Why Does Fullwave Bridge Rectification Raise Voltage by 1.41?

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SUMMARY

Fullwave bridge rectification raises voltage by a factor of 1.41 due to the relationship between RMS voltage and peak voltage, where the equation V_p = \sqrt{2}V_{rms} applies. This factor arises because transformer secondary ratings are specified in RMS voltage rather than peak voltage. Additionally, a bridge rectifier configuration results in a voltage drop of approximately 1.4 volts due to the current passing through two diodes. Understanding these principles clarifies the behavior of voltage in rectification circuits.

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rytmenpinne
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..fullwave bridge rectification raise the voltage *1.41? I scanned the net but I can't find an explanation as to why, only that it does..
 
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Remember, ac voltage is really regarding RMS voltage, and hence,
\sqrt{2}V_{rms}=V_p
where
\sqrt{2}=1.41...
 
Aah, so basically what's going on here is that the secondary ratings on transformers is given in RMS? and not peak?
 
rytmenpinne said:
Aah, so basically what's going on here is that the secondary ratings on transformers is given in RMS? and not peak?
That's correct. Also note that a bridge rectifier will have the current passing through two diodes, so the output voltage will be 1.4 volts less than the peak.
 
Good! thank you both! the world now makes sense again
 

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