hi Vick
your scope is showing you the peak voltage out of the the transformer, where as the transformer is usually rated by its
RMS voltage. That is.... if you transformer has an output of 10VAC... that is its RMS voltage. Its peak voltage is the RMS value x 1.414
your DC voltage out of the rectifier is the RMS value minus the voltage drop in the diodes ( ~ 1.7V for 2 diodes). You need to remember that the diodes are only conducting current in brief bursts. And when you add a capacitor to the output, the current through the diode charges up the capacitor, which then slowly discharges through the load till the next burst of current from one of the diodeson the other 1/2 cycle.
So the capacitor is holding up the voltage during the time that the diodes are not conduction. If you have only a small capacitor, then the voltage is going to drop more ( cuz it holds less charge) than a larger capacitor.
this is where your ripple comes from....
The rectifier diodes will charge up the filter capacitor, to the peak DC value, and between non conducting cycles of the diodes, it will discharge into the load resistor. This creates the sawtooth waveform known more commonly as ripple voltage. The value of the ripple voltage is dependant on load current, power supply frequency and capacitor value. Approximate ripple voltage is calculated using:
Δ V = I / 2fC
where V is ripple voltage (mV), I is DC load current (mA), f is frequency of AC supply and C is smoothing capacitor value (F).
have a ponder on that.... I need to go out for some hours
cheers
Dave