Of course i will not use a 24V transformer, it was just an example
I understand this so far
I need 7V on the input of the 7805, right..
I don't know about he LM78S05 but the LM7805 requires 7.5V.
now i will add 10% ripple voltage meaning 0.7V and aprox 1.4V diode drop resulting 9.1 meaning a 9V AC 3A is good for the project
No, work out the peak voltage, then subtract, the diode losses and the minimum voltage required for the regulator to get the maximum allowable ripple.
Also the diode loss is normally 1.1V per diode at full load, not just 0.6V (look at the datasheet) which makes 2.2V.
[latex]V{peak} = 9 sqrt{2} = 12.73V\\
Ripple = 12.73-7.5-2.2 =3.03V
[/latex]
Now for the capacitor i found this C=5 x i / Vp f
C: Capacitor value.
Vp: Peak voltage. ("Bridge output max voltage")
f: Frequency of the AC supply. 60Hz
i: Load current.
meaning C= 5 x 3 / (9 -1.4) x 60 = 3.300 uF
but I guess I will use a 4.700uF at 16V
by the way i calculated 3A because i will use LM78S05 witch is rated at 3A max.
So a solution for this power supply will be a 9V AC at 3A, 4 diodes 1N4004 (or 4A 50V bridge)
a capacitor elko 4.700uF at 16V and one elko 100uF 16V
To this I will add 2 MKT or ceramics capacitor on the input and output of the regulator.
Any comments are welcome
regards,
That formula is rubbish, iot will grossly undersize the capacitor to size the capacitor use the following approximate formula:
[latex] C = \frac{I}{2f V_{Ripple}}\\
C= \text{Capacitance in Farads}\\
I=\text{Current drawn, in this case 3A}\\
f = \text{Mains frequency, in this case 60Hz}\\
V_{Ripple} =\text{The maximum ripple, in this case 3.03V}\\
C= \frac{3}{2\times 60 \times 3.03}=8.251 \times 10^{-3} = 82501 \mu F [/latex]
So for 3A, I'd use a 10,000µF capacitor to on the safe side.
There is a more accurate formula but it's complicated and this one is fine since it slightly oversizes the capacitor.
EDIT:
Your location says Romainia, I could be wrong but I think you're on 50Hz not 60Hz but you'll still be fine with 10,000µF