Author Topic: Calculating heat of a TO-220 package  (Read 2896 times)

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Offline special_KTopic starter

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Calculating heat of a TO-220 package
« on: January 10, 2025, 11:15:50 pm »
Hello everyone

I've been looking at the datasheet for the LM338 adjustable regulator, which is supposed to be rated for 5A constant (7A peak). The thought occurs that low voltage, high current output would be sinking a ton of heat - 100W or maybe more.

Obviously that kind of use is extreme for a TO-220 package, completely beyond it's real world capability. But how do I calculate how hot it would get?

I'm not going to actually try doing this, I just want to know the maths. Not found a clear explanation on google.
 

Online TimFox

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Re: Calculating heat of a TO-220 package
« Reply #1 on: January 10, 2025, 11:25:23 pm »
There is a max current rating, a max voltage rating, and a max junction temperature rating (or case rating).
You can't run it at max current and max voltage (from input to output) without getting a high power dissipation.
Low voltage and high current may be OK, as may high voltage and low current.
To dissipate lots of heat, you need a heat sink.
To calculate temperature (and compare it to the max temperature rating), you need to know the power dissipation (in W) from the voltage difference and load current and relevant thermal resistance (K/W) to the ambient temperature (which may rise above 25 C inside a box).
Then, you need to know the thermal resistance to ambient, which depends on your heat sink (that information can be hard to find--keep looking).
If the device is specified for case temperature, you only need the heat sink resistance (plus a bit for the interface or a bit more for a mica washer, if you use it) in K/W.
If the device is specified for junction temperature, it should specify the internal thermal resistance (K/W) from junction to case.
 

Offline PGPG

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Re: Calculating heat of a TO-220 package
« Reply #2 on: January 10, 2025, 11:27:05 pm »
In datasheet:
https://www.ti.com/lit/ds/symlink/lm338.pdf
from table 6.3 you read that RQJA for TO220 is 22.9°C/W.
So if you use no radiator than each 1W power dissipated in IC makes its Junction being 22.9°C higher temperature than ambient.
As max operating junction temperature is 125°C than assuming ambient being 25°C you get maximum power dissipated in IC (without radiator) P=100/22.9=4.37W
 
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Offline DavidAlfa

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Re: Calculating heat of a TO-220 package
« Reply #3 on: January 10, 2025, 11:28:40 pm »
Depends on the heatsink it's attached to.
Without heatsink, anything larger than 3-4W will likely trip overtemp protection, much less if inside a sealed enclosure.
Still, the package has thermal transfer limit, beyond it you could attach a truck-sized solid aluminium block that it would overheat anyway.

Some technical details here:
https://www.ti.com/lit/an/slva118a/slva118a.pdf
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Offline special_KTopic starter

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Re: Calculating heat of a TO-220 package
« Reply #4 on: January 11, 2025, 12:14:06 am »
In datasheet:
https://www.ti.com/lit/ds/symlink/lm338.pdf
from table 6.3 you read that RQJA for TO220 is 22.9°C/W.
So if you use no radiator than each 1W power dissipated in IC makes its Junction being 22.9°C higher temperature than ambient.
As max operating junction temperature is 125°C than assuming ambient being 25°C you get maximum power dissipated in IC (without radiator) P=100/22.9=4.37W

Ah, so there's a direct figure I can use, easy enough - it would mean a hundred watts would ramp it up to 2000+ degrees without heatsinking.

Of course, there is no way to sink 100W from a TO-220. No need to explain this; I have a 100W CPU in my computer and the engineering gone into cooling that is an industry in itself.

This does raise another question: An online store I frequent sells heatsinks for this package with the only spec offered (other than dimension) is a "thermal resistivity" figure given in K/W.

One is 10.8K/W, how would that figure be taken into account?
 

Online TimFox

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Re: Calculating heat of a TO-220 package
« Reply #5 on: January 11, 2025, 12:25:47 am »
Like I said, you want to calculate the total power dissipated, which is the voltage difference times the current, in W.
Again, you need the thermal resistance in K/W:  your heat sink value is from the sink to ambient, which may be higher than 25 C.
The total thermal resistance (in K/W) is from the junction or the case, using the device spec and the heat-sink spec.
 

Offline special_KTopic starter

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Re: Calculating heat of a TO-220 package
« Reply #6 on: January 11, 2025, 01:13:06 am »
Like I said, you want to calculate the total power dissipated, which is the voltage difference times the current, in W.

We already have this figure, in both directions. I picked a 100W figure for sake of convenience, wondering how to turn this into degrees C on this component.

PGPG Graciously pointed out the key figure in the datasheet, which I used very quickly to work out the answer to my first question = over 2000C.

PGPG's answer was very good because it demonstrated also how to use the maximum operating temperature and the "RQTA" figure to work out the maximum safe watts to dissipate on the part. = 4.37W. It's an excellent answer because it explained exactly what to do now, so I've learned.

Again, you need the thermal resistance in K/W:  your heat sink value is from the sink to ambient, which may be higher than 25 C.
The total thermal resistance (in K/W) is from the junction or the case, using the device spec and the heat-sink spec.

We have a figure of 22.9 C/W junction to ambient, and a figure of 10.8K/W which you say is heatsink to ambient.

If 10.8K/W is "Kelvin / Watts" then it's no different a measurement than C/W, easy enough. But how do I use these two figures *together*?

It is frustrating to be told to do something without being told how.

« Last Edit: January 11, 2025, 01:27:19 am by special_K »
 

Online TimFox

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Re: Calculating heat of a TO-220 package
« Reply #7 on: January 11, 2025, 01:27:58 am »
22.9 K/W is junction to ambient without heatsink.  You can't dissipate much power from the naked TO-220 package.

You want the junction to case resistance and add it in series to your heat sink thermal resistance to ambient.
In the TI data sheet for the TO-220 LM338, they quote "junction to board" (meaning the interface where the heat goes into the heat sink) of 4.1 K/W.
With your 10.8 K/W heatsink, that would total about 15 K/W, so 100 W is still way too much, giving you 1525 C at the junction.
That tiny heat sink is not very big for this application, since it reduced the total thermal resistance from 22.9 to 15 K/W.
 
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Offline PGPG

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Re: Calculating heat of a TO-220 package
« Reply #8 on: January 11, 2025, 01:28:49 am »
One is 10.8K/W, how would that figure be taken into account?

There is no absolutely clear in datasheet, but I think you should use Junction to Case (bottom) resistance being 0.7 (you attach heatsink to metal (bootom) of TO220 case and not plastic (top) of it.
So you get 10.8+0.7=11.5 K/W thermal resistance Junction to Ambient with this heatsink.
This should be enough for you to follow...
« Last Edit: January 11, 2025, 01:33:07 am by PGPG »
 
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Online TimFox

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Re: Calculating heat of a TO-220 package
« Reply #9 on: January 11, 2025, 01:50:58 am »
My mistake:  the junction-case resistance is 0.7 K/W (last line of section 6.3).
With a perfect heatsink (not practical), 100 W would be 70 C rise above 25 C, or 95 C, compared with the 125 C smoke level for junction temperature.
With your heatsink at 10.8 K/W, for a total of 11.5 K/W, the temperature at 100 W is still way too much.
100 C rise above ambient of 25 C can allow only (100 K) / (11.5 K/W) < 9 W with that small heatsink.
 

Offline special_KTopic starter

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Re: Calculating heat of a TO-220 package
« Reply #10 on: January 11, 2025, 03:38:14 am »
i recall some discussion elsewhere trying to bypass the TO220 power limit by having external BJT

In datasheet for LM317 you find this:



Transistors are not something I understand about, but I have seen someone on youtube use this idea with a 2N2905 in the lower position and a 2N3055 at the top. 2N3055 comes in TO-3.

I do not know if that combination of transistors are appropriate match to LM338 nor really how the circuit works; driving the base of the lower transistor with the voltage unput of the regulator is not intuitive to me.
 

Online TimFox

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Re: Calculating heat of a TO-220 package
« Reply #11 on: January 11, 2025, 04:32:00 am »
Those circuits move the thermal stress from the IC to an external power transistor.
The power dissipation and temperature rise for the transistor are specified similarly:  the headline power rating is for an ideal heat sink, limited only by the internal junction-case thermal resistance, but you need to calculate the allowable power for a practical heat sink.
 

Offline special_KTopic starter

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Re: Calculating heat of a TO-220 package
« Reply #12 on: January 11, 2025, 04:49:49 am »
Of course, but we all already understand that.

What would be interesting is to hear why one transistor is turned on by the input side of the regulator, or why one smaller transistor is used to turn on a larger one, or how the values of the resistors are chosen.
« Last Edit: January 11, 2025, 06:27:25 am by special_K »
 

Offline PGPG

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Re: Calculating heat of a TO-220 package
« Reply #13 on: January 11, 2025, 10:41:03 am »
I do not know if that combination of transistors are appropriate match to LM338 nor really how the circuit works; driving the base of the lower transistor with the voltage unput of the regulator is not intuitive to me.

Like NMOS are 'stronger' than PMOS also NPNs were 'stronger' than PNPs in the time when this schematic was created.
I was doing my power supply (0-2A, 0-40V) around 1975 and the only accessible for me 'strong' transistor was 2N3055.

Two transistors are used just to use NPN in the circuit where PNP can be used.
These transistors are connected as Darlington (pnp+npn behaves as pnp with very high beta). So you can look at them as at single pnp transistor.
This transistor is switched on not by voltage at IC input but by voltage drop at resistor. The higher output current the higher vothage drop. At some point transistor switches on helping IC to supply output load. IC will not output more current because whenever it tries to do it voltage drop at R rises and any extra current is 'taken' by transistor.

This circuit is good except we lose thermal protection.
 

Offline special_KTopic starter

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Re: Calculating heat of a TO-220 package
« Reply #14 on: January 12, 2025, 07:04:20 am »
I used to have a homebuilt bench PSU (I wasn't the one who made it. You could say it came with the house...) which had two TO-3 package parts on the outside of it's case, bolted to rather large heatsinks. I'm sure one of them was a 2N3055. Probably it's creator thought that he could avoid thermal problems by putting it in the open air. I'd say he was successful; it was at least 30 or 40 years old and working fine the last time I saw it.

Something I found on this site:

http://www.reuk.co.uk/wordpress/electric-circuit/lm317-adjustable-power-supply/



What do you think of that? the output of the regulator controls the base through a diode. Seems too good to be true, surely?
 

Offline PGPG

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Re: Calculating heat of a TO-220 package
« Reply #15 on: January 12, 2025, 10:40:39 am »
it was at least 30 or 40 years old and working fine the last time I saw it.

My supply with 2N3055 I have made as child around 1975 (starting from rewinding 100W transformer to get 6+6+6+6+6+1+1+1+1+1 AC) I still have and use when needed. It is in a wooden casing and has plugs on the top that allow you to take any voltage from the transformer from 1V to 35V in 1V steps. It is to limit the power dissipated by the 2N3055.
 

Offline CaptDon

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Re: Calculating heat of a TO-220 package
« Reply #16 on: January 12, 2025, 03:42:16 pm »
Simmed, I wouldn't exactly call that output unregulated. It is sort of a complicated version of the 'one BJT power transistor, one resistor from collector to base, one zener from base to low side' current multiplier circuit that has been used since the 60's. The regulator I.C. is taking the place of the zener. It may not be the best or quickest for transient response but should work as good as the 'zener regulated current multiplier' still in use today in a lot of audio gear.
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Offline special_KTopic starter

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Re: Calculating heat of a TO-220 package
« Reply #17 on: January 12, 2025, 09:04:48 pm »
Problem I expect (as someone who has no idea what he's doing) is that the regulator can't see the voltage coming out of the transistor, because there's nothing from the emitter to the adjust pin. It can only see what the base is drawing.

Switch electronics (UK online supplier) sells 2N2905 for 90p each
 

Offline PGPG

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Re: Calculating heat of a TO-220 package
« Reply #18 on: January 12, 2025, 10:51:00 pm »
It can only see what the base is drawing.

And if transistor is loaded the voltage at emitter is about 0.6..0.7V below base voltage. 0.1V uncertainty is irrelevant in many cases.
 

Offline special_KTopic starter

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Re: Calculating heat of a TO-220 package
« Reply #19 on: January 14, 2025, 10:47:19 pm »
OK so, starting with that LM317 datasheet, decided to try and achieve 24V/3A (70W) linear adjustable supply.

2N3055 transistor can handle it, but then there is the problem that TO-3 is annoying size to work with. Case is conductor instead of third pin, lack of choice in heat sinks available to buy new.

I found TIP3055 which seems to just be 2N3055 equivalent but a TO-247 package, which seems much more capable of dissipating heat watts than the TO-220 and gives a nice flat surface to fit to any generic aluminum heatsinks easily after drilling one screw hole.

My thinking is now to use 2X TIP3055, each with it's own good size (100x69x36mm) heatsink. By using two there is less of a localized hot spot on the transistor, and arranging two heatsinks is more flexible than one giant one.

Circuit diagram currently looks like this:

0.1 resistors are an attempt to balance load between the two TIP3055. Is there a problem with this approach?
« Last Edit: January 14, 2025, 10:50:13 pm by special_K »
 

Offline PGPG

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Re: Calculating heat of a TO-220 package
« Reply #20 on: January 15, 2025, 01:43:44 am »
0.1 resistors are an attempt

Unsuccessful attempt.
Keep tinkering.
 
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Offline special_KTopic starter

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Re: Calculating heat of a TO-220 package
« Reply #21 on: January 15, 2025, 02:21:36 am »
Any hint? Too small resistor? Wrong approach entirely?
 

Offline MrAl

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Re: Calculating heat of a TO-220 package
« Reply #22 on: January 15, 2025, 02:35:05 am »
Any hint? Too small resistor? Wrong approach entirely?

At these power levels a linear regulator is not usually going to be used.  Rather a buck switching regulator would be the choice as there would be much much less power loss at any input or output voltage.  You should look into that really.
 

Offline special_KTopic starter

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Re: Calculating heat of a TO-220 package
« Reply #23 on: January 15, 2025, 02:47:10 am »
At these power levels a linear regulator is not usually going to be used.  Rather a buck switching regulator would be the choice as there would be much much less power loss at any input or output voltage.  You should look into that really.

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You're certainly very opinionated, but that wasn't actually an answer.
« Last Edit: January 15, 2025, 03:07:37 am by special_K »
 

Offline Swainster

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Re: Calculating heat of a TO-220 package
« Reply #24 on: January 15, 2025, 04:06:49 am »
Any hint? Too small resistor? Wrong approach entirely?
I think PGPG is saying that you are on the right track but not quite there yet. Check out this link https://www.allaboutcircuits.com/textbook/semiconductors/chpt-4/bjt-quirks/ and scroll down to the section on thermal mismatch.
 
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