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State Board · Class 12 · Physics

Lesson 16: Semiconductor Devices

7 concepts · 56 practice questions with answers · Free to read

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Rectifier, ripple and filter

From daily life

A mobile charger turns wall AC into steady DC for the phone — a rectifier is inside.

Simple definition

A rectifier uses a diode's one-way conduction to turn AC into DC; a filter reduces the ripple.

Rectifier : input and outputAC inputFull-wave outputV_dc = V_m ÷ π (half) ; 2V_m ÷ π (full)

In the figure : Blue AC, green full-wave output; the red line is V_dc.

Live figure

Rectifier, ripple and filter

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Rule 1: Half-wave rectifier — One diode; current flows in half of each AC cycle. V_dc = V_m ÷ π; V_rms = V_m ÷ 2; ripple factor r = 1.21; efficiency 40.6%; ripple frequency = f.

  • Easy : V_m = 15 V ⇒ V_dc = 15 ÷ π = 4.77 V.
  • Medium : V_rms = 15 ÷ 2 = 7.5 V.
  • Exam (CET) : Half-wave V_dc? (1) V_m/π (2) 2V_m/π (3) V_m/2 (4) V_m → answer 1)

Rule 2: Full-wave rectifier — Two diodes (centre-tap) or four (bridge). V_dc = 2V_m ÷ π; V_rms = V_m ÷ √2; r = 0.48; efficiency 81.2%; ripple frequency = 2f.

  • Easy : V_m = 15 V ⇒ V_dc = 30 ÷ π = 9.55 V.
  • Medium : 50 Hz AC ⇒ ripple at 100 Hz in the output.
  • Exam (CET) : Diodes in a bridge? (1) 4 (2) 2 (3) 1 (4) 3 → answer 1) Trick : bridge = four sides.

Rule 3: Filter (capacitor) — A parallel capacitor charges at the peak and discharges in the dips, reducing ripple. ΔV = I ÷ (f_r C), f_r = ripple frequency.

  • Easy : I = 0.1 A, f_r = 100 Hz, C = 1000 μF ⇒ ΔV = 0.1 ÷ (100 × 10⁻³) = 1 V.
  • Medium : C doubled ⇒ ΔV halves = 0.5 V.
  • Exam (CET) : To reduce ripple? (1) increase C (2) decrease C (3) increase I (4) decrease f → answer 1)

Derivation : Half-wave V_dc = V_m ÷ π

  • Step 1 : V = V_m sinθ only from 0 to π; zero from π to 2π.
  • Step 2 : Average = (1 ÷ 2π) ∫₀^π V_m sinθ dθ.
  • Step 3 : ∫₀^π sinθ dθ = 2.
  • Step 4 : V_dc = 2V_m ÷ 2π = V_m ÷ π; doubled for full-wave.
Remember

Half : V_m/π; full : 2V_m/π; ΔV = I/(f_r C).

Common mistake / CET trap

Do not confuse V_m and V_rms; full-wave ripple frequency is 2f.

Shortcut

Memory clue : "Half = ÷π; full = 2÷π".

Spot the difference

QuantityHalf-waveFull-wave
V_dcV_m/π2V_m/π
V_rmsV_m/2V_m/√2
Ripple r1.210.48
Efficiency40.6%81.2%
Ripple ff2f

Practice questions

Easy
  1. 1What is a rectifier?
  2. 2Diodes in a bridge rectifier?
  3. 3What is ripple?
Medium
  1. 4V_m = 20 V : half-wave V_dc?
  2. 5V_m = 20 V : full-wave V_dc?
  3. 6I = 0.2 A, f_r = 100 Hz, C = 2000 μF : ΔV?
Challenging
  1. 760 Hz AC full-wave : ripple f?
  2. 8Full-wave V_dc is how many times half-wave V_dc?
Show answers (try them yourself first)
  1. 1Device that converts AC to DC
  2. 24
  3. 3Residual AC part in DC
  4. 4V_dc = 6.37 V
  5. 5V_dc = 12.73 V
  6. 6ΔV = 1 V
  7. 7120 Hz
  8. 82 times

Zener diode and voltage regulator

From daily life

Like a pressure valve : beyond a set level it holds the voltage steady.

Simple definition

A Zener diode works in the breakdown region and holds the voltage steady.

Zener diode : I – V curveVI−V_zbreakdownforwardR_s = (V_in − V_z) ÷ (I_z + I_L)

In the figure : In reverse bias the current rises steeply at V_z.

Rule 1: Zener diode — Heavily doped; in reverse bias it breaks down at V_z, current rises sharply while the voltage stays nearly V_z. Hence a voltage regulator.

  • Easy : V_z = 6 V ⇒ ≈ 6 V across it after breakdown.
  • Medium : P_z = V_z I_z = 6 × 0.05 = 0.3 W.
  • Exam (CET) : A Zener is used in? (1) reverse bias (2) forward bias (3) AC (4) any → answer 1)

Rule 2: Zener regulator — Series R_s with a Zener, load R_L across the Zener. I = I_z + I_L; R_s = (V_in − V_z) ÷ (I_z + I_L); I_L = V_z ÷ R_L.

  • Easy : V_in = 12 V, V_z = 6 V, I_L = 20 mA, I_z = 10 mA ⇒ R_s = 6 ÷ 0.03 = 200 Ω.
  • Medium : R_L = 300 Ω ⇒ I_L = 6 ÷ 300 = 20 mA.
  • Exam (CET) : If the load opens, I_z? (1) rises (2) falls (3) zero (4) same → answer 1) Trick : the current goes through the Zener.

Rule 3: Zener power limit — I_z,max = P_max ÷ V_z; with no load all current passes through the Zener. Choose R_s so that I_z < I_z,max.

  • Easy : P_max = 0.5 W, V_z = 6 V ⇒ I_z,max = 83 mA.
  • Medium : R_s = 200 Ω, no load ⇒ I = 6 ÷ 200 = 30 mA < 83 mA, safe.
  • Exam (CET) : I_z,max = ? (1) P/V_z (2) P V_z (3) V_z/P (4) P² → answer 1)

Derivation : Formula for R_s

  • Step 1 : Zener and load are in parallel ⇒ both at V_z.
  • Step 2 : Voltage across R_s = V_in − V_z.
  • Step 3 : Current I = I_z + I_L.
  • Step 4 : R_s = (V_in − V_z) ÷ (I_z + I_L).
Remember

R_s = (V_in − V_z)/(I_z + I_L); I_z,max = P/V_z.

Common mistake / CET trap

Connect a Zener in reverse; if I_L = 0 all current flows through the Zener.

Shortcut

Memory clue : "Zener = steady V_z".

Spot the difference

PropertyNormal diodeZener diode
Dopingnormalheavy
Used inforwardreverse (breakdown)
Userectifierregulator
Voltagevariessteady V_z

Practice questions

Easy
  1. 1Use of a Zener diode?
  2. 2What is Zener breakdown?
  3. 3Why is R_s used in a regulator?
Medium
  1. 4V_in = 15 V, V_z = 9 V, I_z + I_L = 30 mA : R_s?
  2. 5V_z = 5 V, R_L = 250 Ω : I_L?
  3. 6P_max = 1 W, V_z = 10 V : I_z,max?
Challenging
  1. 7V_in = 10 V, V_z = 6 V, R_s = 100 Ω, R_L = 300 Ω : I_z?
  2. 8What changes in the regulator when V_in rises?
Show answers (try them yourself first)
  1. 1Keeping voltage steady
  2. 2Sharp rise of current at a reverse voltage
  3. 3To limit current and drop the extra voltage
  4. 4200 Ω
  5. 520 mA
  6. 6100 mA
  7. 7I_z = 20 mA
  8. 8I_z rises; load voltage stays steady

Photodiode and solar cell

From daily life

A calculator's small solar strip runs on light; a TV remote's receiver is a photodiode.

Simple definition

A photodiode senses light; a solar cell converts light into electricity.

I – V curve of a solar cellI_scV_ocP_max = V_m I_mV_mI_mFF = V_m I_m ÷ (V_oc I_sc)VI

In the figure : I–V in the fourth quadrant; the orange rectangle is P_max.

Rule 1: Photodiode — Connected in reverse bias; when light with E = hν ≥ E_g falls, electron-hole pairs form ⇒ photocurrent proportional to intensity. Uses : light sensors, remote receivers.

  • Easy : Si : E_g = 1.1 eV ⇒ λ_max = 1240 ÷ 1.1 = 1127 nm.
  • Medium : Intensity doubled ⇒ photocurrent doubled.
  • Exam (CET) : A photodiode is operated in? (1) reverse bias (2) forward bias (3) zero bias (4) AC → answer 1)

Rule 2: Solar cell — No external bias; light generates an emf. The I–V curve lies in the fourth quadrant. P_max = V_m I_m; FF = V_m I_m ÷ (V_oc I_sc); η = P_max ÷ P_in. Si cell V_oc ≈ 0.5 – 0.6 V.

  • Easy : V_oc = 0.6 V, I_sc = 0.3 A, FF = 0.75 ⇒ P_max = 135 mW.
  • Medium : Area 10 cm², 1000 W/m² ⇒ P_in = 1 W; η = 13.5%.
  • Exam (CET) : External battery in a solar cell? (1) none (2) yes (3) reverse (4) forward → answer 1)

Rule 3: Comparison — Photodiode : light → current (sensor). Solar cell : light → electric power (generator).

  • Easy : Detecting a remote signal ⇒ photodiode.
  • Medium : Powering a calculator ⇒ solar cell.
  • Exam (CET) : For power generation? (1) solar cell (2) photodiode (3) LED (4) Zener → answer 1)

Derivation : Cut-off wavelength λ = 1240 ÷ E_g

  • Step 1 : An electron-hole pair needs hν ≥ E_g.
  • Step 2 : hc ÷ λ ≥ E_g ⇒ λ ≤ hc ÷ E_g.
  • Step 3 : hc = 1240 eV nm.
  • Step 4 : λ_max (nm) = 1240 ÷ E_g (eV).
Remember

λ = 1240/E_g; P_max = V_m I_m; FF = V_m I_m/(V_oc I_sc); η = P_max/P_in.

Common mistake / CET trap

Photodiode uses reverse bias, a solar cell none; for η use P_in = intensity × area (m²).

Shortcut

Memory clue : "Photo = sensor; solar = generator".

Spot the difference

PropertyPhotodiodeSolar cell
Biasreversezero
Functionsensorpower source
Quadrantthirdfourth
Outputphotocurrentemf and current

Practice questions

Easy
  1. 1How is a photodiode connected?
  2. 2Use of a solar cell?
  3. 3Formula for FF?
Medium
  1. 4E_g = 1.8 eV : λ_max?
  2. 5V_oc = 0.5, I_sc = 0.2 A, FF = 0.8 : P_max?
  3. 6P_max = 90 mW, P_in = 900 mW : η?
Challenging
  1. 7Will 800 nm light work with E_g = 1.1 eV?
  2. 8V when solar cells are in series?
Show answers (try them yourself first)
  1. 1In reverse bias
  2. 2Electricity from light
  3. 3V_m I_m / (V_oc I_sc)
  4. 4689 nm
  5. 580 mW
  6. 610%
  7. 7Yes; E = 1.55 eV > 1.1 eV
  8. 8Voltages add

Light emitting diode (LED)

From daily life

A home LED bulb gives more light for less electricity.

Simple definition

An LED emits light when forward biased.

LED : levels and photonE_cE_vE_ghν = E_gλ (nm) = 1240 ÷ E_g (eV)

In the figure : On falling from E_c to E_v a photon hν = E_g is emitted.

Rule 1: Principle of an LED — In a forward-biased p-n junction, electron-hole recombination emits a photon. hν = E_g, λ = hc ÷ E_g. GaAsP (red), GaP (green), GaN (blue).

  • Easy : E_g = 2.0 eV ⇒ λ = 1240 ÷ 2 = 620 nm (orange-red).
  • Medium : E_g = 1.8 eV ⇒ λ = 689 nm.
  • Exam (CET) : An LED works in? (1) forward bias (2) reverse bias (3) zero bias (4) breakdown → answer 1)

Rule 2: Series resistor — An LED needs a current limit. R = (V_s − V_f) ÷ I; V_f ≈ 1.8 – 3.5 V.

  • Easy : V_s = 5 V, V_f = 2 V, I = 15 mA ⇒ R = 3 ÷ 0.015 = 200 Ω.
  • Medium : Power P = I² R = 0.015² × 200 = 0.045 W.
  • Exam (CET) : LED without R? (1) may burn out (2) brighter safely (3) safe (4) reversed → answer 1)

Rule 3: Advantages — Low voltage, low power, long life, instant on, little heat, compact.

  • Easy : 1 W LED instead of 10 W bulb ⇒ 90% saving.
  • Medium : 1 kWh = 3.6 × 10⁶ J.
  • Exam (CET) : LED colour is decided by? (1) E_g (2) R (3) V_s (4) I → answer 1)

Derivation : LED wavelength

  • Step 1 : Energy released on recombination = E_g.
  • Step 2 : hν = E_g ⇒ hc ÷ λ = E_g.
  • Step 3 : λ = hc ÷ E_g.
  • Step 4 : λ (nm) = 1240 ÷ E_g (eV).
Remember

hν = E_g; λ = 1240/E_g; R = (V_s − V_f)/I.

Common mistake / CET trap

E_g decides the colour, not R; larger E_g = shorter λ.

Shortcut

Memory clue : "Forward = light; reverse = dark".

Spot the difference

PropertyLEDPhotodiode
Biasforwardreverse
Functionelectricity → lightlight → electricity
Outputphotonsphotocurrent

Practice questions

Easy
  1. 1Principle of an LED?
  2. 2Two advantages of an LED?
  3. 3Why a resistor with an LED?
Medium
  1. 4E_g = 2.5 eV : λ?
  2. 5V_s = 9 V, V_f = 2 V, I = 20 mA : R?
  3. 6λ = 620 nm : E_g?
Challenging
  1. 7For a blue LED is E_g larger or smaller?
  2. 8Why does an LED not glow in reverse bias?
Show answers (try them yourself first)
  1. 1Light from recombination
  2. 2Low power; long life
  3. 3To limit current
  4. 4496 nm
  5. 5R = 350 Ω
  6. 62.0 eV
  7. 7Larger (shorter λ)
  8. 8No recombination occurs

Bipolar junction transistor

From daily life

A small turn of a tap controls a large flow; a transistor's base works the same way.

Simple definition

A BJT has three regions; a small base current controls a large collector current.

npn transistor : three regions and currentsn (E)pn (C)EmitterCollectorBase (B)I_EI_CI_BI_E = I_B + I_C ; β = I_C ÷ I_B

In the figure : npn : I_E = I_B + I_C.

Rule 1: Structure and biasing — Three regions : emitter E (heavily doped), base B (thin, lightly doped), collector C (largest). Active region : EB forward, CB reverse biased.

  • Easy : npn : E and C are n-type, B is p-type.
  • Medium : pnp : E and C are p-type, B is n-type.
  • Exam (CET) : The base is? (1) thin and lightly doped (2) thick (3) heavily doped (4) same → answer 1)

Rule 2: Current relations — I_E = I_B + I_C; α = I_C ÷ I_E (< 1, ≈ 0.95 – 0.99); β = I_C ÷ I_B (50 – 300); β = α ÷ (1 − α).

  • Easy : I_B = 40 μA, β = 100 ⇒ I_C = 4 mA; I_E = 4.04 mA.
  • Medium : α = I_C ÷ I_E = 0.990.
  • Exam (CET) : β = ? (1) I_C/I_B (2) I_C/I_E (3) I_B/I_C (4) I_E/I_B → answer 1) Trick : β has B below.

Rule 3: α and β relation — α = 0.98 ⇒ β = 0.98 ÷ 0.02 = 49; α = 0.99 ⇒ β = 99. A small rise in α gives a large rise in β.

  • Easy : α = 0.98 ⇒ β = 49.
  • Medium : β = 99 ⇒ α = 99 ÷ 100 = 0.99.
  • Exam (CET) : α = 0.9 ⇒ β? (1) 9 (2) 0.9 (3) 90 (4) 1 → answer 1)

Derivation : β = α ÷ (1 − α)

  • Step 1 : I_E = I_B + I_C.
  • Step 2 : I_C ÷ I_E = α ⇒ I_E = I_C ÷ α.
  • Step 3 : I_B = I_E − I_C = I_C (1 ÷ α − 1) = I_C (1 − α) ÷ α.
  • Step 4 : β = I_C ÷ I_B = α ÷ (1 − α).
Remember

I_E = I_B + I_C; α = I_C/I_E; β = I_C/I_B; β = α/(1−α).

Common mistake / CET trap

α is always below 1; β is far above 1; I_E is the largest.

Shortcut

Memory clue : "β = C over B".

Spot the difference

QuantityFormulaUnit
I_EI_B + I_CA
αI_C/I_Eunitless
βI_C/I_Bunitless
βα/(1−α)unitless

Practice questions

Easy
  1. 1Three parts of a transistor?
  2. 2Biasing in the active region?
  3. 3Formula for α?
Medium
  1. 4I_B = 25 μA, β = 80 : I_C?
  2. 5I_C = 2 mA, I_E = 2.04 mA : I_B?
  3. 6α = 0.96 : β?
Challenging
  1. 7β = 50 : α?
  2. 8I_C = 4.9 mA, I_B = 0.1 mA : I_E and β?
Show answers (try them yourself first)
  1. 1E, B, C
  2. 2EB forward, CB reverse
  3. 3I_C/I_E
  4. 4I_C = 2 mA
  5. 5I_B = 40 μA
  6. 6β = 24
  7. 7α = 0.980
  8. 8I_E = 5 mA; β = 49

Common-emitter configuration and transistor as amplifier

From daily life

The tiny signal from a mic becomes big at the speaker — an amplifier.

Simple definition

An amplifier enlarges a small signal; in CE the output is inverted by 180°.

CE amplifier : 180° phase shiftA_vv_iv_oA_v = β R_L ÷ R_i

In the figure : Blue input, green output : larger and inverted.

Rule 1: CE configuration and gains — The emitter is common to input and output. β_ac = ΔI_C ÷ ΔI_B (V_CE constant); A_v = Δv_o ÷ Δv_i = β R_L ÷ R_i; A_p = β A_v = β² R_L ÷ R_i.

  • Easy : β = 100, R_i = 1 kΩ, R_L = 4 kΩ ⇒ A_v = 400.
  • Medium : A_p = β A_v = 40000.
  • Exam (CET) : Phase shift in CE? (1) 180° (2) 0° (3) 90° (4) 45° → answer 1)

Rule 2: As an amplifier — A small signal at the base varies I_B, hence I_C, giving a large voltage change across R_L. The output is inverted (180°).

  • Easy : Δv_i = 10 mV, A_v = 400 ⇒ Δv_o = 4 V.
  • Medium : v_o and v_i swing in opposite directions.
  • Exam (CET) : A_v rises with? (1) R_L (2) R_i (3) smaller β (4) V_CC → answer 1)

Rule 3: As a switch — Cut-off : I_B = 0 ⇒ I_C ≈ 0 (OFF). Saturation : large I_B ⇒ V_CE ≈ 0.2 V (ON). Used in digital circuits.

  • Easy : V_BE < 0.7 V ⇒ OFF.
  • Medium : V_BE ≥ 0.7 V with enough I_B ⇒ ON.
  • Exam (CET) : ON as a switch in? (1) saturation (2) cut-off (3) active (4) never → answer 1)

Derivation : A_v = β R_L ÷ R_i

  • Step 1 : Δv_i = ΔI_B × R_i.
  • Step 2 : ΔI_C = β ΔI_B.
  • Step 3 : Δv_o = ΔI_C × R_L = β ΔI_B R_L.
  • Step 4 : A_v = Δv_o ÷ Δv_i = β R_L ÷ R_i.
Remember

A_v = βR_L/R_i; A_p = βA_v; phase 180°.

Common mistake / CET trap

The amplifier does not create energy; it comes from V_CC. A_p = β A_v.

Shortcut

Memory clue : "CE = inverted and bigger".

Spot the difference

GainFormulaUnit
β_acΔI_C/ΔI_Bunitless
A_vβR_L/R_iunitless
A_pβ A_vunitless
Phase180°—

Practice questions

Easy
  1. 1Meaning of CE?
  2. 2Formula for A_v?
  3. 3ON state of a switch?
Medium
  1. 4β = 50, R_i = 500 Ω, R_L = 5 kΩ : A_v?
  2. 5A_v = 200, Δv_i = 5 mV : Δv_o?
  3. 6β = 60, A_v = 300 : A_p?
Challenging
  1. 7ΔI_B = 20 μA, ΔI_C = 1.5 mA : β_ac?
  2. 8A_v if R_L is doubled?
Show answers (try them yourself first)
  1. 1Common emitter
  2. 2βR_L/R_i
  3. 3Saturation
  4. 4A_v = 500
  5. 5Δv_o = 1 V
  6. 6A_p = 18000
  7. 7β_ac = 75
  8. 8Doubles

Digital signals and logic gates

From daily life

A light switch : on (1) or off (0) — the simplest digital signal.

Simple definition

A logic gate gives a fixed output for 0/1 inputs.

Basic logic gatesANDORNOTY = A · BY = A + BY = ĀNAND, NOR : universal

In the figure : Symbols and expressions of AND, OR, NOT.

Live figure

Digital signals and logic gates

In the app you can move this figure with your finger.

Open in the app

Rule 1: OR, AND, NOT — OR : Y = A + B (any 1 ⇒ 1). AND : Y = A · B (both 1 ⇒ 1). NOT : Y = Ā (inverts). Truth table (A B : OR AND) : 00 : 0 0; 01 : 1 0; 10 : 1 0; 11 : 1 1.

  • Easy : A = 1, B = 0 ⇒ OR = 1, AND = 0.
  • Medium : A = 1 ⇒ NOT = 0.
  • Exam (CET) : AND output is 1 when? (1) both 1 (2) any 1 (3) both 0 (4) never → answer 1) Trick : AND = both.

Rule 2: NAND, NOR, XOR — NAND : Y = (A·B)‾; NOR : Y = (A+B)‾; XOR : Y = A ⊕ B (different ⇒ 1). NAND and NOR are universal gates; every gate can be built from them.

  • Easy : A = 1, B = 1 ⇒ NAND = 0, NOR = 0, XOR = 0.
  • Medium : A = 0, B = 1 ⇒ NAND = 1, NOR = 0, XOR = 1.
  • Exam (CET) : XOR is 1 when? (1) inputs differ (2) same (3) both 1 (4) both 0 → answer 1)

Rule 3: Digital signals and binary — The signal has only two levels : 0 and 1. Binary : 1011₂ = 8 + 0 + 2 + 1 = 11.

  • Easy : 25 = 16 + 8 + 1 = 11001₂.
  • Medium : 1101₂ = 8 + 4 + 0 + 1 = 13.
  • Exam (CET) : 101₂ = ? (1) 5 (2) 3 (3) 7 (4) 9 → answer 1)

Derivation : NOT and AND from NAND

  • Step 1 : Tie both inputs to A : (A·A)‾ = Ā ⇒ NOT.
  • Step 2 : Pass a NAND output through a NOT : ((A·B)‾)‾ = A·B ⇒ AND.
  • Step 3 : De Morgan : (A·B)‾ = Ā + B̄.
  • Step 4 : NAND of Ā and B̄ gives A + B ⇒ OR.
Remember

OR : A + B; AND : A·B; NOT : Ā; NAND/NOR universal; XOR : different ⇒ 1.

Common mistake / CET trap

NAND is the inverse of AND; NOR of OR; XOR differs from OR at (1,1).

Shortcut

Memory clue : "NAND = 0 only if all inputs are 1".

Spot the difference

A BORANDNANDNORXOR
0 000110
0 110101
1 010101
1 111000

Practice questions

Easy
  1. 1Boolean expression of OR?
  2. 2Which are the universal gates?
  3. 31010₂ = ?
Medium
  1. 4A = 1, B = 0 : NAND?
  2. 5A = 0, B = 0 : NOR?
  3. 618 in binary?
Challenging
  1. 7How do you make NOT from NOR?
  2. 8A = 1, B = 1, C = 0 : (A·B) + C?
Show answers (try them yourself first)
  1. 1Y = A + B
  2. 2NAND, NOR
  3. 310
  4. 41
  5. 51
  6. 610010₂
  7. 7Tie both inputs to A : (A+A)‾ = Ā
  8. 81

Quick revision of the whole lesson

ConceptKey formula / factUnit / dimensions
RectifierV_dc = V_m/π; 2V_m/πV
FilterΔV = I/(f_r C)V
ZenerR_s = (V_in − V_z)/(I_z + I_L)Ω
Photo/solarλ = 1240/E_g; η = P_max/P_innm; —
LEDhν = E_g; R = (V_s − V_f)/IeV; Ω
BJTI_E = I_B + I_C; β = I_C/I_BA; —
CEA_v = βR_L/R_i—
LogicOR A+B; AND A·B; NOT Ā—

🎯 Most important points for the exam

  • Full-wave : V_dc doubled; ripple f = 2f.
  • Zener in reverse; R_s limits current.
  • Photodiode reverse; solar cell no bias.
  • LED forward; λ = 1240/E_g.
  • BJT : EB forward, CB reverse.
  • β = α/(1−α); α < 1.
  • NAND and NOR are universal.
Study mantra

Identify the device → bias → formula → units / truth table.