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# if i want to rectify 12v AC to 12v DC how much values of diodes i'll use and how can i calculate it(i.e diode 1N4007)?

Like we mostly used (4 x 1N4007) Rectifier bridge to convert 12v AC to 12v DC.why and how we can calculate it?Formula?

The Wikipedia page titled, "Rectifier", does a pretty good job of explaining the full-wave bridge rectifier circuit,

https://en.wikipedia.org/wiki/Rectifier#Full-wave_...

Regarding the question of which, er, what rating, for the diodes, pretty much all you have to do is satisfy 2 conditions:

(1) Each diode must have peak inverse voltage rating (PIV) greater than the largest reverse voltage it will feel.

(2) Each diode must have current rating greater than the largest forward current to flow through it.

Note that the

peakvalue of an AC waveform is greater than theRMSvalue by a factor of square-root-of-2. That is Vpeak = 1.41421*Vrms. So, for example, the peak voltage of 12VAC, is actually 16.97 V peak = (1.41421)*(12.0 V)Although, tt turns out it is very easy to find diodes with PIV rating higher than that. E.g. like every diode in the 1N400x series (see Josehf Murchison's post on this topic) has PIV {50,100, 200, 500, 600, 1000} all greater than 17 volts.

So for a rectifier circuit that does not have to deliver much current (e.g. <1 A), the design is easy, and inexpensive.

If the rectifier circuit must deliver

larger currentthan this, then the diodes start getting more expensive, and physically larger, and also you have start worrying about power, as heat, approximately (0.5)*(0.6 V)*(Iforward) per diode, dissipated by the diodes themselves.By the way, they also make bridge rectifiers as a package. They are typically square shaped (for the 1 phase kind) with four terminals (one on each corner), with a hole through the center so you can bolt it to a heat sink. Also the cost of a such a packaged bridge rectifier is usually the same, or less, than the cost of 4 diodes.

They're even cheaper if can pull one out of some broken piece of junk already in your parts collection.

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I just ball park peak voltage.

RMS is 2/3 peak so 12 volts is just under 18 volts DC ripple and roughly 34 volts peak to peak AC.

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Yeah. I do the same thing actually. Since 1.5 is conveniently just a little bigger than 1.41, and I can multiply by 1.5 without having to reach for a calculator.

Actually the math is kind of reminiscent of calculating tips for waiters, waitresses, (waitpersons?) in restaurants, specifically for tipping 15%, since 15% is just 1.5 divided by 10.

Although for explaining it to somebody for the first time, I thought I'd be pedantic, and use the same formula Wikipedia does.

I dunno. Seems like I'm always linking to some Wikipedia article. Sometimes two or three of 'em in one post.

I don't know if anyone actually follows those links and reads the article, but in the event that happens, I thought it would be reassuring to the reader to see me, and Wikipedia, and almost everyone else, using the same formula.

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Wikipedia is good for something as straight forward as a formula and I do follow and use the links my self on occasion. But some of their articles have been messed with by hackers or have poor references. The downside of being a member and editing.

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That depends on your application. Powering something small, 4 1N4007's would probably be fine. What's most important is that you don't exceed the average current rating of the diode, or bad things will happen. The diodes (and presumably the transformer) should be sized appropriately for the load.

Because 2 diodes will be forward-biased in series with a full bridge during either half of the AC cycle, you will always have 2 'diode' drops. Each individual diode is only used half the cycle, so the average current rating is approximately doubled for the bridge rectifier as a whole (each half of the bridge rectifier does half the work).

http://www.electronics-tutorials.ws/diode/diode_6....

Especially with smaller diodes, make sure that you don't exceed the diode's peak (surge) current rating when it is charging a filter capacitor. Otherwise you are putting a lot of stress on the diodes each time the power supply is turned on. It would not be too difficult to figure out the math for that, just some capacitive reactance + ESR, and plug those numbers into ohm's law.

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I would recommmend simply getting ahold of a full wave bridge rectifer as one unit, with a heatsink tab, like one of these: http://www.alliedelec.com/nte-electronics-inc-nte5... Those are pretty compact, that specific one can handle an average current of up to 25A (plenty for some of the most demanding applications).

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I don't know why you would use a 1000 volt diode for a 12 volt circuit, I would use a 1N4001 in a circuit less than 50 volts.

1N4001 is 50 volts, 1N4002 is 100 volts, 1N4003 is 200 volts, 1N4004 is 400 volts, 1N4005 is 600 volts, 1N4006 is 500 volts, and 1N4007 is 1000 volts, all 1N400x series diodes are 1 amp.

These other diodes are the same voltages just a different amperage for each series.

1N539x series diodes are 1.5 amp.

1N540x series diodes are 3 amp.

For a rectifier current and wattage is more important.

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Can't believe I miss typed that 1N4006 is 800 volts.

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Google is your friend, especially if you specify what you want to calculate.

Same by the way fo Wikipedi - lots of info on AC, DC, transformers and rectifiers.

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The current rating of the diode is all that matters in most cases.

The output voltage is very hard to calculate exactly, because it depends on many factors, but its close to 1.414 x 12V. There is always a ripple voltage, for which you can find calculators online.

To get true 12V DC, it is normal to put a voltage regulator after the reservoir capacitor

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