Semiconductor Electronics — NEET notes

Class 12 Physics · NCERT chapter 14 · 8 topics · free to read

Semiconductors Basics

Semiconductors are materials with conductivity between conductors and insulators. By adding tiny impurities (doping), we can control whether current flows via negative electrons (n-type) or positive holes (p-type). Silicon and germanium — the backbone of every electronic device you own — are semiconductors.

n_e n_h = n_i^2

Energy Bands

When billions of atoms come together in a solid, their discrete energy levels merge into continuous bands. The gap between the valence and conduction bands decides everything.

  • Valence band — highest energy band that is completely or partially filled with electrons at 0 K
  • Conduction band — next higher band, empty or partially filled; electrons here conduct electricity
  • Forbidden energy gap E_g — energy range between valence and conduction bands where NO electron states exist
  • Metals: E_g = 0 (bands overlap or conduction band partially filled)
  • Semiconductors: E_g 1 eV (Si: 1.1 eV, Ge: 0.67 eV)
  • Insulators: E_g > 3 eV (diamond: 5.47 eV)
E_g = 0 (no gap)E_g 1 eV (small gap)
Conductivity: 10^2 to 10^8 S/mConductivity: 10^ -5 to 10^ 0 S/m
Conductivity decreases with temperatureConductivity increases with temperature
Free electrons always availableFree electrons created by thermal excitation

Intrinsic Semiconductors

A pure semiconductor with no impurities. At absolute zero it behaves like an insulator; at room temperature, thermal energy creates electron-hole pairs.

An intrinsic semiconductor is a chemically pure semiconductor (Si or Ge) without any dopant atoms. At 0 K, the valence band is completely full and the conduction band is completely empty — it behaves like an insulator.

n_e = n_h = n_i (intrinsic carrier concentration)

  • At 0 K: perfect insulator (no free carriers)
  • At room temperature: n_i 1.5 10^ 16 m^ -3 for Si
  • Both electrons and holes contribute to current: I = I_e + I_h
  • Conductivity is low because n_i is small compared to metals ( 10^ 28 m^ -3 )
  • Hole movement = valence electrons hopping from one vacancy to the next (like musical chairs)

n-Type Semiconductor

Adding a group 15 element (P, As, Sb) to silicon provides an extra free electron per dopant atom. Electrons become majority carriers; holes become minority carriers.

When a pentavalent atom (5 valence electrons) like phosphorus (P), arsenic (As), or antimony (Sb) replaces a Si atom in the crystal lattice, four of its electrons form bonds with neighboring Si atoms. The fifth electron is loosely bound and easily freed at room temperature — becoming a free conduction electron.

  • Pentavalent dopants (P, As, Sb) are called donor impurities — they donate an extra electron
  • Majority carriers: electrons | Minority carriers: holes
  • n_e n_h but n_e n_h = n_i^2 (mass action law still holds)
  • The material is still electrically neutral — the donor atom becomes a positive ion, but the crystal has no net charge
  • Donor energy level sits just below the conduction band ( 0.01 to 0.05 eV)

n_e n_h = n_i^2 (mass action law)

p-Type Semiconductor

Adding a group 13 element (B, Al, In, Ga) creates a vacancy (hole) in the bonding structure. Holes become majority carriers.

When a trivalent atom (3 valence electrons) like boron (B), aluminium (Al), or indium (In) replaces a Si atom, it can form only 3 bonds. The missing 4th bond creates a hole — a vacancy that can accept an electron from a neighboring bond.

  • Trivalent dopants (B, Al, Ga, In) are called acceptor impurities — they accept an electron, creating a hole
  • Majority carriers: holes | Minority carriers: electrons
  • n_h n_e but n_e n_h = n_i^2 still holds
  • Acceptor energy level sits just above the valence band ( 0.01 to 0.05 eV)
  • Still electrically neutral — acceptor atom becomes a negative ion, balanced by the holes
Dopant: pentavalent (P, As, Sb)Dopant: trivalent (B, Al, In)
Majority: electronsMajority: holes
Minority: holesMinority: electrons
Donor level near conduction bandAcceptor level near valence band
Extra electron is easily freedMissing bond creates a hole

Quick recall

If n_i = 1.5 10^ 16 m ^ -3 and you dope Si to get n_h = 5 10^ 22 m ^ -3 , what is n_e ?

n_e = n_i^2 / n_h = (1.5 10^ 16 )^2 / (5 10^ 22 ) = 4.5 10^ 9 m ^ -3 . Minority carriers are negligible compared to majority carriers.

p-n Junction

When p-type and n-type semiconductors are joined, electrons and holes diffuse across the boundary and recombine, creating a charge-depleted region with a built-in electric field. This depletion region is the heart of every diode, LED, solar cell, and transistor in existence.

V_ barrier 0.7 V (Si), 0.3 V (Ge)

Junction Formation

Majority carriers diffuse across the junction and recombine, leaving behind immobile ions that create a depletion region and an internal electric field.

  1. n-side has excess electrons; p-side has excess holes — a concentration gradient exists across the junction
  2. Electrons diffuse from n → p; holes diffuse from p → n (diffusion current)
  3. Near the junction, electrons and holes recombine and disappear
  4. Left behind: fixed positive donor ions on the n-side and fixed negative acceptor ions on the p-side
  5. These immobile ions create an electric field (E) pointing from n → p (positive to negative ions)
  6. This E-field opposes further diffusion — it drifts minority carriers back across
  7. Equilibrium: diffusion current = drift current → net current = 0
  • Depletion region: the zone near the junction depleted of free charge carriers
  • Width of depletion region: typically 0.1 to 1 m
  • Depletion region extends more into the lightly doped side
  • Barrier potential (built-in voltage): 0.7 V for Si, 0.3 V for Ge
  • The junction is NOT formed by physically pressing two pieces together — it's made in a single crystal by doping different regions

Forward Bias

Connecting p-side to positive terminal and n-side to negative terminal narrows the depletion region and allows significant current to flow.

In forward bias, the external voltage opposes the built-in potential barrier. The p-side is connected to the positive terminal of the battery, and the n-side to the negative terminal. This pushes majority carriers TOWARD the junction.

  • External voltage opposes barrier potential → effective barrier = V_ barrier - V_ external
  • Depletion region narrows (majority carriers pushed toward junction)
  • Diffusion current increases exponentially; drift current stays roughly constant
  • Significant current flows when V > V_ knee ( 0.7 V for Si, 0.3 V for Ge)
  • Forward resistance is very low ( 10 to 100 )

I = I_0 ( e^ eV/k_BT - 1 )

The diode equation I = I_0(e^ eV/k_BT - 1) is called the Shockley diode equation. I_0 is the reverse saturation current (very small, A for Ge, nA for Si). This is in NCERT but rarely tested numerically in NEET.

Reverse Bias

Connecting p-side to negative terminal and n-side to positive terminal widens the depletion region. Only a tiny reverse saturation current flows due to minority carriers.

In reverse bias, the external voltage adds to the built-in barrier. The p-side connects to the negative terminal, pulling holes away from the junction. The n-side connects to the positive terminal, pulling electrons away. The depletion region widens.

  • External voltage adds to barrier potential → effective barrier = V_ barrier + V_ external
  • Depletion region widens (majority carriers pulled away from junction)
  • Diffusion current drops to nearly zero; drift current of minority carriers persists
  • Reverse saturation current I_0 : tiny ( A for Ge, nA for Si), nearly independent of reverse voltage
  • Reverse resistance is very high ( 10^5 to 10^6 )
p → +ve, n → −ve terminalp → −ve, n → +ve terminal
Opposes barrier → barrier decreasesAdds to barrier → barrier increases
Depletion region narrowsDepletion region widens
Large current (mA range)Tiny current ( A/nA range)
Low resistanceVery high resistance

Diode Characteristics

A p-n junction diode is a one-way valve for current: low resistance in forward bias, extremely high resistance in reverse bias. The V-I characteristic curve — with its knee voltage, dynamic resistance, and breakdown region — tells the complete story of how a diode behaves.

r_d = V I

V-I Characteristic Curve

The graph of current vs. voltage for a p-n junction diode shows highly asymmetric behavior — easy flow in one direction, almost zero in the other.

The V-I characteristic of a diode is obtained by varying the applied voltage and measuring the resulting current. The curve has three distinct regions: forward bias (right side), reverse bias (left side), and breakdown (far left).

V-I Characteristic of a p-n Junction DiodeVI0Forwardbias0.7V (Si)0.3V (Ge)Knee voltageReverse saturationcurrent (I0)Breakdownregion-V_BRForward V (+)Reverse V (-)I (mA)I (uA)Highly asymmetric: easy flow forward, blocked in reverse, breakdown at high reverse V
  • Forward bias region: Current is negligible until V V_ knee , then rises sharply (exponentially)
  • Knee voltage (cut-in voltage): 0.7 V for Si, 0.3 V for Ge — below this, current is practically zero
  • Reverse bias region: Only a tiny reverse saturation current I_0 flows (minority carriers)
  • Breakdown region: At high reverse voltage, current suddenly surges — the diode 'breaks down'
  • The curve is NOT symmetric — this asymmetry is what makes diodes useful as rectifiers

Dynamic Resistance

Unlike a fixed resistor, a diode's resistance changes with the operating point. Dynamic (AC) resistance is the slope of the V-I curve at any given point.

A diode does NOT obey Ohm's law — its resistance is not constant. We define two types of resistance:

r_d = V I (dynamic/AC resistance)

R_ static = V I (static/DC resistance)

  • Dynamic resistance r_d : resistance to small AC signals superimposed on a DC bias — given by the slope of V-I curve
  • Static resistance R : simply V/I at any operating point
  • In forward bias: r_d is small (steep curve → small V for large I ) — typically 1 to 100
  • In reverse bias: r_d is very large (flat curve → large V for almost zero I ) — typically 10^5 to 10^6
  • At knee voltage: r_d changes rapidly — the diode transitions from high to low resistance

Breakdown Voltage

At sufficiently high reverse voltage, the diode suddenly allows a large reverse current. Two mechanisms cause this: Zener breakdown and avalanche breakdown.

When the reverse bias voltage exceeds a critical value called the breakdown voltage V_ BR , the reverse current increases dramatically. The diode doesn't actually 'break' — the process is reversible if current is limited.

  • Zener breakdown: occurs in heavily doped junctions at low voltages ( < 6 V). The strong electric field directly pulls electrons out of covalent bonds (field ionization)
  • Avalanche breakdown: occurs in lightly doped junctions at high voltages ( > 6 V). Accelerated minority carriers knock out bonding electrons in chain-reaction collisions
  • Both mechanisms rapidly increase minority carrier concentration → large reverse current
  • Breakdown is reversible — the diode is not damaged if current is limited by an external resistor
  • Ordinary diodes are operated below V_ BR ; Zener diodes are designed to operate AT V_ BR
Heavy doping, narrow depletion regionLight doping, wide depletion region
Low breakdown voltage ( < 6 V)High breakdown voltage ( > 6 V)
Direct field ionization of bondsChain-reaction collisions (impact ionization)
V_ BR decreases with temperatureV_ BR increases with temperature

Diode Applications

The one-way behavior of diodes makes them perfect for converting AC to DC (rectifiers), emitting light (LEDs), detecting light (photodiodes), and harvesting solar energy (solar cells). Every phone charger uses a rectifier; every screen uses LEDs.

V_ dc = V_0 (HWR), V_ dc = 2V_0 (FWR)

Half-Wave Rectifier

A single diode blocks the negative half of AC, converting it into pulsating DC. Simple but wasteful — half the input power is thrown away.

A half-wave rectifier uses a single diode in series with the load. During the positive half-cycle, the diode is forward-biased and current flows. During the negative half-cycle, the diode is reverse-biased and blocks current. The output is a pulsating DC with the negative halves chopped off.

Half-Wave Rectifier CircuitAC InputTrans-formerDR_LV_outInput and Output WaveformsACInputf_in = 50 HzDCOutputf_out = 50 HzNegative half blockedV_dc = V0/pi | Efficiency ~ 40.6% | 1 diode | Heavy ripple
  • Uses 1 diode and a transformer
  • Output frequency = input frequency ( f_ out = f_ in = 50 Hz for Indian mains)
  • DC component: V_ dc = V_0/ 0.318 V_0
  • Efficiency: 40.6 % (theoretical maximum)
  • Ripple is very high — needs heavy filtering for practical use

Full-Wave Rectifier

Two diodes with a center-tapped transformer (or four diodes in a bridge) flip the negative half-cycle to positive, doubling the DC output compared to half-wave.

A full-wave rectifier uses two diodes with a center-tapped transformer. During each half-cycle, one diode conducts while the other blocks. Both halves of the AC input contribute to the output — the negative half is 'flipped' to positive.

Full-Wave Rectifier (Center-Tapped Transformer)Trans-formerCTD1D2R_LV_outInput and Output WaveformsACInputDCOutputf_out = 100 Hz (double)Both halves usedV_dc = 2V0/pi | Efficiency ~ 81.2% | 2 diodes + CT | Less ripple
  • Uses 2 diodes + center-tapped transformer (or 4 diodes in bridge configuration)
  • Output frequency = 2 input frequency ( f_ out = 100 Hz for 50 Hz mains)
  • DC component: V_ dc = 2V_0/ 0.636 V_0 — double the half-wave value
  • Efficiency: 81.2 % (double the half-wave efficiency)
  • Much less ripple → easier to filter into smooth DC
1 diode2 diodes (or 4 in bridge)
f_ out = f_ inf_ out = 2f_ in
V_ dc = V_0/V_ dc = 2V_0/
Efficiency 40.6 %Efficiency 81.2 %
High rippleLower ripple

LED, Photodiode & Solar Cell

LEDs emit light when forward biased, photodiodes detect light in reverse bias, and solar cells generate voltage from sunlight. All are p-n junctions engineered to interact with photons.

LED (Light Emitting Diode): In forward bias, electrons and holes recombine at the junction, releasing energy as photons (light). The wavelength (color) depends on the band gap of the semiconductor material — NOT silicon (Si has indirect band gap).

  • LED materials: GaAs (infrared), GaAsP (red/yellow), GaP (green), InGaN (blue/white)
  • LED forward voltage: 1.5 to 3.5 V (higher than Si diode because of wider band gap)
  • LEDs are more efficient than incandescent bulbs — less heat, more light
  • Photodiode: operated in reverse bias. Incident light creates electron-hole pairs in the depletion region → reverse current increases proportional to light intensity
  • Solar cell: similar to a photodiode but operated at zero bias (no external voltage). Light generates a voltage across the junction (photovoltaic effect)
Forward biasedReverse biased / Zero bias
Electrical energy → lightLight → electrical energy
Wide band gap materials (GaAs, GaN)Si, Ge commonly used
Application: displays, indicatorsApplication: sensors, power generation

Zener Diode

A specially designed diode that operates in reverse breakdown — and loves it. The Zener diode maintains a nearly constant voltage across it regardless of current changes, making it the simplest and most elegant voltage regulator. Think of it as a pressure relief valve for voltage.

V_ out = V_Z = constant

Zener Breakdown Mechanism

In a heavily doped p-n junction, the depletion region is so thin that even a modest reverse voltage creates an enormous electric field — enough to rip electrons directly from covalent bonds.

A Zener diode is a p-n junction diode with heavy doping on both sides. This makes the depletion region extremely thin. Even a small reverse voltage creates a very strong electric field across this thin region — strong enough to directly pull valence electrons out of their bonds (field ionization or tunneling).

  • Heavy doping → very thin depletion region → very strong E -field even at low reverse voltage
  • Zener breakdown voltage V_Z : can be designed from 2 V to 200 V by controlling doping level
  • More doping → thinner depletion → lower V_Z
  • The symbol for a Zener diode has bent edges on the cathode bar (like a Z-shape)
  • Unlike ordinary diodes, Zener diodes are designed to operate in breakdown continuously

Zener as Voltage Regulator

By connecting a Zener diode in reverse bias with a series resistor, any fluctuations in input voltage are absorbed by the resistor while the Zener maintains constant voltage across the load.

A voltage regulator maintains a constant output voltage despite variations in input voltage or load current. The Zener diode does this beautifully: once in breakdown, the voltage across it stays at V_Z regardless of current changes.

Zener Diode Voltage Regulator CircuitRsR_LV_out = V_ZV_in+-Zener(V_Z)I = I_Z + I_LI_ZI_LHow It RegulatesV_in increases: excess V drops across Rs, I_Z increases, V_out stays at V_ZI_L increases: I_Z decreases to compensate, total I through Rs stays sameFormula: V_in = I x Rs + V_Z and I = I_Z + I_L
  1. Zener diode is connected in reverse bias across the load R_L
  2. Series resistor R_s is placed between the unregulated input V_ in and the Zener
  3. When V_ in increases: excess voltage drops across R_s , Zener current I_Z increases, but V_ out = V_Z stays constant
  4. When V_ in decreases: less drop across R_s , Zener current decreases, but V_ out = V_Z remains constant
  5. If load changes: Zener adjusts its current to compensate, keeping V_ out = V_Z

V_ in = I R_s + V_Z and I = I_Z + I_L

  • V_ out = V_Z as long as the Zener remains in breakdown
  • The series resistor R_s absorbs voltage fluctuations: V_ R_s = V_ in
  • Zener current adjusts automatically: if I_L increases, I_Z decreases (and vice versa) to keep total current through R_s roughly constant
  • For regulation to work: V_ in must always be greater than V_Z
  • Zener must carry a minimum current to stay in breakdown — design constraint

Quick recall

A Zener diode with V_Z = 6 V is connected with R_s = 200 to a 9 V supply. If load draws 10 mA, what is I_Z ?

Total current: I = (V_ in - V_Z)/R_s = (9-6)/200 = 15 mA. Since I = I_Z + I_L : I_Z = 15 - 10 = 5 mA.

Transistor

A transistor is three semiconductor layers (npn or pnp) that can amplify weak signals or switch circuits on and off. A tiny base current controls a much larger collector current — like a small tap handle controlling a huge water pipe. Transistors are the building blocks of every computer chip.

I_E = I_B + I_C

Transistor Structure

A transistor has three doped regions (emitter, base, collector) forming two p-n junctions. The base is extremely thin and lightly doped — this is the key to transistor action.

A transistor (transfer + resistor) is a semiconductor device with three alternately doped regions: Emitter (E), Base (B), and Collector (C). It comes in two types: npn (n-p-n sandwich) and pnp (p-n-p sandwich). Two p-n junctions are formed: the emitter-base (EB) junction and the base-collector (BC) junction.

  • Emitter: heavily doped, injects majority carriers into the base. Arrow in symbol shows emitter current direction
  • Base: very thin ( 1 m) and lightly doped — most carriers pass through without recombining
  • Collector: moderately doped, physically larger area to collect carriers that pass through the base
  • In npn: emitter emits electrons → most pass through thin base → collected by collector
  • In pnp: emitter emits holes → most pass through thin base → collected by collector
  • Arrow on emitter: points FROM p TO n (conventional current direction)
Emitter: n, Base: p, Collector: nEmitter: p, Base: n, Collector: p
Majority carriers: electronsMajority carriers: holes
Arrow points away from baseArrow points toward base
More commonly used (electrons are faster)Less common in practice
V_ CE and I_C are positiveV_ CE and I_C are negative

Transistor Action

For active operation: EB junction is forward biased, BC junction is reverse biased. The thin, lightly doped base lets most emitter carriers pass through to the collector.

In the active region (normal amplifier operation), the EB junction is forward biased and the BC junction is reverse biased. This is the fundamental biasing condition for transistor action.

  1. Forward-biased EB junction → emitter injects majority carriers (electrons for npn) into the thin base
  2. Base is very thin and lightly doped → only 2-5 % of injected carriers recombine in the base (this is I_B )
  3. The remaining 95-98 % of carriers diffuse across the base to the BC junction
  4. Reverse-biased BC junction has a strong field that sweeps these carriers into the collector (this is I_C )
  5. Result: a tiny I_B controls a much larger I_C

I_E = I_B + I_C

  • I_E = I_B + I_C (Kirchhoff's current law at the transistor)
  • I_C I_B (typically I_B is 2-5 % of I_E )
  • Current amplification factor = I_C / I_B (typically 20 to 500)
  • = I_C / I_E (typically 0.95 to 0.99)
  • Relation: = / (1 - ) and = / (1 + )

= I_C I_B , = I_C I_E , = 1 -

Quick recall

If = 0.98 for a transistor, what is ?

= /(1- ) = 0.98/0.02 = 49 . A small change in causes a large change in : if = 0.99 , then = 99 .

CE, CB, CC Configurations

Depending on which terminal is common to input and output, we get Common Emitter (CE), Common Base (CB), or Common Collector (CC). CE is the most used because it gives both current and voltage gain.

A transistor has three terminals but four connection points (two for input, two for output). One terminal must be common to both input and output circuits. This gives three configurations:

Common terminalEmitter / Base / Collector
Input signalBase-Emitter / Emitter-Base / Base-Collector
Output signalCollector-Emitter / Collector-Base / Emitter-Collector
Current gain(high, 20-500) / (<1) / (1+ ) (highest)
Voltage gainHigh / Highest / <1 (buffer)
Power gainHighest / Moderate / Low
Phase reversalYes (180°) / No / No
  • CE (Common Emitter): most used — high current gain, high voltage gain, highest power gain, 180° phase shift
  • CB (Common Base): best for high-frequency applications — < 1 (no current gain), highest voltage gain, no phase shift
  • CC (Common Collector): also called emitter follower — current gain (1 + ) , voltage gain 1 , used as impedance buffer
  • NEET almost exclusively tests CE configuration

Transistor as Amplifier & Switch

In the active region, a transistor amplifies — a small input signal at the base produces a large output signal at the collector. In cutoff and saturation, it acts as a switch (OFF and ON). Every computer processor contains billions of transistors working as tiny switches.

A_v = - R_C R_B

CE Amplifier

A common emitter amplifier uses a transistor in active mode to amplify AC signals. A small AC signal at the base produces a much larger (and phase-inverted) AC signal at the collector.

In a CE amplifier, a small AC signal V_ BE is applied to the base-emitter circuit. This causes a small change I_B in base current, which produces a much larger change I_C = I_B in collector current. This amplified current through R_C creates a large voltage change at the output.

CE Amplifier: Input vs Output WaveformSmall AC input at base produces large, phase-inverted AC output at collectorInput(V_BE)SmallsignalOutput(V_CE)LargesignalInput peak= Output trough180 degrees Phase InversionVoltage gain: |Av| = beta x R_C / R_B | Amplified AND inverted
  1. DC bias: V_ CC and R_B set the operating point (Q-point) in the active region
  2. AC input signal v_i causes small variation V_ BE around the Q-point
  3. This produces I_B = V_ BE / r_ in (input resistance)
  4. I_C = I_B (current amplification)
  5. Output voltage: v_o = - I_C R_C = - I_B R_C
  6. Negative sign → output is 180° out of phase with input (CE phase inversion)

A_v = v_o v_i = - R_C R_B

  • Voltage gain: |A_v| = R_C / R_B — can be > 100
  • Current gain: A_i = = I_C / I_B
  • Power gain: A_P = A_v A_i = ^2 R_C / R_B
  • The negative sign means output is inverted (180° phase shift) — when input goes up, output goes down
  • Coupling capacitors block DC but pass AC — they isolate the bias from the signal

Quick recall

A CE amplifier has = 100 , R_C = 5 k , R_B = 500 k . Find the voltage gain.

|A_v| = R_C/R_B = 100 5000/500000 = 1 . Wait — that's only 1! To get higher gain, either increase R_C , decrease R_B , or use a transistor with higher . This shows why component selection matters.

Transistor as Switch

When driven between cutoff (OFF) and saturation (ON), a transistor acts as a fast electronic switch. Billions of these switches form the logic circuits in every processor.

A transistor can operate in three regions: cutoff (both junctions reverse-biased → OFF), active (EB forward, BC reverse → amplifier), and saturation (both junctions forward-biased → ON). For switching, we only use cutoff and saturation.

V_ BE < 0.7 V (below threshold)V_ BE > 0.7 V AND I_B large enough
Both junctions reverse-biasedBoth junctions forward-biased
I_C 0I_C V_ CC /R_C (maximum)
V_ CE V_ CC (full supply)V_ CE 0 (almost zero)
Transistor acts as OPEN switchTransistor acts as CLOSED switch
Output = HIGHOutput = LOW
  • Cutoff: I_B = 0 , I_C = 0 , V_ out = V_ CC → switch is OFF (output HIGH)
  • Saturation: I_B is large, I_C = V_ CC /R_C , V_ out 0 → switch is ON (output LOW)
  • A LOW input produces HIGH output and vice versa — the transistor switch acts as a NOT gate (inverter)!
  • Switching speed: modern transistors switch in nanoseconds — enabling GHz clock speeds in processors
  • Your computer's CPU has billions of transistor switches — each one is either in cutoff or saturation

Logic Gates

Logic gates are electronic circuits that perform Boolean operations — AND, OR, NOT, NAND, NOR. They take binary inputs (0 or 1) and produce binary outputs following strict truth tables. NAND and NOR are 'universal gates' because ANY logic circuit can be built using just one of them. Every computation your phone does is ultimately a combination of these gates.

A B = A + B (De Morgan's)

AND, OR, NOT

These three gates form the basis of all Boolean algebra. AND outputs 1 only when ALL inputs are 1. OR outputs 1 when ANY input is 1. NOT simply flips the input.

AND Gate: Output Y = A B . The output is 1 (HIGH) only when both inputs A and B are 1. If either input is 0, the output is 0.

  • AND truth table: (0,0)→0, (0,1)→0, (1,0)→0, (1,1)→1
  • Boolean expression: Y = A B (also written as A B or simply AB )
  • Real-life AND: car starts only if key is inserted AND brake is pressed

OR Gate: Output Y = A + B . The output is 1 when at least one input is 1. The output is 0 only when both inputs are 0.

  • OR truth table: (0,0)→0, (0,1)→1, (1,0)→1, (1,1)→1
  • Boolean expression: Y = A + B (also written as A B )
  • Real-life OR: alarm rings if door sensor OR motion sensor is triggered

NOT Gate (Inverter): Output Y = A . It simply flips the input. A single transistor in switching mode acts as a NOT gate.

  • NOT truth table: 0→1, 1→0
  • Boolean expression: Y = A (also written as A' or A )
  • This is exactly what the transistor switch does: HIGH in → LOW out, LOW in → HIGH out

NAND & NOR Gates

NAND is AND followed by NOT. NOR is OR followed by NOT. Each alone can implement ANY Boolean function — that's why they're called universal gates. Real-world chips are built almost entirely from NAND gates.

NAND Gate: Output Y = A B . It's an AND gate followed by a NOT gate. Output is 0 only when BOTH inputs are 1 — the exact opposite of AND.

  • NAND truth table: (0,0)→1, (0,1)→1, (1,0)→1, (1,1)→0
  • Opposite of AND: output is 0 only when both inputs are 1
  • NOR truth table: (0,0)→1, (0,1)→0, (1,0)→0, (1,1)→0
  • Opposite of OR: output is 1 only when both inputs are 0
  • NOR expression: Y = A + B
  1. NOT from NAND: Connect both inputs together: Y = A A = A
  2. AND from NAND: NAND + NOT (another NAND as inverter): Y = A B = A B
  3. OR from NAND: NOT each input, then NAND: Y = A B = A + B (by De Morgan's law)

De Morgan's Theorems & Gate Combinations

De Morgan's two theorems let you convert between NAND/NOR and OR/AND expressions. They are essential for simplifying Boolean expressions and identifying equivalent circuits.

A B = A + B (Theorem 1: NAND = OR of NOTs)

A + B = A B (Theorem 2: NOR = AND of NOTs)

  • De Morgan's theorems are used to simplify Boolean expressions and convert between gate types
  • Theorem 1: A B = A + B → a NAND gate is equivalent to an OR gate with inverted inputs
  • Theorem 2: A + B = A B → a NOR gate is equivalent to an AND gate with inverted inputs
  • Can be verified by comparing truth tables of both sides — they always match
  • Extends to more variables: A B C = A + B + C

Quick recall

What gate combination gives the output Y = A + B ?

Apply De Morgan's Theorem 2: A + B = A B = A B . So this combination is equivalent to an AND gate.