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Electronic Devices - Semiconductor Diodes and Rectifiers

Grade 12ICSEPhysics

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

🔑Concepts

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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., Eg≈1.1 eVE_g \approx 1.1 \text{ eV} for SiSi and Eg≈0.7 eVE_g \approx 0.7 \text{ eV} for GeGe).

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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 ne≫nhn_e \gg n_h, and pp-type (trivalent impurities like BB, AlAl) where nh≫nen_h \gg n_e.

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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.

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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).

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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.

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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}).

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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}).

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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+RL≈40.6%\eta_{HWR} = \frac{0.406 R_L}{r_f + R_L} \approx 40.6\%

ηFWR=0.812RLrf+RL≈81.2%\eta_{FWR} = \frac{0.812 R_L}{r_f + R_L} \approx 81.2\%

γ=(IrmsIdc)2−1\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 m−3n_i = 1.5 \times 10^{16} \text{ m}^{-3}.

Solution:

The number of donor atoms ND=1106×5×1028=5×1022 m−3N_D = \frac{1}{10^6} \times 5 \times 10^{28} = 5 \times 10^{22} \text{ m}^{-3}. Since ND≫niN_D \gg n_i, we assume ne≈ND=5×1022 m−3n_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 m−3n_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).