- 2072 Kartik · 2 marks
Unregulated dc supply with bridge rectifier
Short answer
A step-down transformer reduces mains AC. A bridge rectifier of four diodes converts it to full-wave pulsating DC. A filter capacitor smooths the output to an unregulated DC voltage: Vdc ≈ Vm - 1.4 V, with peak-to-peak ripple Vr(p-p) ≈ IL/(2fC).
Draw unregulated dc voltage power supply using bridge rectifier.
230 V ac +---------+ +---------+ +--------+
mains o--->| Step- |--->| Bridge |-->| Filter |--> unregulated
| down | | rectif. | | C | dc output
| T/F | |(4 diode)| +--------+ to load
+---------+ +---------+
Circuit:
T1 D1 D2
230 V o--+ +--A--+---|>|---+---|<|---+--B--+
ac ) || ( | | | |
mains ) || ( | (+) | |
) || ( | | |
o--+ +--+ | (−) | |
| | | | |
| +---|<|---+---|>|---+ |
| D3 D4 |
+----------------------------+
Filter and load, connected to the bridge output:
(+) o----+--------+----o +Vdc (unregulated)
| |
[C] [RL]
| |
(−) o----+--------+----o GND
- A step-down transformer reduces the 230 V mains to the required ac voltage.
- The bridge rectifier (four diodes D1 to D4) converts the ac to pulsating dc (full-wave): D1 and D4 conduct in one half-cycle, and D2 and D3 in the other.
- A filter capacitor C charges to the peak and discharges through the load between peaks, smoothing the output. Ripple voltage: Vr(p-p) ≈ IL/(2fC).
- The output Vdc ≈ Vm − 1.4 V is unregulated: it changes when the mains voltage or the load changes.
✓ Answer checked by Claude Opus 5.5 (high effort)
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