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So im still a bit of an Electronics newbie. and was watching this video: http://www.youtube.com/watch?v=cyhzpFqXwdA

I understood mostly everything, but even at the end.....the voltage ripple was still pretty bad. So powering anything like a Microcontroller would basically be impossible.

What is the simplest way to go from A/C to D/C without much ripple? or would it just involve using a HUGE capacitor? or is the only way is to add voltage regulators (with the schematic listed above in the video IE: a transformer with full bridge rectifier+capacitor) (Schematic is at 10:49 of the video)

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No, I'm not going to watch a 11 minute video just to answer a question here. Put all the important parts of your question here. Give us a schematic and ask specific questions about waveforms of the like. As it is now, we have no way of knowing what the topology of the power supply you are referring to is. – Olin Lathrop Apr 10 '12 at 23:13
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I've not watched the video, but the usual way to do it is to use a stepdown transformer, fully rectify with a diode bridge, add a smoothing cap (2000uF or so should do the job) and feed the result into a voltage regulator. If you don't need more than about 1A from it, a 78xx series fixed linear regulator will do the job nicely. You can also use an LM317 or similar variable regulator if you need more flexibility in your supply voltage. – Polynomial Apr 11 '12 at 0:16
The whole video wasn't neccesary, just the Schematic he uses at the end. 10:49 – Mercfh Apr 11 '12 at 1:04
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OK, so grab the schematic and show it to us. Again, I'm not going to watch that video, certainly not 11 minutes of it just to get the schematic. And no, I'm not going to go there and fiddle with it to get to some time code. It's your job to gather the necessary information and present it here. – Olin Lathrop Apr 11 '12 at 19:04

2 Answers

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There are nomographs for full wave bridges on ripple vs cap value vs load current.. Just get a big cap.but dont exceed the charge current on the diodes charging the cap up. .. depending on esr value.

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I figured just getting a big cap was the answer. But question: How come in wall-warts and such (cheap wall a/c-->d/c converters and such) you see like one semi-large cap. or is that just because the cheapo power supplies you buy just aren't very accurate as far as ripple goes? – Mercfh Apr 11 '12 at 17:00
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Olin is right. We shouldn't have to watch an 11 minute video to get to your question. So I didn't watch it, but I understand your question is about ripple.

enter image description here

The dashed line is the voltage from your rectifier. The solid line is the voltage across the smoothing capacitor. Every half cycle of your AC supply (8.3ms in the US, 10ms in Europe) the input voltage will go higher than the capacitor's value and the capacitor will be charged through the diodes. There's little resistance, so the current peak can be quite large. This charging stops when the input voltage drops below the capacitor's voltage, when the sine is at its peak. From then on the capacitor gets discharged by the load, which makes its voltage sag. The discharging stops when the input voltage goes higher than the capacitor's voltage again, where the sine intersects with the smoothed ripple.

How fast the voltage drops depends on the capacitor's size and the load. A higher load (more current drawn) will make the voltage drop faster, causing a larger ripple. On the other hand a larger capacitor will hold more energy, so that that same load will make the voltage drop slower.

"What is the simplest way to go from A/C to D/C without much ripple?"

Don't load your capacitor and you'll have near zero ripple. :-)
OK, the load will be a given, so there's just the capacitance you can play with. You'll want a reasonably low ripple voltage, but you don't have to go to extremes. A voltage regulator will suppress the remaining ripple quite well. Just make sure you respect the regulator's minimum input voltage.

further reading
Ripple calculator

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