Semiconductors

0.0(0)
Studied by 0 people
call kaiCall Kai
Locked
learnLearn
examPractice Test
spaced repetitionSpaced Repetition
heart puzzleMatch
flashcardsFlashcards
GameKnowt Play
Card Sorting

1/7

encourage image

There's no tags or description

Looks like no tags are added yet.

Last updated 11:47 AM on 8/4/26
Name
Mastery
Learn
Test
Matching
Spaced
Call with Kai
Chat

No analytics yet

Send a link to your students to track their progress

8 Terms

1
New cards

Schrödinger equation

iℏ∂ψ∂t=−ℏ22m∇2ψ+Vψiℏ∂t∂ψ​=−2mℏ2​∇2ψ+Vψ

  • ψ(x,t) = wavefunction; |ψ|² = probability density of finding a particle at (x,t).

  • Stable solutions only exist for fixed energy eigenstates, ψₙ(x,t) → these are atomic orbitals (s, p, d, f).

Molecular orbitals

  • When multiple nuclei are present, atomic orbitals combine into molecular orbitals → chemical bonds.

  • Example: H₂ — two 1s orbitals combine into a bonding (lower energy, in-phase) and antibonding (higher energy, out-of-phase) orbital.

Filling of orbitals (left-step periodic table)

  • Electrons have spin up/down → max 2 electrons per orbital (Pauli exclusion).

  • Atoms are neutral (electrons = protons).

  • Orbitals fill in increasing energy order: 1s, 2s, 2p, 3s, 3p, 4s, 3d, 4p, 5s, 4d, 5p, 6s, 4f, 5d, 6p, 7s...


Quantum Basis of Band Structure

2
New cards
  • In a solid, many atoms → energy levels of eigenstates blur into continuous energy bands.

  • Metals: a band is partially filled at all temperatures → very high conductivity (e.g., Na has a partially filled 3s band).

  • Semiconductors: small band gap between a full valence band and empty conduction band.

    • At normal temps, some valence electrons are thermally excited into the conduction band → become free (current-carrying) electrons.

    • This leaves holes in the valence band, which behave like mobile positive charges (a neighbouring electron hops into the hole).

  • Insulators: same picture as semiconductors but with a much larger band gap (negligible thermal excitation).


Energy Bands

3
New cards
  • Above 0 K, some electrons are randomly knocked loose from Si lattice bonds → electron–hole pairs.

  • Applying an electric field lets both electrons and holes contribute to current flow.

Doping — adding small amounts of impurity atoms changes conductivity dramatically:

Type

Impurity example

Effect

n-type

Antimony (Sb) — 5 valence electrons

Donates an extra free electron

p-type

Boron (B) — 3 valence electrons

Creates an extra hole (acceptor)


Semiconductors and Doping

4
New cards
  • Joining p-type and n-type material creates a p-n junction = basic diode.

  • Current flows easily in one direction (forward biased) and is blocked in the other (reverse biased).

Depletion region

  • At the junction, free electrons from n-side diffuse over and recombine with holes → forms a region of immobile charged ions with no free carriers.

  • Unbiased junction: depletion region ≈ 1 µm.

Forward bias (+ to p-side, – to n-side)

  • Holes pushed toward junction from p-side, electrons pushed toward junction from n-side.

  • Depletion region shrinks; electrons and holes recombine at junction → large continuous current.

Reverse bias (– to p-side, + to n-side)

  • Carriers pulled away from junction → depletion region widens.

  • Current stops except for a tiny thermally-generated reverse current (~10⁻⁸ A) until breakdown voltage (typically –20 V to –700 V).

Diode I–V characteristic

  • Forward turn-on ≈ 0.5 V, significant conduction above ≈ 0.7 V (LEDs: 1.2–3.6 V).

  • Reverse current: tiny and roughly voltage-independent until breakdown.

  • Diode = two-terminal, nonlinear, passive device.


The P-N Junction (Diode)

5
New cards

Half-wave rectifier

  • Single diode; only lets through the half-cycle when source exceeds diode's ~0.6 V forward drop.

  • Output (V_load) is a rectified copy of the AC input's positive half-cycles only.

Full-wave bridge rectifier

  • 4 diodes; current always passes through 2 diodes in series.

  • V_load(max) ≈ V_source(max) – 1.2 V (two diode drops).

  • V_load ≈ 0 whenever –1.2 V ≤ V_source ≤ 1.2 V (dead zone from diode drops).

Smoothing with a capacitor

  • A capacitor in parallel with the load smooths the ripple: charges near the peak, discharges (supplies load current) between peaks.

  • Ripple current relation: i=C dvdt ≈ Vload/Rload​​

  • Ripple voltage:

    Vripple = δv ≈ (Vload/RloadC)δt

    where δt = ripple period = 1/f (half-wave) or 1/(2f) (full-wave).


Diode Circuits — Rectifiers

6
New cards
  • 3 doped regions: emitter, base (very thin, ~10 µm), collector (largest, dissipates most heat).

  • NPN: n-type emitter / p-type base / n-type collector.

  • PNP: p-type emitter / n-type base / p-type collector.

  • Small base current controls a much larger collector current: IC=βIB (β typically 50–200) → current amplification.

  • Base–emitter junction: forward biased (VBE ≈ 0.6 V typical).

  • Base–collector junction: reverse biased; large current still flows because base is thin and field is high.

Modes of operation (NPN)

Mode

Condition

Behaviour

Active

V_C > V_B > V_E (base-emitter forward, base-collector reverse)

Current amplifier: I_C = βI_B

Saturation

V_C < V_B > V_E

Switch ON — high conduction C→E

Cut-off

V_C > V_B < V_E

Switch OFF — no conduction

Reverse-active

V_C < V_B > V_E (rarely used)

Determining the mode — procedure

  1. Is base-emitter forward biased (V_BE > 0.6 V or I_B > 0)?

    • Nocut-off: I_C = I_B = I_E = 0.

    • Yes → active or saturation:

  2. Compute: IC,max=VCC−VBE/RC, IB=VBB−VBE/RB

  3. If IBβ>IC,maxsaturation (I_C = I_C,max, acts as closed switch). If IBβ<IC,maxactive (I_C = βI_B, acts as amplifier).

Worked examples (β=100, V_CC=10 V, V_BB=5 V, R_B=22 kΩ):

  • R_C = 1 kΩ → I_C,max = 9.4 mA; I_B = 200 µA; I_Bβ = 20 mA > 9.4 mA → Saturation, I_C = 9.4 mA (switch ON).

  • R_C = 100 Ω → I_C,max = 94 mA; I_Bβ = 20 mA < 94 mA → Active, I_C = 20 mA (amplifier).

BJT as a current source (LED driver)

  • Choose R_B to set I_B = I_C/β for desired LED current.

  • Example: I_C = 20 mA, V_CC=V_BB=5 V, β=100 → I_B = 200 µA → R_B = (5–0.6)/200µA = 22 kΩ.

  • To stay in active mode: RC<VCC−VLED−VBE/IC​​ (e.g. < 190 Ω).

Emitter follower

  • Output taken at emitter; V_out ≈ V_in – 0.6 V (follows input, one diode drop down).

  • High input impedance, low output impedance → current/power gain but no voltage gain (unity-gain buffer).

Common emitter amplifier

  • DC collector voltage: VC=VCC−ICRC

  • AC input signal capacitively coupled; C chosen so C≥1/2πf(R1∥R2)​ (passes frequencies of interest).

  • AC base perturbation v_in = v_B ≈ v_E (since V_BE ≈ const), and since β is large, i_C ≈ i_E = v_E/R_E.

  • Output: vout=vC=−iCRC= −vinRC/RE

  • Voltage gain = –R_C/R_E (inverting amplifier).


Bipolar Junction Transistor (BJT)

7
New cards
  • 3 terminals: gate, drain, source (+ substrate/body).

  • Gate = insulating oxide layer grown on p- or n-type substrate → forms a planar capacitor.

  • No current flows through the gate (insulated) — voltage-controlled device, unlike current-controlled BJT.

  • n-channel MOSFET: n-type drain/source in p-type substrate. p-channel: reverse.

Gate operation (n-channel)

  • V_GS = 0: no depletion, no channel.

  • 0 < V_GS < V_th: depletion layer forms (holes pushed away, immobile negative ions) — still no conducting channel.

  • V_GS > V_th (threshold voltage): inversion layer forms — high concentration of electrons at the oxide/semiconductor interface → conducting channel exists.

Three modes of operation (n-channel)

Mode

Condition

Behaviour

Cutoff

V_GS < V_th

No conduction — switch OFF

Linear (triode)

V_GS > V_th and V_DS < V_GS – V_th

Acts like a voltage-controlled resistor

Saturation

V_GS > V_th and V_DS > V_GS – V_th

Current flows but only weakly depends on V_DS; mainly set by V_GS. Pinch-off: inversion layer separates from drain as V_DS increases

  • V_th typically 0.5 V – 5 V, device-specific.

  • I_D vs V_DS curves: for each V_GS – V_th, current rises linearly then flattens off (saturates) — family of curves fan out with increasing V_GS – V_th.

MOSFET as a digital switch (LED driver example)

  • V_th = 2 V. Gate driven digitally (0 V / 5 V).

    • V_GS = 0 V → cutoff (LED off).

    • V_GS = 5 V → saturation (LED on), MOSFET behaves as a near-zero-resistance switch.

  • LED current set by supply and series resistor:

    RD=VCC−VLED/ID

    e.g. V_CC = 5 V, I_D = 20 mA → R_D = (5 – 0.6)/0.02 = 220 Ω.


MOSFET (Metal-Oxide-Semiconductor FET)

8
New cards



BJT

MOSFET

Control terminal

Base — current controlled

Gate — voltage controlled

Input current

Significant (I_B)

~Zero (insulated gate)

Key equation

I_C = βI_B

I_D depends on V_GS – V_th

"On" state

Saturation

Linear/Saturation depending on use

Typical use

Analog amplification, switching

Switching (digital logic, power), amplification


BJT vs MOSFET