---
title: "Fermi level position at 310 K"
subject: "Electrical Engineering Material"
chapter: 6
key: "semester_3/electrical_engineering_material:fermi-level-temperature-310k"
source: https://www.bctengineer.com/q/semester_3/electrical_engineering_material/fermi-level-temperature-310k
site: BCT Engineering
---

# Fermi level position at 310 K

Electrical Engineering Material · Chapter 6: Semiconductors

- Asked in: 2078 Kartik, 6 marks
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_3/electrical_engineering_material/fermi-level-temperature-310k
- Whole chapter: https://www.bctengineer.com/md/semester_3/electrical_engineering_material/chapter-6

## Question

In an n-type semiconductor, the Fermi level lies 0.5eV below the conduction band at 300 K, if the temperature is increased to 310 K, find the new position of Fermi level.

## Short answer

For fixed donor concentration in extrinsic range, Ec – EF ∝ T. At 300 K it is 0.5 eV. At 310 K: Ec – EF = 0.5 × (310/300) ≈ 0.517 eV below the conduction band edge.

## Answer

In an n-type semiconductor (non-degenerate, all donors ionised)

$$
n = N_c\,e^{-(E_c-E_F)/kT} \;\Rightarrow\; E_c - E_F = kT\ln\frac{N_c}{N_d}
$$

In the extrinsic range $n = N_d$ is fixed. If the small change of $N_c$ with temperature is neglected, $\ln(N_c/N_d)$ is constant, so $E_c-E_F \propto T$.

#### Calculation

$$
\begin{aligned}
\frac{(E_c-E_F)_{310}}{(E_c-E_F)_{300}} &= \frac{310}{300}\\
(E_c-E_F)_{310} &= 0.5\times\frac{310}{300} \\
&= 0.5167\ \text{eV}
\end{aligned}
$$

**Answer: at 310 K the Fermi level lies about 0.517 eV below the conduction band edge**, i.e. it moves down by about 0.017 eV, towards the middle of the gap.

#### More exact check (including $N_c \propto T^{3/2}$)

$$
\begin{aligned}
(E_c-E_F)_{310} &= \frac{310}{300}(0.5) + k(310)\tfrac{3}{2}\ln\frac{310}{300}\\
&= 0.5167 + 0.0013 = 0.518\ \text{eV}
\end{aligned}
$$

The difference is only about 1 meV, so the simple answer is acceptable.

#### Physical meaning

As temperature rises, more electrons can be thermally excited across the gap and the material moves towards intrinsic behaviour, so the Fermi level drops from near $E_c$ towards $E_{Fi}$ (mid-gap).

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. In the extrinsic range with n = Nd fixed, how does Ec - E_F vary?
   A. Ec - E_F = kT ln(Nc/Nd), proportional to T if Nc change is neglected
   B. It is independent of T
   C. It is inversely proportional to T
   D. It is proportional to T^2
   Answer: A. ln(Nc/Nd) is roughly constant, so Ec - E_F scales with T. (Key point: Governing relation)

2. What is the new Ec - E_F at 310 K?
   A. 0.5 x 300/310 = 0.484 eV
   B. 0.5 x 310/300 = 0.517 eV
   C. 0.5 + 10 = 10.5 eV
   D. 0.500 eV
   Answer: B. Ratio of temperatures gives 0.5167 eV. (Key point: Calculation)

3. Which way does E_F move as T rises?
   A. Up closer to Ec
   B. It stays pinned at Ec - 0.5
   C. Away from Ec toward mid-gap E_Fi, as the material becomes more intrinsic
   D. Into the valence band
   Answer: C. More thermally generated pairs push E_F toward the intrinsic level. (Key point: Direction)

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