bridge rectifier ratings

[m3vuv]

hi all i have a diode bridge rated at 50 amp,if i use one diode from the bridge as a blocking diode for a charging system,would it handle 50a or 1/4 as its only one diode,ie 12.5a?,cheers m3vuv.

[floobydust]

I think you can pass rated 50A out of any one diode. The current rating is the "maximum average forward output rectified current" below 60 ̊C temperature.
Heat is the limiting factor, used as an AC rectifier a pair of diodes is on 1/2 the time.

At 50A and 1.2Vf that's 60W of heat to dissipate in a single diode.
Taitron MB5005 50A bridge datasheet pdf
You could parallel two and use that out of the bridge for less stress on the part. But still 60W.

At those currents, MOSFETS might be more attractive due to their much lower losses and heat generation. 60W requires a huge heatsink.

[m3vuv]

its for the train regulator ive built,it needs a 100 amp blockind diode,they are expensive,im trying to find a way of doing it cheap.

[digsys]

As been pointed out - there is no free meal ! :-)
Any Si diode will incur a 0.8-1.0V drop at that current and temp rise (if it is well heatsinked). You may find lower Vf ones, but they will cost more.
The "proper" 100+A blocks WILL cost a lot more. An ideal diode ie IC and FET pair will likely be cheaper with a LOT less heat, but still cost several $s
If you want "real cheap", just wire 100x 1A diodes in parallel ? A few cents ea. Might look ugly, but it'd work.

[bob91343]

You can parallel two or even four diodes.  Trouble is, diode voltage drop goes down as it warms so unbalance gets worse with higher currents.  You could select the diodes and put them in a series parallel arrangement with a high and a low in each side.

Anyway, 100 A at maybe 1 Volt is a lot of power to dissipate.  You could raise the rating a bit by bolting it to a heat plate and force some air over it.

As has been said, there is no free ride.

[D Straney]

100W+ of heat is a LOT of heat to get rid of, no matter what type of package it's in - it's all about temperature, in the end, and almost everything else ratings-wise is just BS.  Honestly I wouldn't even try to get rid of that much heat without a massive fan, or water-cooling, etc.  In this case, cutting down on the dissipation using some kind of (still very large) MOSFET and ideal-diode-controller IC like digsys suggested seems like the most realistic way to go.

What exactly are you doing with 100A of current?  (Not a model train, I assume?)  Any system that involves that much current is going to have a whole lot of copper, and even ignoring the diode, the rest of the 100A-related stuff is going to be expensive anyways.  Maybe there's a whole different approach that would work better.

Also, for the future, here's how you can make some thermal estimates yourself with simple calculations (for a diode as an example):
1. Figure out your diode's power dissipation.  This will be the forward current multiplied by the forward voltage at that current (look at the forward voltage curves in the datasheet, and choose some reasonable junction temperature like 70-100 C).
2. Find the "junction-to-case thermal resistance" (usually called Rth-JC) in the diode's datasheet.  This will be in units of K/W, or C/W; it shows how many degrees hotter the junction will be than the case when dissipating 1W of power.
3. Find the thermal resistance rating in the heatsink's datasheet.  A good heatsink datasheet will also show you different thermal resistance numbers for different amounts of airflow too, which will let you select a fan (based on the airflow) if you need one.  Obviously if you're using a salvaged heatsink this will be much harder, but in that case find a heatsink with similar geometry on digikey, etc. and use its numbers as a ballpark estimate, then leave lots of headroom.
4. The junction temperature of the diode will now be (ambient air temperature) + (diode power)*(heatsink thermal resistance + diode thermal resistance).  Pretty much all silicon devices top out at 150 C junction temperature, and for leaving headroom, getting decent lifetime, etc. I wouldn't recommend any more than 100 C at very most.

[duak]

I personally don't like to operate a semiconductor at much more than half its rated current or voltage but sometimes you gotta do what you can.

Here's an idea to help spread the current and heat  more evenly among the diodes:  the source circuit connects to one AC terminal and the - terminal.  The load circuit  connects to the other AC terminal and the + terminal.  If the wires are equal lengths they should have similar resistances and should help to balance diode currents somewhat.  This will also spread the current among 4 terminals rather than 3.  Because the diodes are on opposite sides of the package, the temperature profile will be balanced and more spread out.

I've used a rule of thumb that 1 W of power needs one square inch of heat sink area to operate reliably at typical temperatures.  I still do the calculations that D Straney lists to confirm, but it gives me a quick estimate of the heat sink size.

A quick test with the proposed parts will tell you a lot, especially if the bridge overheats and shorts out - it's hard to handwave that away.

Cheers,

[bob91343]

A useful procedure is to measure temperature rise of components.  I have two methods (not counting risking a burned hand).  First, I have an infrared thermometer I got on ebay from China for about $8.  It has an LED to help in pointing it.

Second, I have a cheap thermistor I connect to a digital Ohmmeter and wave it around to find hot spots.  It's not so good for temperature measurement but is great for finding what's getting hot first.  The reading won't be accurate but the temperature variation as you move it around will tell you a lot.

[T3sl4co1l]

Wouldn't recommend more than 35A for a single diode in a FWB.  They're amazingly robust devices, but they're still made like a fullpack device -- plastic molding provides heatsinking.

If your project is so underfunded that you can't afford a proper $10 diode, let alone the hardware needed to heatsink it -- perhaps you should consider more unconventional solutions, like dropping the project altogether... just saying. 

Tim

[floobydust]

A charging system at 2.4kW is big and expensive, I can't see a cheap way out.

OP's diode is for an output diode of a dynamo:
https://www.eevblog.com/forum/projects/steam-train-chargelighting-system-conversion-to-solid-state/msg2467890/#msg2467890

100A alternator rectifiers are forced-air cooled, you'll need a fan. Truck/Caravan battery isolators are closest and have heatsinks with no fan and 100A capabilities. I think the best ones use MOSFETS and need much less heatsinking than diodes.

[David Hess]

You can parallel two or even four diodes.  Trouble is, diode voltage drop goes down as it warms so unbalance gets worse with higher currents.  You could select the diodes and put them in a series parallel arrangement with a high and a low in each side.

If the diodes are mounted to the same substrate like with a bridge rectifier or dual diode, then they can match pretty well over temperature.

Wouldn't recommend more than 35A for a single diode in a FWB.  They're amazingly robust devices, but they're still made like a fullpack device -- plastic molding provides heatsinking.

I would current derate them to 1/2 or 2/3 as well but I do that with all power diodes.  DC operation is actually a little easier on the diode since the average specification takes into account greater losses from a poorer crest factor.

It seems like older bridge rectifiers more commonly used a metal substrate with a plastic encapsulation.  If I had a choice, I would choose that kind of package for adverse operating conditions.  I like the SOD-57 and SOD-64 packages for the same reason.  Glass DO-41 packages used to be available but now they are all plastic.

[exe]

DC operation is actually a little easier on the diode since the average specification takes into account greater losses from a poorer crest factor.

Interesting, I thought the opposite: losses (Vf) grow slower than current: see "Fig. 4 - Typical Instantaneous Forward Characteristics" in https://www.vishay.com/docs/88503/1n4001.pdf , for example. What do you think?

Or did you mean switching losses?

[T3sl4co1l]

I was thinking, because only one diode out of the four is active at a time.  The 50A rating is for alternating pairs, averaging 25A per diode.  This buys a free, about less than two times, margin over DC, at the expense of somewhat higher voltage drop due to the crest factor.

So i was figuring 35A as a compromise between voltage drop and thermal capacity per diode.

FWIW, the two AC pins can be connected in parallel, utilizing half the FWB as a single diode.  Don't expect the rating to be fully double though (i.e., the full ~50A DC from the module, give or take), because of current sharing issues.

Tim

[David Hess]

DC operation is actually a little easier on the diode since the average specification takes into account greater losses from a poorer crest factor.

Interesting, I thought the opposite: losses (Vf) grow slower than current: see "Fig. 4 - Typical Instantaneous Forward Characteristics" in https://www.vishay.com/docs/88503/1n4001.pdf , for example. What do you think?

Or did you mean switching losses?

The forward voltage increases at the typical 60 millivolts per decade of current plus the contribution from series resistance but I^{^2}R losses still apply so for the same average current, a higher crest factor still results in higher losses.

This is reflected in the surge current ratings, if given in enough detail, which allow higher average current at higher duty cycles.

I was thinking, because only one diode out of the four is active at a time.  The 50A rating is for alternating pairs, averaging 25A per diode.  This buys a free, about less than two times, margin over DC, at the expense of somewhat higher voltage drop due to the crest factor.

So i was figuring 35A as a compromise between voltage drop and thermal capacity per diode.

FWIW, the two AC pins can be connected in parallel, utilizing half the FWB as a single diode.  Don't expect the rating to be fully double though (i.e., the full ~50A DC from the module, give or take), because of current sharing issues.

The datasheet is unfortunately not specific enough to give test conditions but I suspect you are right.  Distributing the current over two diodes per cycle halves the junction-to-case thermal resistance.

I suspect using two diodes in parallel and the lower loss at DC more than makes up for the difference and any issues with current sharing.

[calexanian]

Just doing a casual search turned up a whole new crop of reasonably priced Schottkey rectifiers in that current range. There most have been a new process developed recently because some of them are just a few dollars in single piece quantity while others are well in excess of ten dollars and they have similar specs. Leads me to believe the new parts have better fab processes.

[David Hess]

Just doing a casual search turned up a whole new crop of reasonably priced Schottkey rectifiers in that current range. There most have been a new process developed recently because some of them are just a few dollars in single piece quantity while others are well in excess of ten dollars and they have similar specs. Leads me to believe the new parts have better fab processes.

Inexpensive power schottky diodes have been around for a while in modern power packages.  I think what changed is demand in off-line switching power supplies.

[floobydust]

I think 40-55V parts would not be reliable, and we don't have a PIV number.
A problem is a 24VDC vehicle electrical system transients are to ~450V. So a blocking diode would need to have a PIV rating higher than most Schottky's. Could try TVS and clamp to say 100V, but the trains have a bunch of inductive loads as well as lighting and transients- could be high energy. I think it's 300Ah battery.

OP breaks open the piggy bank and uses a real 150A stud-mount diode Vishay VS-45L $27. Cold 25°C Vf=1.0V at 100A, so 100W of heat. On a Wakefield 486K $75, or something smaller with a fan. $100 for a solution and 3/4 of the dollars are for the heatsink.

Schottky, maybe two MBR80100 80A 100V. Cold 25°C Vf=0.75V at 50A, so 75W of heat for a pair assuming good current sharing.

A MOSFET ideal diode needs to be much less than 10mohm and say 100V rated. Most are D2PAK with fake specs and no real way to heatsink them, despite claims of a few milliohm RDS on. I would probably look at using MOSFET array if they could be protected against transients.

[T3sl4co1l]

FWIW, automotive fast transients aren't very severe, they can mostly be handled by a small TVS.  A big diode like this, won't have a problem (and will probably be rated for much more avalanche than required).

That leaves load dump, which is a good reason to have a bit of extra voltage rating.  A 100V schottky probably wouldn't be bad.

Personally, I would use a 200V PN diode, some module thing rated for the current, it'll cost up to maybe $50.  Bolt it to a big chunk of metal (preferably aluminum) and it's done.

Tim