High current pcb

George Jefferson schrieb:

Hello,

there are some pcb manufacturers who can make pcbs with very thick copper, about 300 µm instead of the usual 35 µm. You may even get pcbs with normal and thick copper on both sides. You may need special thermal pads for easy soldering. Just ask some manufacturers about pcbs with thick copper for high currents.

Bye

Methinks you will absolutely need to use a copper busbar - and a lotta FETs.

This could be done on a pcb, but 200 amps is getting serious. The problems will of course be getting rid of the heat from the fets, getting rid of heat from the pcb traces themselves, and getting all that current on and off the board.

Surface-mount fets aren't a good choice here, unless you are willing to use a lot of them. They are hard to keep cool, and there will be current crowding in the pcb traces getting into and out of them.

The best way to get on and off a board like this is to use lots of relatively small wires, again to avoid current crowding in the traces. Fastons soldered to the board, mating with #14 wire maybe, is pretty good. The external leads are mechanically friendly as compared to a single hunk of #4, and the lead resistances help equalize currents on the board.

Is this for production or one-off? If single-unit, it would be a lot easier to build with TO-247s with aluminum heatsink+busbar construction. Expect to blow it all up a couple of times.

Do you need control as well as rectification? If not, mayhe just use schottky diodes on a heat sink.

John

The 200A could be done on 2oz copper with careful layout. Watch for the narrow spots and always calculate the trace resistances. Don't make any assumptions. You may have to reinforce a couple of tough places with bus bars; that could ease up the layout.

Vladimir Vassilevsky DSP and Mixed Signal Design Consultant

George Jefferson schrieb:

Hello,

these fets tolerate 80 W only at an mounting base temperature of 25 °C, see figure 2 in the data sheet. When the pcb should do the job of the heat sink too, there must be enough copper area to get the heat out of the fets and the pcb.

Bye

A small part like this doesn't conduct heat into pcb pours very well. If you stick a part to a relatively thin thermally conductive sheet, theta goes up as the part footprint area goes down [1]. That's a really bad deal when you're working with copper foil and dpak-type parts. Multiple internal pour patches and lots of thermal vias can help increase the part's effective footprint.

A TO-220 on an infinite sheet of 0.062" thick aluminum has a theta of roughly 2K/W. That fact can be used to do some very rough scaling, factor-of-three type stuff maybe. Thermal design like this is very messy. Lateral heat spreading, namely hot-spot effects, are important even on extruded heat sinks; they become dominant on thin copper foil. Now add in current crowding and the positive TCR of copper and convection issues, eventually you have a serious math problem.

ftp://jjlarkin.lmi.net/Infinite_Sheet.jpg

A PCB is an expensive way to make a heat sink, and a bad way to conduct/terminate a lot of current.

John

[1] anybody know the exact relation?

"John Larkin" wrote in = message news: snipped-for-privacy@4ax.com...

Easy to approximate. Assume a circular footprint (cf. spherical = chicken). Assume heat dissipation at the center is zero (fair for an = infinnitessimal segment, blatantly false for an infinite number of = them). If heat diffuses through copper out to infinity, temperature = drops inversely with distance (because cross sectional area increases = linearly). It looks like a point charge in space, and the temperature = is defined by Gauss' law.

Of course, heat diffuses through two or three means, with strange = temperature-dependent coefficients besides. So it's not at all true = that, the device itself, and the little bit of copper surrounding it = that doesn't have a quite circular temperature profile, isn't = dissipating any power. In fact, it could be dissipating a considerable = amount of power. If the heat source were an infinnitessimal point, it = would have infinite temperature, and therefore radiate infinite power = density (power can still be finite, since the area is infinnitessimal).

However, it is true that heat diffuses out, one way or another, so maybe = the power dissipation is just a little higher in the center, and spreads = out in a slightly-steeper-than-inverse relationship, eventually going to = zero at infinity all the same. The trouble is deriving the exponent and = coefficient of that power law.

Tim

--=20 Deep Friar: a very philosophical monk. Website:

Since it's a full wave bridge, average current through each MOSFET is going to be 1/6 or 1/8 of the total, so only 33A or 25A, which looks within the realm of possibility. Power dissipation from the MOSFETs alone might be ~50W though, more if it's a nasty load.

Assume heat dissipation at the center is zero (fair for an infinnitessimal segment, blatantly false for an infinite number of them). If heat diffuses through copper out to infinity, temperature drops inversely with distance (because cross sectional area increases linearly). It looks like a point charge in space, and the temperature is defined by Gauss' law.

temperature-dependent coefficients besides. So it's not at all true that, the device itself, and the little bit of copper surrounding it that doesn't have a quite circular temperature profile, isn't dissipating any power. In fact, it could be dissipating a considerable amount of power. If the heat source were an infinnitessimal point, it would have infinite temperature, and therefore radiate infinite power density (power can still be finite, since the area is infinnitessimal).

power dissipation is just a little higher in the center, and spreads out in a slightly-steeper-than-inverse relationship, eventually going to zero at infinity all the same. The trouble is deriving the exponent and coefficient of that power law.

Nice rant, but still no answer.

Given a perfectly thermally conductive puck attached to an infinite sheet of thin [1] finite-thermal-conductivity material, and assuming conduction cooling only, what is the relationship of puck theta to puck diameter?

This is relevant to situations where you have a choice of, say, SOT89 versus DPAK versus D2PAK and you're heatsinking to copper foil.

John

[1] thin relative to puck diameter

And I'd guess that a battery charger will be a "nasty load." At 200 amps average, unless there's a big inductor available, 600 amp peaks wouldn't shock (punalert) me.

John

In the "one-or-only-a-few-off for lab work" area, we have been known to bend up some heavy copper wire and solder it to a trace to increase current carrying capacity. But the problems of heat dissipation from the transistors mentioned by others probably rule the pile of problems here, and make choosing a serious power package rather than an unsuitable surface mount package the better option. Sometimes smaller is not better.

--
Cats, coffee, chocolate...vices to live by

agenews: snipped-for-privacy@4ax.com...

en). =A0Assume heat dissipation at the center is zero (fair for an infinnit= essimal segment, blatantly false for an infinite number of them). =A0If hea= t diffuses through copper out to infinity, temperature drops inversely with= distance (because cross sectional area increases linearly). =A0It looks li= ke a point charge in space, and the temperature is defined by Gauss' law.

ature-dependent coefficients besides. =A0So it's not at all true that, the = device itself, and the little bit of copper surrounding it that doesn't hav= e a quite circular temperature profile, isn't dissipating any power. =A0In = fact, it could be dissipating a considerable amount of power. =A0If the hea= t source were an infinnitessimal point, it would have infinite temperature,= and therefore radiate infinite power density (power can still be finite, s= ince the area is infinnitessimal).

the power dissipation is just a little higher in the center, and spreads o= ut in a slightly-steeper-than-inverse relationship, eventually going to zer= o at infinity all the same. =A0The trouble is deriving the exponent and coe= fficient of that power law.

Hmm, I think I understand the question. I might try breaking up the sheet into pie shaped pieces. (cylindrical coordinates) The pointed end of the pie piece is cut off becasue it's touching the puck. Now you've got to do the integral for a pie piece with different size bites taken out of the pointy end. But if you insist on an infinte sheet then the answer is infinity no matter what the puck size. So you need a bit better boundary condition for the size of the sheet.

George H.

"John Larkin" wrote in = message news: snipped-for-privacy@4ax.com...

defined by

^ ^ ^ ^ ^

No? Gauss' law is easy. Set up the integral.

Tim

--=20 Deep Friar: a very philosophical monk. Website:

On a sunny day (Fri, 16 Jul 2010 13:57:42 -0500) it happened "Tim Williams" wrote in :

If you want the theoretical side, with calculations, and a lot of integrals, google for ahttv131.pdf, the MIT heat transfer textbook. Only 762 pages... But it gives you that feeling that you know it all :-)

Assume heat dissipation at the center is zero (fair for an infinnitessimal segment, blatantly false for an infinite number of them). If heat diffuses through copper out to infinity, temperature drops inversely with distance (because cross sectional area increases linearly). It looks like a point charge in space, and the temperature is defined by Gauss' law.

temperature-dependent coefficients besides. So it's not at all true that, the device itself, and the little bit of copper surrounding it that doesn't have a quite circular temperature profile, isn't dissipating any power. In fact, it could be dissipating a considerable amount of power. If the heat source were an infinnitessimal point, it would have infinite temperature, and therefore radiate infinite power density (power can still be finite, since the area is infinnitessimal).

power dissipation is just a little higher in the center, and spreads out in a slightly-steeper-than-inverse relationship, eventually going to zero at infinity all the same. The trouble is deriving the exponent and coefficient of that power law.

PC with Core2 series CPU is running >100A regulator 3, 4, or 6 phase around the CPU area, lots of cooling air flow. But they're not trying to get that current off board, just direct it through a couple in^^2 of CPU chip.

OP needs controlled bridge to regulate the charger?

Grant.

2 ounce copper should give you ~250 uohm per square so a two layered board 2 ounce board would dissipate

Except during switching transitions, when "on" the Vds of a full-wave rectifier MOSFET can't really exceed minus one diode drop, nor would it be very useful if it did.

Best regards, Spehro Pefhany

--
"it's the network..."                          "The Journey is the reward"
speff@interlog.com             Info for manufacturers: http://www.trexon.com
Embedded software/hardware/analog  Info for designers:  http://www.speff.com

Of course you're right. I should get near a computer during barbeque saturdays. Vds(on) is actually ~450mV at 100A for 45W which is not that easy to remove from that package. I think based on other parts which I have used, one shouldn't dissipate more than ~10W even with a large piece of copper around the package.

--
Muzaffer Kal

DSPIA INC.
ASIC/FPGA Design Services

http://www.dspia.com

Assume heat dissipation at the center is zero (fair for an infinnitessimal segment, blatantly false for an infinite number of them). If heat diffuses through copper out to infinity, temperature drops inversely with distance (because cross sectional area increases linearly). It looks like a point charge in space, and the temperature is defined by Gauss' law.

temperature-dependent coefficients besides. So it's not at all true that, the device itself, and the little bit of copper surrounding it that doesn't have a quite circular temperature profile, isn't dissipating any power. In fact, it could be dissipating a considerable amount of power. If the heat source were an infinnitessimal point, it would have infinite temperature, and therefore radiate infinite power density (power can still be finite, since the area is infinnitessimal).

power dissipation is just a little higher in the center, and spreads out in a slightly-steeper-than-inverse relationship, eventually going to zero at infinity all the same. The trouble is deriving the exponent and coefficient of that power law.

Really? If the chip requires only 65 Watts how do you get to 100 A even at 1.2 V?

Of course with modern CPU chips that take far less power at idle these issues become very fuzzy.