ARCHETYPE: Semiconductor & Logic Numericals

The six relations you just drilled are, between them, the arithmetic of the electronics industry. Solar cells are sized by equation (1): silicon's $\sim\!1.1$ eV gap is a near-perfect match to the solar spectrum, which is why rooftops are tiled with it. LED lighting and displays run the same relation forwards — engineers dial a compound's gap to hit red, green or blue, from Holonyak's $1.9$-eV red of 1962 to every screen you own. Power supplies in every charger are full-wave bridges chosen for the $2I_m/\pi$ current and $0.48$ ripple of example 3, then filtered smooth. Every amplifier and logic chip is transistors obeying $I_E = I_B + I_C$: a modern processor packs tens of billions of them, and the universality of NAND is why one repeated cell can be printed into any circuit at all. A student's "$\lambda = hc/E_g$" and "$2^n$ rows" underwrite everything from a solar farm to the phone in your pocket.

Bookkeeping first, because in a numerical gym the accounting is the physics. The chapter looks sprawling — bands, doping, junctions, diodes, rectifiers, transistors, gates — but every number an examiner can ask for is generated by one of just six short relations, mapped in the figure below. Learn to name the family the moment you read the question and the arithmetic is almost automatic. Three of these are pure conservation statements you can trust blindly — $n_e n_h = n_i^2$, $I_E = I_B + I_C$, and "$n$ inputs give $2^n$ rows" — and three are formulas with a physical constant baked in. Nothing here needs anything past school algebra: the deeper machinery (energy-band theory, the Shockley diode equation $I = I_0\,(e^{\,V/V_T} - 1)$, and the Fermi–Dirac statistics that set the carrier densities) is quoted by name and used, never derived — obtaining those results is the business of the university seam pointed to at the end.

The forbidden gap $E_g$ is the energy an electron must gain to cross from the valence band into the conduction band — and, run in reverse, it is the energy a recombining electron–hole pair releases as a photon of light. So a single Planck relation ties the material's gap to a wavelength. This is why an LED's colour is set by what it is made of, and why silicon is transparent to far-infrared light it hasn't the gap to absorb:

where: $E_g$ — the band gap in electron-volts; $\lambda$ — the wavelength of the photon absorbed (a valence electron kicked up) or emitted (a pair recombining); $h$, $c$ — Planck's constant and the speed of light, bundled into the exam-friendly product $hc = 1240\ \text{eV}\,\cdot\,\text{nm}$ so a gap in eV gives a wavelength directly in nanometres. Read it as a see-saw: a large gap needs a short-wavelength (high-energy) photon, a small gap an infrared one. Silicon's $1.1$ eV lands at $1127$ nm, just past the red edge of vision — which is why a silicon solar cell harvests visible light but leaks the deep infrared straight through.

A semiconductor carries current with two populations at once — mobile electrons (density $n_e$) in the conduction band and mobile holes (density $n_h$) in the valence band. In a pure crystal thermal energy makes them strictly in pairs, so $n_e = n_h = n_i$, the intrinsic concentration. Doping shatters that symmetry — but not the product. The single most-used relation in the whole chapter is the law of mass action:

where: $n_e$, $n_h$ — the electron and hole concentrations in $\text{m}^{-3}$; $n_i$ — the intrinsic concentration, fixed by temperature and the gap ($n_i(\text{Si}) = 1.5\times10^{16}$, $n_i(\text{Ge}) = 2.4\times10^{19}\ \text{m}^{-3}$ at 300 K); $N_D$, $N_A$ — the donor and acceptor doping densities. Read equation (2) as a hyperbola: raise one carrier by doping and the other must fall by the same factor, because their product is pinned. Add $10^{22}\ \text{m}^{-3}$ phosphorus donors to silicon and the electrons soar to $\sim\!10^{22}$ while the holes are crushed to $2.25\times10^{10}\ \text{m}^{-3}$ — a million-million-fold split from one pinch of dopant.

A rectifier turns AC into pulsating DC by letting a diode pass current one way only. Everything an exam asks about it is captured by two numbers: the average (DC) current the load actually sees, and the ripple factor measuring how much AC "hum" still rides on top. A half-wave rectifier throws away half of every cycle; a full-wave (or bridge) rectifier flips the unwanted half up so both halves drive the load:

where: $I_m$ — the peak current; $I_{dc}$ — the time-average current (what a DC meter reads); $I_{rms}$ — the root-mean-square current ($I_m/2$ half-wave, $I_m/\sqrt2$ full-wave); $r$ — the ripple factor, the ratio of leftover AC to DC. The two circuits split cleanly: ripple $1.21$ (half) against $0.48$ (full), efficiency $40.6\%$ against $81.2\%$. The full-wave circuit wins on every count — twice the DC current, far less ripple, twice the efficiency, and a ripple frequency at twice the line frequency that is far easier to filter smooth.

A bipolar transistor is current-controlled, and its every numerical grows from one conservation law and two definitions of "gain." Charge entering the emitter must leave by the base or the collector, so the three terminal currents form a triangle. The common-base gain $\alpha$ and common-emitter gain $\beta$ are two ways of quoting how much of the emitter current reaches the collector:

where: $I_E$, $I_B$, $I_C$ — the emitter, base and collector currents; $\alpha$ — the fraction of emitter current collected, always just under 1 ($0.95$–$0.99$); $\beta$ — the collector-to-base current gain, a big number ($20$–$200$). Because the base is thin and lightly doped, almost all of $I_E$ sails through to the collector, leaving only a trickle of $I_B$ — which is why $\beta = I_C/I_B$ is large. The two gains are the same fact in two costumes, converted by a one-line algebra bridge (derived below):

where: $A_V$ — the common-emitter voltage gain: it multiplies a small base-side signal by $\beta$ and the resistance ratio $R_L/r_{in}$, and inverts its phase by $180^\circ$. A transistor with $\alpha = 0.98$ has $\beta = 0.98/0.02 = 49$; push $\alpha$ to $0.99$ and $\beta$ leaps to $99$ — the bridge's hallmark, that $\beta$ is hypersensitive to $\alpha$ as $\alpha \to 1$.

At the digital end of the chapter, a gate is defined completely by its truth table — the output for every combination of inputs. With $n$ inputs there are $2^n$ rows, so a 2-input gate has exactly 4. The three basic gates and their two most useful combinations are:

where: $A$, $B$ — binary inputs (0 or 1); $Y$ — the output; the overbar is inversion (NOT); $+$ is OR and $\cdot$ is AND. The headline result — worth memorising — is that NAND and NOR are universal: each of OR, AND and NOT can be built from copies of NAND alone (or NOR alone), so any digital circuit whatsoever comes from a single gate type. That is why real chips are oceans of identical NAND cells, and why an exam loves asking you to reconstruct a NOT or an AND from NANDs.

1 · Board level — band gap to photon wavelength (CBSE and NEET pattern)

Using $hc = 1240\ \text{eV}\,\cdot\,\text{nm}$, find (a) the longest-wavelength light silicon ($E_g = 1.1$ eV) can absorb, (b) the same for germanium ($E_g = 0.7$ eV), and (c) the band gap of a red LED that emits at $653$ nm.

Step 1. Silicon's cutoff, from equation (1): $\lambda = hc/E_g = 1240/1.1 = \boxed{1127\ \text{nm}}$ — in the near-infrared, just beyond visible red. A photon of longer wavelength (lower energy) cannot lift an electron across the $1.1$ eV gap, so silicon lets it pass unabsorbed.

Step 2. Germanium's cutoff: $\lambda = 1240/0.7 = \boxed{1771\ \text{nm}}$. Germanium's smaller gap means it can absorb (and detect) even deeper into the infrared than silicon — one reason Ge photodiodes serve the fibre-optic bands.

Step 3. The red LED, run in reverse (emission): $E_g = hc/\lambda = 1240/653 = \boxed{1.90\ \text{eV}}$. A recombining electron–hole pair drops $1.9$ eV and hands it to a red photon, so the emitting material's gap is about $1.9$ eV.

Step 4. Sense-check the see-saw: larger gap → shorter cutoff → higher photon energy. Ge (0.7 eV) reaches furthest into the infrared, Si (1.1 eV) is intermediate, and the 1.9 eV LED emits visible red.

2 · Board level — minority carriers by mass action (CBSE and JEE-Main pattern)

Silicon has $n_i = 1.5\times10^{16}\ \text{m}^{-3}$ at 300 K. Find the minority-carrier concentration when it is doped with (a) $1.0\times10^{22}\ \text{m}^{-3}$ pentavalent phosphorus donors, and (b) $5.0\times10^{22}\ \text{m}^{-3}$ trivalent boron acceptors.

Step 1. Case (a) is n-type — donors give up electrons — so electrons are the majority carrier: $n_e \approx N_D = 1.0\times10^{22}\ \text{m}^{-3}$.

Step 2. The minority holes follow from mass action, equation (2): $n_h = n_i^2/n_e = (1.5\times10^{16})^2 / 1.0\times10^{22} = 2.25\times10^{32}/10^{22} = \boxed{2.25\times10^{10}\ \text{m}^{-3}}$.

Step 3. Case (b) is p-type — acceptors capture electrons, leaving holes as majority: $n_h \approx N_A = 5.0\times10^{22}\ \text{m}^{-3}$. Minority electrons: $n_e = n_i^2/n_h = 2.25\times10^{32}/5.0\times10^{22} = \boxed{4.5\times10^{9}\ \text{m}^{-3}}$.

Step 4. Sense-check: the majority carrier outnumbers the minority by about $10^{12}$ in each wafer — and the product $n_e n_h$ returns $2.25\times10^{32} = n_i^2$ both times, as it must.

3 · Board level — rectifier DC current and ripple (ICSE and A-Level pattern)

A rectifier feeds a load with a peak current $I_m = 10$ mA. Find the DC (average) load current and the ripple factor for (a) a half-wave and (b) a full-wave rectifier.

Step 1. Half-wave DC current, from equation (3): $I_{dc} = I_m/\pi = 10/\pi = \boxed{3.18\ \text{mA}}$ — only about a third of the peak survives, because half of every cycle is blocked.

Step 2. Full-wave DC current: $I_{dc} = 2I_m/\pi = 20/\pi = \boxed{6.37\ \text{mA}}$ — exactly twice the half-wave value, since both half-cycles now push current through the load.

Step 3. Ripple factors, $r = \sqrt{(I_{rms}/I_{dc})^2 - 1}$. Half-wave: $I_{rms} = I_m/2 = 5$ mA, so $r = \sqrt{(5/3.18)^2 - 1} = \boxed{1.21}$. Full-wave: $I_{rms} = I_m/\sqrt2 = 7.07$ mA, so $r = \sqrt{(7.07/6.37)^2 - 1} = \boxed{0.48}$.

Step 4. Interpret: the full-wave rectifier delivers twice the DC and far lower ripple ($0.48$ vs $1.21$), with efficiency $81.2\%$ against $40.6\%$ — which is why nearly every real power supply uses a full-wave bridge.

4 · Competitive level — the transistor triangle, both gains and the amplifier (JEE-Main and NEET pattern)

An n-p-n transistor in common-emitter has base current $I_B = 20\ \mu\text{A}$ and current gain $\beta = 100$, driving a load $R_L = 5\ \text{k}\Omega$ from a source of input resistance $r_{in} = 1\ \text{k}\Omega$. Find $I_C$, $I_E$, $\alpha$, the voltage gain and the power gain.

Step 1. Collector current from $\beta$, equation (4): $I_C = \beta I_B = 100 \times 20\ \mu\text{A} = 2000\ \mu\text{A} = \boxed{2.0\ \text{mA}}$.

Step 2. Emitter current from the triangle: $I_E = I_B + I_C = 20\ \mu\text{A} + 2000\ \mu\text{A} = 2020\ \mu\text{A} = \boxed{2.02\ \text{mA}}$.

Step 3. Common-base gain, two independent ways that must agree: directly $\alpha = I_C/I_E = 2.0/2.02 = 0.990$, and via the bridge $\alpha = \beta/(1+\beta) = 100/101 = \boxed{0.990}$. The match confirms $I_E$ was formed correctly.

Step 4. Voltage gain, equation (5): $A_V = \beta\,R_L/r_{in} = 100 \times 5000/1000 = \boxed{500}$, with a $180^\circ$ phase inversion. Power gain is current gain times voltage gain: $\beta \times A_V = \boxed{5.0\times10^{4}}$. Timing note: about 90 seconds. Do the currents first ($I_C = \beta I_B$, then $I_E = I_B + I_C$), then the gains — computing $\alpha$ two ways is a free consistency check that catches a slipped microamp. The commonest slip is quoting $\alpha > 1$ or writing $\beta = 1/\alpha$; keep $\alpha < 1 < \beta$ in view.

5 · Competitive level — a NAND truth table and universality (JEE-Main and ICSE pattern)

For the 2-input gate $Y = \overline{A\cdot B}$ (a NAND), (a) write its complete truth table, (b) show that tying the two inputs together turns it into a NOT gate, and (c) explain why NAND is called universal.

Step 1. Build the table — 2 inputs give $2^2 = 4$ rows. Compute $A\cdot B$ (AND), then invert: $(A,B) = (0,0)\!\to\!Y=1$; $(0,1)\!\to\!1$; $(1,0)\!\to\!1$; $(1,1)\!\to\!0$. NAND outputs $1$ everywhere except when both inputs are $1$.

Step 2. Tie the inputs: set $B = A$. Then $Y = \overline{A\cdot A} = \overline{A}$ — the row for $A=0$ gives $Y=1$ and the row for $A=1$ gives $Y=0$. A single NAND with its inputs joined is a NOT gate.

Step 3. Build AND and OR from NAND. AND is a NAND followed by a NAND-inverter: $\overline{\overline{A\cdot B}} = A\cdot B$. OR comes from De Morgan's law, $A + B = \overline{\overline{A}\cdot\overline{B}}$ — invert each input (each with a tied NAND) and feed them into one more NAND.

Step 4. Conclude universality: since NOT, AND and OR — a complete set for any Boolean function — can each be assembled from NAND gates alone, any digital circuit whatsoever can be built from NAND gates only. NOR is universal for the same reason. Timing note: about one minute. Always fill the AND column first and invert it — never write the NAND output straight from memory. For "build gate X from NAND," recall two moves: a tied NAND is a NOT, and De Morgan converts an OR into an AND-of-inverses.

Expecting doping to raise both carriers. Mass action $n_e n_h = n_i^2$ is a fixed hyperbola: doping raises one carrier and lowers the other by the same factor. A minority-carrier answer must come out smaller than $n_i$ (e.g. $2.25\times10^{10}$ against $n_i = 1.5\times10^{16}$). Any "solution" with both carriers above $n_i$ is wrong on sight.

Writing $\beta = 1/\alpha$. They are cousins, not reciprocals: $\beta = \alpha/(1-\alpha)$ and $\alpha = \beta/(1+\beta)$. With $\alpha = 0.98$ the true $\beta$ is $49$, not $1.02$. Keep the ordering $\alpha < 1 < \beta$ on your pen; the tiny denominator $1-\alpha$ is what makes $\beta$ large.

Misusing $hc = 1240\ \text{eV}\,\cdot\,\text{nm}$. Use $\lambda\ [\text{nm}] = 1240/E_g\ [\text{eV}]$ and the wavelength drops out in nanometres directly. Feeding eV into the SI $hc = 1.99\times10^{-25}$ J·m without converting energy to joules is the classic factor-of-$10^{9}$ blunder.

Doubling the wrong rectifier. It is the full-wave rectifier that gives $I_{dc} = 2I_m/\pi$ and ripple $0.48$; the half-wave gives $I_m/\pi$ and ripple $1.21$. Learn them as a pair and quote the efficiencies ($81.2\%$ vs $40.6\%$) as the sanity check — full-wave beats half-wave on every count.

Forgetting the current triangle. $I_E = I_B + I_C$ is not optional: the emitter current is the largest of the three. If you find $I_E < I_C$, you have added wrong — the base current, however small, still counts.

Board scope. These numerical families are firmly examined on CBSE, ICSE, JEE-Main and NEET; Semiconductor Electronics sits in the JEE-Main syllabus but is traditionally excluded from JEE-Advanced. The diode/rectifier family (worked example 3) is also examined on A-Level 9702, but 9702's current syllabus does not carry the common-base/common-emitter α/β conversions or the discrete-transistor amplifier gain of worked example 4. AP Physics 2 does not carry semiconductor electronics at all, so none of this sits on an AP-2 paper. IGCSE 0625 carries only the qualitative diode, LED and logic-gate truth tables — not the mass-action calculation or the $\alpha$/$\beta$ conversions here — and IB has moved most electronics out of its core, so treat the quantitative gym as an Indian-board and A-Level target.

Texts. NCERT Physics Class 12, Ch. 14 ("Semiconductor Electronics"), the worked examples on band gap, doping, rectifiers, transistor gains and logic gates · H. C. Verma, Concepts of Physics Vol. 2, "Semiconductors and Semiconductor Devices" · D. C. Pandey / Irodov modern-physics problem sets for the mass-action and $\alpha$–$\beta$ families · Cambridge A-Level 9702, the rectification and electronics topics.

Provenance. Standard exam-archive problems across the semiconductor families. Recompute: $hc = 1240\ \text{eV}\,\cdot\,\text{nm} \Rightarrow$ Si $1127$ nm, Ge $1771$ nm, red LED ($653$ nm) $1.90$ eV, diamond $225$ nm; mass action ($n_i(\text{Si}) = 1.5\times10^{16}$) $\Rightarrow$ n-Si ($n_e = 10^{22}$) $n_h = 2.25\times10^{10}$, p-Si ($n_h = 5\times10^{22}$) $n_e = 4.5\times10^{9}\ \text{m}^{-3}$; rectifier ($I_m = 10$ mA) $\Rightarrow$ half $I_{dc} = 3.18$ mA (ripple $1.21$, eff $40.6\%$), full $6.37$ mA (ripple $0.48$, eff $81.2\%$); transistor $I_B = 20\ \mu\text{A}$, $\beta = 100 \Rightarrow I_C = 2.0$ mA, $I_E = 2.02$ mA, $\alpha = 0.990$, $A_V = 500$, power gain $5.0\times10^{4}$; gain bridge $\alpha = 0.98 \Rightarrow \beta = 49$, $\beta = 100 \Rightarrow \alpha = 0.990$; NAND table $(0,0),(0,1),(1,0)\to1$, $(1,1)\to0$; $2$ inputs $\to 4$ rows, $3 \to 8$.

Prerequisites. The Junction Diode and its I–V Curve — the concept article that builds the forward/reverse-bias behaviour and the knee voltage this gym rectifies with — and The Bipolar Junction Transistor, which develops the emitter/base/collector structure and $I_E = I_B + I_C$. This archetype assumes both and drills using them at speed. Leads to. the university seam The Fermi–Dirac Distribution (`aq-sm-fermi-dirac-distribution`), which derives the carrier statistics — the actual value of $n_i$ and the temperature dependence quoted here — from quantum statistical mechanics.

Go deeper (UG). Three results are quoted here and derived only at university: the energy-band picture that gives each solid its gap (Bloch's theorem in a periodic crystal, 1928); the Shockley diode equation $I = I_0(e^{\,V/V_T}-1)$ behind the smooth I–V curve; and the intrinsic concentration $n_i$ — the mass-action constant — set by Fermi–Dirac statistics, the subject of the seam above. All three sit above this article's algebra-and-trig ceiling; here you use the results and never derive them.