Power-supply circuits
Most electronics need steady DC, but the wall socket gives AC. A power supply has four blocks:
- Transformer: lowers the AC voltage (for example 230 V to 12 V).
- Rectifier: diodes let current go one way only. A half-wave rectifier keeps only the positive half. A full-wave rectifier (a bridge of 4 diodes) flips the negative half up, so there are twice as many bumps. Output is pulsating DC.
- Filter: a big capacitor charges at each peak and gives current in the dips. The leftover wobble is called ripple. Bigger C or a lighter load means less ripple: ripple voltage ≈ I / (f × C).
- Regulator: a Zener diode or an IC (like 7805) holds the output at a fixed voltage, even if the mains or the load changes.
Ripple frequency: half-wave = 50 Hz, full-wave = 100 Hz (for 50 Hz mains). A diode circuit is explained in PN junction diode.
Oscillator circuits
An oscillator makes a repeating wave by itself, from DC power alone. It uses an amplifier plus positive feedback: a part of the output is sent back to the input in step with it, so the wave grows until it settles at a steady size.
Barkhausen idea: oscillation needs loop gain of at least 1 and the feedback in phase.
- LC oscillator (Colpitts, Hartley): a coil and capacitor set the frequency f = 1 / (2π√(LC)). Used for radio frequencies.
- RC oscillator (phase-shift, Wien bridge): resistors and capacitors set the frequency; used for audio.
- Crystal oscillator: a quartz crystal gives a very exact frequency (watches, computers, radio).
More about the LC pair in LC oscillations.
Pulse circuits
A pulse is a signal that jumps between two levels: HIGH and LOW. Its main numbers:
- Period T = time for one full cycle; frequency f = 1/T.
- Pulse width = time the signal stays HIGH.
- Duty cycle = pulse width / T × 100%.
Multivibrators make pulses with two transistors or a timer IC:
- Astable: keeps pulsing by itself (a clock, a blinking LED). A 555 timer gives f ≈ 1.44 / ((R1 + 2R2) × C).
- Monostable: gives one pulse of fixed length when triggered (a delay).
- Bistable (flip-flop): holds one of two states until told to change (one bit of memory).
A Schmitt trigger cleans a slow or noisy signal into sharp pulses. Changing the duty cycle (PWM) changes the average power: 25% duty gives a quarter of the power.
Modulation and demodulation circuits
A voice signal is slow (about 300 Hz to 3 kHz) and cannot travel far on its own. So we put it on a fast carrier wave (for example 1 MHz). This is modulation.
- AM (amplitude modulation): the message changes the height of the carrier. Modulation index m = message peak / carrier peak (keep m ≤ 1).
- FM (frequency modulation): the message changes the carrier's frequency; it is less affected by noise.
- Digital: ASK, FSK and PSK change the height, frequency or phase in steps for 0s and 1s.
Demodulation takes the message back. A simple AM detector is a diode, a capacitor and a resistor: the diode removes one half, the capacitor follows the peaks (the envelope), and a final filter blocks the carrier. The receiver also has a tuning circuit to choose one station.
Try it: see the waves
In the 3D, make the capacitor smallest and note the ripple, then make it biggest. Predict: which one is better for a radio? Set the pulse duty to 50% and then 20% and say which gives a dimmer lamp. In modulation, raise the depth above the middle and look at where the envelope nearly touches zero: that is the limit before the message gets distorted.
Key formulas and definitions
- Full-wave ripple frequency = 2 × mains frequency (100 Hz for 50 Hz)
- Ripple voltage ≈ I / (f × C)
- LC oscillator: f = 1 / (2π√(LC))
- Pulse: f = 1/T; duty = pulse width / T × 100%
- 555 astable: f ≈ 1.44 / ((R1 + 2R2) × C)
- AM modulation index: m = message peak / carrier peak
Worked examples
1. What is the output ripple frequency of a full-wave rectifier on 50 Hz mains?
2 × 50 = 100 Hz.
2. A power supply gives 100 mA. The ripple frequency is 100 Hz and C = 1000 µF. Find the ripple voltage.
V = I/(fC) = 0.1 / (100 × 0.001) = 1 V.
3. An LC oscillator has L = 1 mH and C = 0.1 µF. Find f.
f = 1/(2π√(10⁻³ × 10⁻⁷)) = 1/(2π × 10⁻⁵) ≈ 15.9 kHz.
4. A pulse is HIGH for 2 ms in a period of 8 ms. Find its frequency and duty cycle.
f = 1/0.008 = 125 Hz. Duty = 2/8 × 100 = 25%.
5. A 555 astable has R1 = 1 kΩ, R2 = 10 kΩ, C = 10 µF. Find f.
f = 1.44 / ((1000 + 20000) × 10×10⁻⁶) = 1.44 / 0.21 ≈ 6.9 Hz.
6. An AM wave has a message peak of 3 V and a carrier peak of 5 V. Find m and say if it is fine.
m = 3/5 = 0.6, which is below 1, so there is no distortion.
Common mistakes
- Saying a rectifier makes pure DC. It makes pulsating DC; a filter and a regulator are still needed.
- Thinking an oscillator needs an input signal. It needs only DC power and feedback.
- Mixing up duty cycle with frequency. Duty is the ON share, not the speed.
- Using a modulation index above 1 in AM, which distorts the message (over-modulation).