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I want to derive an expression of the quasi-Fermi levels as functions of distance in the neutral p and n regions (outside depletion zone) when a forward-bias voltage $V_a$ is applied, as shown in Figure 8.6. (Taken from Neamen's Semiconductor Physics and Devices)

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I already derive expressions for the excess carrier concentrations in the n-region ($x<-x_p$): $$\delta n_{np}(x)=n_p(x)-n_{p_0}=n_{p_0}\left[e^{e\beta Va}-1\right]e^{\frac{x_p+x}{L}},$$ and for quasi-Fermi levels in terms of carrier electron concentration: $$n_p(x)=n_{p_0}+\delta n_{np}(x)=n_i e^{\beta\left(E_{F_n}(x)-E_{F_i}\right)}$$

Since both equations are exponential, the quasi-Fermi levels are then linear functions of distance. To show that, I can equate both equations and then isolate $E_{F_i}(x)$ from: $$n_{p_0}\left[e^{e\beta Va}-1\right]e^{\frac{x_p+x}{L}}+n_{p_0}=n_i e^{\beta\left(E_{F_n}(x)-E_{F_i}\right)},$$ or by collecting all constants terms: $$n_{p_0} \left( C_1 e^{\frac{x}{L}}+1 \right) =C_2 e^{\beta E_{\text{Fn}}}$$

However, the term $n_{p_0}$ messes up any straightforward simplification. Any suggestions on how to obtain a linear equation $E_{F_n}(x) \propto x$?

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  • $\begingroup$ Note that the equilibrium carrier concentration is negligible compared to the excess concentration when the electron quasi-Fermi level is much greater than (i.e. a few $k_B T$ above) the equilibrium Fermi level. $\endgroup$
    – Puk
    Mar 16, 2023 at 4:18

1 Answer 1

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Expression of the Quasi-Fermi levels through a forward-biased pn junction

We focus on the distribution of minority carriers in a steady state when a forward-bias voltage is applied to a PN junction. From [1], in p region we have (please refer to [1] for notation)

$$ \begin{aligned} \delta n_p(x) &=n_p(x)-n_{p 0}=n_{p 0}\left[\exp \left(\frac{e V_a}{k T}\right)-1\right] \exp \left(\frac{x_p+x}{L_n}\right) \\ n_p(x) &=n_{p 0}+\delta n_p=n_i \exp \left(\frac{E_{F n}-E_{F i}}{k T}\right) \end{aligned} \tag{1} $$

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we can determine the quasi-Fermi energy from (1)

$$ E_{F n}=E_{F i}+k T \ln \frac{n_{p 0}}{n_i}+k T \ln \left(\left[\exp \left(\frac{e V_a}{k T}\right)-1\right] \exp \left(\frac{x_p+x}{L_n}\right)+1\right) \tag{2} $$

assume that $\mathrm{exp} (eV_{a}/kT) >> 1$ (for $eV_{a} = 0.4 \ \mathrm{eV}$, $\mathrm{exp} (eV_{a}/kT)$ ~ $5.1 \times 10^{6}$).

$$ E_{F n}=E_{F i}+k T \ln \frac{n_{p 0}}{n_i}+k T \ln \left(\exp \left(\frac{e V_a}{k T} + \frac{x_p+x}{L_n}\right) +1\right) \tag{3} $$

for $\mathrm{exp}(\frac{e V_a}{k T} + \frac{x_p+x}{L_n}) >>1$, we have (the linear region)

$$ \begin{aligned} {{E}_{Fn}}={{E}_{Fi}}+kT\ln \frac{{{n}_{p0}}}{{{n}_{i}}}+kT\left( \frac{e{{V}_{a}}}{kT}+\frac{{{x}_{p}}+x}{{{L}_{n}}} \right) \end{aligned} \tag{4} $$

for $\lvert x \rvert >>L_{n}$ and $\lvert \frac{x_p+x}{L_n}\rvert >>\lvert \frac{e V_a}{k T} \rvert$ (note that $\frac{x_p+x}{L_n}$ is negative in p region), we have

$$ \begin{aligned} {{E}_{Fn}}={{E}_{Fi}}+kT\ln \frac{{{n}_{p0}}}{{{n}_{i}}}\end{aligned} \tag{5} $$

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[1] Neamen D. Semiconductor physics and devices[M]. McGraw-Hill, Inc., 2002.

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