Physics Electronic Devices

Electronic Devices Formula Sheet

All key Electronic Devices formulas for JEE: semiconductors, p-n junction, diodes, rectifiers, Zener, LED, solar cell, transistor & logic gates. Quick revision.

8 min read Updated Jun 2026 #formula sheet#quick revision#jee-main

Last-minute revision sheet for Electronic Devices: every formula, constant, and must-know result from the chapter, grouped by sub-topic for fast scanning.

Semiconductors and Band Theory

Carrier Concentration

QuantityFormulaNotes
Intrinsic balance$n_i = p_i$Pure semiconductor: equal electrons and holes
Intrinsic concentration$n_i^2 = A T^3 e^{-E_g/kT}$$A$ = material constant, increases sharply with $T$
Mass action law$n \times p = n_i^2$Holds for ALL semiconductors at thermal equilibrium
$$\boxed{n \times p = n_i^2}$$

Conductivity

TypeConductivityMajority / Minority
Intrinsic$\sigma_i = n_i e(\mu_e + \mu_h)$$n_i$ (e⁻) / $n_i$ (h⁺)
n-type$\sigma_n = N_D e\mu_e$ (approx)$n_n \approx N_D$ / $p_n = n_i^2/N_D$
p-type$\sigma_p = N_A e\mu_h$ (approx)$p_p \approx N_A$ / $n_p = n_i^2/N_A$

Full (exact) forms before approximation:

$$\sigma_n = n_n e\mu_e + p_n e\mu_h \qquad \sigma_p = p_p e\mu_h + n_p e\mu_e$$
High-yield numbers to memorize

Band gaps: Si $E_g = 1.1$ eV, Ge $E_g = 0.7$ eV, Diamond $E_g = 5.4$ eV.

Thermal energy at 300 K: $kT \approx 0.026$ eV; Boltzmann constant $k = 8.62 \times 10^{-5}$ eV/K.

Si intrinsic concentration: $n_i \approx 1.5 \times 10^{10}$ cm⁻³ at 300 K.

Semiconductors: $\rho \sim 10^{-3}$ to $10^{3}$ Ω·m, resistance DROPS with rising temperature (negative coefficient).

Doping Types

TypeDopant (group)ExamplesMajority carrier
n-typePentavalent (V)P, As, SbElectrons
p-typeTrivalent (III)B, Al, GaHoles

Mnemonic: “5 gives electrons (donor → n-type), 3 takes electrons (acceptor → p-type).”


p-n Junction

Barrier Potential and Depletion Region

$$\boxed{V_0 = \frac{kT}{e}\ln\!\left(\frac{N_A N_D}{n_i^2}\right)}$$

Equivalent form using carrier concentrations ($p_p \approx N_A$, $n_n \approx N_D$):

$$V_0 = \frac{kT}{e}\ln\!\left(\frac{p_p\, n_n}{n_i^2}\right)$$$$\boxed{W = \sqrt{\frac{2\epsilon\epsilon_0 V_0}{e}\left(\frac{1}{N_A} + \frac{1}{N_D}\right)}} \qquad W \propto \sqrt{V_0}$$
QuantityValue / relationNotes
$kT/e$ at 300 K$0.026$ V$\dfrac{1.38\times10^{-23}\times300}{1.6\times10^{-19}}$
Barrier potential $V_0$Si $\approx 0.7$ V, Ge $\approx 0.3$ V, GaAs $\approx 1.2$ VInternal — reads 0 on a voltmeter
Depletion width $W_0$$\sim 10^{-6}$ m (1 µm)Higher doping → narrower $W$
Temperature effect$V_0$ drops $\sim 2$ mV/°C; $I_0$ doubles every 10°C

Diode (Junction) Current

$$\boxed{I = I_0\left(e^{eV/kT} - 1\right)}$$
  • Forward bias ($V > 0.1$ V): $I \approx I_0\, e^{eV/kT}$ (exponential rise)
  • Reverse bias: $I \approx -I_0$ (constant saturation current, $I_0 \sim 10^{-9}$ A for Si)

Bias Behavior

ConditionBarrierDepletion widthCurrent
No bias$V_0$$W_0$$0$
Forward (positive to p)$V_0 - V$ ↓$W$ ↓Large ↑
Reverse (positive to n)$V_0 + \lvert V\rvert$ ↑$W$ ↑$\approx I_0$ (tiny)

Depletion (Junction) Capacitance

$$\boxed{C_j = \frac{\epsilon\epsilon_0 A}{W}} \qquad C_j \propto \frac{1}{\sqrt{V_0 + V_r}}\;\text{(reverse bias)}$$
Conceptual traps that score marks

An ideal voltmeter across an unbiased junction reads ZERO — contact potentials cancel $V_0$.

Reverse current is NOT zero: minority carriers give $I_0$.

Knee/threshold voltage $\approx$ barrier potential numerically (0.7 V Si, 0.3 V Ge) but they are conceptually different.


Diode Characteristics and Rectifiers

Diode Equation and Resistance

$$\boxed{I = I_0\left(e^{eV/\eta kT} - 1\right)} \xrightarrow{\text{300 K}} I \approx I_0\, e^{V/0.026}$$

($\eta$ = ideality factor: 1 ideal, 1–2 real diodes.)

ResistanceFormulaNotes
Static (DC)$r_{dc} = \dfrac{V}{I}$Changes with operating point
Dynamic (AC)$r_{ac} = \dfrac{dV}{dI} = \dfrac{\eta kT}{eI}$At 300 K: $r_{ac} = \dfrac{0.026}{I}$ Ω
$$\boxed{r_{ac} = \frac{0.026}{I}\ \Omega \quad (\text{at } 300\text{ K})}$$

Rectifier Output Formulas

ParameterHalf-WaveFull-Wave
$V_{dc}$$\dfrac{V_m}{\pi} = 0.318\,V_m$$\dfrac{2V_m}{\pi} = 0.636\,V_m$
$V_{rms}$$\dfrac{V_m}{2}$$\dfrac{V_m}{\sqrt{2}}$
Efficiency $\eta$$\dfrac{4}{\pi^2} = 40.6\%$$\dfrac{8}{\pi^2} = 81.2\%$
Ripple factor $r$$1.21$$0.48$
Ripple frequency$f$$2f$
PIV (bridge)$V_m$$V_m$
PIV (center-tap)$2V_m$
$$\boxed{V_{dc}^{\text{half}} = \frac{V_m}{\pi}, \quad V_{dc}^{\text{full}} = \frac{2V_m}{\pi}}$$

Efficiency: $\eta = \dfrac{P_{dc}}{P_{ac}} = \dfrac{V_{dc}^2}{V_{rms}^2}$, so $\eta_{\text{full}} = 2\,\eta_{\text{half}}$ exactly.

Ripple Factor and Capacitor Filter

$$\boxed{r = \sqrt{\left(\frac{V_{rms}}{V_{dc}}\right)^2 - 1}}$$$$V_r = \frac{I_{dc}}{fC} = \frac{V_{dc}}{fC R_L} \qquad r_{\text{full,filtered}} = \frac{1}{2\sqrt{3}\,f C R_L}$$
Rectifier shortcuts

Quick DC estimate: Half-wave $\approx 0.3 \times$ peak; Full-wave $\approx 0.6 \times$ peak.

Real Si diode drops 0.7 V; bridge has TWO conducting diodes per half cycle → subtract 1.4 V.

Lower ripple factor = smoother DC. Larger $C$ or higher $f$ → lower ripple.


Zener Diode and Voltage Regulation

Operated in REVERSE breakdown; cathode to the positive rail.

$$\boxed{V_{out} = V_Z\ (\text{constant in breakdown})}$$$$\boxed{I_s = I_Z + I_L} \qquad I_s = \frac{V_{in} - V_Z}{R_s}, \quad I_L = \frac{V_Z}{R_L}$$$$\boxed{I_Z = \frac{V_{in} - V_Z}{R_s} - \frac{V_Z}{R_L}}$$
QuantityFormulaNotes
Series resistor$R_s = \dfrac{V_{in} - V_Z}{I_s}$$I_s = I_Z + I_L$
Power dissipation$P_Z = V_Z I_Z$Must stay below $P_{Z,max}$
Max current$I_{Z,max} = \dfrac{P_{Z,max}}{V_Z}$Power-limited
Regulation window$I_{Z,min} < I_Z < I_{Z,max}$Both bounds must hold
Output with real $r_z$$V_{out} = V_Z + I_Z r_z$$r_z$ ≈ 5–50 Ω; assume 0 if not given

Minimum load resistance for regulation (worst case, $V_{in} = V_{min}$):

$$R_{L,\min} = \frac{V_Z R_s}{V_{min} - V_Z - I_{Z,min} R_s}$$

Breakdown types: Zener effect (field ionization, $< 5$ V, heavy doping) vs Avalanche (collision chain, $> 5$ V, light doping).

Zener exam checks

Always verify $I_Z > I_{Z,min}$ (else regulation fails) and $P_Z < P_{Z,max}$ (else it burns out).

Common trap: forgetting that $I_s$ adds BOTH $I_Z$ and $I_L$.


Special Diodes: LED, Photodiode, Solar Cell

LED (Light Emitting Diode)

Forward biased; emits a photon per electron-hole recombination.

$$\boxed{E = h\nu = \frac{hc}{\lambda} = E_g} \qquad \boxed{\lambda(\text{nm}) = \frac{1240}{E_g(\text{eV})}}$$$$\boxed{R = \frac{V_s - V_f}{I_f}}\ \text{(series current-limiting resistor)}$$
QuantityValueNotes
Forward voltage $V_f$Red ~1.8 V, Green ~2.0 V, Blue ~3.0 V, White ~3.2 VHigher energy → higher $V_f$
Typical current10–20 mA$P = V_f I$
Planck constant $h$$6.63 \times 10^{-34}$ J·s
Speed of light $c$$3 \times 10^8$ m/s

Si and Ge make poor LEDs (indirect band gap → energy lost as lattice vibration, not light).

Photodiode

Reverse biased; current rises with light intensity.

$$\boxed{I = I_0 + I_L} \qquad \boxed{I_L \propto \Phi} \qquad \boxed{R = \frac{I_L}{P_{opt}}}$$

$I_0$ = dark current (~nA); $I_L$ = photo-current; responsivity $R \approx 0.4$–0.6 A/W.

Solar Cell (Photovoltaic)

Self-powered (no external bias).

$$\boxed{FF = \frac{V_m I_m}{V_{oc} I_{sc}}} \qquad \boxed{\eta = \frac{V_m I_m}{P_{in}} = \frac{FF \cdot V_{oc} \cdot I_{sc}}{P_{in}}}$$
QuantityValueNotes
Open-circuit voltage $V_{oc}$Si ~0.5–0.7 V (≈0.6 V)Set by band gap, ~constant with intensity
Short-circuit current $I_{sc}$$\propto$ area × intensity~30–40 mA/cm² in full sun
Fill factor $FF$0.7–0.85“Squareness” of I-V curve
Commercial Si efficiency15–20%Shockley-Queisser limit (Si) ~33%
Light-device essentials

Master formula: $\lambda(\text{nm}) = 1240 / E_g(\text{eV})$. Example: $E_g = 2$ eV → 620 nm (red).

More light on a solar cell raises CURRENT ($I_{sc}$), not voltage. For higher voltage, wire cells in series.

Photodiode needs an external battery; a solar cell generates its own power.


Transistor (BJT) as Amplifier and Switch

Active mode: base-emitter forward biased, base-collector reverse biased. $V_{BE} \approx 0.7$ V (Si).

Current Relations

$$\boxed{I_E = I_B + I_C} \qquad \boxed{\beta = \frac{I_C}{I_B}} \qquad \boxed{\alpha = \frac{I_C}{I_E}}$$$$\boxed{\beta = \frac{\alpha}{1-\alpha}} \qquad \boxed{\alpha = \frac{\beta}{\beta + 1}}$$

Typical: $\beta = 50$–300, $\alpha = 0.95$–0.99, $I_E > I_C \gg I_B$.

Common-Emitter Amplifier

$$\boxed{A_v = -\frac{R_C}{r_e}} \quad\text{where}\quad r_e = \frac{26\ \text{mV}}{I_E(\text{mA})}$$$$\boxed{A_i = \beta} \qquad \boxed{A_p = A_v \times A_i}$$

The negative sign means 180° phase inversion (only CE inverts; CB and CC do not).

DC bias (voltage-divider): $V_B = \dfrac{R_2}{R_1 + R_2}V_{CC}$, $\;V_E = V_B - V_{BE}$, $\;I_E = \dfrac{V_E}{R_E}$, $\;I_C \approx I_E$, $\;V_C = V_{CC} - I_C R_C$.

Transistor as a Switch

StateCondition$V_{CE}$$I_C$
Cut-off (OFF)$I_B = 0$$V_{CE} = V_{CC}$$\approx 0$
Saturation (ON)$I_B$ large$V_{CE,sat} \approx 0.2$ V$I_{C,sat} = \dfrac{V_{CC}}{R_C}$
$$\boxed{I_{C,sat} = \frac{V_{CC}}{R_C}} \qquad \boxed{R_B < \frac{(V_{in} - V_{BE})\,\beta}{I_{C,sat}}}$$
Transistor exam reminders

In SATURATION do NOT use $I_C = \beta I_B$; use $I_{C,sat} = V_{CC}/R_C$ instead.

Always subtract $V_{BE} = 0.7$ V: $V_E = V_B - 0.7$.

A small change in $\alpha$ causes a large change in $\beta$ ($\alpha: 0.98\to0.99$ gives $\beta: 49\to99$).


Logic Gates and Boolean Algebra

Truth Table (all seven gates)

ABANDORNANDNORXORXNOR
00001101
01011010
10011010
11110001

Boolean Expressions

GateBooleanOutput = 1 when
NOT$Y = \overline{A}$Input is 0
AND$Y = A \cdot B$ALL inputs are 1
OR$Y = A + B$ANY input is 1
NAND$Y = \overline{A \cdot B}$Not all inputs are 1
NOR$Y = \overline{A + B}$All inputs are 0
XOR$Y = A\overline{B} + \overline{A}B = A \oplus B$Inputs are different
XNOR$Y = AB + \overline{A}\,\overline{B} = \overline{A \oplus B}$Inputs are same

NAND and NOR are universal gates (any function can be built from either alone).

De Morgan’s Theorems

$$\boxed{\overline{A + B} = \overline{A} \cdot \overline{B}} \qquad \boxed{\overline{A \cdot B} = \overline{A} + \overline{B}}$$

Boolean Laws

LawIdentity
Identity$A + 0 = A$, $\;A \cdot 1 = A$
Null$A + 1 = 1$, $\;A \cdot 0 = 0$
Idempotent$A + A = A$, $\;A \cdot A = A$
Complement$A + \overline{A} = 1$, $\;A \cdot \overline{A} = 0$
Double negation$\overline{\overline{A}} = A$
Distributive$A(B + C) = AB + AC$
Absorption$A + AB = A$, $\;A(A + B) = A$

XOR Properties

$$A \oplus 0 = A, \quad A \oplus 1 = \overline{A}, \quad A \oplus A = 0, \quad A \oplus \overline{A} = 1$$

Combinational Circuits

CircuitSumCarry
Half adder$S = A \oplus B$$C = A \cdot B$
Full adder$S = A \oplus B \oplus C_{in}$$C_{out} = AB + BC_{in} + AC_{in}$
Logic-gate shortcuts

NAND = flip AND’s output column; NOR = flip OR’s output column.

XOR output: count the 1s in the inputs — odd number of 1s → output 1.

De Morgan’s: “Break the bar, change the operator.”