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Question about diode ratings in a rectifier bridge

I'm not expert on electronics, but I think what he is getting at is that theoretically yes - but that doesn't mean you should do it.

Some things to remember, device ratings are rarely 100% accurate, there is varience in everything. If you assume that a device is 100% precise then you're running a risk when a real part is slightly out. Running devices at their absolute maximum limit can reduce their life time because you're placing more stress on them. As mentioned before - devices typically take on different properties with temperature.... so how it reacts are 5 celsius could be different 30 celsius. Also don't forget that your circuit should be tollerent of outside changes beyond your control. For example, what happens if your supply voltage surges briefly or if the alternating frequency changes or stops alternating (because someone connected it to an odd supply etc). You don't want the diodes going bang as a result.

The main point is that while a lot of things can work, that doesn't mean its advisable to cut things that close - especially when you're dealing with power circuits.

Mike

Reply to
mjt
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Sure. Don't forget you'll be burning about a kilowatt in that bridge.

Wait: "balanced 60Hz AC"? As in a center-tapped winding? Cut your losses by half and use a single pair of diodes!

Tim

-- Deep Fryer: a very philosophical monk. Website:

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Reply to
Tim Williams

And don't forget that if you are doing something evil and immoral, like using your diode bridge to charge up a big reservoir capacitor, the current flows through the diode for only a few milliseconds in each cycle, so the peak current through the diode - and the peak voltage drop - are higher than the numers you first thought of.

Diodes are relatively forgiving of peaky currents, so the bottom line is usually the peak junction temperature of the diode under a sustained worse case load, and answering that not only requires that you know curent and voltage over the cycle, but also the thermal reistance to ambient from the diode junction through any heat-sink you use to mount the diodes to the air around the heat-sink.

People have been known to worry about the heat-capacity of the junction.

In some semiconductors - laser diodes are the obvious example - the junction is so small, and the heat capacity so low, that a few microseconds of overload can heat the junction to the point of re-annealing it into something useless for the intended purpose.

The data sheets for well-specified rectifier diodes do give a series of absolute maximum currents for a series of progressively longer pulses, which provides pretty much this information.

-------------- Bill Sloman, Nijmegen

Reply to
bill.sloman

All other things being equal (like ambient air temperature around the diode or amount of heat sink per diode, etc.) then, yes. Each diode in a full bridge carries half of the average output current.

Reply to
John Popelish

Yes. If each diode is heat sinked to meet the requirements of Ifav=240 amps, half wave, then the bridge would be capable of 480 amps.

Reply to
John Popelish 

--
But then you need twice as much wire in the secondary (but only half
the cross sectional area) as in the case of a full bridge, and the
amount of reverse voltage the rectifiers have to stand off increases
by a factor of two.
Reply to
John Fields

Suppose that I build a rectifier bridge to rectify 60 Hz AC.

Each diode has a rating of Ifav = X amps.

Ifav is described as the maximum mean on-state current.

Since in a full rectifier bridge, every diode conducts only half the time, would it be proper to say that I can make the bridge to rectify

2X current, out of four such diodes?

Thanks

i
Reply to
Ignoramus361

That's great. So, to paraphrase your answer, I could build a 480 amp rectifier bridge for balanced 60 Hz AC, out of four Ifav=240V diodes, provided that I satisfy your conditions on equal heating etc.

Am I correct?

thanks

i
Reply to
Ignoramus361

Thanks. Yes, it would be a lot!

Well, by balanced I meant that 50% of time it is positive and 50% it is negative.

Good point on the center tapped winding.

i
Reply to
Ignoramus361

Sure. Real life and being careful demands some degree of derating. So that, say, a 240 amp diode is only used for 200 amps. But, if I use the same degree of derating, could I use diodes that could be used to conduct 200 amps after derating, to form a full rectifier bridge to conduct 400 amps with the same degree of safety?

i
Reply to
Ignoramus361

Thank you! At 1.4 voltage drop on each diode, that would mean

1.4*2*480 = 1344 watts dissipated. Quite a bit! i
Reply to
Ignoramus361

It seems that higher power draw of the full bridge, vs. a bigger transformer and higher reverse voltage, is a tradeoff depending of investment cost vs. use cost, in other words depends on how much the project item would be used in real life. Though the extra tw power diodes would also add $60 or so to the cost.

i
Reply to
Ignoramus361

Thanks. That makes perfect sense. For now, my interest is mainly theoretical.

i

Reply to
Ignoramus361

On Mon, 05 Dec 2005 21:04:11 GMT, Ignoramus361 wrote in Msg.

Why did I just know the OP would start pulling numbers like these out of his hat? ;-) robert

Reply to
Robert Latest

Bill is correct...

this is a very simple circuit but very complex to analyze

If you are using a cap input filter after the bridge rectifier then the diodes do NOT conduct 1/2 the time, they conduct much LESS then 1/2 the time. The cap delivers the AVERAGE load current and the diodes have to recharge the AVERAGE load current... but they only conduct for a few ms each so the PEAK current is MUCH higer then the average and the RMS current is also much higher than the average.

This is what power factor correction is all about.

Four diodes rated at 10 AMPs RMS used in a bridge with a cap input filter and a small conduction angle can proably deliver considerably less than 10 Amps DC.

This is NOT a simple question.

By the way, this low conduciton angle and high RMS current also impacts the transformer rating...

Cap input filter power supplies are not simple to analyze.

Do some reaserch on power factor correction, conduction anlge and RMS vs AVERAGE values.

Enjoy..

Mark

Reply to
Mark

How would one correct for it, add an inductor between the caps and diodes?

i

Reply to
Ignoramus21666

ther eis no simple answer..

theoretically, an inductor will help but such an inductor is usually not practical for low vltage high current supplies

you can use a smaller output C which increases the ouput ripple so the diodes conduct longer but then you need to filter or reulate the output to get rid of the ripple..

or you can add some series R which reduces the peak current but hurts efficency and regulation...

probably the best idea in practice is to make sure the diodes are well over-rated

...build a breadboard and get a current probe and a scope and look at some voltage AND CURRENT waveforms...that is the best way to get a feel for this...maybe make a spreadsheet and take some points off the current waveform and convert it to RMS ....try that for some simple rectangular waveform...say a 100 Amp pulse that is 1% duty factor... the peak current is 100 Amps,, the average current is 1 Amp....you calculate the RMS current... the RMS current is what determines how hot the diode will get...

Mark

Reply to
Mark

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