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View Full Version : DC SSR heat sink question (is it really necessary?)



steve
04-19-2012, 02:21 AM
I'm working out a PCB layout design for a bank of DC SSRs similar to wjohn and labrat's, using the FQPF13N06L MOSFET. I'm trying to pack them in pretty tightly into the space I have available, and it would be great if I didn't have to have a heat sink. This will be my first time working with DC SSRs and low-voltage LED loads, so I'm wondering what others' experiences have been in this area. Do the MOSFETs actually get too warm on their own, from what you've seen? Can I get away without a heatsink?

One of the things I'm thinking of is to turn them 90 degrees so they're stacked together in a row like this: ||||||| instead of being in a long line like this: _ _ _ _ _.

Thanks!
steve

David_AVD
04-19-2012, 05:48 AM
You'll need to get the specs, do the calculations and work out the theoretical dissipation. Then you'll need to confirm that with some real testing.

LabRat
04-19-2012, 07:19 AM
Yep. It's all going to depend on how much current you are expecting to draw through the MOSFET.

steve
04-19-2012, 12:54 PM
Yeah, that's pretty much what I was thinking too, just looking for any more data points from others' experience I could add.
As it turns out, I'm able to rearrange the design to have two parallel rows of MOSFETs so I can put a heat sink into the design anyway. Being generally paranoid, my tendency is to use a heat sink anyway, just in case.

jklingert
04-19-2012, 03:23 PM
Maybe you could use one of the other style heatsinks from Mouser (http://www.mouser.com/access/?pn=532-577102B00)

Penfold
04-20-2012, 12:52 PM
Perhaps if you stacked the DCSSRs on top of each other with standoff like this: http://www.mouser.com/ProductDetail/Keystone-Electronics/8411/?qs=1eFRaVyeUDW%2bYNr2X1zwCVU5iu2xmGqCtCGUXHjAeww% 3d. And then put them on their side like you have shown, but you will probably need to find a two inch standoff. Maybe a nylon collar with a 4/40 screw that you can get at Home Depot might be a good alternative.

g2ktcf
04-21-2012, 10:39 AM
Gents,

MOSFETs are not like TRIACs...they have a VERY low internal resistance. There is no need to use a heat sink unless you are pulling MASSIVE amps which I cannot see happening here.

steve
04-21-2012, 11:57 AM
I was able to rearrange things (barely) to fit all the MOSFETs in a row so I could still bolt them all to an aluminum bar or something if I need a heatsink in there, although I think some misunderstood which axis I was rotating them on. The idea of stacking them is interesting, but I was leaving them standing up, just arranged so their wide sides were facing each other rather than their narrow edges, so I could pack them in more tightly.

These are rated at 10A maximum current load, and I was figuring I'd label the channel outputs at 5A max considering the trace widths and so forth, but in practice the loads I have in mind to run from these would be either 200mA max or 1A max per MOSFET. So I'm not convinced heat would really be an issue, but better safe than sorry.

I'm not used to reading the thermal ratings on the datasheets here... when it indicates the degrees per watt, is that just the simple calculation of watts flowing through the device (supply voltage * current), or do you have to manually figure in internal resistances or other factors? I'm wondering because if I'm reading it right it would indicate they would heat up a lot more than I've ever actually experienced to be the case.

David_AVD
04-21-2012, 05:58 PM
Gents,

MOSFETs are not like TRIACs...they have a VERY low internal resistance. There is no need to use a heat sink unless you are pulling MASSIVE amps which I cannot see happening here.

I did not read the spec sheet for the intended device, but I never assume that people will pick a device with a low rdson figure. Often people choose on lowest price without looking at the actual specs.

Also, the MOSFET's rdson is not the only determining factor. The PWM frequency and drive type can have a marked effect on the overall losses. Sometimes more than the steady on-state loss.

P. Short
04-21-2012, 06:21 PM
Steve: I don't know what datasheet that you're looking at, but typically the C/W number will indicate how much the temperature of the junction on the internal semiconductor die will increase above either the temperature of metal tab on the transistor (ΘJC) or above the ambient temperature (ΘJA) of the transistor (i.e. no heatsink).

In the former case you would add up the ΘJC of the transistor and the ΘCA of the heatsink to determine an overall ΘJA. In either case you then calculate the junction temperature of the transistor (TJ = ΘJA * P + TA), where TJ is the junction temp, TA is the ambient temp, and P is the power dissipated), and compare this with the maximum allowed junction temp from the data sheet.

Bear in mind that all of these calculations are very rough, because they don't really take into account any cooling effects (or not) due to air circulation, nor how efficiently heat is transferred from the transistor case to the heat sink, etc.

steve
04-25-2012, 07:15 PM
Ok, thanks. This is probably a dumb newbie question, but just so I'm clear what I'm calculating here, let me try an example and see if I'm right here (none of my designs have had parts other than voltage regs and in one design triacs which would get any significant heat, and in those cases I just generously applied heat sinks :)

Say I have a diode rated at 10A. It has forward voltage Vf=1V, RΘja=10K/W. I assume that we're just working on the power dissipated as heat, which would be calculated by the amps and the voltage drop (1V * 10A = 10W, ergo 100K or 180F above ambient temperature).

For the MOSFETs, I see RθJC=6.2C/W and RθJA=62.5C/W maximum. RDS(on) ranges from 0.088 to 0.14Ω so let's take the larger value, so if I run a load of say 5A on it, the dissipated power would be P=Id**2 * Rds(on) = 25 * 0.14 = 3.5W, so the RθJC=6.2C/W gives a rise on the case of 21.7C or 39F, but the RθJA=62.5C/W gives an answer of 218.75C or 393F above ambient temperature. I can see 3.5W heating up the part by 22C, but I'm not sure why the RθJA is so much higher.


Datasheet from http://www.datasheetcatalog.org/datasheet/diodes/ds28010.pdf and http://www.fairchildsemi.com/ds/FQ/FQPF13N06L.pdf

P. Short
04-26-2012, 04:07 AM
Would you willingly touch a 7W light bulb (i.e. an incandescent night-light) that had been on for a while? I think not. And if the exterior is that hot, think about how much hotter the tiny heat source (filament or junction) must be. The 393F number that you give doesn't seem unreasonable for the circumstances...

steve
04-26-2012, 11:22 AM
Granted, but what am I misunderstanding about what RθJC and RθJA actually represent, then? If that's the amount the internal junction heats, and how much the part heats the outside air, I'd expect the junction temp to be a lot hotter than the outside, but the numbers seem to say the opposite, so I'm assuming that I'm misinterpreting this somehow.

P. Short
04-26-2012, 05:25 PM
RθJA is a measure of the temperature difference between the ambient environment and the junction. This is usually relevant when you do not have any heatsink. The transistor is either in open air without any flow restrictions (in which case the air temperature would be the ambient temperature), or in an enclosure (in which case the temperature inside the enclosure is the ambient temperature, which is probably much higher than the room temperature).

RθJC is a measure of the temperature difference inside the component between some specified spot on the component case and the junction. You use this number when you can figure out what the temperature is at that specific spot on the case, either by measuring it or by calculating it based on the thermal resistance of a heatsink of some sort and the thermal resistance of the mounting scheme (including thermal goop, any insulator, etc). The temperature of that location on the package is probably a lot higher than the ambient (air) temperature.

steve
04-26-2012, 05:46 PM
Okay, that makes sense now. Thanks.