Minority vs majority carriers: Exercises — Hardware deep dive

Intuition How to use this page

Work each problem before opening its solution. The problems climb a ladder: from recognising which carrier is which (L1) up to mastering the exact quadratic and compensated cases (L5). Every solution is inside a collapsible callout so you can self-test. All numeric answers are machine-checked.

Throughout: $n$ = electron concentration, $p$ = hole concentration, $n_{i}$ = intrinsic carrier concentration, $N_{D}$ = donor density, $N_{A}$ = acceptor density. Two master tools: $np = n_i^2$ and the charge-neutrality condition $p + N_{D} = n + N_{A}$ . Unless told otherwise, assume $T = 300 K$ , silicon with $n_{i} = 1.0 \times 1 0^{10} cm^{- 3}$ , and full ionization.

(A note on the highlighting below: words wrapped in ==double equals== are the key answer terms — in Obsidian they render as a coloured highlight. Read them simply as the emphasised answer.)

Level 1 — Recognition

Can you read the doping and name the carriers?

Recall Solution L1.1

(a) A donor donates a spare electron. Extra electrons → n-type. What we did: matched the dopant group to the carrier it releases. Why: donors sit just below the conduction band and give up electrons at room temperature. (b) In n-type the abundant carrier is electrons (majority); the rare one is holes (minority).

Recall Solution L1.2

Boron is an acceptor — it grabs an electron, leaving a hole. Extra holes → p-type. Majority = holes ( $p$ ), minority = electrons ( $n$ ).

Recall Solution L1.3

False. By the law of mass action $n p = n_{i}^{2}$ , and $n_{i}$ depends only on temperature (and the material), not on doping. Doping shifts $n$ and $p$ in opposite directions but their product is pinned.

Level 2 — Application

Plug into $n \approx N_{D}$ and $p = n_{i}^{2} / N_{D}$ .

Recall Solution L2.1

Step 1 (majority). Check $N_{D} ≫ n_{i}$ : $2 \times 1 0^{16} ≫ 1 0^{10}$ ✓. So $n \approx N_{D} = 2.0 \times 1 0^{16} cm^{- 3}$ . Why: with so many donors, the tiny hole term in neutrality is negligible, leaving $n \approx N_{D}$ . Step 2 (minority). Mass action forces the product to stay $n_{i}^{2}$ : $p = \frac{n _{i}^{2}}{n} = \frac{( 1.0 \times 1 0 ^{10} ) ^{2}}{2.0 \times 1 0 ^{16}} = \frac{1.0 \times 1 0 ^{20}}{2.0 \times 1 0 ^{16}} = 5.0 \times 1 0^{3} cm^{- 3} .$

Recall Solution L2.2

Majority: $p \approx N_{A} = 4.0 \times 1 0^{17} cm^{- 3}$ (since $N_{A} ≫ n_{i}$ ). Minority: $n = \frac{n _{i}^{2}}{p} = \frac{1 0 ^{20}}{4.0 \times 1 0 ^{17}} = 2.5 \times 1 0^{2} cm^{- 3}$ .

Recall Solution L2.3

$\frac{n}{p} = \frac{2.0 \times 1 0 ^{16}}{5.0 \times 1 0 ^{3}} = 4.0 \times 1 0^{12} .$ Roughly twelve to thirteen orders of magnitude more electrons than holes — the majority utterly dominates the count, even though the minority runs the junction.

Level 3 — Analysis

When approximations break, and compensated material.

Recall Solution L3.1

Step 1 — net dopant. Start from neutrality $p + N_{D} = n + N_{A}$ . Subtract $n$ from both sides and subtract $N_{D}$ from both sides: $p + N_{D} = n + N_{A} ⟹ p - n = N_{A} - N_{D} .$ (Explicitly: moving $n$ left gives $p - n + N_{D} = N_{A}$ ; then moving $N_{D}$ right gives $p - n = N_{A} - N_{D}$ .) The right-hand side is the net acceptor $= 6.0 \times 1 0^{16} - 1.0 \times 1 0^{16} = 5.0 \times 1 0^{16} cm^{- 3} > 0$ → holes exceed electrons → p-type. Why they cancel: each ionized donor supplies an electron that fills an acceptor's hole; only the excess acceptors leave free holes. Step 2 — majority (explicit "why"). We have exactly $p - n = N_{A} - N_{D}$ . The minority $n$ will turn out tiny (next step confirms $n \sim 1 0^{3}$ ), while $N_{A} - N_{D} = 5.0 \times 1 0^{16}$ . Since $n ≪ N_{A} - N_{D}$ , we neglect $n$ in $p - n = N_{A} - N_{D}$ , giving $p \approx N_{A} - N_{D} = 5.0 \times 1 0^{16} cm^{- 3} .$ (This neglect is self-consistent: we drop $n$ , compute $p$ , then verify $n$ really is negligible.) Step 3 — minority. $n = \frac{n _{i}^{2}}{p} = \frac{1 0 ^{20}}{5.0 \times 1 0 ^{16}} = 2.0 \times 1 0^{3} cm^{- 3}$ . Indeed $2.0 \times 1 0^{3} ≪ 5.0 \times 1 0^{16}$ , so neglecting $n$ in Step 2 was justified. ✓

Recall Solution L3.2

Here $N_{D}$ is comparable to $n_{i}$ , so $n \approx N_{D}$ would be wrong. Step 1 — set up two equations. Neutrality (no acceptors): $n = p + N_{D}$ . Mass action: $n p = n_{i}^{2} \Rightarrow p = n_{i}^{2} / n$ . Step 2 — substitute to get one equation in $n$ : $n = \frac{n _{i}^{2}}{n} + N_{D} \Rightarrow n^{2} - N_{D} n - n_{i}^{2} = 0.$ Step 3 — quadratic formula (take the + root, since $n > 0$ ): $n = \frac{N _{D}}{2} + (\frac{N _{D}}{2})^{2} + n_{i}^{2} .$ Step 4 — numbers. $\frac{N _{D}}{2} = 1.5 \times 1 0^{13}$ , and $(1.5 \times 1 0^{13})^{2} + (2.4 \times 1 0^{13})^{2} = 2.25 \times 1 0^{26} + 5.76 \times 1 0^{26} = 8.01 \times 1 0^{26}$ , so $\cdot = 2.83 \times 1 0^{13}$ . $n = 1.5 \times 1 0^{13} + 2.83 \times 1 0^{13} = 4.33 \times 1 0^{13} cm^{- 3} .$ $p = \frac{n _{i}^{2}}{n} = \frac{( 2.4 \times 1 0 ^{13} ) ^{2}}{4.33 \times 1 0 ^{13}} = \frac{5.76 \times 1 0 ^{26}}{4.33 \times 1 0 ^{13}} = 1.33 \times 1 0^{13} cm^{- 3} .$ Sanity: the shortcut would have given $n = 3.0 \times 1 0^{13}$ — off by ~44%. The exact answer is larger because the intrinsic sea of pairs adds carriers on top of the donors.

Read the figure below. The curves plot $n$ versus $N_{D}$ for germanium. The teal curve is the exact quadratic result; the dashed orange line is the shortcut $n = N_{D}$ ; the dotted plum line is the intrinsic floor $n_{i}$ . The black dot marks this problem ( $N_{D} = 3.0 \times 1 0^{13}$ , $n = 4.33 \times 1 0^{13}$ ). Notice how, at small $N_{D}$ , the exact teal curve bends away above the orange shortcut and flattens onto the $n_{i}$ floor — that vertical gap between teal and orange is the 44% error. Only far to the right, where $N_{D} ≫ n_{i}$ , do teal and orange merge.

Level 4 — Synthesis

Combine mass action, neutrality, and physical reasoning.

Recall Solution L4.1

Reverse the minority formula. From $p_{n} = n_{i}^{2} / N_{D}$ , solve for $N_{D}$ : $N_{D} = \frac{n _{i}^{2}}{p _{n}} = \frac{( 1.0 \times 1 0 ^{10} ) ^{2}}{1.0 \times 1 0 ^{2}} = \frac{1 0 ^{20}}{1 0 ^{2}} = 1.0 \times 1 0^{18} cm^{- 3} .$ Check the shortcut is valid: $N_{D} = 1 0^{18} ≫ n_{i} = 1 0^{10}$ ✓, so $n \approx N_{D}$ holds and the minority formula is legitimate.

Recall Solution L4.2

Sample A: $n_{A} = 1.0 \times 1 0^{15}$ (majority electrons), $p_{A} = n_{i}^{2} / N_{D} = 1 0^{20} /1 0^{15} = 1.0 \times 1 0^{5}$ (minority holes). $n_{A} p_{A} = (1.0 \times 1 0^{15}) (1.0 \times 1 0^{5}) = 1.0 \times 1 0^{20} = n_{i}^{2} . ✓$ Sample B: $p_{B} = 1.0 \times 1 0^{17}$ (majority holes), $n_{B} = 1 0^{20} /1 0^{17} = 1.0 \times 1 0^{3}$ (minority electrons). $n_{B} p_{B} = (1.0 \times 1 0^{3}) (1.0 \times 1 0^{17}) = 1.0 \times 1 0^{20} = n_{i}^{2} . ✓$ Punchline: wildly different carrier makeups, but the product is the same $n_{i}^{2}$ — the fingerprint of thermal equilibrium at fixed $T$ .

Recall Solution L4.3

Minority on the n-side: $p_{n} = n_{i}^{2} / N_{D} = 1 0^{20} /1 0^{16} = 1.0 \times 1 0^{4}$ . Minority on the p-side: $n_{p} = n_{i}^{2} / N_{A} = 1 0^{20} /1 0^{16} = 1.0 \times 1 0^{4}$ . They are equal — factor of 1. Equal and opposite doping gives equal minority reservoirs. (The diffusion currents then differ only through the different mobilities/diffusion lengths of electrons vs holes, not through the concentrations.)

Level 5 — Mastery

Full exact treatment, degenerate cases, limiting behaviour.

Recall Solution L5.1

Step 1 — net dopant. $N_{A} - N_{D} = 2.0 \times 1 0^{10} - 0.5 \times 1 0^{10} = 1.5 \times 1 0^{10} cm^{- 3}$ (net acceptor → p-type). Call $N \equiv N_{A} - N_{D}$ . Step 2 — general neutrality (explicit derivation). Start from $p + N_{D} = n + N_{A}$ . Subtract $n$ from both sides, then subtract $N_{D}$ from both sides: $p + N_{D} = n + N_{A} ⟹ p - n = N_{A} - N_{D} \equiv N .$ This is the direct route — no sign juggling: the carriers land on the left, the net dopant on the right. Now substitute mass action $n = n_{i}^{2} / p$ : $p - \frac{n _{i}^{2}}{p} = N \Rightarrow p^{2} - N p - n_{i}^{2} = 0.$ Step 3 — solve (+ root): $p = \frac{N}{2} + (\frac{N}{2})^{2} + n_{i}^{2} .$ Step 4 — numbers. $N /2 = 7.5 \times 1 0^{9}$ ; $(7.5 \times 1 0^{9})^{2} + (1.0 \times 1 0^{10})^{2} = 5.625 \times 1 0^{19} + 1.0 \times 1 0^{20} = 1.5625 \times 1 0^{20}$ , $\cdot = 1.25 \times 1 0^{10}$ . $p = 7.5 \times 1 0^{9} + 1.25 \times 1 0^{10} = 2.0 \times 1 0^{10} cm^{- 3} .$ $n = \frac{n _{i}^{2}}{p} = \frac{1 0 ^{20}}{2.0 \times 1 0 ^{10}} = 5.0 \times 1 0^{9} cm^{- 3} .$ Note $p$ and $n$ are the same order as $n_{i}$ — this is nearly-intrinsic material, "barely p-type."

Recall Solution L5.2

(a) Heavy net doping, $N ≫ n_{i}$ . Why we may approximate: inside the radical the two terms are $(N /2)^{2}$ and $n_{i}^{2}$ ; when $N ≫ n_{i}$ we have $(N /2)^{2} ≫ n_{i}^{2}$ , so the $n_{i}^{2}$ term is negligible next to $(N /2)^{2}$ and we may drop it. Then $(N /2)^{2} + n_{i}^{2} \approx (N /2)^{2} = N /2$ , giving $p \approx N /2 + N /2 = N$ . → recovers the shortcut $p \approx N_{A} - N_{D}$ . ✓ (b) Undoped, $N = 0$ . Then $p = 0 + 0 + n_{i}^{2} = n_{i}$ . And $n = n_{i}^{2} / p = n_{i}$ . → recovers the intrinsic case $n = p = n_{i}$ . ✓ What this proves: the single quadratic is the universal law; the shortcut and the intrinsic result are just its two ends.

Read the figure below. The plot shows both carriers versus the net donor density $N$ for silicon: the teal curve is the majority $n$ , the orange curve is the minority $p$ , and the dotted plum line is $n_{i}$ . Trace it left-to-right to see both limits proven above. At the far left ( $N \to 0$ , left of the black dashed line at $N = n_{i}$ ) the teal and orange curves merge onto the plum $n_{i}$ line — that meeting point is limit (b), $n = p = n_{i}$ . Moving right into heavy doping, the teal majority climbs as a straight $4 5^{\circ}$ log-log line ( $n \approx N$ , limit (a)) while the orange minority mirrors it downward as $p = n_{i}^{2} / N$ — the two curves stay symmetric about the plum line because their product is always $n_{i}^{2}$ . The visual symmetry is the law of mass action.

Recall Solution L5.3

Net dopant zero means $N = 0$ , i.e. perfectly compensated. From L5.2(b), $n = p = n_{i} \neq = 0$ . Neither carrier is "minority" or "majority" — the sample behaves intrinsic. The claim confuses "net doping = 0" with "carriers = 0." Thermal generation of pairs keeps both populations at $n_{i} > 0$ . The product $n p = n_{i}^{2}$ can never be zero at $T > 0$ , so no carrier population can ever hit zero in equilibrium.

Recall Solution L5.4

Minority: $p_{n} = n_{i}^{2} / N_{D} \propto n_{i}^{2}$ . Doubling $n_{i}$ multiplies $p_{n}$ by $2^{2} = 4$ .

Before: $p_{n} = (1 0^{10})^{2} /1 0^{16} = 1.0 \times 1 0^{4}$ .
After: $p_{n} = (2 \times 1 0^{10})^{2} /1 0^{16} = 4.0 \times 1 0^{20} /1 0^{16} = 4.0 \times 1 0^{4}$ . Factor = 4. Majority: $n \approx N_{D} = 1 0^{16}$ , still $≫ n_{i} = 2 \times 1 0^{10}$ , so it is essentially unchanged. This is exactly why leakage (minority-driven) current in diodes is so temperature-sensitive: it rides on $n_{i}^{2}$ .

The 80/20 core of this exercise set

Connections

Yeh exercises Hinglish mein →
Law of mass action — the product constraint used in every problem.
Charge neutrality condition — the second equation, source of the quadratic.
Intrinsic carrier concentration $n_i$ — the temperature-set floor for both carriers.
Doping — donors and acceptors — decides which carrier is majority.
pn junction diode — where the minority reservoir controls the current.
Diffusion and drift currents — how injected minority carriers move.
Fermi level position vs doping — the same light/heavy crossover in energy terms.