Target Exam

Quick Revision

🎯 Perfect for Last 24 Hours

This page contains EVERYTHING you need for quick revision: formulas, concepts, diagrams, and memory tricks. Print this page if needed.

One-Page Concept Summary

Energy Bands

Conductor: E_g ≈ 0 (bands overlap)

Semiconductor: E_g = 0.7-1.5 eV

Insulator: E_g > 3 eV

Si: 1.1 eV | Ge: 0.7 eV

Intrinsic vs Extrinsic

Intrinsic: Pure, n = p = n_i

n-type: Pentavalent (P, As, Sb)

→ Majority: Electrons

p-type: Trivalent (B, Al, In, Ga)

→ Majority: Holes

p-n Junction

Depletion region: No mobile carriers

Forward bias: Width decreases, conducts

Reverse bias: Width increases, blocks

V_th: Si ≈ 0.7V, Ge ≈ 0.3V

Special Diodes

LED: Forward bias, emits light

Photodiode: Reverse bias, detects light

Solar Cell: No bias, generates voltage

Zener: Reverse breakdown, voltage regulation

Formula Dump (Memorize These)

Carrier Concentration

n × p = n_i²

Valid for intrinsic AND extrinsic

LED Wavelength (CRITICAL)

λ(nm) = 1240 / E_g(eV)

Most asked in NEET/JEE Main

Half-Wave Rectifier

V_DC = V_m/π

Efficiency: 40.6% | Ripple: 1.21

Full-Wave Rectifier

V_DC = 2V_m/π

Efficiency: 81.2% | Ripple: 0.48

RMS to Peak

V_m = √2 × V_RMS

√2 ≈ 1.414

Dynamic Resistance

r_d = ΔV/ΔI

NOT V/I (diode is non-ohmic)

Zener Current

I_Z = (V_in - V_Z)/R_S

For voltage regulation

Photon Energy

E = hc/λ = hν

h = 6.63×10^-34 J·s, c = 3×10⁸ m/s

🔬 Memory Trick

1240 Rule: For ALL LED/photodiode problems, use λE = 1240 (λ in nm, E in eV)
π ≈ 3.14: For rectifier calculations
Full = 2× Half: All full-wave values are exactly double of half-wave

Interactive Flashcards

Click card to flip. Use for rapid-fire revision.

Pentavalent impurity creates which type?

n-type semiconductor

Majority carriers: Electrons

LED with E_g = 2 eV emits which wavelength?

620 nm (Red light)

λ = 1240/2 = 620 nm

Full-wave rectifier efficiency?

81.2%

Exactly double of half-wave (40.6%)

NAND(1, 1) = ?

0 (LOW)

NAND gives 0 only when both inputs are 1

Photodiode works in which bias?

Reverse Bias

Faster response, light generates current

Silicon band gap energy?

1.1 eV

Germanium: 0.7 eV

Logic Gate Truth Tables (NEET Must-Know)

AND Gate

A	B	Y
0	0	0
0	1	0
1	0	0
1	1	1

OR Gate

A	B	Y
0	0	0
0	1	1
1	0	1
1	1	1

NOT Gate

A	Y
0	1
1	0

NAND Gate

A	B	Y
0	0	1
0	1	1
1	0	1
1	1	0

NOR Gate

A	B	Y
0	0	1
0	1	0
1	0	0
1	1	0

XOR Gate

A	B	Y
0	0	0
0	1	1
1	0	1
1	1	0

🎯 Memory Tricks

AND: Output 1 only when ALL inputs are 1
OR: Output 1 when ANY input is 1
NAND: Opposite of AND (0 only when both 1)
NOR: Opposite of OR (1 only when both 0)
XOR: Output 1 when inputs are DIFFERENT

Final Checklist: Avoid These Mistakes

❌ Before Exam, Check This List

☐ Remember: n-type and p-type are BOTH electrically neutral
☐ For LED: Use λ = 1240/E_g (NOT E = hc/λ directly)
☐ Check RMS vs Peak voltage (V_m = √2 × V_RMS)
☐ Half-wave vs Full-wave (2× difference in V_DC, η, frequency)
☐ Photodiode works in REVERSE bias (NOT forward)
☐ Dynamic resistance: r_d = ΔV/ΔI (NOT V/I)
☐ Silicon V_th = 0.7 V (NOT 0.3 V, which is Germanium)
☐ NAND(1,1) = 0, NOR(0,0) = 1
☐ In intrinsic: n = p always
☐ Forward bias → depletion width DECREASES
☐ Energy band diagram: Label E_g, CB, VB
☐ Show ALL steps in numerical problems
☐ Write units in final answer

Perfect for last-minute revision

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