Semiconductor Electronics formulas

Class 12 physics formula sheet for NEET and JEE: the key equations of NCERT chapter 14, the special cases questions are built on, and diagrams where they help.

27 formulas6 sectionsClass 12 · Chapter 142 of 6 sections free

By Sreeraj P, M.Sc Physics · 10+ years teaching NEET and JEE

Most used formulasOther formulas and cases

1Bands and carriers

Solid	Band gap $E_g$	Resistivity (Ω m)
Conductor	0 (bands overlap / CB half filled)	$10^{-2}$–$10^{-8}$
Semiconductor	< 3 eV (Ge 0.7, Si 1.1, GaAs 1.4 eV)	$10^{-5}$–$10^{6}$
Insulator	> 3 eV (diamond 5.4 eV)	$10^{11}$–$10^{19}$

$$n_en_h=n_i^2,\qquad n_i\ (\text{intrinsic}):\ n_e=n_h=n_i$$

$$\begin{array}{l}\displaystyle n_en_h=n_i^2\\[6pt]\displaystyle n_i\ (\text{intrinsic}):\ n_e=n_h=n_i\end{array}$$

Mass-action law (holds for doped samples too). n-type (pentavalent donor: P, As, Sb): $n_e\approx N_D$, $n_h=n_i^2/N_D$. p-type (trivalent acceptor: B, Al, In, Ga): $n_h\approx N_A$. Both dopants: majority $=|N_D-N_A|$. Doped crystal is still neutral.

$$\sigma=\frac1\rho=e(n_e\mu_e+n_h\mu_h),\qquad I=eA(n_ev_e+n_hv_h),\quad \mu=\frac{v_d}{E}$$

$$\begin{array}{l}\displaystyle \sigma=\frac1\rho=e(n_e\mu_e+n_h\mu_h)\\[6pt]\displaystyle I=eA(n_ev_e+n_hv_h),\quad \mu=\frac{v_d}{E}\end{array}$$

$\mu_e>\mu_h$. Intrinsic: $\sigma=en_i(\mu_e+\mu_h)$.

$$n_i\propto T^{3/2}e^{-E_g/2kT},\qquad \frac{\sigma_2}{\sigma_1}\approx e^{\frac{E_g}{2k}\left(\frac1{T_1}-\frac1{T_2}\right)}$$

$$\begin{array}{l}\displaystyle n_i\propto T^{3/2}e^{-E_g/2kT}\\[6pt]\displaystyle \frac{\sigma_2}{\sigma_1}\approx e^{\frac{E_g}{2k}\left(\frac1{T_1}-\frac1{T_2}\right)}\end{array}$$

Resistance of a semiconductor falls with temperature (negative temperature coefficient). Photon can create an e–h pair if $\dfrac{hc}\lambda\ge E_g$, i.e. $\lambda_{\max}(\text{nm})=\dfrac{1240}{E_g(\text{eV})}$. $kT\approx0.026$ eV at 300 K.

Donor level lies just below CB (~0.01 eV for Ge, 0.05 eV for Si); acceptor level just above VB.

2p–n junction diode

Diffusion of majority carriers → depletion layer of immobile ions → barrier $V_0$ (Ge ≈ 0.3 V, Si ≈ 0.7 V); field points n → p. Diffusion current (majority) balanced by drift current (minority) at equilibrium.
Forward bias (p at higher potential): barrier $V_0-V$, depletion width ↓, current in mA. Reverse bias: barrier $V_0+V$, width ↑, tiny µA current of minority carriers (saturation, rises with $T$, not with $V$).

$$E=\frac{V_0}{d},\qquad I=I_0\left(e^{eV/kT}-1\right),\qquad r=\frac{\Delta V}{\Delta I}$$

$$\begin{array}{l}\displaystyle E=\frac{V_0}{d}\\[6pt]\displaystyle I=I_0\left(e^{eV/kT}-1\right)\\[6pt]\displaystyle r=\frac{\Delta V}{\Delta I}\end{array}$$

Field in the depletion layer; diode equation; dynamic resistance from the curve. Carrier crossing the barrier: needs KE $\ge e(V_0\mp V)$ (− forward, + reverse).

$$I=\frac{E-V_0}{R+r_f}$$

Real diode in series with $R$. Ideal diode: forward → short circuit (0 V), reverse → open circuit. In circuits: first decide each diode's bias (p higher than n → ON), replace, then solve. Parallel Ge and Si: Ge (lower $V_0$) conducts, Si stays off.

4 more sections and 16 formulas in the full chapter

3Rectifiers1 case table · 1 diagram
4Special diodes4 formulas · 1 case table · 1 diagram
5Transistor10 formulas · 1 case table
6Logic gates2 formulas · 1 case table · 1 diagram

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