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The rms value of an alternating quantity which is the input to a half wave rectifier is Imax divided by square root of two.

Then the rms value of output should be IMAX by 2*sqrt two. But it Isht given every where that it is IMAX by two. Can somebody help me?

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  • \$\begingroup\$ StackExchange has support \$L_AT_EX\$. If you write you equations better, it would make your questions clearer. \$\endgroup\$ – Nick Alexeev Feb 23 '14 at 7:03
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    \$\begingroup\$ I have no clue what you are asking. Please elaborate. \$\endgroup\$ – jippie Feb 23 '14 at 7:03
  • \$\begingroup\$ Related electronics.stackexchange.com/questions/72130/… \$\endgroup\$ – Dan D. Feb 23 '14 at 8:49
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Assuming sinusoide waveform:

enter image description here

Diode drop voltage shorten the duty cycle δ. Diode drop voltage depends on Iavg (see diode datasheet Iavg vs Vf curve)

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Convert to power then halve the power (because only one half of the waveform gets thru a diode) then convert back to RMS.

You can use 1 ohm as the resistance when converting to power to make the math easier.

It's all about halving the power and not about halving the RMS value.

Consider the much simpler case of a square wave of amplitude +1v and -1v. It has an RMS value of 1V - positive half and negative half will heat a 1 ohm load resistor and dissipate 1W equally. The diode stops the negative voltage and therefore reduces the power to 0.5W.

If you convert this back to an RMS voltage remembering power = \$\dfrac{V^2}{1 ohm}\$,

Voltage = \$\sqrt{0.5}\$ = 0.7071 V\$_{RMS}\$ (and not 0.5V)

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Looks like you just want to know the result: Vrms_with_diode = 0.7071 * Vrms_sine_wave

Simply it is not 50% of the full value but 70% of the full value.

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