Built-in potential of a junction: Exercises — Hardware deep dive

Before we start, three plain-language reminders so no symbol is used unexplained:

Definition The concentrations, in words

$N_{A}$ = acceptor doping on the p-side: how many "hole factories" per cm³ we deliberately added. Units cm⁻³ (things per cubic centimetre).
$N_{D}$ = donor doping on the n-side: how many "electron factories" per cm³.
$n_{i}$ = intrinsic carrier concentration: how many electrons (and equally holes) a pure undoped crystal makes by itself from heat alone. See Intrinsic carrier concentration.
$n$ = electron concentration (free electrons per cm³) at some spot; $p$ = hole concentration (holes per cm³) at that spot. A "hole" is a missing electron that behaves like a positive mobile charge.
When we want the electron count specifically on the n-side we write $n_{n}$ ; specifically on the p-side we write $n_{p}$ . (Same letter $n$ , small subscript telling you which side.) Likewise $p_{n}$ , $p_{p}$ for holes if needed.
$V_{T} = k T / q$ = thermal voltage, the natural voltage scale set by temperature. At room temperature it is $\approx 25.85$ mV. Read " $k$ " as Boltzmann's constant (energy per degree), " $T$ " as absolute temperature in kelvin, " $q$ " as the charge on one electron.

Level 1 — Recognition

Can you pick the right formula and read off the pieces?

Problem 1.1

State, in one sentence each, what physical quantity $N_{A}$ , $N_{D}$ , and $n_{i}$ each represent, and give the value of $V_{T}$ at 300 K.

Recall Solution 1.1

$N_{A}$ = acceptor (p-side) doping density, in cm⁻³.
$N_{D}$ = donor (n-side) doping density, in cm⁻³.
$n_{i}$ = intrinsic carrier concentration of the pure semiconductor, in cm⁻³.
$V_{T} = k T / q \approx 0.02585$ V $= 25.85$ mV at 300 K.

Which of these is the correct built-in potential formula? (a) $V_{bi} = V_{T} \frac{N _{A} N _{D}}{n _{i}^{2}}$ (b) $V_{bi} = V_{T} ln (\frac{N _{A} N _{D}}{n _{i}^{2}})$ (c) $V_{bi} = V_{T} ln (\frac{n _{i}^{2}}{N _{A} N _{D}})$

Recall Solution 1.2

(b) is correct.

(a) drops the logarithm — that would make $V_{bi}$ astronomically large (a bare ratio like $1 0^{13}$ volts). Physically wrong.
(c) flips the ratio, giving a negative $V_{bi}$ (the log of a number below 1 is negative). The larger concentrations $N_{A}, N_{D}$ belong on top because carriers are more concentrated than intrinsic. The picture: the log measures how many "e-foldings" of concentration difference the carriers had to climb; $V_{T}$ converts each e-folding into volts.

Level 2 — Application

Plug numbers in cleanly and get the number out.

Problem 2.1

Silicon at 300 K, $N_{A} = 1 0^{18} cm^{- 3}$ , $N_{D} = 1 0^{15} cm^{- 3}$ , $n_{i} = 1.0 \times 1 0^{10} cm^{- 3}$ . Find $V_{bi}$ .

Recall Solution 2.1

Argument $= \frac{1 0 ^{18} \times 1 0 ^{15}}{( 1 0 ^{10} ) ^{2}} = \frac{1 0 ^{33}}{1 0 ^{20}} = 1 0^{13}$ . $V_{bi} = 0.02585 \times ln (1 0^{13}) = 0.02585 \times 13 \times 2.302585 = 0.02585 \times 29.93 \approx 0.774 V$ Squarely in the 0.6–0.8 V silicon window ✓. Notice the answer is identical to the parent note's Example 1 even though the individual dopings differ — because only the product $N_{A} N_{D} = 1 0^{33}$ matters.

Problem 2.2

Same silicon, but now $N_{A} = N_{D} = 5 \times 1 0^{16} cm^{- 3}$ . Find $V_{bi}$ .

Recall Solution 2.2

Product $= 2.5 \times 1 0^{33}$ . Divide by $n_{i}^{2} = 1 0^{20}$ : argument $= 2.5 \times 1 0^{13}$ . $ln (2.5 \times 1 0^{13}) = ln (2.5) + 13 ln (10) = 0.9163 + 29.934 = 30.85$ $V_{bi} = 0.02585 \times 30.85 \approx 0.798 V$

Problem 2.3

A germanium junction ( $n_{i} = 2.4 \times 1 0^{13} cm^{- 3}$ ) has $N_{A} = N_{D} = 1 0^{17} cm^{- 3}$ . Find $V_{bi}$ .

Recall Solution 2.3

Product $= 1 0^{34}$ ; $n_{i}^{2} = (2.4 \times 1 0^{13})^{2} = 5.76 \times 1 0^{26}$ . Argument $= 1 0^{34} /5.76 \times 1 0^{26} = 1.736 \times 1 0^{7}$ . $ln (1.736 \times 1 0^{7}) = ln (1.736) + 7 ln 10 = 0.5516 + 16.118 = 16.67$ $V_{bi} = 0.02585 \times 16.67 \approx 0.431 V$ Higher than the parent's Example 3 (0.312 V) because we doped 10× harder on both sides — but still far below silicon's ~0.7 V, thanks to Ge's big $n_{i}$ .

Level 3 — Analysis

Reason about how $V_{bi}$ responds to changes; the log is the whole story.

Problem 3.1

Starting from Problem 2.1 ( $V_{bi} = 0.774$ V), you increase both $N_{A}$ and $N_{D}$ by a factor of 10. Without a calculator, predict the new $V_{bi}$ , then confirm.

Recall Solution 3.1

Multiplying each doping by 10 multiplies the product by 100, i.e. adds two powers of ten to the argument. Each power of ten adds $V_{T} ln 10 = 0.02585 \times 2.302585 = 0.0595$ V $= 59.5$ mV. $Δ V_{bi} = 2 \times 59.5 mV = 119 mV$ $V_{bi}^{new} = 0.774 + 0.119 = 0.893 V$ Check directly: argument $= 1 0^{15}$ , $V_{bi} = 0.02585 \times 15 \times 2.302585 = 0.893$ V ✓.

Problem 3.2

Look at Figure 1. The blue curve is $V_{bi}$ against $lo g_{10} (N_{A} N_{D})$ for silicon, with a dashed red line drawn at $V_{bi} = 1.0$ V. Using only the graph (a ruler, not the formula): (i) explain why "doping harder" is a losing strategy, and (ii) read off, straight from the horizontal axis, the log-doping value where the blue curve crosses the red 1.0 V line, and convert that to a doping product.

Figure — Built-in potential of a junction

Recall Solution 3.2

(i) Why it is a losing strategy — read the slope off the picture. Pick any two points one gridline apart on the horizontal axis (one decade of doping product). The blue curve rises by only a tiny vertical step — the green label on the figure marks it: $\approx 59.6$ mV per decade. Because the curve is a straight line against log-doping, that step never grows. So to climb even 0.1 V you must slide $\sim 2$ decades to the right, i.e. multiply the doping product by nearly $100 \times$ . Each extra sliver of voltage costs an exponential amount of doping — that is the losing strategy, visible as the graph's stubbornly shallow slope.

(ii) Read the crossing point directly. Follow the dashed red $V_{bi} = 1.0$ V line rightward until it meets the blue curve, then drop straight down to the horizontal axis. The foot lands near $lo g_{10} (N_{A} N_{D}) \approx 36.8$ . Converting: $N_{A} N_{D} \approx 1 0^{36.8} \approx 6 \times 1 0^{36} cm^{- 6}$ — e.g. both sides near $2.5 \times 1 0^{18}$ cm⁻³.

Cross-check (only after reading): algebra gives $ln (arg) = 1.0/0.02585 = 38.7$ , arg $= e^{38.7} \approx 6 \times 1 0^{16}$ , and with $n_{i}^{2} = 1 0^{20}$ that is $N_{A} N_{D} = 6 \times 1 0^{36}$ — the same value the ruler gave. This doping is already brushing the degenerate limit where the formula breaks (see L5).

Problem 3.3

A silicon junction has $V_{bi} = 0.72$ V at 300 K with $n_{i} = 1 0^{10}$ cm⁻³. What is the doping product $N_{A} N_{D}$ ?

Recall Solution 3.3

Invert the formula. From $V_{bi} = V_{T} ln (N_{A} N_{D} / n_{i}^{2})$ : $\frac{N _{A} N _{D}}{n _{i}^{2}} = exp (\frac{V _{bi}}{V _{T}}) = exp (\frac{0.72}{0.02585}) = e^{27.85} = 1.24 \times 1 0^{12}$ $N_{A} N_{D} = 1.24 \times 1 0^{12} \times (1 0^{10})^{2} = 1.24 \times 1 0^{32} cm^{- 6}$ Why exponentiate? $ln$ and $exp$ are inverse operations — $exp$ "undoes" the log to release the ratio, exactly like $arctan$ undoes $tan$ .

Level 4 — Synthesis

Combine $V_{bi}$ with a second concept: temperature, the mass-action law, or band diagrams.

Problem 4.1 (temperature)

For a silicon junction with $N_{A} = N_{D} = 1 0^{16}$ cm⁻³, $V_{bi} = 0.714$ V at 300 K. Silicon's $n_{i}$ roughly doubles for every 11 °C rise. Estimate $V_{bi}$ at 322 K (a 22 °C rise) using this rule. (Take $V_{T}$ scaling with $T$ into account too.)

Recall Solution 4.1

Two things change: $V_{T}$ (rises with $T$ ) and $n_{i}$ (rises fast). We handle both.

Step A — new $V_{T}$ . $V_{T} = k T / q \propto T$ , so $V_{T} (322) = 0.02585 \times \frac{322}{300} = 0.02775$ V.

Step B — new $n_{i}$ . A 22 °C rise is two doublings ( $22/11 = 2$ ), so $n_{i}$ grows by $2^{2} = 4 \times$ . Then $n_{i}^{2}$ grows by $16 \times$ , shrinking the argument by $16 \times$ . Original argument $= 1 0^{12}$ → new argument $= 1 0^{12} /16 = 6.25 \times 1 0^{10}$ . $ln (6.25 \times 1 0^{10}) = ln 6.25 + 10 ln 10 = 1.8326 + 23.026 = 24.86$

Step C — combine. $V_{bi} (322) = 0.02775 \times 24.86 \approx 0.690 V$ So a 22 °C rise dropped $V_{bi}$ from 0.714 to ≈0.690 V — about $- 24$ mV, i.e. $\approx - 1.1$ mV/°C. Same direction as the textbook " $- 2$ mV/°C" rule (the shrinking- $n_{i}$ effect beats the rising $V_{T}$ ), a little gentler because our doubling rule is approximate. Note the rising $V_{T}$ prefactor did not rescue $V_{bi}$ : the log shrinks faster than $V_{T}$ grows.

Problem 4.2 (mass-action law)

On the p-side of Problem 2.1 ( $N_{A} = 1 0^{18}$ , $n_{i} = 1 0^{10}$ ), compute the minority-electron concentration $n_{p}$ (recall $n_{p}$ = electrons per cm³ on the p-side). Then reconstruct $V_{bi}$ from the ratio $n_{n} / n_{p}$ , where $n_{n}$ = electrons per cm³ on the n-side, and confirm it equals the doping-form answer.

Recall Solution 4.2

The Mass-action law says $n p = n_{i}^{2}$ everywhere in equilibrium, where $n$ is the local electron concentration and $p$ the local hole concentration. On the p-side holes are the majority carrier, so the hole concentration there is $p_{p} \approx N_{A} = 1 0^{18}$ . Applying the law on that side: $n_{p} = \frac{n _{i}^{2}}{p _{p}} = \frac{n _{i}^{2}}{N _{A}} = \frac{( 1 0 ^{10} ) ^{2}}{1 0 ^{18}} = \frac{1 0 ^{20}}{1 0 ^{18}} = 1 0^{2} cm^{- 3}$ On the n-side electrons are the majority carrier, so $n_{n} \approx N_{D} = 1 0^{15}$ . $V_{bi} = V_{T} ln (\frac{n _{n}}{n _{p}}) = 0.02585 ln (\frac{1 0 ^{15}}{1 0 ^{2}}) = 0.02585 ln (1 0^{13}) = 0.774 V$ Identical to Problem 2.1 ✓. This shows the two forms of the formula are the same equation — the doping form is just the carrier form with $n_{n} = N_{D}$ and $n_{p} = n_{i}^{2} / N_{A}$ substituted in.

Problem 4.3 (band diagram)

The built-in potential also equals the energy separation of the two Fermi levels (before contact) divided by $q$ . In Figure 2 the left panel shows the two crystals apart; the right panel shows them joined. Using the left panel's numbers — the n-side Fermi level sits $0.15$ eV below the conduction band, the p-side Fermi level sits $0.20$ eV above the valence band, and the silicon bandgap is $E_{g} = 1.12$ eV — find $V_{bi}$ , then check that it matches the vertical slide drawn in the right panel.

Recall Solution 4.3

Read the left panel first (Energy band diagrams). The green double-arrow between the two dashed Fermi levels is the quantity we want, $q V_{bi}$ . To turn the pictured energies into a number, walk down the full bandgap and subtract the two "already-used" slivers marked on the figure: $q V_{bi} = 1.12 E_{g} - 0.15 (n-side E_{c} - E_{F}) - 0.20 (p-side E_{F} - E_{v})$ $q V_{bi} = 1.12 - 0.15 - 0.20 = 0.77 eV \Rightarrow V_{bi} = 0.77 V$ Now check the right panel. When the crystals touch, the Fermi level must become flat (equilibrium), so the entire n-side band structure slides down. The red down-arrow in the right panel labels that slide as $q V_{bi} = 0.77$ eV — exactly the number we computed from the left panel. Two panels, one equilibrium: the band picture and the diffusion-drift picture agree, and both land near our ~0.7 V doping answers.

Level 5 — Mastery

Edge cases, degenerate limits, and traps the formula silently hides.

Problem 5.1 (upper bound)

Show that $V_{bi}$ can never reach the bandgap voltage $E_{g} / q$ . For silicon ( $E_{g} / q = 1.12$ V, $n_{i} = 1 0^{10}$ ), what doping product would formally be required to hit exactly 1.12 V, and why is that unphysical?

Recall Solution 5.1

Set $V_{bi} = 1.12$ : $\frac{N _{A} N _{D}}{n _{i}^{2}} = e^{1.12/0.02585} = e^{43.33} = 6.6 \times 1 0^{18}$ $N_{A} N_{D} = 6.6 \times 1 0^{18} \times 1 0^{20} = 6.6 \times 1 0^{38} cm^{- 6}$ That requires each side near $6.6 \times 1 0^{38} \approx 2.6 \times 1 0^{19}$ cm⁻³. But silicon has only $\sim 5 \times 1 0^{22}$ atoms/cm³, and beyond $\sim 1 0^{19}$ cm⁻³ the doping becomes degenerate: the Fermi level enters the band, the mass-action relation $n p = n_{i}^{2}$ fails, and the majority carrier count saturates below $N_{D}$ . So the formula's own assumptions ( $n_{n} \approx N_{D}$ , non-degenerate) break before $V_{bi}$ can reach $E_{g} / q$ . The gap voltage is a physical ceiling: the two Fermi levels can slide apart by at most $E_{g}$ (one hits $E_{c}$ , the other hits $E_{v}$ ).

Problem 5.2 (degenerate limit — the trap side)

A student plugs $N_{A} = N_{D} = 1 0^{20}$ cm⁻³ silicon into the formula. Carry out the correct plug-in, then explain in one paragraph why the resulting number is not trustworthy.

Recall Solution 5.2

Naive plug-in: argument $= \frac{1 0 ^{20} \times 1 0 ^{20}}{( 1 0 ^{10} ) ^{2}} = \frac{1 0 ^{40}}{1 0 ^{20}} = 1 0^{20}$ , so $V_{bi} = 0.02585 \times ln (1 0^{20}) = 0.02585 \times 20 \times 2.302585 = 0.02585 \times 46.05 \approx 1.19 V .$ (If a student reports something smaller like 0.95 V, they slipped an arithmetic step — but even the correct 1.19 V is untrustworthy, and that is the real lesson.) Here is why: at $1 0^{20}$ cm⁻³ the doping is far above silicon's degenerate threshold ( $\sim 1 0^{19}$ cm⁻³). In that regime three of the derivation's pillars collapse at once — (1) $n_{n}$ no longer equals $N_{D}$ , because band-filling caps the number of free electrons the conduction band can hold; (2) the mass-action relation $n p = n_{i}^{2}$ no longer holds, because it too assumes non-degenerate Boltzmann statistics; and (3) the Boltzmann $ln$ we derived must be replaced by Fermi–Dirac integrals plus a bandgap-narrowing correction. So the clean $V_{T} ln (\cdot)$ formula, built on non-degenerate, fully-ionised assumptions, produces a 1.19 V figure the real device never actually reaches. The number is a mirage the formula cannot warn you about.

Problem 5.3 (zero-doping / intrinsic limit)

What does the formula predict if one side is undoped (intrinsic), i.e. $N_{D} \to n_{i}$ and $N_{A} \to n_{i}$ ? Interpret the result physically.

Recall Solution 5.3

If both sides approach intrinsic, $N_{A} \to n_{i}$ and $N_{D} \to n_{i}$ , so $\frac{N _{A} N _{D}}{n _{i}^{2}} \to \frac{n _{i} \cdot n _{i}}{n _{i}^{2}} = 1, V_{bi} \to V_{T} ln (1) = 0$ Physically: with no doping there is no concentration difference across the join, so no net diffusion, so no exposed ion charge, so no field, so no potential. $V_{bi} = 0$ is exactly right — the built-in potential is powered entirely by the doping contrast. This is also the "degenerate input" sanity check the contract demands: at the trivial (equal-concentration) input the formula correctly returns zero, not a division-by-zero or a negative value.

Problem 5.4 (asymmetry synthesis)

Two silicon diodes have the same product $N_{A} N_{D} = 1 0^{33}$ but different splits: Diode A is $N_{A} = N_{D} = 1 0^{33} \approx 3.16 \times 1 0^{16}$ ; Diode B is $N_{A} = 1 0^{18}, N_{D} = 1 0^{15}$ . Do they have the same $V_{bi}$ ? Which has the wider depletion region on the lightly-doped side?

Recall Solution 5.4

Same $V_{bi}$ : the formula depends only on the product $N_{A} N_{D}$ , which is $1 0^{33}$ for both. So both give $V_{bi} = 0.02585 ln (1 0^{33} /1 0^{20}) = 0.02585 ln (1 0^{13}) = 0.774 V .$ Identical. Depletion width differs: the Depletion region width pushes further into the lightly-doped side (fewer ions there, so it must expose a thicker slab to balance the charge). Diode B, with its very light $N_{D} = 1 0^{15}$ n-side, has a much wider n-side depletion region than symmetric Diode A. So: same built-in voltage, very different internal geometry. $V_{bi}$ alone cannot tell you the shape of the junction.

Recall One-line self-test before you leave

Q: A junction's doping product goes up by exactly $1000 \times$ . By how much does $V_{bi}$ rise? Verify: $1000 = 1 0^{3}$ → three decades → $3 \times 59.6 mV \approx 178.7$ mV. (Independent of the starting value — pure logarithm.)

Connections

Built-in potential of a junction — parent
Einstein relation, Mass-action law, Intrinsic carrier concentration — tools used above
Depletion region width, PN junction under bias, Diode I-V equation, Energy band diagrams — where $V_{bi}$ goes next

Exercises — Built-in potential of a junction