Average output voltage of a full-wave bridge rectifier
BlinkNBuild practice problem · GATE standard, authored and verified in-house
A full-wave bridge rectifier is driven by a sinusoid of $20\ \text{V}$ peak. All four diodes are silicon with a forward drop of $0.7\ \text{V}$ each. The load is resistive and the output is unfiltered.
The average (DC) output voltage, in volts, is ________.
Show the step-by-step derivation
Answer
11.84 V (accept 11.7 to 12.0) V (accept 11.7 to 12.0)
Step-by-step derivation
- 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.
- Subtract both drops from the input peak to get the peak at the load: $$V_{m,\text{out}} = 20 - 2(0.7) = 20 - 1.4 = 18.6\ \text{V}.$$
- For an unfiltered full-wave rectifier the output is $|V_m \sin\theta|$, so the average over one half cycle is $$V_{DC} = \frac{1}{\pi}\int_0^{\pi} V_{m,\text{out}}\sin\theta\, d\theta.$$
- Evaluate the integral: $$\int_0^{\pi}\sin\theta\, d\theta = \big[-\cos\theta\big]_0^{\pi} = -(-1) - (-1) = 2,$$ giving the standard result $$V_{DC} = \frac{2V_{m,\text{out}}}{\pi}.$$
- Substitute: $$V_{DC} = \frac{2 \times 18.6}{\pi} = \frac{37.2}{3.1416} = \mathbf{11.84\ V}.$$
- 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.
The trap this question is built around
Subtracting only one diode drop, or using the half-wave average $V_m/\pi$. The bridge has two diodes conducting per half cycle and delivers both half cycles, so it is $2(V_m - 1.4)/\pi$ - the factor of two appears once for the diodes and once for the waveform, in opposite directions.