Dissipating 700mW through a D2PAK 7805 regulator? Copper pour heat-sink design

[incf]

Due to certain constraints in a commercial design, under a hypothetical worst case scenario, I am looking to dissipate 700mW (30mA @ 28V) of heat in a D2PAK 7805 regulator (link to digikey).

I am seeking guidance on designing the copper pour heatsink.

Worst case I intend to operate at an air temperature of 70C inside of an enclosure. I desire to hit a 30C temperature rise (40C/W Theta-j-a) worst case.
I've read that the best case junction to ambient thermal resistance is 15C/W and a typical one under nonideal conditions is 63C/W.

Supposedly a 1" PCB with 2oz copper can supply 40C/W (Theta-j-a) if it is designed correctly.

I use Altium, but do not have the PDN analysis tool. I have a 4 layer PCB with 1oz copper on all layers.

As far as I can tell "all I have to to" is make at least a 1" square copper pour on all 4 layers, and stitch it with ~0.3mm filled vias on a 1mm pitch. (I know that I'm using thinner copper than what most datasheets use)
Supposedly convection of hot air over the PCB will do the rest if the enclosure is no too close to the PCB surface?

This datasheet claims 15C/W under idealized conditions.
It also provides this nifty graph for a single sided copper pour heatsink.



edit: switch from DPAK to D2PAK

[tom66]

You can get D2PAK heatsinks which are SMD placeable if that is an option.

https://www.mouser.co.uk/ProductDetail/Aavid/573300D00000G?qs=HIKTYMuL%252B9N4XRiNzvrYsg%3D%3D

Given this heatsink achieves 16C/W, I suggest that a copper area to achieve 15C/W would be challenging.  However, nothing stopping a belt-and-braces approach and going for both.

Another option I have used in the path is a power resistor in series with the regulator, aiming to dissipate half the power in the resistor.   Resistors tend to tolerate higher temperatures than semiconductors.

And of course you have the "XY problem" question:  is there a really good reason a buck converter wouldn't do?

[incf]

You can get D2PAK heatsinks which are SMD placeable if that is an option.

https://www.mouser.co.uk/ProductDetail/Aavid/573300D00000G?qs=HIKTYMuL%252B9N4XRiNzvrYsg%3D%3D

Given this heatsink achieves 16C/W, I suggest that a copper area to achieve 15C/W would be challenging.  However, nothing stopping a belt-and-braces approach and going for both.

Another option I have used in the path is a power resistor in series with the regulator, aiming to dissipate half the power in the resistor.   Resistors tend to tolerate higher temperatures than semiconductors.

And of course you have the "XY problem" question:  is there a really good reason a buck converter wouldn't do?

• I only intend to achieve 40C/W (or better if I am lucky)
• The PCB in question is a effectively a specialized "high power" switching power supply. (passively cooled)
• The regulator powers the control circuitry for said SMPS.
• I feel that the part count on my design is too high, so I'd like to keep things as simple as possible.
• For various reason I would like to avoid mechanical heatsinking - I have plenty of space for a copper pour heatsink.
• I fear that the box containing this device will be dropped down concrete staircases and from loading docks with some frequency (don't ask).

[floobydust]

I did a design with those Aavid D2PAK heatsinks and found they are not the saviour they purport to be.
The problem is poor heat transfer from the D2PAK to the heatsink - it's only the PCB copper which had too much thermal resistance- sideways.
Second, you will never really install/remove one manually. It just takes too much heat.

OP I think you can pull it off but it's not an ideal approach. Inner layers can't do much with heat as they are covered (insulated) by FR4.
I would drop the 28V down using another part to split up the heat dissipated between two components. Or conduct some of the heat to the enclosure, depending on how the board is mounted or what wiring/terminal blocks there are, these can also be part of a "heatsink". The heat will cook nearby electrolytic capacitors so try keep them away.

[incf]

... The heat will cook nearby electrolytic capacitors so try keep them away.

I plan on using ceramic capacitors for the linear 28V->5V @ 30mA supply.

I have taken steps to protect my expensive 5000h 105C rated electrolytics (used for a high power SMPS on the same PCB). They are located far away. Also they are switched by relays so that they only connected to power for 1 or 2 hours every few days. They should last at least 20 years under worst case conditions.

[tom66]

Will the 28V supply ever not be 28V?  For instance, will the product have to operate with a 10V input?

[incf]

Will the 28V supply ever not be 28V?  For instance, will the product have to operate with a 10V input?

Yes, the input supply voltage fluctuates wildly. I'd really prefer this to be a particular supply rail to be a "plain old" single stage linear 7805 regulator.

*for context, one of several different power inputs for this system is a direct connection to a solar panel. It's effectively a MPPT unit.

[PCB.Wiz]

Due to certain constraints in a commercial design, under a hypothetical worst case scenario, I am looking to dissipate 700mW (30mA @ 28V) of heat in a D2PAK 7805 regulator

• I only intend to achieve 40C/W (or better if I am lucky)
• The PCB in question is a effectively a specialized "high power" switching power supply. (passively cooled)
• For various reason I would like to avoid mechanical heatsinking - I have plenty of space for a copper pour heatsink.

It sounds like there are other sources of heat, you forgot to mention in the OP.

If you have room, just throw as much spread copper as is available, and measure it in a box.
Your copper spreads the heat, but it still needs to escape the box. If you have other heat sources, the combination is what matters.

A 7805 can take up to 8mA just for the regulator, so you could look to a better alternative to lower the total power ?

[incf]

...

It sounds like there are other sources of heat, you forgot to mention in the OP.

If you have room, just throw as much spread copper as is available, and measure it in a box.
Your copper spreads the heat, but it still needs to escape the box. If you have other heat sources, the combination is what matters.

A 7805 can take up to 8mA just for the regulator, so you could look to a better alternative to lower the total power ?

Its 5.8 watts on a roughly 4x4" PCB in an open ended plastic extrusion in a 18x24x4" plastic box.

I might need to buy a box of 10k thermistor assemblies and epoxy.

[Someone]

in an open ended plastic extrusion in a 18x24x4" plastic box.

Thermal dissipation, plastic enclosure...
you're already fighting physics from the outset.

[PCB.Wiz]

Its 5.8 watts on a roughly 4x4" PCB in an open ended plastic extrusion in a 18x24x4" plastic box.

I might need to buy a box of 10k thermistor assemblies and epoxy.

Below is an indication of how many watts a similar area PCB is benchmarked to - 5.8 watts sounds a lot, inside a box.

(Four-layer Board)  Board Dimensions 76.2 mm × 114.3 mm × 0.8 mm

[Siwastaja]

in an open ended plastic extrusion in a 18x24x4" plastic box.

Thermal dissipation, plastic enclosure...
you're already fighting physics from the outset.

But that plastic box is just HUGE (if those really were inches and it was not a typo)?

Similar power dissipation is normal through plastic enclosures of laptop power supplies, which have one hundredth of the surface area. Granted, they have to pull tricks like use internal aluminum heatsinks or graphite spreader pads to use as much of the plastic for cooling as possible.

It's not like plastic is a thermal insulator; with conduction of ~0.3 W/mK you just need a lot more of surface area than an open design. Which this huge box does have. The question is just about spreading the heat inside of the box. Which, given the sheer size, would work through convection, unless the airflow is somehow seriously blocked.

Realistically, 700mW for D2PAK is not much at all. Typical 40K/W can be achieved with a simple 1-sided copper pour. But because 78xx regulator conveniently uses GND as tab, the OP would have at very least full uniterrupted ground plane on at least one of the four(!) layers and it would be probably trivial to sacrifice pretty large area of either of the external layers for a ground plane, too. And for EMC reasons alone the grounds would be stitched together everywhere. And with 4 layers, prepreg is so thin that especially given D2PAKs large area, it thermally couples to the ground layer below additionally through the FR4 (not just thermal vias). In this kind of arrangement dissipating 3-4W out of D2PAK is perfectly normal. Sure, OP's operational temperature requirement is a tad high, but 700mW is still very little for this package size.

But I agree the real question is everything else dissipating 5.1W, not the 0.7W dissipated by this 7805.

[incf]

in an open ended plastic extrusion in a 18x24x4" plastic box.

Thermal dissipation, plastic enclosure...
you're already fighting physics from the outset.

But that plastic box is just HUGE (if those really were inches and it was not a typo)?

...

But I agree the real question is everything else dissipating 5.1W, not the 0.7W dissipated by this 7805.

The 24" x 18" x 4" size is correct, the box contains a few batteries and a lot of air (least 1" free space between all the components, often a "fair bit" more).