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Electronics begins with one extraordinary class of materials: semiconductors. Insulators do not conduct because their valence electrons are tightly bound; metals conduct freely because their valence electrons form a delocalised sea. Semiconductors sit precisely in the middle, and that middle position is what makes them useful. A pure crystal of silicon at room temperature has only a tiny number of free charge carriers, but by deliberately introducing impurity atoms at a level of one part per million we can engineer the conductivity over many orders of magnitude. By placing differently engineered slabs of silicon side by side we can produce devices — diodes, transistors, MOSFETs — that selectively rectify, amplify, switch, and gate electrical signals. The whole of modern computing, communications, and signal processing is built on the controlled doping of semiconductor crystals. This lesson develops the underlying physics from the band-gap picture upwards: intrinsic vs extrinsic semiconductors, n-type and p-type doping, the p–n junction, the diode current–voltage characteristic, bipolar junction transistors as amplifiers and switches, and a brief survey of the MOSFETs that now dominate large-scale integrated circuits.
Spec mapping (AQA 7408 §3.13.1 — Discrete semiconductor devices, option E): This lesson covers intrinsic and extrinsic semiconductors, n-type and p-type doping, the formation of the p–n junction and the depletion region, the diode I–V characteristic in forward and reverse bias, the bipolar junction transistor (npn and pnp) and its three regions of operation, transistor action as a small-signal amplifier and as a switch, and a brief contextual mention of MOSFETs as the dominant technology used in modern integrated circuits. (Refer to the official AQA specification document for exact wording.)
Synoptic links: Semiconductor band gaps connect directly to the quantum physics of energy levels in solids (§3.2.2 photoelectric effect and electron energy levels) — the band gap of silicon, around 1.1 eV, is precisely the energy scale of visible-light photons, which is why silicon photodiodes work and why solar cells are made of doped silicon. The transistor-as-amplifier connects to the operational amplifier (Lesson 2 of this course) because an op-amp is essentially many transistors arranged on a single chip; the same negative-feedback ideas that stabilise op-amp gain are first seen here in the emitter-resistor stabilisation of a transistor amplifier. Finally, the transistor as a digital switch connects to logic gates (Lesson 4) — every NAND, NOR, AND and OR gate inside a CMOS chip is ultimately a small array of transistors operating in their saturation and cut-off regions.
When isolated atoms come together to form a crystal, the discrete atomic energy levels broaden into continuous energy bands. The two bands that matter for electrical behaviour are the highest band that is fully occupied at absolute zero — the valence band — and the next band above it, which is empty at absolute zero — the conduction band. Whether electrons can flow under an applied electric field depends on whether they can enter the conduction band.
Pure (intrinsic) semiconductor: a perfectly pure crystal of Si or Ge in which the only charge carriers are thermally generated electron–hole pairs. At room temperature the carrier density in intrinsic Si is roughly 1.5 × 10¹⁰ cm⁻³ — a remarkable 10¹² times smaller than in a typical metal, which is why pure silicon is a poor conductor and never used directly in devices.
A crucial property of intrinsic semiconductors is that the carrier density rises rapidly with temperature, because more thermal energy lifts more electrons across the gap. This temperature sensitivity is exploited in thermistors (Lesson 6) — a semiconductor's resistance falls as temperature rises, the opposite of a metal.
The trick that turns a useless lump of pure silicon into a useful electronic device is doping: deliberately introducing a controlled amount of impurity atoms, typically at the parts-per-million level. There are two families of dopants, depending on which group of the periodic table the impurity comes from.
Silicon sits in Group IV: each Si atom forms four covalent bonds with its neighbours. If we replace a tiny fraction of Si atoms with Group V atoms — typically phosphorus (P), arsenic (As) or antimony (Sb) — each impurity atom brings five valence electrons. Four of them slot into the covalent bonds; the fifth is loosely bound to its parent atom and easily ionised. At room temperature essentially all the donor atoms are ionised, releasing their fifth electron into the conduction band.
The resulting material has:
Because the mobile charges that respond to an applied field are negatively charged electrons, this is called n-type silicon. A typical doping level might be 10¹⁶ to 10¹⁸ donors per cm³ — still only one impurity for every 10⁴ to 10⁶ silicon atoms, but enormous compared with the intrinsic carrier density of 10¹⁰ cm⁻³.
Replace a tiny fraction of Si atoms with Group III atoms — typically boron (B), aluminium (Al) or gallium (Ga) — and each impurity brings only three valence electrons. The fourth covalent bond is left "missing" an electron — a hole. At room temperature a neighbouring valence-band electron can hop in to fill the hole, leaving a hole behind in the new location, and so on. The hole behaves as a positively charged mobile carrier moving through the valence band.
The resulting material has:
Because the mobile carriers are positively charged holes, this is called p-type silicon.
Exam Tip: When asked to identify majority and minority carriers, always state both. "n-type silicon: majority carriers are electrons; minority carriers are holes." Examiners frequently test whether you remember that minority carriers exist at all — they are the reason for reverse-bias leakage current.
| Property | Intrinsic Si | n-type Si | p-type Si |
|---|---|---|---|
| Dopant | None | Group V (P, As, Sb) | Group III (B, Al, Ga) |
| Majority carrier | electron = hole | electron | hole |
| Minority carrier | — | hole | electron |
| Fixed ion cores | — | positive (donor) | negative (acceptor) |
| Typical resistivity | ~2300 Ω m | 10⁻⁴ to 10⁻¹ Ω m | 10⁻⁴ to 10⁻¹ Ω m |
The single most important device in electronics is the p–n junction: a single silicon crystal in which one side has been doped n-type and the adjacent side p-type. The doping is not done by joining two separate crystals (which would create surface defects) but by diffusing or implanting dopants into selected regions of a single wafer.
At the moment of formation, the n-side has a huge excess of electrons and the p-side has a huge excess of holes. The concentration gradient at the junction drives:
After a small fraction of carriers has crossed, the region either side of the metallurgical junction is depleted of mobile carriers — but the fixed donor and acceptor ions remain. The result is a thin layer (typically 0.1 to 1 μm wide) containing exposed positive donor ions on the n-side and exposed negative acceptor ions on the p-side. This produces an internal electric field pointing from n to p across the junction, and a corresponding potential difference — the built-in potential V_bi.
For silicon at room temperature, V_bi ≈ 0.7 V. For germanium it is roughly 0.3 V. The built-in potential opposes any further diffusion and the junction reaches equilibrium with zero net current.
graph LR
A["p-type<br/>Holes (majority)<br/>Fixed negative<br/>acceptor ions"] --> B["Depletion<br/>region<br/>(no mobile<br/>carriers)"]
B --> C["n-type<br/>Electrons (majority)<br/>Fixed positive<br/>donor ions"]
style A fill:#e74c3c,color:#fff
style B fill:#95a5a6,color:#fff
style C fill:#3498db,color:#fff
Connect the positive terminal of a battery to the p-side and the negative terminal to the n-side. The applied field opposes the built-in field and narrows the depletion region. Once the applied voltage exceeds the built-in potential of ~0.7 V, the depletion region collapses and majority carriers flow freely across the junction: holes from p to n, electrons from n to p. Current rises sharply. The junction is a conductor in this direction.
Connect the positive terminal of the battery to the n-side and the negative terminal to the p-side. The applied field now reinforces the built-in field, widening the depletion region. Majority carriers are pushed away from the junction; the only current that can flow is a small reverse-bias leakage current driven by thermally generated minority carriers, typically a few nanoamps to microamps in silicon. The junction is an insulator in this direction.
If the reverse voltage is increased far beyond ~50 V (depending on the diode), eventually the field is large enough to cause avalanche breakdown — the carriers accelerate enough between collisions to ionise lattice atoms, multiplying the current catastrophically. Special Zener diodes are designed to operate in this breakdown region as voltage regulators, but for a general-purpose diode breakdown is destructive.
Question: A silicon diode is forward-biased by a 5.0 V battery in series with a 220 Ω resistor. Assuming the diode forward voltage drop is fixed at 0.7 V, calculate the current through the diode.
Solution: The voltage across the resistor is 5.0 − 0.7 = 4.3 V. Ohm's law gives I = V/R = 4.3 / 220 = 0.0195 A ≈ 19.5 mA. This is a typical operating point for a small-signal silicon diode like the 1N4148.
Plotting the current through a diode against the voltage across it gives the iconic diode curve. In forward bias, the current is exponentially related to voltage by the Shockley equation:
I = I_s (exp(V/V_T) − 1)
where V_T = k_B T / e ≈ 25 mV at room temperature and I_s is the reverse saturation current. For practical purposes:
In reverse bias the current is approximately −I_s, a tiny constant leakage. At the breakdown voltage the current rises catastrophically (downward and to the left on the I–V plot).
graph TD
A["Apply forward voltage V"] --> B{"V > 0.7 V?"}
B -->|Yes| C["Diode ON<br/>I follows Ohm's law in external circuit<br/>V_diode ≈ 0.7 V fixed"]
B -->|No| D["Diode mostly off<br/>I exponentially small"]
E["Apply reverse voltage V"] --> F{"V < V_breakdown?"}
F -->|Yes| G["Diode OFF<br/>I ≈ tiny leakage I_s"]
F -->|No| H["Avalanche breakdown<br/>I rises catastrophically"]
style C fill:#27ae60,color:#fff
style G fill:#3498db,color:#fff
style H fill:#e74c3c,color:#fff
A single diode connected in series with an AC source and a load resistor conducts only on the positive half-cycle. The output is a series of half-sinusoid pulses — a half-wave rectified version of the AC input. With a smoothing capacitor in parallel with the load, the pulses are integrated to produce a near-DC voltage. This is the basis of every plug-top wall adapter ever made (now usually upgraded to a four-diode bridge rectifier for full-wave operation).
A bipolar junction transistor is a three-layer sandwich of doped silicon. Two layouts exist:
The three terminals are emitter (E), base (B) and collector (C). In normal "active-mode" operation:
In an npn transistor in active mode:
The defining property is that I_C is large and approximately proportional to I_B:
I_C = β I_B (common-emitter current gain; β typically 50 to 500)
A small change in base current produces a much larger change in collector current — this is current amplification. By placing a load resistor in series with the collector, the amplified current produces an amplified voltage at the collector. This is the heart of transistor action.
A typical common-emitter audio amplifier stage looks like:
+Vcc
│
R_C
│
─────┼─────→ V_out
│
C ────┐
│ │
B ─┐ │ │
│ │ │
───┤[T] │ (load)
│ │ │
E ─┘ │ │
│ │
R_E │
│ │
GND GND
The signal is applied to the base (with a DC bias network setting the quiescent operating point), and the amplified output is taken from the collector. The voltage gain at small signals is approximately G_v = −R_C/R_E (the negative sign indicates a 180° phase inversion). The emitter resistor R_E provides negative feedback that stabilises the gain against variations in temperature and transistor parameters — exactly the same negative-feedback trick used in op-amp circuits in Lesson 2.
Question: A common-emitter amplifier has R_C = 4.7 kΩ, R_E = 470 Ω, V_CC = 12 V, and a transistor with β = 200. The DC base current is set to 20 μA. Calculate (a) the DC collector current, (b) the DC voltage at the collector, (c) the small-signal voltage gain.
Solution:
(a) I_C = β I_B = 200 × 20 × 10⁻⁶ = 4.0 × 10⁻³ A = 4.0 mA.
(b) Voltage drop across R_C = I_C × R_C = 4.0 × 10⁻³ × 4700 = 18.8 V. But this exceeds V_CC = 12 V, so the transistor is actually saturated — the collector pinned at ≈ 0.2 V above the emitter. In a real bias design we would reduce I_B until V_C sat in the linear region; for this example, a sensible quiescent point would be I_C ≈ 1.5 mA giving V_C = 12 − 1.5 × 4.7 = 4.9 V.
(c) Small-signal gain G_v = −R_C/R_E = −4700/470 = −10. A 100 mV peak input produces a 1 V peak output, inverted.
When a BJT is driven hard, it operates in only two regions:
Between cut-off and saturation lies the active region used for analogue amplification. For digital switching we deliberately drive the transistor into saturation to minimise the on-state voltage drop, and into cut-off to minimise the off-state leakage current. This binary operation (saturation = "1", cut-off = "0") is the foundation of every logic gate.
The same npn transistor running as an amplifier in an audio circuit can run as a switch in a logic gate. The difference is the operating point — analogue circuits sit in the middle of the active region; digital circuits hop between the two extremes.
Question: A 5.0 V logic signal is used to drive a BJT switch controlling an LED with forward voltage 2.0 V and current 15 mA. The transistor has β = 100 and the supply is 5.0 V. Calculate the value of (a) the collector resistor needed to set the LED current, (b) the base resistor needed to ensure saturation, assuming the logic high is 5.0 V and we want at least 2 × the calculated I_B to guarantee saturation.
Solution:
(a) The collector circuit: V_CC = V_LED + I_C R_C + V_CE(sat). So R_C = (V_CC − V_LED − V_CE(sat)) / I_C = (5.0 − 2.0 − 0.2) / 0.015 = 2.8 / 0.015 = 187 Ω — round up to 220 Ω (standard value).
(b) Minimum base current I_B(min) = I_C/β = 15 mA / 100 = 0.15 mA. Use 2× this to guarantee saturation: I_B = 0.30 mA. With V_BE ≈ 0.7 V and the logic high at 5.0 V, R_B = (5.0 − 0.7) / 0.30 × 10⁻³ = 4.3 / 0.00030 = 14.3 kΩ — round down to 10 kΩ (standard) to guarantee saturation across temperature.
While the bipolar junction transistor (invented 1947 at Bell Labs by Bardeen, Brattain and Shockley) was the dominant device for the first quarter-century of solid-state electronics, modern integrated circuits use a different family: the MOSFET (metal–oxide–semiconductor field-effect transistor). MOSFETs come in two complementary types — NMOS and PMOS — wired together in CMOS (complementary MOS) logic, which is the technology behind essentially every microprocessor, memory chip, and microcontroller manufactured today.
The MOSFET differs from the BJT in a fundamental way. Where a BJT is current-controlled (the base current sets the collector current), a MOSFET is voltage-controlled — the voltage on a thin, insulated gate electrode controls the conductivity of a channel underneath. Because the gate is insulated from the channel by a layer of silicon dioxide, no DC gate current flows. This makes MOSFETs vastly more power-efficient in digital logic than BJTs.
A-Level coverage of MOSFETs is intentionally light: you need to know that they exist, that they have three terminals (source, drain, and gate), that the gate voltage controls a conductive channel between source and drain, and that they dominate large-scale digital integration because of their near-zero static power consumption. The BJT remains widely used for high-current power switching, analogue amplification, and discrete circuit work.
| Property | BJT (npn) | MOSFET (n-channel enhancement) |
|---|---|---|
| Terminals | E, B, C | S, G, D |
| Control mechanism | I_B (current) | V_GS (voltage) |
| Input impedance | Low (~kΩ) | Very high (~10¹² Ω) |
| Switching speed | Fast | Very fast |
| Dominant use | Power, analogue, discrete | Digital, integrated |
Specimen question modelled on the AQA paper format.
A silicon p–n junction diode is connected in series with a 100 Ω resistor and a variable voltage source. The diode is approximately ideal except for a fixed forward voltage drop of 0.70 V and a reverse breakdown voltage of −50 V.
(a) Sketch the current–voltage characteristic of this diode for the range −60 V ≤ V ≤ +1.5 V, labelling the values at the breakdown and the turn-on. [3 marks]
(b) Calculate the current that flows when the source is set to +5.0 V in the forward direction. [2 marks]
(c) The source is reversed to −10 V. Calculate the current that flows. [2 marks]
(d) Explain, in terms of carriers and the depletion region, why almost no current flows in (c) but a substantial current flows in (b). [4 marks]
This 11-mark item tests three Assessment Objectives. AO1 (knowledge) is rewarded for the I–V sketch shape and for the correct identification of majority/minority carriers. AO2 (application) is rewarded for the numerical calculations in (b) and (c) using Ohm's law with the assumed 0.7 V forward drop. AO3 (analysis) is the bulk of (d) — the candidate must explain why the depletion region widens under reverse bias and why the only available current is the minority-carrier drift, not the majority-carrier diffusion. Candidates who simply name "depletion region" without explaining its width-and-field role typically lose 2 of the 4 marks.
Grade C response (about 200 words):
(a) The forward branch is approximately zero current for V < 0.7 V, rising rapidly at V = 0.7 V. The reverse branch is approximately zero until V = −50 V, when breakdown occurs and the current rises sharply. [Sketch shown with both axes labelled.]
(b) V_R = 5.0 − 0.7 = 4.3 V. I = V_R / R = 4.3 / 100 = 43 mA.
(c) Reverse bias of 10 V is well below the −50 V breakdown, so almost no current flows. I ≈ 0.
(d) In forward bias the depletion region narrows, allowing majority carriers (electrons in the n-side, holes in the p-side) to cross the junction freely once the applied voltage exceeds 0.7 V. In reverse bias the depletion region widens, blocking majority-carrier flow, and only a small minority-carrier leakage remains.
Examiner commentary: This earns 9 of 11 marks. The sketch and numerical work are correct (3 + 2 + 1 = 6). The explanation in (d) earns 3 of 4 — it identifies the depletion-region widening and majority-vs-minority carrier role but does not mention the direction of the built-in field or that minority carriers in reverse bias are thermally generated. A clean, correct, minimal answer — typical Grade C border.
Grade A response (about 320 words):*
(a) The I–V curve: for V ≥ 0.7 V the current rises sharply (forward conduction); for −50 V < V < 0.7 V the current is essentially zero (a few μA of leakage at most); at V = −50 V the curve drops sharply downward as avalanche breakdown begins. [Sketch with horizontal axis labelled "V_diode (V)" and vertical axis labelled "I (A)", with the turn-on at 0.7 V and the breakdown at −50 V both annotated.]
(b) Apply Kirchhoff's voltage law: V_source = V_diode + V_R. With V_diode = 0.7 V (fixed-drop model), V_R = 4.3 V, so I = 4.3 / 100 = 43 mA. The current is comfortably below the diode's forward-current rating and the diode operates safely in the "on" region.
(c) At −10 V the diode is reverse-biased but well short of the −50 V breakdown. The only current is the thermally generated reverse-saturation leakage I_s, typically tens of nA for a small-signal silicon diode. I ≈ 0 to within the resolution of the question.
(d) In equilibrium the built-in potential V_bi ≈ 0.7 V across the depletion region prevents net diffusion. Forward bias opposes V_bi, narrowing the depletion region; once the applied voltage exceeds V_bi, majority carriers from each side (electrons from the n-side, holes from the p-side) can diffuse freely across, producing a large current proportional to the load. Reverse bias reinforces V_bi, widening the depletion region and sweeping mobile carriers away from the junction. Only the thermally generated minority carriers (holes in n, electrons in p) drift in the field, giving the tiny reverse-saturation current. This asymmetry — large forward current, negligible reverse current — is the defining property of a diode.
Examiner commentary: Full marks. The sketch is annotated with the turn-on and breakdown values (M3 in (a)). The numerical work in (b) and (c) is correct and uses the fixed-drop model explicitly (M2 + M2). Part (d) earns all four marks: identification of V_bi and the depletion region (M1), the bias-direction effect on the depletion width (M1), the role of majority vs minority carriers (M1), and the thermal-generation explanation of the leakage current (M1). The explicit naming of V_bi and the assertion of the carrier asymmetry are the A*-band moves.
University electronics introduces the band diagram for the p–n junction — a plot of conduction-band edge E_c and valence-band edge E_v across the junction. The Fermi level is constant across the device in equilibrium and the bending of E_c and E_v captures both the built-in potential and the asymmetry under bias. A Cambridge admissions interview might ask you to draw the band diagram for a forward-biased and a reverse-biased junction and explain where the diffusion and drift currents come from. The Shockley equation I = I_s (exp(eV/k_BT) − 1) emerges naturally from this band picture; the factor e/k_BT ≈ 39 V⁻¹ at room temperature is why a small change in V produces a huge change in I.
This content is aligned with the AQA A-Level Physics (7408) specification, option E — Electronics, section 3.13.1.