I am designing a circuit that needs to output 5VDC @ 1A. I'm trying to use a wall transformer to step the voltage down to 12VAC. The next step is the diode bridge and ripple capacitor.
The ripple voltage equation is:
$$V_{ripple} = \frac{I}{2fC}$$
I = load current (1A)
f = AC frequency (60Hz)
C = Filter Capacitor (? uF)
If I choose a C of 1000 uF, the ripple voltage is 8.3 V! Do I really need to put more capacitance to lower the ripple voltage? Is there another method of converting AC to DC?
Answer
1000 µF at this voltage isn't terribly big. Are you limited by size or something?
To completely get rid of the ripple and produce 5 V, you need to add a voltage regulator after the capacitor.
12 VRMS = 17 VPeak, which, minus the two diode drops, is the peak DC voltage you'll see at the output of the rectifiers: 17 - 1.1 - 1.1 = 14.8 V. So there's no threat of exceeding the input limits of the regulator (35 V input).
If the ripple is 8.3 V, then the DC voltage will be varying from 6.5 V to 15 V. This is just barely high enough to feed into the regulator without dropping out of regulation, since the 7805 has about 1.5 V dropout at 1 A (depending on temperature). So yes, you should use a slightly higher capacitor (or multiple capacitors in parallel, if space is an issue).
(Here's a guide to each stage of the power supply circuit.
Also:
Real life power line voltages vary from one outlet to the next, and the frequency varies by country. You need to calculate the low line/high load condition to make sure it doesn't drop below regulation, as well as the high line/low load condition to make sure it doesn't exceed the regulator's input voltage limit. These are the generally recommended values:
- JP: 85 VAC to 110 VAC (+10%, -15%), 50 and 60 Hz
- US: 105 VAC to 132 VAC (+10%), 60 Hz
- EU: 215 VAC to 264 VAC (+10%), 50 Hz
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