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)

enter image description here

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

  • $\begingroup$ 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


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)

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

enter image description here

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

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

enter image description here

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


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.