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Hello, and happy new year!

Question about IRF, does it need to isolate IRF from ground, from metal enclosure, i want to use another kind of heatsink,?

thx you!

73

Yes, you will need the mica insulators.? I can speak from experience that the heatsink is at PA potential (i inadvertently sorted one to ground while taking a measurement).? Thankfully only blew the fuse.

73
Evan
AC9TU

If Evan says it, recommend you take his advice to the bank.?

On the other hand, I didn't insulate the heat sinks under the misguided motivation to maximize heat transfer. Given I Chernobyl'ed my PAs due to "operator error", I got more sensitive to heat (mis) management.
Note the big ugly hacked up heat sinks, with no insulation.? Though big and ugly (like my teenage self...), they work like gangbusters. And they should not touch each other (yea, I learned that as well...).?

Now, they are just waiting for a probe at ground potential, or a errant wire, or a fat finger, or a stray random unplanned event. Like a time bomb with no expiration date. Then I will regret my choices.

I did order some alumina ceramic insulators that have a high heat transfer rates.? So when they come in, I will put them in and lower the risk of smoke and excitement. But for now...we live dangerously...and cool...



Note that with 24V+ to the finals, the big heat sinks, the 40mx40mx20 fan, the rig makes 20 to 30 watts without a fuss. And no spectral complaints. Beyond the kitchen speakers that buzz when the rig hits 80m TX for 2 full minutes doing WSPR duties...I learned how to turn off the kitchen speakers because my WSPR operations are much more important....hahah.

Don
km4udx

Don,??

Ted and Co...this insulator has a Thermal conductivity: 29.3w/m.k.? And that number is bigger than the silicon type insulator (but I can't find the number for that....suspicious? sure...)

Yes I have too much time on my hands...but these look like the bomb.? And given that the net heat transfer is a bunch of terms multiplied (or divided), then a higher number improves heat transfer.

Note, I don't actually know what that means really, but the math looks like a bunch of linear operations, so if one term changes by half, then the end thing changes by the same relative amount. Clearly I'm at the end of my technical understanding. But...



Don

Don,

The math of heat transfer is not just too bad.
Here's how I would calculate it:

What does 29.3 w/m*k mean?
Assume you have a huge cube of this material that is 1 meter on a side.
Apply heat to one side of the cube, and have a heat sink at a fixed temperature
on the opposite side of the cube.  We assume no heat is otherwise lost to the environment.
If 29.3 watts of power is applied, that side of the cube will (eventually) be one degree Kelvin
hotter than the heat sink.   

A change of one degree Kelvin is the same as a change of one degree Celsius,
so I will just use Celsius from here on out.


Assume your insulator is 0.025 mm thick, or 0.000025 meters.  (About a thousandth of an inch.)
The surface area of the IRF510 from which we conduct the heat is around 1 cm by 1 cm, or 0.0001 square meters.
You are getting 10 Watts of RF out of that single IRF510 at 50% efficiency, so it is dissipating 10 Watts as heat.
How hot will the IRF510 tab be if the heatsink can be kept down at 30 degrees Celsius with a fan?

We divide the thickness into the area and reduce those dimensions to a single value of 0.0001/0.000025 = 4 meters.
Not terribly intuitive, but that's all we need to know about the geometry to do this heat conductivity calculation.
As the area increases, the amount of heat that can be transferred increases proportionally.
But as the thickness is increased, the heat has a proportionally harder time moving through it.


The material will have a 1 C temp rise with 29.3 Watts applied and a 1 meter figure of merit for the geometry.
Our 10 Watts is less than their 29.3 Watts,  so that temp rise to be (1 degree C) * 10/29.3 = 0.341 degrees C
Our geometry has a large area for the thickness, so the temperature rise is even less: 0.341 / 4 = 0.085 K

So the IRF510 side of the insulator is 0.085 degrees Celsius hotter than the heatsink side.
And is at 30.085 degrees C.
That is less of a temperature difference than I expected, perhaps someone will see an error in my math?


Inside the IRF510 it's a different story.
Page 2 of the datasheet says thermal resistance from silicon junction to the tab of the IRF510 case 
is 3.5 degrees C per Watt, plus an additional 0.5 degrees C per Watt from tab to a well greased heat sink.
    
So the silicon die inside the IRF510 is at (10 Watts * (3.5+0.5)) + 30.085 = 70.085 degrees C.
The datasheet also states that the IRF510 will work OK with a die as hot as 175 C
(though it likely won't last very long).


The interface from IRF510 tab to the insulating material and from the insulating material to the heat sink
is surprisingly problematic, given the 0.5 degrees C per Watt figure from the IRF510 datasheet.
There will likely be another 0.5 degrees C per Watt temp rise between the insulator and the heat sink.
Without heat sink grease, these temperature differences will be considerably greater.
Using appropriate heat sink grease appears to be far more consequential than choosing the correct insulating material.

Jerry, KE7ER

Looking again at the ebay listing, those guys are much thicker than the TO-220 insulators I am used to.
Listing says they are 1 mm thick, so 40 times thicker than I was calculating for.
So the temperature rise across that insulator is 40 times what I calculated, or 0.085*40 = 3.4 degrees C
for 10 Watts of heat out of a TO-220 tab.

Makes sense, they are ceramic, anything too thin would fracture easily.
There are much thinner insulators, but they are made of material (such as mica) with much lower heat conductivity.
Given that these are ceramic and will not conform to metal surfaces when compressed,?
an appropriate heat sink grease is especially important.

Jerry

Here is my thermal grease....5.15w/m.k
Is this stuff up for the job?

Jerry, many thanks for the walk through.? As normal, there is much more to everything than we think. Or at least I think.? So thanks a million (degrees C).

That 5.15 W/m*K means the grease is almost 6 times worse at conducting heat?
than your ceramic insulators, which were 29.3 W/m*K
.
But if that grease winds up being more than 6 times thinner than your 1 mm thick insulators,
the temperature drop across the grease will be less than that across the insulator.

Once you torque down the bolts holding the TO-220 IRF510 to your heatsink,?
that grease will be very thin indeed.?
I'd guess the grease will be more than 100 times thinner than your insulators.

I bet 99% of uBitx users just leave the IRF510 heatsinks as they found them.
No insulation (or grease) between the IRF510 and heatsink.
But they are careful not to have the two heatsinks touch each other
or anything else conductive such as a metal enclosure.

Jerry, KE7ER

So I got the ceramic insulators.? The included bushing fits the screw just fine.? And the screw, without the bushing fits the wafer just fine. But put the bushing on the screw, and ain't nothing going in the ceramic hole...

They are (relatively) thicker than thin silicon wafers/pads. Which would not be a problem if...

The hole were not too small for the insulating bushing. Which would not be a problem if...

The ceramic material were not so hard and brittle that I couldn't drill or sand/grind it bigger. Which wouldn't be a problem if...

I didn't want the heat sink and screw electrically isolated from the IFR510 case.? Which wouldn't be a problem if...

I could return them. Which as a practical matter I can't.? Does anyone want a few of these to play with??

hahah.

On Tue, 21 Jan 2020, Don - KM4UDX wrote:

So I got the ceramic insulators.? The included bushing fits the screw just fine.? And the screw, without the bushing fits the wafer just fine. But put the bushing on the screw, and ain't
nothing going in the ceramic hole...

It looks to me like it was intended to be assembled thus:

1) Enlarge the hole in the heat sink to accomodate the bushing.
2) Insert the screw in the bushing, and the bushing in the BACK of the heat sink.
3) Place the wafer over the screw on the front of the heat sink.
4) Install the IRF510 on the screw protruding from the wafer.
5) Secure the entire stack with a lockwasher and nut.

This will insulate the IRF510 from the heat sink, but the screw head on the back will still be at the same potential as the IRF510 tab. At least, its a much smaller target for a distracted errant screwdriver ;-)

Rick Green

They are (relatively) thicker than thin silicon wafers/pads. Which would not be a problem if...
The hole were not too small for the insulating bushing. Which would not be a problem if...
The ceramic material were not so hard and brittle that I couldn't drill or sand/grind it bigger. Which wouldn't be a problem if...
I didn't want the heat sink and screw electrically isolated from the IFR510 case.? Which wouldn't be a problem if...
I could return them. Which as a practical matter I can't.? Does anyone want a few of these to play with??
hahah.[IMAGE]

--
Rick Green

We, the People of the United States of America, reject the U.S. Supreme Court's
Citizens United ruling, and move to amend our Constitution to firmly establish
that money is not speech, and that human beings, not corporations, are persons
entitled to constitutional rights.

Simple,? so long as the metal scre does not contact the IRF510 tab its good.
The insulators are non conducting so if they touch the screw no harm or foul.

Allison
---------------------------------
No direct email, it goes to bit bucket due address harvesting in groups.IO

hummm...I confess I never thought of using the bushing on the back side against the heat sink....that assumes in torquing down the screw that the bushing is not distorted or flattened...in my case the hole in the heatsinks looks like there were drilled with a bad stick, a twisted file, and a blow torch for good measure. In short, they are a mess and bigger than the bushing already.? I use a washer to cover up my metallurgical butchery.

But....the notion of using the bushing on the back of the heatsink strikes me as very workable, in more competent hands than my own...hahahah.? Thank you very very much...I'm going to think about this approach....the old bean is on the case...hahah.

Don

On Wed, Jan 1, 2020 at 12:05 PM, Anthony F4HUY wrote:

Question about IRF, does it need to isolate IRF from ground, from metal enclosure, i want to use another kind of heatsink,?

I used a thermal pad between the MOSFETs and heat sinks and used nylon M3 screws and nuts so I didn't have to deal with bushings.
?
--

Without the long story...

A vendor did that using plastic (Nylon, later Delrin)?screws and it didn't work out well.

Seems the screws would get warm and stretch allowing the mounted device to
loose solid contact and get very much hotter then the screw would melt and fail.
The device aware of its over heated condition would notify the user by releasing
smoke and ceasing operation, promptly and permanently.

The field kit included new transistor, metal screw, nut and a fiber shoulder washer.
The instructions admonished the service person to use the parts and directed,
or else, in bold print.

Allison
---------------------------------
No direct email, it goes to bit bucket due address harvesting in groups.IO