14. Semiconductor Electronics: Materials, Devices And Simple Circuits

A semiconductor device turns the microscopic behaviour of electrons and holes into a useful circuit function. Follow carriers, field direction and the complete current path. Numerical constants and model voltage drops in examples are rounded classroom values.

The development of semiconductor physics connected crystal bonding and energy bands with practical electronic devices. Controlled doping and junction fabrication made reproducible diodes possible and later enabled large numbers of components to be integrated on one chip.

A small change in carrier concentration can turn the same base material into a useful electronic device.

Electrical components work by controlling charge motion. A metal wire provides a conducting path. A semiconductor lets material composition, temperature, light and applied voltage change the available carriers and their motion. This controllability is the reason semiconductor devices can act as switches, rectifiers and sensors.

A diode has two terminals and a junction that conducts much more readily in one bias direction than the other over its normal operating range. Rectification uses that asymmetry to obtain a unidirectional load current from an alternating source. The output need not be steady to qualify as rectified.

Begin with the energy-band picture and electron–hole bookkeeping. Then follow how diffusion creates a junction barrier, how an external voltage changes that barrier, and how the diode behaves in a complete circuit. Isolated facts about “positive” and “negative” sides are not enough.

A circuit needs current to pass mainly during one polarity of an alternating input. What device property is useful?

The required function distinguishes the two voltage directions. A symmetric resistor cannot do this by itself.

Result: The asymmetric current–voltage behaviour of a diode can select the conducting polarity.

1. Why are semiconductors useful even when a metal is a better conductor?
2. Does a simple diode need three terminals?

Answers

1. Their carrier population and transport can be controlled in useful ways.
2. No. A basic junction diode has two terminals.

A material conducts when carriers and nearby available states permit a sustained response to an electric field.

Conductivity $\sigma$ measures current response to electric field in the simple relation $J=\sigma E$. Resistivity is its reciprocal, $\rho=1/\sigma$. Conductors have large conductivity, insulators have very small conductivity, and semiconductors occupy a controllable intermediate range. These labels depend on temperature, purity and the physical conditions.

When many atoms form a crystal, their interacting electron states produce closely spaced groups of allowed energies called bands. A forbidden energy gap separates some bands. The valence band is the highest occupied band in the ideal zero-temperature semiconductor; the conduction band is the next band available for mobile electron carriers.

In a metal, partially occupied bands or overlapping bands provide nearby available states, so electrons can respond to a modest field. In an ideal intrinsic semiconductor at absolute zero, a full valence band and empty conduction band do not provide ordinary conduction carriers. At higher temperature, some electrons gain enough energy to reach the conduction band, leaving holes behind. A larger gap makes thermal excitation much less likely at the same temperature.

The simple distinction between a semiconductor and an insulator is a matter of gap scale and operating conditions, not a universal sharp gap value. Ordinary metals often become more resistive as temperature rises because scattering increases. In a typical intrinsic semiconductor, carrier creation can dominate and conductivity increases strongly with temperature. Doped materials can have additional temperature regimes.

A sample has resistivity $0.25\,\Omega\,\mathrm m$. Find conductivity.

Use reciprocal resistivity. The unit is siemens per metre, because siemens is reciprocal ohm.

Result: $\sigma=4\,\mathrm{S\,m^{-1}}$.

1. Why is a completely full band not by itself a source of ordinary conduction?
2. What happens to valence-band occupancy when an electron is thermally excited across the gap?

Answers

1. Nearby state occupancy and symmetry prevent a simple net current response within that full band.
2. A vacancy, or hole, is left in the valence band.

In a pure semiconductor, creating a mobile conduction electron also creates a mobile valence-band hole.

Silicon and germanium form covalent bonding networks. In the simple local picture, each atom shares valence electrons with four neighbours. At absolute zero in the ideal intrinsic model, bonding states are filled and conduction states are empty. At a finite temperature, thermal excitation can create electron–hole pairs.

A hole is an unoccupied state in an otherwise nearly full valence band. A neighbouring electron can fill the vacancy and leave another vacancy behind. The resulting hole motion can be treated as the motion of a positive charge carrier. The hole is not a positive proton moving through the solid.

In intrinsic thermal equilibrium, electron density n and hole density p are equal: $n=p=n_i$. Pair generation and recombination continue, but their average rates balance. The conductivity is $\sigma=q_{\rm e}(n\mu_n+p\mu_p)$, where the mobilities $\mu_n$ and $\mu_p$ relate average drift speed to field. Electron and hole mobilities need not be equal.

Under an electric field, electron drift is opposite to the field, while hole drift is along it. Conventional current contributions from both carrier types point along the field. Random thermal motion exists in addition to the much smaller mean drift.

An intrinsic sample has $n_i=2.0\times10^{16}\,\mathrm{m^{-3}}$, $\mu_n=0.14$ and $\mu_p=0.05\,\mathrm{m^2\,V^{-1}\,s^{-1}}$. Use $q_{\rm e}=1.602\times10^{-19}\,\mathrm C$ to find conductivity.

Set n=p=n_i. Add mobilities before multiplying by the carrier density and elementary charge.

Result: $\sigma=0.0006088\,\mathrm{S\,m^{-1}}$.

1. What is left behind when a valence electron reaches the conduction band?
2. If an intrinsic sample has electron density nᵢ, what is its hole density?

Answers

1. A hole in the valence band.
2. The same density nᵢ, at thermal equilibrium.

A small concentration of suitable impurity atoms changes which mobile carrier is most common.

Doping means adding a controlled amount of impurity to a semiconductor crystal. A group-V donor in silicon has one more valence electron than needed for the simple local bonding pattern. That extra electron can be excited into the conduction band relatively easily. A group-III acceptor introduces a state that can accept a valence electron, leaving a mobile hole.

An ionised donor is a fixed positive ion after releasing an electron. An ionised acceptor is a fixed negative ion after accepting an electron. The mobile carriers and fixed ions together preserve approximate bulk electrical neutrality. Thus n-type does not mean the entire crystal is negatively charged, and p-type does not mean the entire crystal is positively charged.

For a nondegenerate semiconductor at thermal equilibrium and fixed temperature, $np=n_i^2$. If donors are fully ionised, greatly outnumber the intrinsic carrier density and are not compensated by acceptors, then $n\approx N_D$ and $p\approx n_i^2/N_D$. For an analogous p-type sample, $p\approx N_A$. These are approximations with stated conditions, not universal doping formulas.

In a band picture, a shallow donor level lies close below the conduction band, while a shallow acceptor level lies close above the valence band. Only a modest energy is needed for the associated excitation compared with crossing the full intrinsic gap. Donors do not inject free protons into the crystal, and acceptors do not make holes by removing nuclei.

At a specified temperature, let $n_i=1.5\times10^{10}\,\mathrm{cm^{-3}}$ and donor density $N_D=5.0\times10^{15}\,\mathrm{cm^{-3}}$. Assume full ionisation and negligible compensation. Estimate n and p.

Take n approximately equal to donor density. Then use the equilibrium product $np=n_i^2$. Keep both densities in the same volume unit.

Result: $n\approx5.0\times10^{15}\,\mathrm{cm^{-3}}$; $p\approx45000\,\mathrm{cm^{-3}}$.

1. What is the sign of a donor ion after donating its mobile electron?
2. Does an n-type semiconductor contain holes?

Answers

1. Positive, and fixed in the lattice.
2. Yes. They are normally the minority carriers under the stated doping conditions.

A junction barrier grows naturally when carriers diffuse across an initially uneven concentration profile.

A p–n junction is made within a semiconductor structure, not by merely pressing two unrelated wires together. Initially, electrons are more abundant on the n side and holes on the p side. Their concentration gradients drive diffusion across the junction. Electrons and holes that meet can recombine.

Near the junction, mobile carriers are depleted. Ionised donors remain as fixed positive charge on the n side; ionised acceptors remain as fixed negative charge on the p side. This depletion region therefore contains fixed space charge rather than being completely empty of all matter or all charge.

The resulting electric field points from the positive donor region on the n side towards the negative acceptor region on the p side. It opposes further majority-carrier diffusion. It also produces drift currents. At thermal equilibrium the opposing drift and diffusion contributions balance, so the net current through the junction is zero even though microscopic carrier motion continues.

Explain how zero equilibrium current can coexist with electrons diffusing across a junction.

Diffusion is driven by carrier-density gradients. The space-charge electric field drives drift in the opposing direction. At equilibrium their total contributions cancel.

Result: Zero net current does not mean all microscopic motion has stopped.

1. Which fixed ion signs appear on the n side of a depleted junction?
2. What is meant by the depletion region?

Answers

1. Positive donor ions.
2. A region strongly depleted of mobile majority carriers, containing fixed space charge.

An external voltage changes the junction barrier and therefore changes the balance of carrier currents.

Connect the p side to the positive terminal and n side to the negative terminal. This forward bias opposes the built-in barrier and narrows the depletion region in the usual abrupt-junction picture. Majority carriers are injected across the junction, so appreciable conventional current can flow from p to n through the device.

A forward diode does not have a universal exact “on” voltage. Its current rises continuously and steeply with voltage in its useful forward range. A constant drop, often chosen as about 0.7 V for a silicon classroom model, is an approximation that must be stated. The actual drop depends on current, temperature and device design.

Connect the p side to the negative terminal and n side to the positive terminal. The barrier and depletion width increase. The majority-carrier flow is strongly suppressed, but a small reverse current associated with minority carriers and generation remains. It is not a perfect open circuit.

At sufficiently large reverse voltage, breakdown can produce a large current. Uncontrolled current can damage an ordinary diode. Devices designed for breakdown operation still need appropriate current limiting. Ordinary rectifier analysis assumes the rated reverse voltage is not exceeded.

In any model, combine the diode relation with the rest of the circuit. With a forward supply $V_s$, a series resistor R and assumed constant diode drop $V_D$, Kirchhoff’s loop rule gives $I=(V_s-V_D)/R$ when the forward-conduction assumption is self-consistent. If it predicts a negative current, reconsider the assumed state rather than accepting the contradiction.

A forward-connected diode and $1.0\,\mathrm{k\Omega}$ resistor are connected to $6.0\,\mathrm V$. Assume a constant diode drop of $0.7\,\mathrm V$.

The resistor receives the supply voltage minus the diode drop. Apply Ohm’s law to the resistor, not to the diode as though its resistance were constant.

Result: $I=5.3\,\mathrm{mA}$.

To measure a diode characteristic, vary the applied voltage while measuring both diode voltage and series current. Use a resistor to limit current. Forward current may be measured in a much larger range than reverse current, so label axes and units carefully when drawing both branches.

1. Which way does ordinary conventional forward current pass through a p–n diode?
2. Why is a series resistor useful when testing a diode?

Answers

1. From p to n.
2. It limits current and prevents a steep current rise from overloading the device.

A rectifier routes alternating input so that load current keeps one direction.

In a half-wave rectifier, a series diode conducts during the polarity that forward biases it. During the opposite polarity the diode is reverse biased and the ideal model gives zero current. The resistor voltage therefore contains separated pulses of one sign. Energy is delivered during only one half of each ideal input cycle.

In a full-wave arrangement, a centre-tapped transformer with two diodes or a bridge with four diodes directs both input half-cycles through the load in the same direction. In a bridge, one diode pair conducts during one half-cycle and the other pair during the next. Each conducting path contains two forward drops in the simple constant-drop model.

The full-wave load pulses repeat twice per input period, while a half-wave circuit has one load pulse per input period. This makes the full-wave ripple frequency twice the supply frequency. A capacitor placed across the load can charge near peaks and discharge through the load between them, reducing ripple. It does not produce perfectly steady output under every load.

For ideal sinusoidal input of peak $V_m$ and a resistor load, average output is $V_m/\pi$ for half-wave and $2V_m/\pi$ for full-wave rectification. These follow by averaging the respective sine-wave pulses over a full input period. The averages change if a smoothing capacitor changes the waveform.

An ideal rectifier receives a sine wave of peak $12\,\mathrm V$ at $50\,\mathrm{Hz}$. Compare half-wave and full-wave average outputs and pulse frequencies, with no capacitor.

Use the unsmoothed resistive-load averages. Half-wave pulses repeat once per input cycle; full-wave pulses repeat twice.

Result: Half-wave mean $=3.8197\,\mathrm V$ at $50\,\mathrm{Hz}$ pulse rate; full-wave mean $=7.6394\,\mathrm V$ at $100\,\mathrm{Hz}$.

When drawing a full-wave exam circuit, trace the complete conducting loop separately for each source polarity. In a centre-tapped design, the two secondary ends alternate their polarity relative to the centre tap. In a bridge, the diode orientations must keep the same load terminal positive for both source polarities. A waveform alone does not prove that a drawn circuit is wired correctly.

1. Does full-wave mean current reverses through the load every half-cycle?
2. For 60 Hz input, what is the ideal full-wave ripple frequency?

Answers

1. No. The circuit maintains the same load-current direction.
2. 120 Hz.

• Intrinsic equilibrium gives $n=p=n_i$; electron and hole contributions both enter $\sigma=q_{\rm e}(n\mu_n+p\mu_p)$.
• Donors create n-type behaviour and acceptors create p-type behaviour without making the entire bulk charged.
• A depletion region has few mobile majority carriers but contains fixed ion charge.
• Forward bias lowers the effective barrier; reverse bias raises it until breakdown behaviour becomes relevant.
• Rectification keeps load current in one direction. Full-wave output has two pulses per input cycle.

• An energy gap is an energy interval, not a spatial gap.
• The diode model and complete circuit must be solved together; a guessed conducting state must be checked.
• A rectified waveform can remain strongly time-dependent. Smoothing and regulation are additional functions.

Try these newly written questions before reading the answers. Show a diagram, the governing equation, substitutions and units where appropriate.

14.1 Why does a semiconductor provide more control than a simple metal wire?

14.2 Find conductivity for resistivity $2.0\,\Omega\,\mathrm m$.

14.3 What does a band gap represent?

14.4 Why does heating a typical intrinsic semiconductor often increase conductivity?

14.5 What is a hole, and why is it not a proton?

14.6 In intrinsic equilibrium, if electron density is 4×10¹⁶ m⁻³, what is hole density?

14.7 Compare electron drift and conventional electron-current direction in an applied field.

14.8 Name one donor and one acceptor dopant for silicon.

14.9 Why is an n-type crystal usually electrically neutral in its bulk?

14.10 At a specified temperature let $n_i=2\times10^{10}\,\mathrm{cm^{-3}}$ and n approximately $4\times10^{15}\,\mathrm{cm^{-3}}$. Find equilibrium p.

14.11 Where are shallow donor and acceptor levels placed relative to the main bands?

14.12 What drives diffusion when a p–n junction is first formed?

14.13 Give the signs of fixed ions on the p and n sides of the depletion region.

14.14 Which direction does the built-in junction field point?

14.15 Why is there no net junction current at thermal equilibrium?

14.16 State the external polarity for forward bias and its main effect.

14.17 Why is reverse current small but not exactly zero?

14.18 A diode with assumed $0.7\,\mathrm V$ drop is in series with $2.0\,\mathrm{k\Omega}$ across $8.7\,\mathrm V$. Find current when forward conducting.

14.19 Why is 0.7 V not an exact universal silicon turn-on threshold?

14.20 Sketch the sign of ideal half-wave output during each input half-cycle.

14.21 How many diodes conduct at a time in an ordinary bridge rectifier?

14.22 For $60\,\mathrm{Hz}$ input, compare half-wave and full-wave pulse repetition frequencies.

14.23 For an ideal full-wave rectifier with input peak $9.0\,\mathrm V$, find unsmoothed mean output.

14.24 How does a capacitor across the rectifier load reduce ripple?

14.25 Which assumptions must be stated before using Vm/π or 2Vm/π?

Answers

14.1 Doping and operating conditions can strongly alter carrier populations and junction behaviour.

14.2 $\sigma=0.5\,\mathrm{S\,m^{-1}}$.

14.3 A range of forbidden electron energies, not a physical empty space between crystal layers.

14.4 It creates more electron–hole pairs; the carrier increase can dominate mobility changes.

14.5 It is an unoccupied valence-band state whose motion acts as a positive carrier; the nuclei remain in the lattice.

14.6 The same density, 4×10¹⁶ m⁻³.

14.7 Electrons drift opposite the field; their conventional current contribution is along the field.

14.8 Phosphorus is a donor; boron is an acceptor.

14.9 Mobile electrons balance positive ionised donors along with the other charges.

14.10 $p=n_i^2/n=100000\,\mathrm{cm^{-3}}$.

14.11 Donor levels lie near below the conduction band; acceptor levels near above the valence band.

14.12 The carrier-concentration gradients across the junction.

14.13 Negative acceptor ions on the p side; positive donor ions on the n side.

14.14 From the n-side positive ion region towards the p-side negative ion region.

14.15 Opposing drift and diffusion current contributions balance.

14.16 p positive relative to n; the barrier is reduced and majority-carrier injection increases.

14.17 Minority carriers and carrier generation can still contribute.

14.18 $I=(8.7-0.7)/2000=4\,\mathrm{mA}$.

14.19 The current–voltage relation is continuous and depends on current, temperature and device construction.

14.20 It follows the conducting positive half-cycle and is zero during the reverse-biased negative half-cycle for the chosen orientation.

14.21 Two, with the conducting pair changing on the next input half-cycle.

14.22 $60\,\mathrm{Hz}$ and $120\,\mathrm{Hz}$, respectively.

14.23 $2V_m/\pi=5.7296\,\mathrm V$.

14.24 It charges near voltage peaks and supplies the load between peaks; its discharge still causes some ripple.

14.25 Ideal diodes, sinusoidal input, a resistive load, no smoothing capacitor, and the intended half-wave or full-wave topology.