chief
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power-transistors as rectifiers ?
Since I have burned my 2 bridge-rectifiers (33 A each), now AC passing ...
might I use power transistors for rectification ?
I have a lot of (2 N 3055), on cooling-aluminum, from a power-DC-AC inverter; each of the (2 N 3055) is rated for 100 W, at (10 A * 10 V) maximum.
Since I have them, already wired ... might I use them ?
Does the B-E permit the 10 A or only the C-E ? In the latter case I would have to feed something into the B, to get the the thing conducting, right ?
The other way I think about is to use such a welding-inverter:
http://cgi.ebay.de/Einhell-Inverter-Schweissger%E4t-ISG-1000...
Only: They typically claim the folowing:
-----------------------------------------------------
Schweißstrom I2 35-80 A
Leerlaufspannung V0 85 V
Netzspannung 230 V ~ 50 Hz
Absicherung 10 A
Aufnahmestrom I1max 16,1 A
Einschaltdauer bei 80 A 15%
Einschaltdauer bei 40 A 60 %
Einschaltdauer bei 35 A 100 %
-----------------------------------------------
That would mean: The duration at 40 A would e only 60% of the time on ..; but on the other hand: They claim Vo == 85 V (@40 A == 3.4 kW); as long as
the electrolytic cell uses only 5-10 V (200 - 400 W), what happens: Are the additional (max.) 3 kW just burned into heat (boiling the water or
whatever), or does the thing run cool (does it adjust the voltage to the 10 V) despite their 60%-time-claim ??
[Edited on 6-11-2008 by chief]
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12AX7
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2N3055 is rated at 10A, and doesn't work real well at that. The best you could do is tie B and E together, but since most of the current will flow
through the base, you're limited to about 5A. GBPC3502 will do better, it's much beefier -- no to mention purpose made. If you need more current,
parallel the bridges -- making sure to add a small resistance on each AC pin, a bit of stainless steel wire will do. This ensures current is shared
evenly. If you'd rather have monolithic elements, try a TO-247 or (DO-??) stud package, or a diode module. 50-100A is pedestrian for industrial
diodes.
Welders burn extra VAs in reactance so they don't have to dissipate it. Inverter welders may not even need that, as they can regulate VAs cycle to
cycle.
Tim
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tentacles
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I can understand wanting to use what you've got on hand, but big rectifier diodes aren't exactly expensive - Digikey has 15V 65A schottky diodes for
$3.30 each (http://search.digikey.com/scripts/DkSearch/dksus.dll?Detail?...). Or 50A rectifier bridges for $4.80 - http://search.digikey.com/scripts/DkSearch/dksus.dll?Detail?...
I'd suggest buying the proper tool for the job here, and buy it overspec, the price difference between something that's "enough" and something that'll
handle big current surges and overvoltage (which may be quite a lot less desirable) is insignificant. Plus, a rectifier diode/bridge won't be
dissipating tens of watts as heat.
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gregxy
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If you want to use them as diodes, tie the base to the collector. But since you burned out a 33 amp bridge and have only 10amp BJTs then you would
need to put 4 or 5 in parallel. They won't share current equally and you will soon have a bunch of burned out BJTs.
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12AX7
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| Quote: | Originally posted by gregxy
If you want to use them as diodes, tie the base to the collector. |
That gets you a little more current capacity (when Vce(sat) = Vbe and Ib = Ib(max), you're basically at the limit), but the reverse voltage rating is
completely useless, Veb = 7V or so.
Tim
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chief
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I don't have 5 such Transistors, but 16 Aluminum-cooloings with 3 transistors on each, wired with 0.15-Ohm-resistors; they are already paralleled to 2
banks of 8*3 == 2 * 24 transistors. basically the base, emitter and collector are parallel, but I didn't figure out yet, if 1 on each coolin is used
to drive the other 2
It was a 1000 VA-DC-AC converter, before I took the powerful transformer out (no problem to let it run at 100 A continuous (10-15 V): Doesn't even get
warm from that).
Also the mentioned transformer has the 15-V-winding double, as 30 V-winding with a tap in the middle. So I could use the rectification-method with 2
rectifiers instead of four ...
Besides I have another such Transistor-array: 50 Transistors of "6065"-type, 150 Watt each, at max. 175 volts (datasheet wasn't freely available on
the net). Thats why I just would like to get along with what I have, instead of getting new stuff ...
Besides I have also maybe 10 switching-power-supplys: Do they contain anything worth considering for that purpose ??
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gregxy
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If you short B-C you can get 10A/transistor, but as 12AX7 pointed out, the reverse voltage will be too low.
If you short B-E you can get 7A/transistor so for your 40A you will need 6 in parallel, (i.e. 2 groups of 3) however you
should put a small resistor in series with each one so
that they share the current evenly. This is what the .15ohm
resistor is for, but since it is tied to the emitter it does not
help you now.
You may be able to get some high current schottky diodes out of one of the other power supplies. But best is to spend
$10 and get the right component for the job.
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chief
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Now I have obtained 8 * MBR2545CT Schottky-Diodes, each rated at 30 A/(45V) (1.23 EUR a piece), datasheet here: http://www.datasheetcatalog.org/datasheet/irf/mbr2535ct.pdf
I was going for larger ones (60 A), but they were not available; how is it with the paralleling of such Schottky-diodes ? Resistor nexessary ?
And also I have no real Idea of how to cooling-mount them, never did such a thing before ...
Could these let be run under oil ? In an oil-reservoir ? Or would it damage (leak into) the package over time ?
I plan to parallel 2*4, using the center-tap of the transformer (2-Diode-full-wave rectification, instead of 4-diode-bridge)
[Edited on 8-11-2008 by chief]
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bquirky
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Gday Cheif,
Generally speaking diodes should not be disapating a lot of heat and therefor not require anything as radical as oil cooling in 'normal' applications
if your diode is dropping say 0.7volts (about average for a PN junction) at 30 amps thats about 20 watts an old computer CPU heat sink should be
enough for that. as a general rule for fire and finger safety the heat sink shouldn't get to hot to touch 
If your swimming in bags of old Transistors/FET's or stuck on a desert island. Many power Switching devices designed for switching inductive loads
have a built in diode designed to protect the transistor from the back emf generated by the Coil/magnet/motor when the device turns off.
In many cases this diode is working just dandy even if the transistor is dead the diode that is in the these transistors is internaly connected in
paralell to the transistor between the Colector and Emmitor or the Source and the Drain.
Running diodes in paralell can work ok as long as you keep the legs an evan length But running high power transistors or fets in paralell under load
is a black art becuse of slight variations between each device.
But if it was me id just order a packaged bridge rectifyer of twice the current rating required from your local electronics supplyer. and get on with
my project
Good Luck !
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chief
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I soldered everything and measured the voltage-difference of each diode:
==> 5 were at 204 (+/-1) mV
==> 3 were at 220 (+/-2) mV
With my power-transistors I verified: It are 2N3055, not 2N3065 . They are parallel, 0.15 Ohm at the Emitter of each one. A while ago I was thinking
about an induction-furnace ...
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12AX7
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Ah, except that normal diodes, like 1N4001, aren't passing thirty amperes.
Yes, you should have current-spreading resistors, and you MUST have a heatsink. Each one will dissipate perhaps 20A * 1V = 20W and the TO-220 package
only handles 1W in air, and that's being optimistic.
Why did you get TO-220 packaged diodes? TO-247 (or TO-3P, nearly identical) are much more common and much more robust (such diodes are in every
computer power supply I've taken apart).
What is your voltage supply? You understand you cannot safely use these to rectify more than 10VDC, right?
Tim
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chief
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Well, the voltage supply is 30 V with center-tap, and that's what I want: 2 * 12-15 V, full wave-rectification, for some electrolysis ...
I originally was going for some 60A-Diodes, but those they hadn't either. I assumed these (that I got) to have the best amperage-Price-ratio, and
anyhow: They are rated at 45 Volts. Do you really thing they couldn't be used for eg. 30 V ? (I'm gonna have them at max. 15 anyhow).
I also ,mounted them onto some coolin Aluminum: Its flat (no extra rips), but 4 mm thick. The screws were in place,in fact i demounted 4 * 2N3055
Transistors and used the 8 screws then free ..
So each of the diodes has its own 0.5-1 inch material-circumference, approximately, of 4 mm Al.
Would this give enough cooling ?
I also very carefully heavy-soldered copper-cable to each of the pins, so _that_ is not gonne be source of any extra heat. The soldering was done by
dipping the diodes with the cabling in place into molten colophonium, and then immediately into a molten solder-bath (the legs only, of course).
Immediately after that it was cooled in alcohol, so the temp. probably never reached the interieur of the elements, since the whole hot-soldering was
done in less than 1 second.
My plan so far is to run it tomorrow at 60 A (but it's rated at 120 A, since the rectification will be 2 * 4 elements, each at max. 30 A rated) ...
What do you think ? Will I still need some resistors ? At the mentioned forward-voltages from above ??
Edit:
===============
Shouldn't be the power-dissipation calculated from the forward-voltage (200 mV) * Amperage ?
Then for 30 A it would be (30 A) * (0.2 V) == (6 W) ??? That wouldn't be too much ?
[Edited on 9-11-2008 by chief]
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12AX7
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| Quote: | Originally posted by chief
Well, the voltage supply is 30 V with center-tap, and that's what I want: 2 * 12-15 V, full wave-rectification, for some electrolysis ...
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That's a bit high for schottky. You should have regular silicon rectifiers for that range (100V, however many amps).
So you've got 30VCT and FWCT (full wave, but not full wave bridge) rectification? That's the way to go with low volts and high amps. Obviously too
much for one cell, more like three, maybe two with ballast.
| Quote: | | I also ,mounted them onto some coolin Aluminum: Its flat (no extra rips), but 4 mm thick. |
Cooling comes from area, not volume, so we'll see. Not a big deal, if it gets too hot too quickly, turn it off and bolt on more aluminum until it
stays cool!
Incidentially, test the case temperature of the diodes, using the back side of your finger (after checking with the moistened front side of your
finger that it's not already boiling  ), more sensitive that way and if it's
live, your finger (and hand) will jump away from the shock. Case temperature is more important than heatsink temperature, for obvious reasons.
| Quote: | | Immediately after that it was cooled in alcohol, so the temp. probably never reached the interieur of the elements, since the whole hot-soldering was
done in less than 1 second. |
That's not a problem. Actually quenching is probably worse, due to thermal stress on the die. Everything silicon is rated for soldering at 250C for
10 seconds and I've desoldered plenty of [working!] parts with worse than that. Heat's not really a concern as long as it's under 250C.
| Quote: | | What do you think ? Will I still need some resistors ? At the mentioned forward-voltages from above ?? |
Forward voltage isn't what matters, it's how it changes. If one chip is heating up a lot more than the others, its Vf falls, so it hogs current,
heating up, etc.... Your choice of 4mm aluminum is a good step to preventing this. Paralelling silicon ~~usually~~ works okay, particularly on a
stocky heatsink, so you're probably fine. And they're only diodes, so if they burn, who cares -- although I hope you have a fuse on your transformer
in case they do.
| Quote: |
Shouldn't be the power-dissipation calculated from the forward-voltage (200 mV) * Amperage ? |
Yes, that's what I did.
Refer to Figure 1 in the IR datasheet you linked. Vf = 0.8 and some change, at 30A. Your figure of 200mV was measured at 1mA or so, which is 10^3
off the scale of this plot. (Notice the voltage drops to a hair over 0.7V up at 175C. A matched diode still at 25C would be conducting only 20A at
the same voltage, so for this rather extreme temperature difference and a total current of 50A, you'd get 20A in one and 30A in the other -- a small
enough difference that you really can wave your hand and say it's okay, as long as heatsinking is good anyway.)
As for current, 30A total is more like 15A per diode. But peak is more like 20-25A (more like 50-200A if you have a lot of capacitance on there -- no
need, I think, to filter the voltage just for a cell, so save the diodes the trouble!). And power isn't from averages, it's from RMS, so what's more,
you really ought to take the sine wave, pass it through the diode's graph (which is logarithmic plus resistive), multiply the instantaneous
voltage and current, and integrate over a half wave. Then halve that because each diode is on only half the time. (And this is all possible between
calculus, pad and pen, but that doesn't mean you want, or need to do it...) This will get you the actual value (within whatever parts tolerances),
but such abusive precision is unnecessary; it's sufficient to guess within a factor of 2 that the voltage drop and current are so-and-so and therefore
the average power is this, and we'll just cover the uncertainty by using a heatsink twice as big, which will give us nice low operating temperatures
most of the time.
Tim
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IrC
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You should find a couple old junk alternators and knock out the 6 diodes in each. Drill proper size holes in an empty heat sink and using a socket tap
them into the holes. The sink would be one terminal so mount accordingly. Tremendous ampere handling ability at near zero cost.
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Texium
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Thread Moved
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