Electrical properties — conductors, semiconductors, insulators; doping (n-type, p-type): Foundations

Before you touch the parent note Electrical Properties, you need every symbol it throws at you to already feel obvious. This page builds each one from nothing, in the order they depend on each other.

1. Energy — the height of a ball on a hill

Every symbol on this page is measured against energy, so we start there.

Look at Figure 1. That shelf is where an electron can finally roll sideways freely — for now just hold onto the picture of "a high empty shelf you must climb to". In section 3 we give that shelf its proper name.

2. The electron, the hole, and charge

3. From one atom to bands — where energy levels come from

4. The room's "kick" — temperature and $k_{B} T$

5. The exponential — why $e^{- E_{g} /2 k_{B} T}$ appears

Where the $\frac{1}{2}$ in $e^{- E_{g} /2 k_{B} T}$ comes from

Formula Carrier count, the

$\frac{1}{2}$ factor, and the hidden $T^{3/2}$ With the Fermi level $E_{F}$ near the middle of the gap, an electron reaching the CB must climb $E_{C B} - E_{F} \approx E_{g} /2$ . The full statistical-mechanics result for the intrinsic carrier density is $n_{i} = C T^{3/2} e^{- E_{g} /2 k_{B} T}$ where $C$ is a constant and the $T^{3/2}$ factor comes from the density of states — roughly, "how many seats exist in each band" grows as temperature rises. Two things to notice:

The exponential dominates completely: as $T$ rises the $e^{- E_{g} /2 k_{B} T}$ term explodes far faster than the gentle $T^{3/2}$ , so we usually just write $n \propto e^{- E_{g} /2 k_{B} T}$ .
That leftover $T^{3/2}$ does not vanish — it quietly rides along inside the prefactor $σ_{0}$ of the conductivity law (section 8), which is why $σ_{0}$ is called "slowly varying" rather than truly constant.

6. Counting carriers — concentrations and their symbols

Formula Conductivity

$σ$ — both carriers count In a semiconductor both electrons and holes carry current, so the total conductivity is the sum of an electron part and a hole part: $σ = n e μ_{e} + p e μ_{h}$ Reading it: each term is (how many carriers) × (magnitude of charge each carries $e$ ) × (how slippery they are $μ$ ). The left term is the electrons' contribution, the right term the holes'. Why $e$ and not $- e$ here? Conductivity is always a positive quantity — it measures how much current flows for a given push, regardless of direction. Electrons (charge $- e$ ) drift against the field but carry current along it; holes (charge $+ e$ ) drift along the field. Both effects add to the current in the same direction, so we use the magnitude $e$ and add the two terms. The sign of the carrier charge only matters when you track the direction of drift (see Hall Effect), not the size of $σ$ . Special cases:

In a metal or strongly n-type material, electrons dominate ( $n ≫ p$ ) and it reduces to $σ \approx n e μ_{e}$ .
In a strongly p-type material, holes dominate ( $p ≫ n$ ) and it reduces to $σ \approx p e μ_{h}$ .
The units are siemens per metre ( $S/m$ ) — using $n, p$ in $m^{- 3}$ .

Formula Law of mass action — where

$n \cdot p = n_{i}^{2}$ comes from $n \cdot p = n_{i}^{2}$ Plain words: in any silicon at fixed temperature, boosting electrons automatically suppresses holes so their product stays fixed at $n_{i}^{2}$ . Where it comes from (derivation sketch): electron and hole densities each follow a Boltzmann factor measured from the Fermi level $E_{F}$ : $n = N_{C} e^{- (E_{C B} - E_{F}) / k_{B} T}, p = N_{V} e^{- (E_{F} - E_{V B}) / k_{B} T}$ where $N_{C}, N_{V}$ are the effective "seat counts" of the two bands. Multiply them: the $E_{F}$ terms cancel, leaving $n \cdot p = N_{C} N_{V} e^{- (E_{C B} - E_{V B}) / k_{B} T} = N_{C} N_{V} e^{- E_{g} / k_{B} T} .$ The right-hand side contains no $E_{F}$ — it depends only on the material and temperature, so it is the same whether the crystal is pure or doped. In a pure crystal $n = p = n_{i}$ , so that fixed value equals $n_{i}^{2}$ . Hence $n \cdot p = n_{i}^{2}$ always (at fixed $T$ ). Picture: a seesaw pinned so that $n \times p$ never changes; push $n$ up and $p$ drops to compensate.

7. The Fermi level — the "water line" of electrons

8. Putting the conductivity law together

Formula The intrinsic conductivity law and its prefactor

$σ_{0}$ Combining "carrier count grows like $e^{- E_{g} /2 k_{B} T}$ " (section 5) with "conductivity is proportional to carrier count" (section 6) gives: $σ = σ_{0} e^{- E_{g} /2 k_{B} T}$ What is $σ_{0}$ ? It is the prefactor — a slowly-varying quantity that collects everything except the dominant exponential: the charge $e$ , the carrier mobilities $μ_{e}, μ_{h}$ , the constant $C$ , and crucially the $T^{3/2}$ density-of-states factor from section 5. So $σ_{0}$ is not a true constant — it drifts gently with temperature — but because the exponential changes so violently, we treat $σ_{0}$ as "roughly constant" over a modest temperature range. You can read it as the scale conductivity would have if the exponential barrier were switched off. Why $ln$ helps: taking logs (using the natural logarithm from section 5), $ln σ = ln σ_{0} - \frac{E _{g}}{2 k _{B}} \cdot \frac{1}{T}$ is a straight line if you plot $ln σ$ (vertical axis) against $1/ T$ (horizontal axis). Its slope is $- \frac{E _{g}}{2 k _{B}}$ , which hands you the band gap.

Worked example Extracting

$E_{g}$ from a real $ln σ$ vs $1/ T$ plot Problem: A silicon sample is measured at two temperatures:

At $T_{1} = 300 K$ : $σ_{1} = 4.0 \times 1 0^{- 4} S/m$
At $T_{2} = 350 K$ : $σ_{2} = 8.7 \times 1 0^{- 3} S/m$

Find the band gap $E_{g}$ .

Step 1 — write the line at both points. Since $ln σ = ln σ_{0} - \frac{E _{g}}{2 k _{B}} \cdot \frac{1}{T}$ , subtract the two equations so the unknown $ln σ_{0}$ cancels: $ln σ_{2} - ln σ_{1} = - \frac{E _{g}}{2 k _{B}} (\frac{1}{T _{2}} - \frac{1}{T _{1}}) .$ Why subtract? It kills the prefactor we don't know, leaving only $E_{g}$ .

Step 2 — put in numbers. $ln (σ_{2} / σ_{1}) = ln (8.7 \times 1 0^{- 3} /4.0 \times 1 0^{- 4}) = ln (21.75) = 3.08$ . The temperature factor: $\frac{1}{350} - \frac{1}{300} = - 4.76 \times 1 0^{- 4} K^{- 1}$ .

Step 3 — solve for $E_{g}$ . With $k_{B} = 8.62 \times 1 0^{- 5} eV/K$ : $E_{g} = - 2 k_{B} \cdot \frac{l n ( σ _{2} / σ _{1} )}{\frac{1}{T _{2}} - \frac{1}{T _{1}}} = - 2 (8.62 \times 1 0^{- 5}) \cdot \frac{3.08}{- 4.76 \times 1 0 ^{- 4}} \approx 1.12 eV .$ Conclusion: $E_{g} \approx 1.1 eV$ — exactly the accepted band gap of silicon. The slope of the $ln σ$ vs $1/ T$ line is the band-gap measurement.

How these foundations feed the topic

Equipment checklist

Cover the right side and answer out loud before revealing.

What is the prefactor $\s

What is the band gap

E_{g}

in plain words, and in symbols?

The height of forbidden empty space between the top of the valence band (

E_{V B}

) and the bottom of the conduction band (

E_{C B}

);

E_{g} = E_{C B} - E_{V B}

Why can a completely full band carry no current?

Every energy seat is occupied, so electrons have nowhere new to shift into when pushed by a voltage.

What is a hole, and what charge does it carry?

A missing electron in an otherwise-full band; it behaves as a mobile positive charge.

The letter

e

means two different things here — what are they?

e = 1.6 \times 1 0^{- 19} C

(magnitude of electron charge) when multiplying a concentration;

e \approx 2.718

(Euler's number) when it appears in an exponent

e^{(\dots)}

What does

k_{B} T

represent, and its value at room temperature?

The typical random thermal energy kick a particle gets; about

0.026 eV

300 K

Why is the exponent

E_{g} /2

and not

E_{g}

The Fermi level sits near mid-gap, so an electron climbs only

E_{g} /2

to the CB and the hole falls

E_{g} /2

to the VB — each carrier pays half the gap.

Why does the intrinsic carrier count carry a hidden

T^{3/2}

factor, and where does it go?

It comes from the density of states (how many band seats exist); it rides along inside the "slowly varying" prefactor

σ_{0}

Derive where

n \cdot p = n_{i}^{2}

comes from.

Write

n = N_{C} e^{- (E_{C B} - E_{F}) / k_{B} T}

and

p = N_{V} e^{- (E_{F} - E_{V B}) / k_{B} T}

; multiplying cancels

E_{F}

, giving

n p = N_{C} N_{V} e^{- E_{g} / k_{B} T}

, independent of doping, which in pure Si equals

n_{i}^{2}

Why does the conductivity formula use

+ e

(magnitude) for both carriers?

Electrons and holes drift opposite ways but both push current the same way; conductivity is a positive quantity, so we use the charge magnitude

e

and add both terms.

Convert

5 \times 1 0^{16} cm^{- 3}

to SI units.

5 \times 1 0^{16} cm^{- 3} \times 1 0^{6} = 5 \times 1 0^{22} m^{- 3}