changers abruptly from acceptor impurities \( N_A \) to donor impurities \( N_D \), as shown in Fig. 1a, one obtains an abrupt junction. In particular, if \( N_A \gg N_D \) (or vice versa), one obtains a one-sided abrupt \( p^{+}-n \) (or \( n^{+}-p \) ) junction.
We first consider the thermal equilibrium condition, that is, one without applied voltage and current flow. From the current equation of drift and diffusion (Eq. 156a in Chapter 1), \[ J_n = 0 = q \mu_n (n \mathcal{E} + \frac{k T }{q} \frac{dn}{dx}) = \mu_n n \frac{dE_{F}}{dx} \tag{1}\] or \[ \frac{dE_F}{dx} = 0 \tag{2} \] Similarly, \[ J_p = 0 = \mu_p p \frac{dE_F}{dx}. \tag{3} \]
Thus the condition of zero net electron and hole currents requires that the Fermi level must be constant throughout the sample. The build-in potential \( \Psi_{bi} \), or diffusion potential, as shown in Fig. 1b, c, and d, is equal to \[ q \Psi_{bi} = E_{g} - (q \psi_{n} + q \psi_{p}) = q \Psi_{Bn} + q \Psi_{Bp}. \tag{4} \]
For nodegenerate semiconductors, \[ \Psi_{bi} = \frac{k T}{q} \ln(\frac{n_{n0}}{n_i}) + \frac{k T}{q} \ln(\frac{p_{p0}}{n_{i}}) \tag{5} \] \[ \approx \frac{kT}{q} \ln( \frac{N_{D} N_{A}}{ n_{i}^2} ) \]. Since at equilibrium \( n_{n0} p_{n0} = n_{p0} p_{p0} = n_{i}^2 \), \[ \Psi_{bi} = \frac{k T}{q} \ln(\frac{p_{p0}}{n_{n0}}) = \frac{k T}{q} \ln(\frac{n_{n0}}{n_{p0}}) \tag{6} \] This given the relationship between carrier densities on either side of the junction.
If one of both sides of the junction are degenerate, care has to be taken in calculating the Fermi-levels and build-in potential. Equation 4 has to be used since Boltzmann statistics cannot be used to simplify the Fermi-Dirac integral. Furthermore, incomplete ionization has to be considered, i.e., \( n_{n0} \neq N_D\) and/or \( n_{p0} \neq N_A\) (Eqs. 34 and 35 of Chapter 1).
Next, we proceed to calculate the field and potential distribution inside the depletion region. To simplify the analysis, the depletion approximation is used which assumes that the depleted charge has a box profile. Since in the thermal equilibrium the electric field in the neutral regions (far from the junction at either side) of the semiconductor must be zero, the total negative charge per unit area in the p-side must be precisely equal to the total positive charge per unit area in the n-side: \[ N_A W_{Dp} = N_D W_{Dn}. \tag{7} \] From the Poisson equation we obtain \[ -\frac{d^2 \Psi_i}{dx^2} = \frac{d \mathcal{E}}{dx} = \frac{\rho(x)}{\varepsilon_s} = \frac{q}{\varepsilon_s} [ N_D^{+}(x) - n(x) - N_A^{-}(x) + p(x)]. \tag{8}\] Inside the depletion region, \(n(x) \approx p(x) \approx 0\), and assuming complete ionization, \[ \frac{d^2 \Psi_i}{dx^2} \approx \frac{q N_A}{\varepsilon_s} \qquad \mathrm{for} \quad -W_{Dp} \leq x \leq 0, \tag{9a} \] \[ -\frac{d^2 \Psi_i}{dx^2} \approx \frac{q N_D}{\varepsilon_s} \qquad \mathrm{for} \quad 0 \leq x \leq W_{Dn} . \tag{9b} \] The electric field is then obtained by integrating the above equations, as shown in Fig. 1b: \[ \mathcal{E} = -\frac{q N_A(x + W_{Dp})}{\varepsilon_s} \qquad \mathrm{for} \quad -W_{Dp} \leq x \leq 0, \tag{10} \] \begin{align} \mathcal{E} &= -\mathcal{E_m} + \frac{q N_D x}{\varepsilon_s} \\ &= -\frac{q N_D}{\varepsilon_s}(W_{Dn} - x) \qquad \mathrm{for} \quad 0 \leq x \leq W_{Dn} \tag{11} \end{align} where \( \mathcal{E_m} \) is the maximum field that exists at \(x=0\) and is given by \[ |\mathcal{E_m} | = \frac{q N_D W_{Dn}}{\varepsilon_s} = \frac{q N_A W_{Dp}}{\varepsilon_s}. \tag{12} \] Integrating Eqs. 10 and 11 once again gives the potential distribution \(\Psi_{i}(x)\) (Fig. 1c) \[ \Psi_{i}(x) = \frac{q N_A}{2 \varepsilon_s}(x + W_{Dp})^2 \qquad \mathrm{for} \quad -W_{Dp} \leq x \leq 0, \tag{13} \] \[ \Psi_{i}(x) = \Psi_{i}(0) + \frac{q N_D}{\varepsilon_s}(W_{Dn} - \frac{x}{2})x \qquad \mathrm{for} \quad 0 \leq x \leq W_{Dn}. \tag{14} \] With these, the potentials across different regions can be found as: \begin{align} \Psi_p &= \frac{q N_A W_{Dp}^2}{2 \varepsilon_s}, \tag{15a} \ |\Psi_n| &= \frac{q N_D W_{Dn}^2}{2 \varepsilon_s}, \tag{15b} \end{align}