Bridge rectifier output and ripple factor

Bridge rectifier circuit

                    A
                    o
                  /   \
       D4 (N -> A)     D1 (A -> P)
                /       \
   N (-) o-----+         +-----o P (+)
                \       /
       D3 (N -> B)     D2 (B -> P)
                  \   /
                    o
                    B

 ac input across A and B; (X -> Y means anode at X, cathode at Y)

   P (+) o----+---------+
              |         |
             [RL]     === C (smoothing)
              |         |
   N (-) o----+---------+

Diodes D1, D2, D3 and D4 form a bridge; ac is applied to two corners (A and B) and RL (with C in parallel) is connected across the other two (P and N). D1 and D3 conduct in the positive half cycle, and D2 and D4 in the negative half cycle.

Output waveform with RL (no capacitor)

 vo   /\  /\  /\  /\
     /  \/  \/  \/  \
 ---------------------

The output is a train of positive half-sine pulses at 2f, with Vdc = 2Vm/π and ripple factor 0.482.

With a smoothing capacitor C

The capacitor charges to Vm during each rising pulse and discharges through RL between pulses. The ripple is approximately triangular. The time between charging peaks is T/2 = 1/(2f), so:

• Peak-to-peak ripple: Vr(pp) = Idc/(2fC) = Vdc/(2fCRL)
• RMS ripple (triangular wave): Vr(rms) = Vr(pp)/(2√3) = Vdc/(4√3 f C RL)

Ripple factor: r = Vr(rms)/Vdc = 1/(4√3 f C RL)

A larger C, a larger RL or a higher f reduces the ripple. For example, f = 50 Hz, C = 1000 µF and RL = 100 Ω give r = 1/(4 × 1.732 × 50 × 10⁻³ × 100) = 0.029 (2.9%).