The DC output Voltage of a Centre Tapped Full Wave Rectifier

[@ovi_4]

But in your example you don't show any smoothing caps after the rectification. Does this mean that once you put some cap in you would measure an increased voltage up to the level of the initial "peak voltage" of 7.49V (no load situation) ? And here I have my own "real life problem" I came across and try to solve. I only have access to the schematic diagram of a "device" where there is a center tap transformer but... NO secondary voltage values are given !!! at all. All I see, is the DC voltage value marked as + 385v DC ( between the positive DC output and the center tap). The rectification is the same as in your video (2 silicone diodes with I assume 0.7v drop) and a 100uF cap rated at 450v !!! So, the problem is: how would I work out the theoretical secondary (end to end winding) voltage . In other words, what are the needed formulas in this case. Also, another question is: let us say that the secondary (end to end) is rated at 100mA of AC current, what would be the OUTPUT's DC current available after the rectification? and , also, are there any significant losses @ the rectification diodes worth considering or they are insignificant , so we can simply ignore those ???

[@saikiransarihaddu1238]

Why is not DC Output voltage calculated in RMS which V peak/ square root of 2 ?

[@TrevorHeadley]

To answer the question as to why DC output voltage is not calculated using the RMS value. In order to take into account the voltage drop of 0.7V across the diode, the peak voltage in is used. Then the peak voltage out would be the peak voltage in minus 0.7 V. The to calculate the average DC voltage the formula is 2 times Vpeak out, divided by pi.

[@gideonkorir4201]

I am kinda confused is it RMS=Vp/√2. Or RMS=Vp/0.7071

[@yourplctutor9431]

RMS = Vp x 0.707 , therefore Vp = RMS divided by 0.707

[@aimi526]

can I ask, where did the 0.707 came from?

[@TrevorHeadley]

The Peak value of an AC wave is the RMS value divided by 0.707. The RMS is the stated value of an AC voltage from the mains.

[@timmeke_8213]

peak = rms * square root of 2 or peak = rms / 0.707, it's the same