simple diode full wave rectification!

[yellowmania]

Just a simple question, for a 4 diode bridge full wave rectification circuit with input of 12Vac supply i have measured voltage on scope to be 19.2V, when voltage measured across load was taken it measured 12.2volts on my DC multimeter! now i know that the meter is reading the average voltage for the output and by looking around online i have seen the average power is equal to 0.637V of the max! This number agrees when inputting my numbers but i was just wondering where the 0.637 number originates from???????

 

[btbass]

The diodes rectify the peak of the AC waveform, which is 1.414.
So for 12 volts AC in, the rectified DC voltage should be 12 * 1.414 = 16.968.

I doubt if the regulation of the transformer is 100% so the AC voltage will rise with no load and fall under load.

If you are measuring the output with a DC multimeter, surly the measurement is a DC measurement and not an average measurement?

 

[yellowmania]

''As the current flowing through the load is unidirectional, so the voltage developed across the load is also unidirectional therefore the average DC voltage across the load is 0.637Vmax. However in reality, during each half cycle the current flows through two diodes instead of just one so the amplitude of the output voltage is two voltage drops ( 2 x 0.7 = 1.4V ) less than the input VMAX amplitude. ''

i read this online so naturally it confused me straight away, have done about the two voltage drops (1.4V) in lessons but then i saw that .637Vmax and that fitted with my results. Ill have to go back to the drawing board and check my results again i think! i've probably missed something basic

 

[dave9000]

The 0.637 value you sometimes found around should be 2/pi, which is the mean value of a rectified sine wave - note well: the "mean value" not the "mean squared value" (that is 1/2).
If you perfectly rectify a sine wave and low-pass filter the result you get 2/pi of the peak value.