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Oct 16, 2016 at 5:59 comment added EM Fields @JoãoPedro: With a perfect voltage source, perfect diodes, a perfect load and a perfect capacitor, if the input voltage is given in volts, RMS, the peak output voltage will be, simply, \$ Vin \times \sqrt 2\$
Oct 15, 2016 at 20:53 comment added João Pedro Can I get the peak voltage at Vo by applying KVL? Or that value depends on the capacitance, period of the wave, etc?
Oct 15, 2016 at 18:31 history answered EM Fields CC BY-SA 3.0