Average output voltage of a full-wave bridge rectifier

Step-by-step derivation

1. Count the conducting diodes. In a bridge, current flows through two diodes in series on each half cycle - one on the way to the load and one on the return path.
2. Subtract both drops from the input peak to get the peak at the load: $$V_{m,\text{out}} = 20 - 2(0.7) = 20 - 1.4 = 18.6\ \text{V}.$$
3. For an unfiltered full-wave rectifier the output is $|V_m \sin\theta|$, so the average over one half cycle is $$V_{DC} = \frac{1}{\pi}\int_0^{\pi} V_{m,\text{out}}\sin\theta\, d\theta.$$
4. Evaluate the integral: $$\int_0^{\pi}\sin\theta\, d\theta = \big[-\cos\theta\big]_0^{\pi} = -(-1) - (-1) = 2,$$ giving the standard result $$V_{DC} = \frac{2V_{m,\text{out}}}{\pi}.$$
5. Substitute: $$V_{DC} = \frac{2 \times 18.6}{\pi} = \frac{37.2}{3.1416} = \mathbf{11.84\ V}.$$
6. Compare the topologies. A centre-tapped full-wave rectifier has only one diode in the path, so it would give $2(20-0.7)/\pi = 12.29$ V. The bridge trades that extra $0.7$ V for not needing a centre-tapped transformer.