Unit 09: Semiconductor Electronics • Section 14.3
Rectifiers (Half-Wave & Full-Wave) & Optoelectronics
Complete working principles, mathematical efficiencies, ripple factors, live circuit current path animations, filter circuits, and optoelectronic junction devices.
A. Half-Wave Rectifier (HWR) Single Diode Design
Half-wave rectifier AC voltage ki sirf ek aadhi cycle (positive half-cycle) ko load tak pahunchata hai aur doosri aadhi cycle (negative half-cycle) ko completely block kar deta hai.
1. Positive Half-Cycle ($0 \to T/2$)
Secondary coil ka upper terminal $A$ positive $(+)$ hota hai aur lower terminal $B$ negative $(-)$ hota hai. Diode ka p-side positive se judta hai $\implies$ Diode Forward Biased hota hai (low resistance). Current load resistor $R_L$ mein top-to-bottom flow karta hai aur output voltage develop hota hai.
2. Negative Half-Cycle ($T/2 \to T$)
Polarity reverse hoti hai: terminal $A$ negative $(-)$ aur terminal $B$ positive $(+)$ ho jata hai. Diode ka p-side negative se judta hai $\implies$ Diode Reverse Biased ho jata hai (infinite resistance). Circuit mein koi current flow nahi karta ($I = 0$), aur output voltage zero rehta hai.
B. Full-Wave Center-Tapped Rectifier (FWR) Dual Diode + Center-Tap Transformer
Full-wave rectifier AC input ki dono cycles (positive aur negative) ko utilize karke load resistor $R_L$ mein hamesha ek hi disha mein current banaye rakhta hai.
1. Positive Half-Cycle ($0 \to T/2$)
Center-tap $T$ ($0\text{V}$) ke respect mein terminal $A$ positive $(+)$ aur terminal $B$ negative $(-)$ hota hai.
• Diode $D_1$ is Forward Biased (ON).
• Diode $D_2$ is Reverse Biased (OFF).
Current path: $A \to D_1 \to X \to R_L \to Y \to T$.
2. Negative Half-Cycle ($T/2 \to T$)
Center-tap $T$ ($0\text{V}$) ke respect mein terminal $A$ negative $(-)$ aur terminal $B$ positive $(+)$ ban jata hai.
• Diode $D_1$ is Reverse Biased (OFF).
• Diode $D_2$ is Forward Biased (ON).
Current path: $B \to D_2 \to X \to R_L \to Y \to T$.
Crucial Symmetry: Dono half-cycles mein load resistor $R_L$ ke andar current hamesha ek hi direction ($X \to Y$) mein behta hai. Isliye continuous pulsating DC output milta hai.
How Does a Shunt Capacitor Filter Convert Pulsating DC into Steady DC?
1. Rectifier ka direct output pure DC nahi hota; usme bohot saara AC ripple ($50\text{ Hz}$ ya $100\text{ Hz}$) maujood hota hai.
2. Capacitor ki reactance $X_C = \frac{1}{2\pi f C}$ hoti hai. DC ke liye ($f = 0$) capacitor infinite resistance offer karta hai, jabki AC ripples ke liye low impedance path offer karta hai.
3. Charging Phase: Jab pulsating DC wave $0$ se peak $V_m$ ki taraf badhti hai, diode forward biased rehta hai aur capacitor instantaneously charge hokar maximum voltage $V_m$ store kar leta hai.
4. Discharging Phase: Jab input wave peak $V_m$ se neeche girti hai, diode reverse biased hokar switch off ho jata hai. Ab load resistor $R_L$ ko current transformer nahi balki discharge hota hua capacitor supply karta hai. Agar time constant $\tau = C R_L \gg T/2$ ho, toh capacitor bohot dheere discharge hota hai. Agli cycle ka agla pulse aane se pehle voltage barely gir pata hai, jisse load par lagbhag smooth steady DC voltage milta hai!
| Device | Biasing Mode | Physical Working Principle | I-V Operating Quadrant | Key Material / Condition |
|---|---|---|---|---|
| Photodiode | Reverse Bias | Incident light ($h\nu \ge E_g$) generates electron-hole pairs inside depletion layer. Built-in electric field separates them, causing reverse saturation current $\Delta I \propto \text{Intensity}$. | 3rd Quadrant (Negative V, Negative I) | Operated in reverse bias because fractional change in minority carrier current ($\Delta p/p$) is vastly higher and easier to measure than in majority carriers. |
| Light Emitting Diode (LED) | Forward Bias | Injected electrons and holes recombine near junction. Conduction band electrons fall into valence band holes, releasing energy as photons of frequency $\nu \approx E_g/h$. | 1st Quadrant (Standard diode curve) | Silicon/Germanium cannot be used (indirect band gap releases heat). Direct bandgap semiconductors like $\text{GaAs}_{1-x}\text{P}_x$ with $E_g > 1.8\text{ eV}$ are used for visible light. |
| Solar Cell | Zero External Bias | Three-step generation: (i) Generation of e-h pairs by light, (ii) Separation by built-in field $E_0$ (electrons to n-side, holes to p-side), (iii) Collection at metal electrodes producing photovoltage. | 4th Quadrant (Positive V, Negative I — Delivers Power) | Silicon ($E_g \approx 1.1\text{ eV}$) or $\text{GaAs}$ ($1.43\text{ eV}$) preferred because their absorption spectra match the peak solar photon emission ($1.5\text{ eV}$). |
Trap 1: Output Ripple Frequency for 50 Hz AC Line
*Board Question:* An input alternating current of $50\text{ Hz}$ is rectified. What is the frequency of output ripple?
• For Half-Wave Rectifier: $f_{\text{out}} = f_{\text{in}} = \mathbf{50\text{ Hz}}$. (One pulse per period $T$).
• For Full-Wave Rectifier: $f_{\text{out}} = 2f_{\text{in}} = 2 \times 50 = \mathbf{100\text{ Hz}}$. (Two pulses per period $T$).
Trap 2: Peak Inverse Voltage (PIV) Danger
*Exam Comparison:*
• Half-Wave Rectifier: $\text{PIV} = V_m$.
• Center-Tapped Full-Wave Rectifier: $\mathbf{\text{PIV} = 2V_m}$!
When upper diode conducts, terminal $A$ is at $+V_m$ and lower terminal $B$ is at $-V_m$. Non-conducting diode $D_2$ faces a total reverse voltage of $(+V_m) - (-V_m) = 2V_m$. Diode must be rated for at least $2V_m$ to prevent breakdown!
Trap 3: Why Silicon & Germanium are Useless for LEDs
*Why can't Silicon ($1.1\text{ eV}$) or Germanium ($0.7\text{ eV}$) emit light?*
1. They have indirect band gaps; electron-hole recombination requires phonon momentum conservation, releasing energy as lattice vibration (heat), not light photons.
2. Their band gaps are $< 1.8\text{ eV}$; even if photons were emitted, they would fall in the invisible infrared spectrum.
Trap 4: Direction of Current in Load Resistor
Students often incorrectly assume current in $R_L$ reverses during the negative half-cycle of full-wave rectification.
In reality, diode $D_1$ guides current from $X \to Y$, and diode $D_2$ ALSO guides current from $X \to Y$. The load current is strictly unidirectional!
Problem: A center-tapped full-wave rectifier uses two ideal diodes. The AC voltage across the entire secondary winding of the transformer is $50\text{ V (rms)}$ at $50\text{ Hz}$. If load resistance is $R_L = 500\text{ }\Omega$, determine: (i) Peak load voltage, (ii) DC output voltage, (iii) DC load current, and (iv) Output ripple frequency.
Solution: The total secondary voltage is $50\text{ V (rms)}$. In a center-tapped configuration, each diode sees only half of the secondary winding:
1. Peak load voltage ($V_m$):
2. DC output voltage ($V_{\text{dc}}$):
3. DC load current ($I_{\text{dc}}$):
4. Output ripple frequency:
Problem: An LED is constructed from a $\text{GaAs}_{1-x}\text{P}_x$ semiconductor material with a bandgap of $E_g = 1.9\text{ eV}$. (a) Calculate the wavelength of light emitted by the LED. (b) Identify the color of the emitted light. (c) What minimum forward voltage is needed to turn this LED on?
Solution: In an LED, recombination of an electron and hole releases energy equal to the bandgap: $E = h\nu = \frac{hc}{\lambda} \approx E_g$.
1. Wavelength of emitted light:
2. Color identification:
3. Minimum forward voltage (knee voltage):
HWR Output Freq
$f_{\text{out}} = f_{\text{in}}$ ($\eta = 40.6\%$)
FWR Output Freq
$f_{\text{out}} = 2f_{\text{in}}$ ($\eta = 81.2\%$)
FWR PIV Rating
$\text{PIV} = 2V_m$
Filter Condition
$C R_L \gg T/2$