Why Does Fullwave Bridge Rectification Raise Voltage by 1.41?

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Fullwave bridge rectification increases voltage by a factor of 1.41 due to the relationship between RMS voltage and peak voltage, where V_peak equals √2 times V_rms. The secondary ratings on transformers are specified in RMS, not peak voltage. Additionally, a bridge rectifier causes the output voltage to drop by approximately 1.4 volts due to the current passing through two diodes. Understanding these principles clarifies the behavior of voltage in rectification processes. This knowledge is essential for accurate electrical engineering applications.
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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 basicly what's going on here is that the secondary ratings on transformers is given in RMS? and not peak?
 
rytmenpinne said:
Aah, so basicly 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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