Apply electrical laws to analyze the operation of semiconductor devices including direct/indirect band gap semiconductors, solar cells, PN junction diodes, rectifiers, special diodes (Zener, LED, photodiode), and bipolar junction transistors (BJTs) in practical circuits.
Semiconductors are classified based on the alignment of the conduction band minimum (CBM) and valence band maximum (VBM) in k-space (momentum space).
In a direct band gap semiconductor, the CBM and VBM occur at the same k-value (usually \( k = 0 \)). Electrons can transition directly between bands by absorbing or emitting a photon.
| Property | Description |
|---|---|
| Examples | GaAs, GaN, InP, CdTe |
| Light emission | Efficient (fast radiative recombination) |
| Absorption coefficient | High (good for solar cells) |
| Applications | LEDs, laser diodes, solar cells |
In an indirect band gap semiconductor, the CBM and VBM occur at different k-values. Electron transitions require a phonon (lattice vibration) to conserve momentum — making light emission inefficient.
| Property | Description |
|---|---|
| Examples | Si, Ge |
| Light emission | Very inefficient (poor LEDs) |
| Absorption coefficient | Lower (needs thicker material) |
| Applications | Solar cells (Si), transistors, ICs |
| Feature | Direct Band Gap | Indirect Band Gap |
|---|---|---|
| CBM and VBM alignment | Same k-value | Different k-values |
| Photon emission | Efficient | Inefficient (needs phonon) |
| Examples | GaAs, GaN, InP | Si, Ge |
| LEDs / Lasers | Excellent | Poor |
| Solar cells | Thin-film possible | Thicker material needed |
| Cost | Higher | Lower (Si is abundant) |
Silicon has an indirect band gap. When an electron recombines with a hole in Si, the energy is released mostly as heat (phonons) rather than light. This is why LEDs are made from GaAs, GaN, and other direct band gap materials.
| Device | Preferred Material | Reason |
|---|---|---|
| LEDs | GaAs, GaN | Direct band gap → efficient light emission |
| Laser diodes | GaAs, InP | Direct band gap → stimulated emission |
| Solar cells | Si (indirect), GaAs (direct) | Si is cheap; GaAs is efficient |
| Photodetectors | Si, Ge, InGaAs | Depends on wavelength range |
Problem: A GaAs LED emits light at 850 nm. If the band gap of GaAs is 1.42 eV, is the emission possible via a direct transition? Calculate the photon energy.
Solution:
\[ E_{photon} = \frac{hc}{\lambda} = \frac{(6.626 \times 10^{-34})(3 \times 10^8)}{850 \times 10^{-9}} \] \[ E_{photon} = \frac{1.988 \times 10^{-25}}{8.5 \times 10^{-7}} = 2.34 \times 10^{-19} \text{ J} \] \[ E_{photon} = \frac{2.34 \times 10^{-19}}{1.6 \times 10^{-19}} \approx 1.46 \text{ eV} \]Since \( E_{photon} \approx E_g \) and GaAs has a direct band gap, the emission is highly efficient.
Problem: Silicon has \( E_g = 1.1 \) eV. Find the maximum wavelength of light that silicon can absorb.
Solution:
\[ \lambda_{max} = \frac{hc}{E_g} = \frac{1240 \text{ eV·nm}}{1.1 \text{ eV}} \approx 1127 \text{ nm} \]Silicon absorbs light with wavelengths shorter than 1127 nm (visible + near-IR).
A solar cell (photovoltaic cell) is a large-area PN junction that converts sunlight directly into electrical energy via the photovoltaic effect.
| Parameter | Symbol | Description |
|---|---|---|
| Open-circuit voltage | \( V_{oc} \) | Voltage when no current flows (≈ 0.5–0.6 V per Si cell) |
| Short-circuit current | \( I_{sc} \) | Current when terminals are shorted (proportional to light) |
| Fill factor | FF | Measure of squareness of I-V curve (0.7–0.85) |
| Efficiency | \( \eta \) | \( \eta = \dfrac{P_{max}}{P_{in}} = \dfrac{V_{oc} I_{sc} FF}{P_{in}} \) |
| Type | Material | Efficiency | Cost |
|---|---|---|---|
| Monocrystalline Si | Single-crystal Si | 18–22% | High |
| Polycrystalline Si | Multi-crystal Si | 15–18% | Moderate |
| Thin-film | CdTe, CIGS, a-Si | 10–15% | Low |
| Perovskite | Hybrid organic-inorganic | 20–25% | Emerging |
| Advantages | Limitations |
|---|---|
| Renewable, clean energy | Intermittent (weather-dependent) |
| Low maintenance | High initial cost |
| No moving parts | Requires large area |
| Modular (scalable) | Efficiency limited by band gap |
| Long lifespan (20–25 years) | Energy storage needed |
Problem: A solar cell has \( V_{oc} = 0.6 \) V, \( I_{sc} = 3 \) A, and \( FF = 0.8 \). If the incident light power is 2 W, find the efficiency.
Solution:
\[ \eta = \frac{V_{oc} I_{sc} FF}{P_{in}} = \frac{(0.6)(3)(0.8)}{2} = \frac{1.44}{2} = 0.72 = 72\% \](Note: This is unusually high for a real cell; typical Si cells are 15–22%.)
Problem: For a solar cell with \( V_{oc} = 0.62 \) V, \( I_{sc} = 2.5 \) A, and \( FF = 0.78 \), find the maximum power output.
Solution:
\[ P_{max} = V_{oc} \cdot I_{sc} \cdot FF = (0.62)(2.5)(0.78) = 1.21 \text{ W} \]Problem: A solar panel needs to deliver 12 V. If each Si cell provides 0.6 V, how many cells must be connected in series?
Solution:
\[ N = \frac{12}{0.6} = 20 \text{ cells in series} \]Ohm's Law states that the current through a conductor is directly proportional to the voltage across it, provided physical conditions (temperature, etc.) remain constant.
Where:
Components connected end-to-end. Current is the same through all components.
Components connected across the same two points. Voltage is the same across all components.
| Quantity | Series | Parallel |
|---|---|---|
| Current (I) | Same through all | Divides among branches |
| Voltage (V) | Divides across components | Same across all |
| Equivalent Resistance | \( R_{eq} = \sum R_i \) | \( \dfrac{1}{R_{eq}} = \sum \dfrac{1}{R_i} \) |
Problem: A resistor of \( 220\,\Omega \) is connected across a 12 V battery. Find the current and power dissipated.
Solution:
\[ I = \frac{V}{R} = \frac{12}{220} = 0.0545 \text{ A} = 54.5 \text{ mA} \] \[ P = VI = 12 \times 0.0545 = 0.654 \text{ W} \]The algebraic sum of currents entering and leaving a node (junction) is zero. Equivalently: total current entering a node = total current leaving it.
Basis: Conservation of charge — charge cannot accumulate at a node.
The algebraic sum of all voltages around any closed loop in a circuit is zero. Equivalently: sum of voltage rises = sum of voltage drops.
Basis: Conservation of energy — a charge returning to its starting point has no net energy change.
| Traversal Direction | Voltage Source | Resistor |
|---|---|---|
| From − to + (rise) | +V | — |
| From + to − (drop) | −V | — |
| Along current direction | — | −IR (drop) |
| Against current direction | — | +IR (rise) |
Problem: A series circuit has a 12 V battery and two resistors \( R_1 = 4\,\Omega \), \( R_2 = 2\,\Omega \). Find the current and voltage across each resistor.
Solution:
Total resistance: \( R_{eq} = 4 + 2 = 6\,\Omega \)
Current: \( I = V/R_{eq} = 12/6 = 2\,\text{A} \)
Voltage drops: \( V_1 = IR_1 = 2 \times 4 = 8\,\text{V} \), \( V_2 = IR_2 = 2 \times 2 = 4\,\text{V} \)
KVL check: \( 12 - 8 - 4 = 0 \) ✓
Problem: At a node, \( I_1 = 5 \) A enters, \( I_2 = 2 \) A leaves, and \( I_3 \) leaves. Find \( I_3 \).
Solution:
\[ \sum I_{\text{in}} = \sum I_{\text{out}} \Rightarrow 5 = 2 + I_3 \Rightarrow I_3 = 3 \text{ A} \]In a series circuit, the voltage across each resistor is proportional to its resistance. This is a direct application of Ohm's Law and KVL.
In a parallel circuit, the current through each branch is inversely proportional to its resistance. This is a direct application of Ohm's Law and KCL.
Voltage divider: The resistor you want the voltage across goes in the numerator.
Current divider: The other resistor goes in the numerator (opposite rule).
Problem: A 12 V source is connected across \( R_1 = 3\,\Omega \) and \( R_2 = 6\,\Omega \) in series. Find the voltage across each resistor.
Solution:
\[ V_{R_1} = 12 \cdot \frac{3}{3+6} = 12 \cdot \frac{3}{9} = 4\,\text{V} \] \[ V_{R_2} = 12 \cdot \frac{6}{3+6} = 12 \cdot \frac{6}{9} = 8\,\text{V} \]KVL check: \( 4 + 8 = 12 \) ✓
Problem: A 6 A current splits between \( R_1 = 2\,\Omega \) and \( R_2 = 4\,\Omega \). Find \( I_1 \) and \( I_2 \).
Solution:
\[ I_1 = 6 \cdot \frac{4}{2+4} = 6 \cdot \frac{4}{6} = 4\,\text{A} \] \[ I_2 = 6 \cdot \frac{2}{2+4} = 6 \cdot \frac{2}{6} = 2\,\text{A} \]KCL check: \( I_1 + I_2 = 4 + 2 = 6\,\text{A} \) ✓
Problem: A circuit has a 10 V source, \( R_1 = 2\,\Omega \) in series with a parallel combination of \( R_2 = 6\,\Omega \) and \( R_3 = 3\,\Omega \). Find the total current and the current through each resistor.
Solution:
Parallel combination of \( R_2 \) and \( R_3 \):
\[ R_p = \frac{R_2 R_3}{R_2 + R_3} = \frac{6 \times 3}{6+3} = 2\,\Omega \]Total resistance:
\[ R_{total} = R_1 + R_p = 2 + 2 = 4\,\Omega \]Total current:
\[ I = \frac{V}{R_{total}} = \frac{10}{4} = 2.5\,\text{A} \]Current through \( R_2 \) and \( R_3 \) (current divider):
\[ I_{R_2} = 2.5 \times \frac{3}{6+3} = 0.833\,\text{A} \qquad I_{R_3} = 2.5 \times \frac{6}{6+3} = 1.667\,\text{A} \]Check: \( I_{R_2} + I_{R_3} = 0.833 + 1.667 = 2.5 \) A ✓
Problem: A loop has a 12 V source, a 3 V source opposing it, and resistors \( R_1 = 2\,\Omega \) and \( R_2 = 3\,\Omega \). Find the current.
Solution:
Net voltage: \( V_{net} = 12 - 3 = 9 \) V
\[ I = \frac{V_{net}}{R_1 + R_2} = \frac{9}{5} = 1.8 \text{ A} \]A PN junction is formed when a p-type semiconductor is brought into intimate contact with an n-type semiconductor. At the boundary, a depletion region forms due to diffusion of carriers.
| Semiconductor | Barrier Potential (V₀) |
|---|---|
| Silicon (Si) | ≈ 0.7 V |
| Germanium (Ge) | ≈ 0.3 V |
| Gallium Arsenide (GaAs) | ≈ 1.2 V |
External voltage applied such that the p-side connects to the positive terminal and the n-side to the negative terminal.
External voltage applied such that the p-side connects to the negative terminal and the n-side to the positive terminal.
Where:
When \( V \gg V_T \):
When \( V \) is negative and large:
Ratio of DC voltage to DC current at the operating point.
Slope of the V-I curve at the operating point — much smaller than \( R_{DC} \).
| Parameter | Ideal Diode | Practical Diode |
|---|---|---|
| Forward resistance | Zero | Small (few Ω to kΩ) |
| Reverse resistance | Infinite | Very high but finite |
| Forward voltage drop | Zero | 0.7 V (Si), 0.3 V (Ge) |
| Reverse saturation current | Zero | nA (Si), µA (Ge) |
| Breakdown | Never | Occurs at rated V_BR |
| Mechanism | Description | Occurs In |
|---|---|---|
| Zener breakdown | Direct rupture of covalent bonds by strong electric field | Heavily doped junctions (V_BR < 6 V) |
| Avalanche breakdown | Carrier multiplication by impact ionization | Lightly doped junctions (V_BR > 6 V) |
Problem: A silicon diode has \( I_0 = 10 \) nA and \( \eta = 2 \). Find the current at \( V = 0.5 \) V. (\( V_T = 26 \) mV)
Solution:
\[ I = I_0 \left( e^{V / \eta V_T} - 1 \right) \] \[ I = 10 \times 10^{-9} \left( e^{0.5 / (2 \times 0.026)} - 1 \right) \] \[ I = 10 \times 10^{-9} \left( e^{9.615} - 1 \right) \approx 10 \times 10^{-9} \times 1.5 \times 10^4 \] \[ I \approx 1.5 \times 10^{-4} \, \text{A} = 0.15 \, \text{mA} \]Problem: A diode operates at \( I = 2 \) mA with \( \eta = 2 \). Find its dynamic resistance at room temperature.
Solution:
\[ r_d = \frac{\eta V_T}{I} = \frac{2 \times 0.026}{2 \times 10^{-3}} = 26 \,\Omega \]A rectifier is a circuit that converts AC (alternating current) into pulsating DC (direct current). Diodes are the key components because they conduct in only one direction.
A single diode conducts during one half-cycle of the input AC.
| Parameter | Formula | Value |
|---|---|---|
| Average (DC) output | \( V_{dc} = V_m / \pi \) | 0.318 \( V_m \) |
| RMS output | \( V_{rms} = V_m / 2 \) | 0.5 \( V_m \) |
| Ripple factor | \( r = 1.21 \) | High ripple |
| Efficiency | \( \eta = 40.6\% \) | Low |
| PIV | \( V_{PIV} = V_m \) | — |
| Output frequency | \( f_{out} = f_{in} \) | Same as input |
Two diodes used with a center-tapped transformer. Each diode conducts during alternate half-cycles.
| Parameter | Formula | Value |
|---|---|---|
| Average (DC) output | \( V_{dc} = 2V_m / \pi \) | 0.636 \( V_m \) |
| RMS output | \( V_{rms} = V_m / \sqrt{2} \) | 0.707 \( V_m \) |
| Ripple factor | \( r = 0.482 \) | Lower ripple |
| Efficiency | \( \eta = 81.2\% \) | High |
| PIV | \( V_{PIV} = 2V_m \) | High |
| Output frequency | \( f_{out} = 2f_{in} \) | Double input |
Four diodes arranged in a bridge configuration. This is the most commonly used rectifier — no center-tapped transformer is needed.
| Parameter | Formula | Value |
|---|---|---|
| Average (DC) output | \( V_{dc} = 2V_m / \pi \) | 0.636 \( V_m \) |
| RMS output | \( V_{rms} = V_m / \sqrt{2} \) | 0.707 \( V_m \) |
| Ripple factor | \( r = 0.482 \) | Lower ripple |
| Efficiency | \( \eta = 81.2\% \) | High |
| PIV | \( V_{PIV} = V_m \) | Lower than center-tap |
| Output frequency | \( f_{out} = 2f_{in} \) | Double input |
| Parameter | Half-Wave | Center-Tap Full-Wave | Bridge Full-Wave |
|---|---|---|---|
| Number of diodes | 1 | 2 | 4 |
| Transformer | Simple | Center-tapped | Simple |
| \( V_{dc} \) | \( V_m/\pi \) | \( 2V_m/\pi \) | \( 2V_m/\pi \) |
| Ripple factor | 1.21 | 0.482 | 0.482 |
| Efficiency | 40.6% | 81.2% | 81.2% |
| PIV | \( V_m \) | \( 2V_m \) | \( V_m \) |
| Output frequency | \( f \) | \( 2f \) | \( 2f \) |
Problem: A full-wave bridge rectifier has an input of 12 V (RMS). Find \( V_{dc} \), ripple factor, and PIV. Assume ideal diodes.
Solution:
Peak voltage: \( V_m = 12 \times \sqrt{2} = 16.97 \) V
\[ V_{dc} = \frac{2V_m}{\pi} = \frac{2 \times 16.97}{\pi} = 10.8 \text{ V} \]Ripple factor = 0.482 (from table).
\[ V_{PIV} = V_m = 16.97 \text{ V} \]A filter is a circuit that removes the AC ripple component from the rectifier output, producing a smoother DC voltage.
The smaller the ripple factor, the better the filter performance.
A large capacitor is connected in parallel with the load. It charges during the rising part of the rectified waveform and discharges through the load during the falling part, smoothing the output.
Where \( f \) = input frequency, \( C \) = filter capacitance, \( R_L \) = load resistance.
| Filter Type | Components | Application |
|---|---|---|
| Inductor (L) filter | Series inductor | High-current loads |
| L-C filter | Inductor + capacitor | Better smoothing |
| π (C-L-C) filter | Capacitor + inductor + capacitor | Best smoothing, low ripple |
| RC filter | Resistor + capacitor | Low-cost, low-current |
After filtering, a Zener diode can be used to maintain a constant output voltage despite variations in input voltage or load current.
Problem: A Zener diode with \( V_Z = 5.1 \) V is used to regulate a 12 V input. If the load current is 20 mA and Zener current is 10 mA, find the series resistance \( R_s \).
Solution:
\[ R_s = \frac{V_{in} - V_Z}{I_L + I_Z} = \frac{12 - 5.1}{(20 + 10) \times 10^{-3}} = \frac{6.9}{0.03} = 230 \,\Omega \]Problem: A capacitor filter has \( C = 100 \) µF and \( R_L = 1 \) kΩ. For a 50 Hz input, find the ripple factor.
Solution:
\[ r \approx \frac{1}{4\sqrt{3} \times 50 \times 100 \times 10^{-6} \times 1000} \] \[ r \approx \frac{1}{4 \times 1.732 \times 50 \times 0.1} = \frac{1}{34.64} \approx 0.029 \]Ripple factor ≈ 2.9%.
A heavily doped PN junction designed to operate in the reverse breakdown region. It maintains a constant voltage across it.
| Application | Description |
|---|---|
| Voltage regulation | Maintains constant output voltage |
| Voltage reference | Provides stable reference voltage |
| Overvoltage protection | Clamps voltage to safe level |
A LED is a forward-biased PN junction that emits light when electrons and holes recombine. Made from direct band gap semiconductors (GaAs, GaN, GaP).
| Color | Material | Forward Voltage |
|---|---|---|
| Red | GaAsP | 1.8 – 2.2 V |
| Green | GaP | 2.0 – 2.4 V |
| Blue | GaN | 3.0 – 3.6 V |
| White | GaN + phosphor | 3.0 – 3.6 V |
Advantages: Low power, long life, fast switching, small size.
A photodiode is a reverse-biased PN junction that converts light into electrical current. When photons strike the depletion region, they generate electron-hole pairs, increasing reverse current.
Where \( \eta \) = quantum efficiency, \( \Phi \) = incident photon flux.
A solar cell is a large-area PN junction that converts sunlight directly into electrical energy via the photovoltaic effect.
| Parameter | Description |
|---|---|
| Open-circuit voltage (\( V_{oc} \)) | ≈ 0.5 – 0.6 V per cell (Si) |
| Short-circuit current (\( I_{sc} \)) | Proportional to light intensity |
| Fill factor (FF) | ≈ 0.7 – 0.85 |
| Efficiency | 15% – 22% (commercial Si) |
Both use direct band gap semiconductors. A LED converts electrical energy into light; a solar cell converts light into electrical energy. They are essentially reverse processes!
| Diode Type | Biasing | Key Feature | Application |
|---|---|---|---|
| Zener | Reverse (breakdown) | Constant voltage | Voltage regulator |
| LED | Forward | Light emission | Displays, lighting |
| Photodiode | Reverse | Light detection | Sensors, optocouplers |
| Solar cell | None (self-generating) | Light → electricity | Solar panels |
Problem: An LED with forward voltage 2 V and forward current 20 mA is connected to a 5 V supply. Find the series resistor required.
Solution:
\[ R = \frac{V_{supply} - V_F}{I_F} = \frac{5 - 2}{20 \times 10^{-3}} = \frac{3}{0.02} = 150 \,\Omega \]Problem: A photodiode has quantum efficiency \( \eta = 0.8 \) and receives \( 10^{17} \) photons per second. Find the photocurrent.
Solution:
\[ I_{ph} = \eta q \Phi = 0.8 \times 1.6 \times 10^{-19} \times 10^{17} \] \[ I_{ph} = 1.28 \times 10^{-2} \text{ A} = 12.8 \text{ mA} \]A BJT consists of three doped semiconductor regions separated by two PN junctions:
By KCL at the transistor.
Problem: An NPN transistor has \( \alpha = 0.98 \). If \( I_E = 10 \) mA, find \( I_C \), \( I_B \), and \( \beta \).
Solution:
\[ I_C = \alpha I_E = 0.98 \times 10 = 9.8 \, \text{mA} \] \[ I_B = I_E - I_C = 10 - 9.8 = 0.2 \, \text{mA} \] \[ \beta = \frac{\alpha}{1-\alpha} = \frac{0.98}{0.02} = 49 \]Problem: A transistor has \( \beta = 100 \). Find \( \alpha \).
Solution:
\[ \alpha = \frac{\beta}{1+\beta} = \frac{100}{101} \approx 0.990 \]| Region | Emitter-Base | Collector-Base | Application |
|---|---|---|---|
| Active | Forward | Reverse | Amplifier |
| Saturation | Forward | Forward | Switch (ON) |
| Cutoff | Reverse | Reverse | Switch (OFF) |
| Inverse Active | Reverse | Forward | Rarely used |
In the active region, a small change in base current produces a large change in collector current. This is the basis of amplification.
| Parameter | Common Base (CB) | Common Emitter (CE) | Common Collector (CC) |
|---|---|---|---|
| Input terminal | Emitter | Base | Base |
| Output terminal | Collector | Collector | Emitter |
| Current gain | \( \alpha \) (< 1) | \( \beta \) (high) | \( \gamma \) (highest) |
| Voltage gain | High | High | < 1 |
| Power gain | Moderate | Highest | Moderate |
| Phase shift | 0° | 180° | 0° |
| Applications | High-frequency | Amplifiers | Impedance matching |
Plot of \( I_E \) vs \( V_{EB} \) at constant \( V_{CB} \).
Plot of \( I_C \) vs \( V_{CB} \) at constant \( I_E \).
Plot of \( I_B \) vs \( V_{BE} \) at constant \( V_{CE} \). Looks like a forward-biased diode curve.
Plot of \( I_C \) vs \( V_{CE} \) at constant \( I_B \).
Also called emitter follower. Voltage gain ≈ 1, current gain high. Used for impedance matching.
| Region | Condition | Behavior |
|---|---|---|
| Cutoff | \( I_B = 0 \) | \( I_C \approx 0 \), transistor OFF |
| Saturation | \( V_{CE} < 0.2 \) V | \( I_C \) maximum, transistor fully ON |
| Active | \( V_{CE} > 0.2 \) V | \( I_C = \beta I_B \), linear amplification |
Problem: An NPN transistor has \( \beta = 100 \), \( V_{CC} = 12 \) V, \( R_C = 1 \) kΩ, \( R_B = 500 \) kΩ. Find the Q-point (\( I_C, V_{CE} \)).
Solution:
\[ I_B = \frac{V_{CC} - V_{BE}}{R_B} \approx \frac{12 - 0.7}{500 \times 10^3} = 22.6 \text{ µA} \] \[ I_C = \beta I_B = 100 \times 22.6 \times 10^{-6} = 2.26 \text{ mA} \] \[ V_{CE} = V_{CC} - I_C R_C = 12 - (2.26 \times 10^{-3})(1000) = 9.74 \text{ V} \]Biasing establishes the proper DC operating point (Q-point) so the transistor operates in the active region for amplification.
\( I_B = (V_{CC} - V_{BE}) / R_B \), \( I_C = \beta I_B \).
Disadvantage: Poor stability — \( I_C \) varies with temperature and β.
Feedback resistor from collector to base improves stability.
Most widely used — excellent stability against temperature and β variations.
Smaller \( S \) → better stability. Voltage divider bias has the lowest \( S \).
| Application | Configuration | Description |
|---|---|---|
| Audio Amplifier | CE | High gain, used in audio equipment |
| RF Amplifier | CB | High-frequency applications |
| Switch | CE (cutoff/saturation) | Digital logic, relay drivers |
| Oscillator | CE with feedback | Signal generation |
| Impedance Matching | CC | Buffer amplifier |
| Concept | Key Formula / Rule |
|---|---|
| Photon energy | \( E = hc/\lambda = 1240/\lambda \text{ (eV·nm)} \) |
| Solar cell efficiency | \( \eta = \dfrac{V_{oc} I_{sc} FF}{P_{in}} \) |
| Diode equation | \( I = I_0(e^{V/\eta V_T} - 1) \) |
| Dynamic resistance | \( r_d = \eta V_T / I \) |
| Barrier potential (Si) | ≈ 0.7 V |
| Barrier potential (Ge) | ≈ 0.3 V |
| Concept | Formula |
|---|---|
| Ohm's Law | \( V = IR \) |
| Power | \( P = VI = I^2R = V^2/R \) |
| Series resistance | \( R_{eq} = R_1 + R_2 + \cdots \) |
| Parallel resistance | \( 1/R_{eq} = 1/R_1 + 1/R_2 + \cdots \) |
| KCL | \( \sum I_{\text{in}} = \sum I_{\text{out}} \) |
| KVL | \( \sum V_{\text{loop}} = 0 \) |
| Voltage divider | \( V_x = V \cdot R_x / R_{eq} \) |
| Current divider | \( I_x = I \cdot R_{\text{other}} / R_{eq} \) |
| Parameter | Half-Wave | Full-Wave |
|---|---|---|
| \( V_{dc} \) | \( V_m/\pi \) | \( 2V_m/\pi \) |
| Ripple factor | 1.21 | 0.482 |
| Efficiency | 40.6% | 81.2% |
| PIV | \( V_m \) | \( V_m \) or \( 2V_m \) |
| Output frequency | \( f \) | \( 2f \) |
| Concept | Formula |
|---|---|
| Current relation | \( I_E = I_B + I_C \) |
| Common-base gain | \( \alpha = I_C / I_E \) |
| Common-emitter gain | \( \beta = I_C / I_B \) |
| Relation | \( \beta = \alpha/(1-\alpha) \) |
| Common-collector gain | \( \gamma = 1 + \beta \) |
| Active region | EB forward, CB reverse |
| Saturation region | Both junctions forward |
| Cutoff region | Both junctions reverse |
Explain why silicon is not used for LEDs while GaAs is. Include band structure in your answer.
A solar cell has \( V_{oc} = 0.62 \) V, \( I_{sc} = 2.5 \) A, and \( FF = 0.78 \). If incident power is 3 W, find the efficiency.
A circuit has a 10 V source, \( R_1 = 2\,\Omega \) in series with a parallel combination of \( R_2 = 6\,\Omega \) and \( R_3 = 3\,\Omega \). Find the total current and the current through each resistor.
A 24 V source is connected across \( R_1 = 4\,\Omega \) and \( R_2 = 8\,\Omega \) in series. Find the voltage across each. Then, if the same resistors are in parallel across a 6 A source, find the current through each.
A silicon diode with \( I_0 = 20 \) nA and \( \eta = 2 \) is forward biased at 0.6 V. Calculate the diode current. (\( V_T = 26 \) mV)
A full-wave bridge rectifier has an input of 12 V (RMS). Find \( V_{dc} \), ripple factor, and PIV. Assume ideal diodes.
A Zener diode with \( V_Z = 5.1 \) V is used to regulate a 12 V input. If the load current is 20 mA and Zener current is 10 mA, find the series resistance \( R_s \).
A transistor has \( \beta = 100 \). If \( I_B = 20 \) µA, find \( I_C \), \( I_E \), and \( \alpha \).
Explain why voltage divider bias is preferred over fixed bias for amplifier circuits. Include stability factor in your answer.
An NPN transistor has \( \beta = 120 \), \( V_{CC} = 15 \) V, \( R_C = 2 \) kΩ, \( R_B = 800 \) kΩ. Find the Q-point (\( I_C, V_{CE} \)).
Silicon has an indirect band gap: the conduction band minimum and valence band maximum occur at different k-values. Electron-hole recombination requires a phonon to conserve momentum, making radiative recombination very inefficient (energy released as heat). GaAs has a direct band gap: CBM and VBM are at the same k-value, allowing efficient photon emission. This is why GaAs is used for LEDs and laser diodes while Si is used for transistors and solar cells.
Parallel combination of \( R_2 \) and \( R_3 \):
\[ R_p = \frac{R_2 R_3}{R_2 + R_3} = \frac{6 \times 3}{6+3} = 2\,\Omega \]Total resistance:
\[ R_{total} = R_1 + R_p = 2 + 2 = 4\,\Omega \]Total current:
\[ I = \frac{V}{R_{total}} = \frac{10}{4} = 2.5\,\text{A} \]Current through \( R_2 \) and \( R_3 \) (current divider):
\[ I_{R_2} = 2.5 \times \frac{3}{6+3} = 0.833\,\text{A} \qquad I_{R_3} = 2.5 \times \frac{6}{6+3} = 1.667\,\text{A} \]Check: \( I_{R_2} + I_{R_3} = 0.833 + 1.667 = 2.5 \) A ✓
Voltage divider (series):
\[ V_{R_1} = 24 \times \frac{4}{4+8} = 8\,\text{V} \qquad V_{R_2} = 24 \times \frac{8}{4+8} = 16\,\text{V} \]Current divider (parallel):
\[ I_{R_1} = 6 \times \frac{8}{4+8} = 4\,\text{A} \qquad I_{R_2} = 6 \times \frac{4}{4+8} = 2\,\text{A} \]Check: \( I_{R_1} + I_{R_2} = 6 \) A ✓
Peak voltage: \( V_m = 12 \times \sqrt{2} = 16.97 \) V
\[ V_{dc} = \frac{2V_m}{\pi} = \frac{2 \times 16.97}{\pi} = 10.8 \text{ V} \]Ripple factor = 0.482.
\[ V_{PIV} = V_m = 16.97 \text{ V} \]Voltage divider bias is preferred because it provides excellent stability against temperature variations and transistor parameter (β) variations. The stability factor S is much smaller (closer to 1) compared to fixed bias (where S = 1 + β, which is very large). This means the Q-point remains stable even if β changes due to temperature or device replacement. Fixed bias is simple but has poor stability because \( I_C = \beta I_B \) depends directly on β.
| Ref | Title | Author | Publisher |
|---|---|---|---|
| T-1 | Principles of Electronics | V. K. Mehta and Rohit Mehta | S. Chand & Company |
| R-1 | Electronic Devices and Circuit Theory | Robert L. Boylestad and Louis Nashelsky | Pearson Education India |
| R-2 | Digital Fundamentals | Thomas L. Floyd | Pearson Education India |
| Ref | Web Address | Feature |
|---|---|---|
| RW-1 | eia.gov/energyexplained/solar | Solar cell basics |
| RW-2 | geeksforgeeks.org/computer-networks | Optical fiber and wireless communication |
| Ref | Topic |
|---|---|
| AV-3 | PN junction diode (working and characteristics) and its applications (rectifiers and switch) |
| AV-4 | BJT basic operations |
| CO | Description | Sections Covered |
|---|---|---|
| CO1 | Explain principles of solid-state physics and semiconductor materials | I, II |
| CO2 | Apply electrical laws to analyze the operation of semiconductor devices | III, IV, V, VI, VII, VIII, IX, X, XI, XII |
Semiconductor Devices & Circuit Laws · Diodes · Rectifiers · BJT · Solar Cells
PHY 175 · Modern Physics and Electronics
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