Formation of a PN junction: Worked examples — Hardware deep dive

This page drills the formation of a PN junction with worked examples that hit every case: symmetric doping, asymmetric doping, temperature extremes, the degenerate " $n_{i} \to$ everything" limit, real-world word problems, and an exam twist. Each example is tagged with the matrix cell it covers.

Everything here uses two boxed results from the parent note. Let me re-state them once, in plain words, so no symbol is unearned.

The scenario matrix

Cell	Case class	What is stressed	Example
A	Symmetric doping $N_{A} = N_{D}$	Baseline $V_{bi}$ , sanity	Ex 1
B	Asymmetric doping $N_{A} \neq = N_{D}$	Which side is wider, $x_{p}$ vs $x_{n}$	Ex 2
C	Doping ratio changes only	Log dependence (slow growth)	Ex 3
D	Temperature swing (cold + hot)	Two competing $T$ effects	Ex 4
E	Degenerate limit $N_{A} N_{D} \to n_{i}^{2}$	$V_{bi} \to 0$ (no junction)	Ex 5
F	Real-world word problem	Reading a spec sheet → $V_{bi}$	Ex 6
G	Exam twist: reverse the formula	Given $V_{bi}$ , find the doping	Ex 7
H	Extreme asymmetry ( $p^{+} n$ )	One-sided depletion approximation	Ex 8

The figures below carry the geometry for cells B and H.

Cell A — Symmetric baseline

Worked example

Ex 1 (Cell A). Silicon at 300 K, $N_{A} = N_{D} = 1 0^{16} cm^{- 3}$ , $n_{i} = 1.5 \times 1 0^{10} cm^{- 3}$ . Find $V_{bi}$ . Forecast: Guess the number before computing. Silicon diodes "turn on" near 0.7 V — so bet on something in the 0.6–0.7 V range.

Step 1. Write the ratio inside the log. $\frac{N _{A} N _{D}}{n _{i}^{2}} = \frac{( 1 0 ^{16} ) ( 1 0 ^{16} )}{( 1.5 \times 1 0 ^{10} ) ^{2}} = \frac{1 0 ^{32}}{2.25 \times 1 0 ^{20}} = 4.44 \times 1 0^{11}$ Why this step? The formula only cares about how many dopant ions there are compared to the intrinsic background; that comparison is a pure ratio.

Step 2. Take the natural log. $ln (4.44 \times 1 0^{11}) \approx 26.82$ Why this step? Diffusion balances drift exponentially (from the $d n / n$ integral in the parent note), so undoing that exponential means taking a log.

Step 3. Multiply by $V_{T}$ . $V_{bi} = 0.0259 \times 26.82 \approx 0.695 V$ Why this step? $V_{T}$ converts the dimensionless log into an actual voltage.

Verify: 0.695 V sits squarely in the forecast 0.6–0.7 V band. Units: $[\frac{k T}{q}] = V$ times a dimensionless log $= V$ . ✅

Cell B — Asymmetric doping, which side is wider

Worked example

Ex 2 (Cell B). Now $N_{A} = 1 0^{15}$ (light p-side), $N_{D} = 1 0^{17}$ (heavy n-side), $n_{i} = 1.5 \times 1 0^{10}$ , 300 K. Find $V_{bi}$ and the ratio $x_{p} / x_{n}$ . Forecast: The product $N_{A} N_{D} = 1 0^{15} \cdot 1 0^{17} = 1 0^{32}$ is the same as Ex 1. So predict $V_{bi}$ is identical to Ex 1 (≈0.695 V) — the log only sees the product. For width: the lightly doped side (p) should be wider.

Step 1. Product and log: $\frac{N _{A} N _{D}}{n _{i}^{2}} = \frac{1 0 ^{32}}{2.25 \times 1 0 ^{20}} = 4.44 \times 1 0^{11}, V_{bi} = 0.0259 \times 26.82 \approx 0.695 V$ Why this step? Confirms the forecast — $V_{bi}$ depends only on the product, not the split.

Step 2. Use charge balance $N_{A} x_{p} = N_{D} x_{n}$ : $\frac{x _{p}}{x _{n}} = \frac{N _{D}}{N _{A}} = \frac{1 0 ^{17}}{1 0 ^{15}} = 100$ Why this step? Equal hidden charge on each side; the side with fewer ions per volume must stretch further to match. This is the geometry in the figure.

Verify: $x_{p} / x_{n} = 100 > 1$ , so the p-side (lightly doped) is 100× wider — matches the forecast. Sanity: putting numbers back, $N_{A} x_{p} = 1 0^{15} \cdot 100 x_{n} = 1 0^{17} x_{n} = N_{D} x_{n}$ . ✅

Cell C — Doping ratio, the slow log

Cell D — Temperature swing (two competing effects)

Worked example

Ex 4 (Cell D). For $N_{A} = N_{D} = 1 0^{16}$ , compare $V_{bi}$ using the front factor only (i.e. hold $n_{i}$ fixed at its 300 K value) at $T = 250$ K (cold) and $T = 350$ K (hot). $V_{T} = k T / q$ scales linearly, so $V_{T} (T) = 0.0259 \times \frac{T}{300}$ . Forecast: With $n_{i}$ artificially frozen, only $V_{T}$ moves. It is linear in $T$ , so predict cold < 0.695 V < hot, spread by the ratio $250/300$ and $350/300$ .

Step 1. Cold: $V_{T} (250) = 0.0259 \times \frac{250}{300} = 0.02158$ V. $V_{bi} (250) = 0.02158 \times 26.82 \approx 0.579 V$ Why this step? Same log (frozen $n_{i}$ ), scaled front factor.

Step 2. Hot: $V_{T} (350) = 0.0259 \times \frac{350}{300} = 0.03022$ V. $V_{bi} (350) = 0.03022 \times 26.82 \approx 0.810 V$ Why this step? Same log, larger front factor.

Verify: $0.579 < 0.695 < 0.810$ — ordered as forecast, and the ratio $0.810/0.579 = 1.40 = 350/250$ confirms the pure linear scaling. Reality caveat: once $n_{i} (T)$ is allowed to grow, the true $V_{bi}$ trend reverses and falls with heat — see the [!mistake] callout. This example isolates one factor on purpose. ✅

Cell E — Degenerate limit: the junction disappears

Worked example

Ex 5 (Cell E). Push doping down until $N_{A} N_{D} = n_{i}^{2}$ exactly (barely-doped, nearly intrinsic material — see Doping of Semiconductors). What is $V_{bi}$ ? Forecast: If p-side and n-side become almost identical to plain silicon, there is no concentration difference to drive diffusion — so guess $V_{bi} \to 0$ .

Step 1. Substitute $N_{A} N_{D} = n_{i}^{2}$ : $V_{bi} = V_{T} ln \frac{n _{i}^{2}}{n _{i}^{2}} = V_{T} ln (1)$ Why this step? This is the degenerate corner of the matrix — the doping ratio collapses to 1.

Step 2. $ln (1) = 0$ , so $V_{bi} = V_{T} \times 0 = 0 V$ Why this step? The log of 1 is zero because $e^{0} = 1$ ; no exponential imbalance means no barrier.

Verify: Zero built-in potential = no functioning junction. Physically correct: two identical materials touching build no field. This is the boundary case that tells you doping must differ for a diode to exist. ✅

Cell F — Real-world word problem

Worked example

Ex 6 (Cell F). A datasheet lists a germanium diode: p-region $N_{A} = 5 \times 1 0^{16} cm^{- 3}$ , n-region $N_{D} = 2 \times 1 0^{16} cm^{- 3}$ , and germanium's $n_{i} = 2.4 \times 1 0^{13} cm^{- 3}$ at 300 K. Estimate $V_{bi}$ . Forecast: Germanium has a much larger $n_{i}$ than silicon (that big $n_{i}$ sits in the denominator). Larger denominator → smaller ratio → smaller log → predict a lower barrier than silicon's 0.7 V — germanium diodes famously turn on near 0.3 V.

Step 1. Product / $n_{i}^{2}$ : $\frac{N _{A} N _{D}}{n _{i}^{2}} = \frac{( 5 \times 1 0 ^{16} ) ( 2 \times 1 0 ^{16} )}{( 2.4 \times 1 0 ^{13} ) ^{2}} = \frac{1 0 ^{33}}{5.76 \times 1 0 ^{26}} \approx 1.736 \times 1 0^{6}$ Why this step? Same recipe; the only villain is germanium's big $n_{i}$ .

Step 2. Log: $ln (1.736 \times 1 0^{6}) \approx 14.37$ Why this step? Undo the exponential imbalance as always.

Step 3. Times $V_{T}$ : $V_{bi} = 0.0259 \times 14.37 \approx 0.372 V$ Why this step? Convert to volts.

Verify: ≈0.37 V — in the classic germanium 0.3 V neighbourhood, well below silicon's 0.7 V, exactly as the forecast reasoned from the large $n_{i}$ . ✅

Cell G — Exam twist: run the formula backwards

Worked example

Ex 7 (Cell G). A silicon junction ( $n_{i} = 1.5 \times 1 0^{10}$ , 300 K) is measured to have $V_{bi} = 0.75$ V. It is symmetric ( $N_{A} = N_{D} = N$ ). Find $N$ . Forecast: 0.75 V is a bit above the $1 0^{16}$ baseline (0.695 V), so predict $N$ a little above $1 0^{16}$ .

Step 1. Invert the formula. Since $V_{bi} = V_{T} ln \frac{N ^{2}}{n _{i}^{2}}$ , $\frac{N ^{2}}{n _{i}^{2}} = e^{V_{bi} / V_{T}} = e^{0.75/0.0259} = e^{28.96}$ Why this step? $ln$ and $e^{(\cdot)}$ are inverse operations — to free the doping we apply $e$ to both sides. This is the "which number has this log?" move.

Step 2. Solve for $N$ : $N = n_{i} e^{V_{bi} / (2 V_{T})} = 1.5 \times 1 0^{10} \times e^{14.48}$ Why this step? $N^{2} = n_{i}^{2} e^{28.96}$ , so take the square root, which halves the exponent.

Step 3. Evaluate $e^{14.48} \approx 1.95 \times 1 0^{6}$ : $N \approx 1.5 \times 1 0^{10} \times 1.95 \times 1 0^{6} \approx 2.9 \times 1 0^{16} cm^{- 3}$ Why this step? Multiply out the numbers.

Verify: Feed $N = 2.9 \times 1 0^{16}$ forward: $\frac{N ^{2}}{n _{i}^{2}} = \frac{8.5 \times 1 0 ^{32}}{2.25 \times 1 0 ^{20}} = 3.8 \times 1 0^{12}$ , $ln = 29.0$ , $\times 0.0259 = 0.751$ V ✅. And $2.9 \times 1 0^{16} > 1 0^{16}$ — matches the forecast that we'd land just above baseline.

Cell H — Extreme asymmetry ( $p^{+} n$ ), one-sided depletion

Worked example

Ex 8 (Cell H). A $p^{+} n$ junction: $N_{A} = 1 0^{19}$ (very heavy p-side), $N_{D} = 1 0^{15}$ (light n-side), $n_{i} = 1.5 \times 1 0^{10}$ , 300 K. Find $V_{bi}$ and the fraction of the depletion width that lies in the n-side. Forecast: Product $= 1 0^{34}$ , larger than Ex 1's $1 0^{32}$ , so $V_{bi}$ a bit above 0.695 V. For geometry: the heavy p-side barely depletes; almost the entire depletion region sits in the light n-side — so predict the n-fraction ≈ 100%.

Step 1. $V_{bi}$ : $\frac{N _{A} N _{D}}{n _{i}^{2}} = \frac{1 0 ^{34}}{2.25 \times 1 0 ^{20}} = 4.44 \times 1 0^{13}, ln = 31.4, V_{bi} = 0.0259 \times 31.4 \approx 0.813 V$ Why this step? Standard recipe; product is 100× Ex 1, so $V_{bi}$ rises by $V_{T} ln 100 \approx 0.119$ V → $0.695 + 0.119 = 0.814$ V. Consistent.

Step 2. From $N_{A} x_{p} = N_{D} x_{n}$ , the n-side reach fraction: $\frac{x _{n}}{x _{n} + x _{p}} = \frac{x _{n}}{x _{n} + \frac{N _{D}}{N _{A}} x _{n}} = \frac{1}{1 + \frac{N _{D}}{N _{A}}} = \frac{1}{1 + 1 0 ^{- 4}} \approx 0.9999$ Why this step? $x_{p} = \frac{N _{D}}{N _{A}} x_{n}$ from charge balance; substitute and simplify.

Verify: ≈99.99% of the depletion width is in the lightly doped n-side — the "one-sided junction" approximation used in device design. Forecast confirmed. ✅

Recall Quick self-test

$V_{bi}$ depends on doping through which mathematical function? ::: The natural logarithm of the product $N_{A} N_{D}$ , so it grows very slowly. Same product, different split $N_{A}$ vs $N_{D}$ — does $V_{bi}$ change? ::: No. $V_{bi}$ depends only on the product $N_{A} N_{D}$ , not the split. Why does real $V_{bi}$ fall when the diode heats up? ::: Because $n_{i}$ (in the denominator) grows with $T$ faster than the front factor $k T / q$ grows. In a $p^{+} n$ junction, where does almost all the depletion width sit? ::: In the lightly doped n-side. What is $V_{bi}$ when $N_{A} N_{D} = n_{i}^{2}$ ? ::: Zero — no doping contrast, no junction.

See also: Diffusion and Drift Currents, PN Junction under Bias, Diode I-V Characteristics, Einstein Relation.