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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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