krit.club logo

Electronic Devices - Semiconductor Diodes and Rectifiers

Grade 12ICSEPhysics

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

Energy Bands in Solids: In semiconductors, the valence band is completely filled and the conduction band is empty at 0 K0 \text{ K}. The forbidden energy gap EgE_g is small (e.g., Eg1.1 eVE_g \approx 1.1 \text{ eV} for SiSi and Eg0.7 eVE_g \approx 0.7 \text{ eV} for GeGe).

Intrinsic and Extrinsic Semiconductors: Pure semiconductors are intrinsic (ne=nh=nin_e = n_h = n_i). Extrinsic semiconductors are formed by doping: nn-type (pentavalent impurities like AsAs, PP) where nenhn_e \gg n_h, and pp-type (trivalent impurities like BB, AlAl) where nhnen_h \gg n_e.

Mass Action Law: In thermal equilibrium, the product of the concentrations of electrons and holes is constant: nenh=ni2n_e n_h = n_i^2.

pp-nn Junction Formation: When pp and nn type materials are joined, diffusion of charge carriers creates a depletion region (devoid of mobile carriers) and a barrier potential VBV_B (approx. 0.7 V0.7 \text{ V} for SiSi).

Biasing: Forward bias (p to positive, n to negative) decreases the depletion layer width and allows current flow. Reverse bias (p to negative, n to positive) increases the depletion layer width and prevents current flow, except for a tiny leakage current IsI_s.

Half-Wave Rectifier: Uses a single diode to conduct during only one half-cycle of the ACAC input. The output frequency is equal to the input frequency (fout=finf_{out} = f_{in}).

Full-Wave Rectifier: Uses two diodes (center-tapped) or four diodes (bridge) to conduct during both half-cycles. The output frequency is twice the input frequency (fout=2finf_{out} = 2f_{in}).

Zener Diode: A heavily doped diode designed to operate in the reverse breakdown region (Zener breakdown) to act as a voltage regulator.

📐Formulae

nenh=ni2n_e n_h = n_i^2

I=Ie+Ih=eA(neve+nhvh)I = I_e + I_h = e A (n_e v_e + n_h v_h)

σ=e(neμe+nhμh)\sigma = e(n_e \mu_e + n_h \mu_h)

rd=ΔVΔIr_d = \frac{\Delta V}{\Delta I}

ηHWR=0.406RLrf+RL40.6%\eta_{HWR} = \frac{0.406 R_L}{r_f + R_L} \approx 40.6\%

ηFWR=0.812RLrf+RL81.2%\eta_{FWR} = \frac{0.812 R_L}{r_f + R_L} \approx 81.2\%

γ=(IrmsIdc)21\gamma = \sqrt{\left(\frac{I_{rms}}{I_{dc}}\right)^2 - 1}

💡Examples

Problem 1:

A pure Silicon crystal has 5×1028 atoms/m35 \times 10^{28} \text{ atoms/m}^3. It is doped by 1 ppm1 \text{ ppm} concentration of pentavalent Arsenic. Calculate the number of electrons and holes given ni=1.5×1016 m3n_i = 1.5 \times 10^{16} \text{ m}^{-3}.

Solution:

The number of donor atoms ND=1106×5×1028=5×1022 m3N_D = \frac{1}{10^6} \times 5 \times 10^{28} = 5 \times 10^{22} \text{ m}^{-3}. Since NDniN_D \gg n_i, we assume neND=5×1022 m3n_e \approx N_D = 5 \times 10^{22} \text{ m}^{-3}. Using Mass Action Law: nh=ni2ne=(1.5×1016)25×1022=2.25×10325×1022=4.5×109 m3n_h = \frac{n_i^2}{n_e} = \frac{(1.5 \times 10^{16})^2}{5 \times 10^{22}} = \frac{2.25 \times 10^{32}}{5 \times 10^{22}} = 4.5 \times 10^9 \text{ m}^{-3}.

Explanation:

Doping with pentavalent atoms makes it an nn-type semiconductor where the electron concentration is approximately equal to the donor concentration.

Problem 2:

A full-wave rectifier is fed with a sinusoidal input of frequency 50 Hz50 \text{ Hz}. What is the frequency of the output ripple?

Solution:

For a Full-Wave Rectifier (FWR), the output frequency is given by fout=2×finf_{out} = 2 \times f_{in}. Therefore, fout=2×50 Hz=100 Hzf_{out} = 2 \times 50 \text{ Hz} = 100 \text{ Hz}.

Explanation:

In a full-wave rectifier, both the positive and negative halves of the ACAC input cycle are converted into unidirectional output pulses, effectively doubling the frequency of the fluctuations (ripples).

Semiconductor Diodes and Rectifiers Revision - Class 12 Physics ICSE