7812 datasheet
http://www.datasheetcatalog.org/datasheets2/52/529144_1.pdf
That aside, if the power source is 1watt and 1 amp then the rms voltage
of the source is 12vac rms. Am I correct so far?
The peak voltage available is 12x1.414=16.968vac, yes?
If using a 7812, he needs a capacitor that can hold the input to the
regulator above 15 vdc at a load of 1amp.
Best to use a full wave bridge.
Considering bridge voltage drop of 0.7v there is now a 16.268 peak ac voltage @60hz
going to the cap.
What is the formula to select the capacitor to keep the voltage above 15vdc at a
current of 1 amp?
I found a site that said "current is roughly Vripple*2*pi*F*C"
Our acceptable ripple is 16.268v - 15v = 1.268v
1 = 1.268 * 2 * 3.14 * 60 * C
C = 477.7824 farads can't be right.
If C is in microfarads then this makes more sense.
Am I completely lost?
I also found this link to a post by Mr Al
https://www.electro-tech-online.com/threads/power-supply-design.86669/
"What helps a bit is to estimate the peak to peak ripple voltage,
which is approximately:
Vpp=6000*I/C
where
Vpp is the peak to peak ripple voltage in volts
I is the load current in amps
C is the capacitance in microfarads (uf).
The frequency is 60 Hz and it is full wave rectified.
For example, a 10v rms 60Hz input to a full wave bridge when filtered
with a 6000uf cap and 1 amp load will mean there will be about 1v peak
to peak ripple. You can compute ripple factor from that if you wish."
He has more to say in the above link.