But there will still be a small timeframe in which all the diodes are nonconducting(Using constant voltage drop model), i.e. when 0 < Vs < 1.4V and 0 > Vs > -1.4V, correct?In this case you don't need to calculate anything to work out the diode conduction cycle, since this an ideal full-wave bridge rectifier with a purely resistive load and without any capacitive or inductive filtering applied. Each diode will conduct for one half the AC cycle or 180° electrical.
Wouldn't the RMS for a sine wave be amplitude*sqrt(2)?Just model the voltage across the diodes with a sine function, and then use an inverse sine to find the value in your calculator. Do note that rms for a sine wave is amplitude/sqrt(2), and don't forget to account for the voltage drop across *both* diodes.
In the actual output of a regulator, this will look like gaps between the peaks- normally not a concern because a capacitor is typically used.
D1, and D2 conduct in the first half cycle, D3 and D4 in the second half cycle, it doesn't specify which diodes in particular so if we consider all the diodes they are conducting 2(47.37%) = 94.74%, right?The question actually asks for what fraction of a cycle each diode conducts.
A cycle presumably means a complete AC cycle. Therefore I would say the fraction is in relation to a full AC cycle period rather than a half period.
The correct answer would then be ~47.37%.
by Jake Hertz
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by Aaron Carman
by Aaron Carman