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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..
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.rytmenpinne said:Aah, so basicly what's going on here is that the secondary ratings on transformers is given in RMS? and not peak?
Fullwave bridge rectification is a process used to convert alternating current (AC) into direct current (DC) by using a diode bridge circuit. This circuit consists of four diodes arranged in a specific configuration to allow the flow of current in only one direction.
Fullwave bridge rectification raises voltage by 1.41 because it utilizes the full peak-to-peak voltage of the AC input. The four diodes in the bridge circuit work together to rectify both the positive and negative halves of the AC waveform, resulting in a higher overall voltage output.
The voltage is raised by exactly 1.41 due to the root-mean-square (RMS) value of the AC waveform. The RMS value is a mathematical calculation that takes into account the average voltage and the peaks of the waveform. In a fullwave bridge rectifier, the RMS value is multiplied by the square root of 2, which results in a voltage increase of 1.41.
Fullwave bridge rectification offers several advantages, including higher voltage output, more efficient use of the AC input, and a smoother DC output. It also eliminates the need for a center-tapped transformer, which can be costly and bulky.
One potential disadvantage of fullwave bridge rectification is the increased complexity of the circuit compared to a halfwave rectifier. This can make troubleshooting and repairs more difficult. Additionally, the four diodes used in the bridge circuit may introduce more resistance and cause some power loss.