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14.3 Electronics: p-n Junction, Rectifiers, Transistors, Op-Amps & Logic Gates

Key Takeaways

  • A p-n junction forms a depletion region with a barrier potential of about 0.7 V for silicon and 0.3 V for germanium; it conducts when forward biased and blocks (apart from a tiny leakage current) when reverse biased.

  • A half-wave rectifier uses one diode and passes alternate half-cycles; a full-wave bridge rectifier uses four diodes so both half-cycles drive current the same way through the load, and a capacitor smooths the output.

  • In a transistor IE=IB+ICI_E = I_B + I_C and the current gain is β=ICIB\beta = \frac{I_C}{I_B}; for amplification the base–emitter junction is forward biased and the collector–base junction reverse biased.

  • An ideal op-amp has very high open-loop gain and input resistance; the inverting amplifier gain is −RfR1-\frac{R_f}{R_1} and the non-inverting gain is 1+RfR11 + \frac{R_f}{R_1}.

  • NAND and NOR gates are universal: any logic function can be built from either one alone, and the NAND output is 0 only when all inputs are 1.

Last updated: October 2026

Electronics: p-n Junction, Rectifiers, Transistors, Op-Amps & Logic Gates

Electronics builds directly on the n-type and p-type semiconductors introduced in Section 14.2. Most questions test either a definition (biasing, majority carriers, gate outputs) or a short calculation (transistor currents, amplifier gain), so clarity on the basic rules is enough to score well.


1. The p-n Junction

When p-type and n-type material are joined, electrons diffuse from the n-side into the p-side and holes diffuse the other way. They recombine near the boundary, leaving a thin depletion region of fixed ions (negative on the p-side, positive on the n-side) with no free carriers. The resulting electric field creates a potential barrier:

  • about 0.7 V0.7\text{ V} for silicon
  • about 0.3 V0.3\text{ V} for germanium

Biasing

BiasConnectionDepletion regionCurrent
Forwardp-side to positive terminalNarrowsLarge current once the applied voltage exceeds the barrier
Reversep-side to negative terminalWidensVery small leakage current from minority carriers

The diode's I–V curve is therefore highly non-linear: almost no current below about 0.7 V (Si) in forward bias, then a steep rise. A very large reverse voltage causes breakdown.


2. Rectification

A rectifier converts AC into DC.

  • Half-wave rectifier: one diode in series with the load conducts only during the half-cycles that forward-bias it; the other half-cycles are cut off. Output is pulsating DC at the supply frequency.
  • Full-wave bridge rectifier: four diodes arranged in a bridge. In each half-cycle, two diodes conduct and route the current through the load in the same direction. The output pulses at twice the supply frequency (100 Hz from 50 Hz mains).
  • Smoothing: a capacitor connected across the load charges near each peak and discharges slowly between peaks, reducing the ripple.

3. Special-Purpose Diodes

DevicePrincipleUse
Light-emitting diode (LED)Forward-biased; electrons and holes recombine and emit photons. Colour depends on the band gap of the material (for example gallium arsenide phosphide)Indicators, displays, lighting
PhotodiodeReverse-biased; light creates electron–hole pairs that increase the reverse currentLight detection, optical communication, counters
Photovoltaic (solar) cellUnbiased junction; light creates carriers that are separated by the junction field, producing an emfSolar panels, calculators

4. The Transistor

A bipolar junction transistor (BJT) is a thin, lightly doped base sandwiched between an emitter (heavily doped) and a collector (moderately doped, largest). It may be npn or pnp; in the circuit symbol the arrow on the emitter points in the direction of conventional current.

Currents

IE=IB+IC,β=ICIBI_E = I_B + I_C, \qquad \beta = \frac{I_C}{I_B}

Because the base is thin and lightly doped, most carriers from the emitter cross into the collector, so IBI_B is only a small fraction of IEI_E and β\beta is typically 50 to a few hundred.

Example: with IB=20 μAI_B = 20\,\mu\text{A} and β=150\beta = 150, IC=3.0 mAI_C = 3.0\text{ mA} and IE=3.02 mAI_E = 3.02\text{ mA}.

Transistor as an Amplifier

For amplification, the base–emitter junction is forward biased and the collector–base junction is reverse biased. In the common-emitter amplifier, a small change in base current produces a large change in collector current; the voltage across the collector load resistor RCR_C changes correspondingly. The output voltage is inverted (180∘180^\circ out of phase) relative to the input.

Transistor as a Switch

  • Cut-off (OFF): base current is zero, collector current is (almost) zero, and the transistor behaves like an open switch.
  • Saturation (ON): a large base current drives the collector current to its maximum, and the voltage across the transistor is very small, like a closed switch.

Switching circuits are used in automatic lighting (with a light-dependent resistor) and in digital electronics.


5. The Operational Amplifier (Op-Amp)

An op-amp is an integrated-circuit amplifier with an inverting input (−-), a non-inverting input (++) and one output. Its open-loop gain AOLA_{OL} is very high (around 10510^5 or more), its input resistance is very high and its output resistance is low. Without feedback, the output is

Vo=AOL(V+−V−)V_o = A_{OL}(V_+ - V_-)

and it saturates near the supply voltage for any appreciable difference between the inputs.

CircuitFeedbackGain
Inverting amplifierRfR_f from output to −- input; input through R1R_1G=−RfR1G = -\frac{R_f}{R_1}
Non-inverting amplifierRfR_f from output to −- input; R1R_1 from −- input to ground; signal at ++G=1+RfR1G = 1 + \frac{R_f}{R_1}
ComparatorNo feedbackOutput switches to +Vsat+V_{\text{sat}} or −Vsat-V_{\text{sat}} depending on which input is larger

Example: an inverting amplifier with R1=10 kΩR_1 = 10\text{ k}\Omega and Rf=100 kΩR_f = 100\text{ k}\Omega has G=−10G = -10, so a 0.2 V input gives −2.0 V-2.0\text{ V}. A non-inverting amplifier with the same resistors has G=11G = 11.

A comparator with a thermistor or light-dependent resistor in a voltage divider forms a night switch or temperature alarm: the output changes state when the sensor voltage crosses a reference.


6. Logic Gates

Digital circuits use two voltage levels, written 1 (high) and 0 (low).

ABANDORNANDNORXORXNOR
00001101
01011010
10011010
11110001

The NOT gate (inverter) has one input and outputs its complement: Aˉ\bar{A}.

GateBoolean expressionRule of thumb
ANDX=A⋅BX = A \cdot B1 only if all inputs are 1
ORX=A+BX = A + B1 if any input is 1
NANDX=A⋅B‾X = \overline{A \cdot B}0 only if all inputs are 1
NORX=A+B‾X = \overline{A + B}1 only if all inputs are 0
XORX=A⊕B=ABˉ+AˉBX = A \oplus B = A\bar{B} + \bar{A}B1 if the inputs differ
XNORX=A⊕B‾X = \overline{A \oplus B}1 if the inputs are equal

Universal gates: NAND alone (or NOR alone) can build every other gate. Joining both inputs of a NAND gate makes a NOT gate; a NAND followed by a NOT gives AND.

De Morgan's theorems for logic, A+B‾=Aˉ⋅Bˉ\overline{A + B} = \bar{A}\cdot\bar{B} and A⋅B‾=Aˉ+Bˉ\overline{A \cdot B} = \bar{A} + \bar{B}, mirror the set-theory laws in Section 2.1. Logic gates are used in alarms (for example, sounding if a door is open AND the system is armed), in arithmetic circuits and in computer memory.

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Choosing the Operating Mode of a Transistor
Test Your Knowledge

In a common-emitter circuit the base current is 40 μA40\,\mu\text{A} and the collector current is 6.0 mA6.0\text{ mA}. What are the current gain and the emitter current?

A

β=150\beta = 150 and IE=6.04 mAI_E = 6.04\text{ mA}

B

β=150\beta = 150 and IE=5.96 mAI_E = 5.96\text{ mA}

C

β=0.0067\beta = 0.0067 and IE=6.04 mAI_E = 6.04\text{ mA}

D

β=240\beta = 240 and IE=6.40 mAI_E = 6.40\text{ mA}

Test Your Knowledge

An op-amp is used as a non-inverting amplifier with R1=2 kΩR_1 = 2\text{ k}\Omega and Rf=18 kΩR_f = 18\text{ k}\Omega. What is the output voltage for an input of 0.3 V?

A

2.7 V

B

-2.7 V

C

3.0 V

D

-3.0 V

Test Your Knowledge

A two-input logic gate gives output 1 for every input combination except A = 1, B = 1, when its output is 0. Which gate is it?

A

NOR

B

XOR

C

AND

D

NAND

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