These high current bridges have a large voltage drop and at 50A my first thought is that's an insane amount of power to dissipate! Has anyone done this? Is it possible even with a large heat sink?

For example, here is the the datasheet for KBPC5010 50A Bridge Rectifier.

http://pdf1.alldatasheet.com/datasheet-pdf/view/34010/WTE/KBPC5010.html

It is rated at 50A. According to the datasheet, the voltage drop is about 1.2V "per element". Per element means per diode, correct?

Further, the junction to case thermal resistance is 1.5K/W also "per element". So it is like 3K/W. Plus whatever heat sink resistance, let's say 10K/W. So at the Vf=2.4V that's not much current at all before things get very hot. But even worse, according to the data sheet the rated current is only achievable at 50C and then it drops sharply and pretty much goes to zero at higher temps.

So how do you cool it to 50C? Even with an indefinitely large heat sink you still have 3K/W of the Rtjc .

Is this just a marketing trick or my calculations are wrong?

 

This looks bogus to me.

I have used 50 amp bridges with no difficulty, so I am leaning toward calling this low quality.

I have provided datasheets so other people can examine them.

 

I got another datasheet off a trusted site and it also looks like the front page ratings are bogus.

 

This looks bogus to me.

I have used 50 amp bridges with no difficulty, so I am leaning toward calling this low quality.

I have provided datasheets so other people can examine them.

Your second datasheet is the same as mine for the same device. Your other datasheet is very similar. When you say bogus, do you mean the numbers I quoted? Like I said perhaps my calculations or understanding of these parameters are wrong.

 

Ah, sorry, I thought you provided the datasheets of the devices you used...

 

I examined all 3 datasheets and ran the numbers myself. The best I can get is a 75C rise at 50 amps, and that is outside the derating curve.

If these things only meet their ratings inside a freezer, the Mfg should say so.

 

Here is another device with very similar

I examined all 3 datasheets and ran the numbers myself. The best I can get is a 75C rise at 50 amps, and that is outside the derating curve.

If these things only meet their ratings inside a freezer, the Mfg should say so.

Disregarding the derating curve, I am just curious how did you arrive at 75C rise?


Here is a different device by a different manufacturer, very similar numbers:

http://www.radioshack.com/nte-nte53016-silicon-bridge-rectifier-200v-50a/55052170.html#.VNGJ0i4TD_I
Datasheet here:

http://www.nteinc.com/specs/53000to53099/pdf/nte53016_20.pdf

 

Even without the derating curve, the numbers I get don't make any sense. Can you see an error?

Vfmax = 1.1V per element, so about 2V total for two diodes in series. Rtjc = 1.5k/W, again times two, so 3k/W total. Rt heat sink about 7K/W (that is very optimistic). So assuming the total thermal resistance 10K/W, at 50A we would need to dissipate about 50A*2V = 100W??? That sounds nuclear hot. I must be wrong somewhere.

Perhaps another factor is the Vfmax is given at 25C, so it might decrease or increase at higher temps based on the temperature coefficient. Since it is a silicon bridge, the Ic where the temperature coefficient goes from negative to positive can be quite low. So it is likely that the voltage drop will be higher that 2V.

 

I used the best graph I could find and read 1.18 Volts across one diode and multiplied that by 1.5 C/W. That got me 88.5 C temperature rise, which I promptly wrote down as 75 C.

Then I looked on the derating graph and saw that the temperature rise above 25 C is unacceptable.

I gave every spec the bast value I could find, even if it was wrong, and the bridge rectifier still failed to meet the advertised claims.

 

Here is how I read it:
Since it is per element each diode is only on 1/2 the time so average current is 25 amps. Times 2.2 volts is 55watts.
Jc= 1.5X55=83 C. The heat sink ia 10 X 10 inch (about) = 1.6C/W = 132C plus 25C ambient = 157C, which is close to 150C max spec.

 

My bad it's 2.4 volts drop. So the easy way is the graph wit Tc = 55C (big heat sink)
2.4X25x1.5+55=145

 

That looks the same as 1.2V x 50 A x 1.5 =90C
That's the temperature rise of the silicon if the case is locked at 55 C.
So the heat sink has to get rid of 60 watts into 25 C air.
The heat sink will be rated as, "1/3 C/W"
That's a lot of heat sink.
Probably needs a fan.

 

We must be missing something. The hs is not big enough.

 

Here is how I read it:
Since it is per element each diode is only on 1/2 the time so average current is 25 amps. Times 2.2 volts is 55watts.
Jc= 1.5X55=83 C. The heat sink ia 10 X 10 inch (about) = 1.6C/W = 132C plus 25C ambient = 157C, which is close to 150C max spec.

So 25A flowing through 2 diodes at the same time? Then both of them are dissipating 55W at 1.5C/W? Wouldn't that double the Jc?

 

That's the problem with a datasheet that says, "per element". Is that, "per diode"? Is it, "per conducting pair"? Is it, "per bridge rectifier"? Those guys need to learn the word, "diode".

 

That looks the same as 1.2V x 50 A x 1.5 =90C
That's the temperature rise of the silicon if the case is locked at 55 C.
So the heat sink has to get rid of 60 watts into 25 C air.
The heat sink will be rated as, "1/3 C/W"
That's a lot of heat sink.
Probably needs a fan.

Correct me if I am wrong but I read that the total heat resistance is the sum of the junction to case and then case to ambient since they can be thought of as connected in series. So I thought that even if you have an indefinitely huge heat sink you still have the junction to case rise of 1.5K/W. So my interpretation is 1.5K/W per element is the best you can do assuming the device is mounted on a heat sink (presumably a very large one). So you can't go lower than that only higher.
I know that in some datasheets they just give you the R junction to ambient with the heat sink resistance already factored in for an application specific sinking method (such as a copper PCB trace).

So to me it sounds like no matter what you do you get at least a 90C rise at the max rated current. And that's "per element". Ouch.

 

Junction temperature max is 125C according to the third datasheet.
Room temperature is assumed to be 25C. At 1.2 V per, "element"
100C/60W = 1.666 C/W maximum, and the case already uses up 1.5 of those.

If you assume a 125 C allowable rise and use Rons 55 watt figure, 125C/55W = 2.27 C/W and the case uses up 1.5 of those.

Either way, you have less than 1 C/W to work with.

 

Junction temperature max is 125C according to the third datasheet.
Room temperature is assumed to be 25C. At 1.2 V per, "element"
100C/60W = 1.666 C/W maximum, and the case already uses up 1.5 of those.

If you assume a 125 C allowable rise and use Rons 55 watt figure, 125C/55W = 2.27 C/W and the case uses up 1.5 of those.

Either way, you have less than 1 C/W to work with.

Yeah, and considering the derating graph, sounds like it is basically a 10A device. With a huge heat sink No wonder it is only 99c on Ebay... Reminds of those 6000 mah UltraFire batteries.

 

Yes, I think so.
Here is a more honest one (25 amp) that gives Jc values of .6C/watt. So if you look at the curve it almost works out to a 1C/W heat sink. 5 amps without it.
http://www.diodes.com/datasheets/ds21221.pdf

 

If these things only meet their ratings inside a freezer, the Mfg should say so.

Thats very common...
Look at the "current rating" of Mosfets,etc... 100-200A.. yeah right..