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

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)

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 $$E_{F_n}(x) \propto x$$?

• 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.
– Puk
Mar 16, 2023 at 4:18

# 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}

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}

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