Discussion
In the filtered half wave rectifier circuit as shown below, the value of the DC output voltage is ............ .
Answer: B
The DC output voltage is 16 V.
The transformer steps 240 V rms down by 20 : 1, and the capacitor charges to the peak of the secondary voltage less one diode drop:
\[ \begin{aligned} V_{2,\text{rms}} &= 240/20 = 12\ \text{V} \\ V_{2,\text{peak}} &= 12\sqrt{2} \approx 16.97\ \text{V} \\ V_{dc} &\approx 16.97 - 0.7 \approx 16.3\ \text{V} \end{aligned} \]
With 500 mF across 1 kΩ the ripple is negligible (the time constant is 500 s against a 20 ms cycle), so the output sits at about 16 V.
17 V is the ideal-diode peak, which ignores the diode drop. 12 V is the rms value, and 18 V exceeds the peak of the secondary altogether.
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