High current pcb

[George Jefferson]

I want to replace a CT full wave rectifier for a 200A+ battery charger with
a active rectifying system.

http://www.nxp.com/documents/data_sheet/PSMN2R5-30YL.pdf

Are some 100A 80W logic level fets which means I can parallel 3 or 4 of them
and increase the efficiency of the charger(there are other reasons for doing
it too).

I'll be driving the mosfets using a driver circuit instead of a uC of
course(just to reducing cross conduction problems).

In any case the real issue I have is how to do this on a pcb, if it is even
possible. Not only are the leads to the packages somewhat small I can't
think of any way how this can be done on a pcb(30oz pcb's?) or large solder
traces in any decent fashion.

Are thee any tricks to defeat mother nature or will I have to design some
special module that uses a copper bus bar like thing with the mosfets
soldered directly to it?

 

[Uwe Hercksen]

In any case the real issue I have is how to do this on a pcb, if it is
even possible. Not only are the leads to the packages somewhat small I
can't think of any way how this can be done on a pcb(30oz pcb's?) or
large solder traces in any decent fashion.

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

 

[Uwe Hercksen]

I want to replace a CT full wave rectifier for a 200A+ battery charger
with a active rectifying system.

http://www.nxp.com/documents/data_sheet/PSMN2R5-30YL.pdf

Are some 100A 80W logic level fets which means I can parallel 3 or 4 of
them and increase the efficiency of the charger(there are other reasons
for doing it too).

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

 

[Tim Williams]

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].

[1] anybody know the exact relation?

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

 

[Spehro Pefhany]

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

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.

 

[Ecnerwal]

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.

 

[Grant]

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].

[1] anybody know the exact relation?

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

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.

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.

John

[1] thin relative to puck diameter

 

[Spehro Pefhany]

2 ounce copper should give you ~250 uohm per square so a two layered
board 2 ounce board would dissipate <5W over the etch which is
certainly acceptable. I think your problem is more basic than that. If
you look at Figure 3 of the datasheet you refered to, you would notice
that at 10V Vds, DC current carrying capacity of the device is less
than 5A!

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

 

[[email protected]]

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].
<snip>
[1] anybody know the exact relation?

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

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.

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.

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.

OP needs controlled bridge to regulate the charger?

Grant.

John

[1] thin relative to puck diameter

 

[JW]

[...]

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.

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

Some of the Quad cores have TDPs of 105W:
http://ark.intel.com/ProductCollection.aspx?familyID=28398&MarketSegment=DT

 

[Spehro Pefhany]

[...]

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.

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

Some of the Quad cores have TDPs of 105W:
http://ark.intel.com/ProductCollection.aspx?familyID=28398&MarketSegment=DT

The i7 (6 core) has a TDP of 130W. The core temperatures _rapidly_
shoot up when you give it something significant to do.

 

[Grant]

On Fri, 16 Jul 2010 10:44:40 -0500, "Tim Williams"

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].
<snip>
[1] anybody know the exact relation?

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

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.

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.

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

Q6600 is 105W, I got one

Grant.

 

[Grant]

[...]

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.

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

Some of the Quad cores have TDPs of 105W:
http://ark.intel.com/ProductCollection.aspx?familyID=28398&MarketSegment=DT

The i7 (6 core) has a TDP of 130W. The core temperatures _rapidly_
shoot up when you give it something significant to do.

They do some fancy power saving plus (over) clocking on-chip to improve
windoze performance? Measure temp on chip in places and go fast as they
can in the current (!) circumstance.

Grant.

 

[[email protected]]

[...]

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.

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

Some of the Quad cores have TDPs of 105W:
http://ark.intel.com/ProductCollection.aspx?familyID=28398&MarketSegment=DT

The i7 (6 core) has a TDP of 130W. The core temperatures _rapidly_
shoot up when you give it something significant to do.

The original dual core G5 sucked up 130W at 1.25V. That was five years ago.

The big boss had a codename that chip contest, which I won with the codename
"Antares", with the series theme "Bright Stars". He made the comment in an
email that he hoped it wasn't a comment about the power density. I replied
(in an email) that the dual-core G5 had a power density of 1E9 times that of
Sol. ANother time I didn't make any points. ;-)

 

[JosephKK]

[...]

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.

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

Some of the Quad cores have TDPs of 105W:
http://ark.intel.com/ProductCollection.aspx?familyID=28398&MarketSegment=DT

Not a big surprise. Just the same a lot of power goes to the 1.8 V to
2.5 V north bridge interface. 1066 MHz memory interface is power
hungry.

 

[JosephKK]

On Fri, 16 Jul 2010 10:44:40 -0500, "Tim Williams"

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].
<snip>
[1] anybody know the exact relation?

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

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.

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.

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

Q6600 is 105W, I got one

Grant.

Does it take the rather typical 3 or 4 supply voltages at various
currents? A lot of the high power is getting on/off chip (high speed
busses).

 

[Grant]

On Fri, 16 Jul 2010 10:44:40 -0500, "Tim Williams"

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].
<snip>
[1] anybody know the exact relation?

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

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.

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.

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

Q6600 is 105W, I got one

Grant.

Does it take the rather typical 3 or 4 supply voltages at various
currents? A lot of the high power is getting on/off chip (high speed
busses).

I'm sure you can go find and read the specs as easily as I can.

Grant.

 

[Uwe Hercksen]

The big boss had a codename that chip contest, which I won with the codename
"Antares", with the series theme "Bright Stars". He made the comment in an
email that he hoped it wasn't a comment about the power density. I replied
(in an email) that the dual-core G5 had a power density of 1E9 times that of
Sol.

Hello,

did you calculate the power density per surface area of the chip, or per
volume of the chip?

Bye

 

[[email protected]]

Hello,

did you calculate the power density per surface area of the chip, or per
volume of the chip?

Volume.

 

[JosephKK]

On Mon, 19 Jul 2010 03:19:26 -0700, [email protected] wrote:



On Fri, 16 Jul 2010 10:44:40 -0500, "Tim Williams"

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].
<snip>
[1] anybody know the exact relation?

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

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.

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.

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

Q6600 is 105W, I got one

Grant.

Does it take the rather typical 3 or 4 supply voltages at various
currents? A lot of the high power is getting on/off chip (high speed
busses).

I'm sure you can go find and read the specs as easily as I can.

Grant.

Oops, my bad. I guessed from the way you were talking you already had
the datasheet and knew already.