Target Exam

Formula Bank & Dimensional Analysis

🎯 Formula Mastery Strategy

Don't just memorize formulas. Understand when they fail. Every formula here includes: derivation hint, dimensional check, application context, and common misuse scenarios.

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Intrinsic Carrier Concentration

1. Intrinsic Carrier Concentration (Temperature Dependent)

n_i = AT^3/2 e^-E_g/2kT

Where: n_i = intrinsic carrier concentration, A = material constant, T = absolute temperature (K), E_g = band gap energy, k = Boltzmann constant (1.38 × 10^-23 J/K)

Dimensional Check: [n_i] = [L^-3] ✓ (Number per unit volume)
Key Insight: n_i increases exponentially with temperature. Si has lower n_i than Ge at same T.

2. Mass Action Law (Universal for Semiconductors)

n × p = n_i²

Where: n = electron concentration, p = hole concentration, n_i = intrinsic carrier concentration

Critical Application: Valid for both intrinsic AND extrinsic semiconductors at thermal equilibrium.
JEE Trick: If n increases (doping), p must decrease to keep product constant.

🔬 Exam Usage Pattern

NEET/JEE Main loves: "In an n-type semiconductor, electron concentration is 10¹⁶ cm^-3. If n_i = 10¹⁰ cm^-3, find hole concentration."
Solution: p = n_i²/n = (10¹⁰)²/10¹⁶ = 10⁴ cm^-3

Diode Current-Voltage Relations

3. Ideal Diode Equation (Shockley Equation)

I = I₀(e^eV/kT - 1)

Where: I = diode current, I₀ = reverse saturation current (~10^-12 to 10^-9 A), V = applied voltage, e = electron charge, k = Boltzmann constant, T = temperature

Forward Bias (V > 0): e^eV/kT >> 1, so I ≈ I₀e^eV/kT (exponential growth)
Reverse Bias (V < 0): e^eV/kT ≈ 0, so I ≈ -I₀ (constant saturation current)

4. Dynamic Resistance of Diode

r_d = dV/dI = kT/eI

Where: r_d = dynamic resistance (AC resistance), I = forward bias current

At Room Temperature (T = 300 K): r_d ≈ 26 mV / I
Key Point: Higher current → Lower resistance (non-linear device)

❌ Diode Equation Misuse

Students write: "Diode resistance = V/I (Ohm's law)" WRONG for diodes.
Correct: Use r_d = dV/dI (dynamic resistance) or approximate using ΔV/ΔI from I-V curve.

Rectifier Performance Metrics

5. Half-Wave Rectifier - DC Output Voltage

V_DC = V_m/π

Where V_m = peak AC voltage

6. Half-Wave Rectifier - RMS Output Voltage

V_RMS = V_m/2

7. Half-Wave Rectifier - Efficiency

η = (P_DC/P_AC) × 100% = 40.6%

Maximum theoretical efficiency (assumes ideal diode)

8. Half-Wave Rectifier - Ripple Factor

γ = √((V_RMS/V_DC)² - 1) = 1.21

121% ripple - Very poor DC quality

🔬 JEE Main Pattern

Questions ask: "Peak voltage is 220 V. Find DC output."
Answer: V_DC = 220/π ≈ 70 V
Always remember π ≈ 3.14 (don't use 22/7 unless specified).

9. Full-Wave Rectifier - DC Output Voltage

V_DC = 2V_m/π

Exactly double of half-wave (both half-cycles utilized)

10. Full-Wave Rectifier - RMS Output Voltage

V_RMS = V_m/√2

11. Full-Wave Rectifier - Efficiency

η = (P_DC/P_AC) × 100% = 81.2%

Double the efficiency of half-wave

12. Full-Wave Rectifier - Ripple Factor

γ = 0.48

48% ripple - Much better DC quality than half-wave

🎯 Comparison Trick (Never Forget This)

Full-wave is ALWAYS better than half-wave:
• V_DC: 2× higher
• Efficiency: 2× higher (81.2% vs 40.6%)
• Ripple: 2.5× lower (0.48 vs 1.21)
• Output frequency: 2× input frequency

If exam asks "Which is better?" → Always Full-Wave.

Optoelectronic Device Formulas

13. LED Wavelength from Band Gap

λ = hc/E_g = 1240/E_g (nm)

Where: λ = wavelength (nm), E_g = band gap (eV), h = Planck's constant, c = speed of light
Simplified form: Use 1240 nm·eV for quick calculation

Dimensional Check: [E_g] = [Energy], [λ] = [Length] ✓
Example: For GaAs (E_g = 1.43 eV), λ = 1240/1.43 ≈ 867 nm (Infrared)

14. Photon Energy to Wavelength

E = hν = hc/λ

Where: E = photon energy, ν = frequency, λ = wavelength
Numerical values: h = 6.626 × 10^-34 J·s, c = 3 × 10⁸ m/s

🔬 NEET Favorite Question

Q: "An LED emits red light of wavelength 620 nm. Find its band gap energy in eV."
Solution: E_g = 1240/λ = 1240/620 = 2 eV
This appears in NEET almost every alternate year. Memorize: λE = 1240

LED Color	Wavelength (nm)	Band Gap (eV)	Material
Infrared	900-1000	1.2-1.4	GaAs
Red	620-750	1.6-2.0	GaAsP
Orange	590-620	2.0-2.1	GaAsP
Yellow	570-590	2.1-2.2	GaAsP
Green	500-570	2.2-2.5	GaP
Blue	450-500	2.5-2.8	GaN

Zener Diode Voltage Regulation

15. Series Resistance for Zener Regulation

R_S = (V_in - V_Z)/I_Z

Where: V_in = input voltage, V_Z = Zener voltage, I_Z = Zener current

Application: Zener diode maintains constant output voltage V_Z despite input fluctuations.
Condition: Zener must operate in breakdown region (reverse biased with V > V_Z)

🧠 Reasoning: Why Zener Works as Regulator

In breakdown region, Zener maintains nearly constant voltage across a wide range of currents. This property makes it ideal for voltage regulation. Series resistor R_S drops excess voltage, protecting the load.

Quick Reference: All Key Formulas

Formula	Expression	Unit	Exam Importance
Intrinsic carrier conc.	n × p = n_i²	cm^-3	⭐⭐⭐⭐⭐
Half-wave V_DC	V_m/π	V	⭐⭐⭐⭐
Full-wave V_DC	2V_m/π	V	⭐⭐⭐⭐⭐
Half-wave efficiency	40.6%	%	⭐⭐⭐⭐
Full-wave efficiency	81.2%	%	⭐⭐⭐⭐⭐
LED wavelength	λ = 1240/E_g	nm	⭐⭐⭐⭐⭐
Diode dynamic resistance	kT/eI	Ω	⭐⭐⭐

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