Formation of a PN junction: Exercises — Hardware deep dive

Throughout, we use the constants from the parent note at $T = 300 K$ :

Level 1 — Recognition

L1.1

State, for a p-type region before contact, which of these carries the fixed (immobile) charge and what its sign is: (a) holes, (b) acceptor ions, (c) free electrons.

Recall Solution

WHAT we're doing: sorting mobile vs fixed charge.

(a) Holes are mobile positive carriers — they can wander (diffuse/drift).
(b) ==Acceptor ions $N_{A}^{-}$ are the fixed, negative charge==. They are dopant atoms locked into the crystal lattice; they cannot move.
(c) Free electrons are mobile (minority carriers on the p-side).

Answer: (b), and its sign is negative. WHY it matters: only the fixed ions can build the permanent field, because mobile carriers get swept away.

L1.2

The built-in electric field $E$ inside the depletion region points from ____ to ____. Fill the blanks and say what it does to a stray electron sitting in that region.

Recall Solution

The field points from the n-side to the p-side (from the exposed $+$ donor ions toward the exposed $-$ acceptor ions — field lines run from $+$ to $-$ ). A stray electron feels force $F = - q E$ , i.e. opposite to $E$ , so it is pushed from p back toward n. The field is a restoring force that undoes diffusion.

Level 2 — Application

L2.1

A silicon junction has $N_{A} = N_{D} = 1 0^{16} cm^{- 3}$ . Compute $V_{bi}$ at 300 K.

Recall Solution

Why this formula? $V_{bi}$ measures how big the potential hill must be so that drift exactly cancels diffusion — that is exactly what $V_{T} ln (N_{A} N_{D} / n_{i}^{2})$ encodes. $V_{bi} = 0.0259 ln \frac{( 1 0 ^{16} ) ( 1 0 ^{16} )}{( 1.5 \times 1 0 ^{10} ) ^{2}} = 0.0259 ln (4.444 \times 1 0^{11})$ $= 0.0259 \times 26.82 \approx 0.695 V$ Why it makes sense: the classic silicon barrier ≈ 0.6–0.7 V — the reason a Si diode "turns on" near 0.7 V.

L2.2

Keep $N_{D} = 1 0^{16}$ but raise $N_{A}$ to $1 0^{17} cm^{- 3}$ . By how much does $V_{bi}$ change compared to L2.1?

Recall Solution

WHAT changes: the argument of the log is multiplied by $10$ (since $N_{A}$ went up $10 \times$ ). WHY: $ln (10 \cdot X) = ln X + ln 10$ , so the extra piece is $Δ V_{bi} = V_{T} ln 10 = 0.0259 \times 2.3026 \approx 0.0596 V .$ New value: $V_{bi} \approx 0.695 + 0.060 = 0.755 V$ . Why it makes sense: doping enters only through a logarithm, so a $10 \times$ change in doping shifts the barrier by only $\approx 60 mV$ . The barrier is stubborn.

Level 3 — Analysis

L3.1

A one-sided junction has $N_{A} = 1 0^{15} cm^{- 3}$ (lightly doped p) and $N_{D} = 1 0^{18} cm^{- 3}$ (heavily doped n). Using $N_{A} x_{p} = N_{D} x_{n}$ , find the ratio $x_{p} / x_{n}$ . Which side is wider, and does that match the forecast in the parent note?

Recall Solution

Why this equation? The total exposed negative ionic charge on the p-side must equal the total positive ionic charge on the n-side (the depletion region is neutral overall). With uniform doping, charge $\propto$ (doping) × (width), giving $N_{A} x_{p} = N_{D} x_{n}$ . $\frac{x _{p}}{x _{n}} = \frac{N _{D}}{N _{A}} = \frac{1 0 ^{18}}{1 0 ^{15}} = 1000 .$ So the lightly doped p-side is 1000× wider. See the figure: few ions on the light side means you must expose them over a long stretch to match the charge.

Matches the forecast? Yes — the parent note predicted the lightly doped side is wider. ✅

L3.2

For that same junction, what fraction of the total depletion width $W = x_{p} + x_{n}$ lies on the n-side?

Recall Solution

WHAT: we want $x_{n} / (x_{p} + x_{n})$ . Use $x_{p} = 1000 x_{n}$ from L3.1: $\frac{x _{n}}{x _{p} + x _{n}} = \frac{x _{n}}{1000 x _{n} + x _{n}} = \frac{1}{1001} \approx 0.000999 \approx 0.0999% .$ Why it makes sense: almost the entire depletion region sits in the lightly doped p-side. This is the defining feature of a one-sided junction — the heavily doped side barely depletes at all.

Level 4 — Synthesis

L4.1

A silicon junction has $N_{A} = 5 \times 1 0^{16}$ and $N_{D} = 2 \times 1 0^{17} cm^{- 3}$ . (a) Find $V_{bi}$ at 300 K. (b) You now heat the same junction. Ignoring the (weak) change in $n_{i}$ for a moment, does the prefactor $V_{T} = k T / q$ push $V_{bi}$ up or down as $T$ rises? (Answer directionally, prefactor only.)

Recall Solution

(a) Plug in: $V_{bi} = 0.0259 ln \frac{( 5 \times 1 0 ^{16} ) ( 2 \times 1 0 ^{17} )}{( 1.5 \times 1 0 ^{10} ) ^{2}} = 0.0259 ln \frac{1 0 ^{34}}{2.25 \times 1 0 ^{20}}$ $= 0.0259 ln (4.444 \times 1 0^{13}) = 0.0259 \times 31.43 \approx 0.814 V .$ (b) The prefactor $V_{T} = k T / q$ grows linearly with $T$ , so by itself it pushes $V_{bi}$ up. BUT — the honest full picture: $n_{i}^{2}$ in the denominator grows exponentially with $T$ , which shrinks the log fast. The exponential wins, so real silicon $V_{bi}$ actually decreases with temperature (about $- 2 mV/K$ ). The question asked prefactor-only, so the answer to (b) is "up," with this caveat noted. This is exactly why PN Junction under Bias and Diode I-V Characteristics are temperature-sensitive.

L4.2

Two diodes A and B are made of the same silicon at the same $T$ . Diode A has $V_{bi, A} = 0.70 V$ . Diode B is doped so that both $N_{A}$ and $N_{D}$ are $100 \times$ larger than in A. Find $V_{bi, B}$ without re-plugging the full formula — reason from logs.

Recall Solution

Why we can shortcut: the product $N_{A} N_{D}$ multiplies by $100 \times 100 = 1 0^{4}$ . Inside the log, $Δ V_{bi} = V_{T} ln (1 0^{4}) = 0.0259 \times 4 ln 10 = 0.0259 \times 9.210 \approx 0.2385 V .$ $V_{bi, B} = 0.70 + 0.2385 \approx 0.938 V .$ Why it makes sense: even a $1 0^{4}$ -fold doping increase adds under a quarter volt. The log keeps $V_{bi}$ tightly bounded — this is why silicon barriers cluster near 0.6–0.9 V no matter how you dope.

Level 5 — Mastery

L5.1

Design task. You want a silicon junction at 300 K with $V_{bi} = 0.75 V$ exactly, using a symmetric doping $N_{A} = N_{D} = N$ . Solve for $N$ .

Recall Solution

WHAT we do: invert the formula. With $N_{A} = N_{D} = N$ , the product is $N^{2}$ : $V_{bi} = V_{T} ln \frac{N ^{2}}{n _{i}^{2}} .$ Step 1 — divide out $V_{T}$ : $\frac{V _{bi}}{V _{T}} = \frac{0.75}{0.0259} = 28.96 = ln \frac{N ^{2}}{n _{i}^{2}} .$ Step 2 — undo the log with $exp$ (the log's inverse — that's why we use it): $\frac{N ^{2}}{n _{i}^{2}} = e^{28.96} = 3.79 \times 1 0^{12} .$ Step 3 — solve for $N$ : $N = n_{i} 3.79 \times 1 0^{12} = (1.5 \times 1 0^{10}) (1.947 \times 1 0^{6}) \approx 2.9 \times 1 0^{16} cm^{- 3} .$ Sanity check: plug back — $V_{T} ln ((2.9 \times 1 0^{16})^{2} / n_{i}^{2}) \approx 0.750 V$ . ✅

L5.2

Full chain. A one-sided junction has $N_{A} = 1 0^{15}$ (p, light) and $N_{D} = 1 0^{19} cm^{- 3}$ (n, heavy) at 300 K. (a) Compute $V_{bi}$ . (b) State the ratio $x_{p} / x_{n}$ . (c) In one sentence, explain physically why almost the entire built-in voltage drop and depletion width live on the p-side, tying the two results together.

Recall Solution

(a) $V_{bi} = 0.0259 ln \frac{( 1 0 ^{15} ) ( 1 0 ^{19} )}{( 1.5 \times 1 0 ^{10} ) ^{2}} = 0.0259 ln \frac{1 0 ^{34}}{2.25 \times 1 0 ^{20}}$ $= 0.0259 ln (4.444 \times 1 0^{13}) = 0.0259 \times 31.43 \approx 0.814 V .$ (b) $\frac{x _{p}}{x _{n}} = \frac{N _{D}}{N _{A}} = \frac{1 0 ^{19}}{1 0 ^{15}} = 1 0^{4}$ — the p-side is $10000 \times$ wider. (c) Because charge must balance ( $N_{A} x_{p} = N_{D} x_{n}$ ), the sparsely-doped p-side must stretch enormously to expose enough ions; the field and voltage therefore develop almost entirely across that long p-side stretch, while the heavily-doped n-side barely depletes. This is precisely the "one-sided junction" approximation used when analysing real diodes in PN Junction under Bias.

Recall One-line self-check before you leave

Bigger doping ::: bigger $V_{bi}$ , but only logarithmically (barrier is stubborn). Lighter-doped side ::: wider depletion region (from $N_{A} x_{p} = N_{D} x_{n}$ ). Real $V_{bi}$ vs temperature ::: decreases (the $n_{i}^{2}$ exponential beats the $V_{T}$ prefactor). To invert $V_{bi} = V_{T} ln (\cdot)$ ::: divide by $V_{T}$ , then exponentiate, then solve.