Author Topic: Calculating heat of a TO-220 package  (Read 2889 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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Offline special_KTopic starter

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Re: Calculating heat of a TO-220 package
« Reply #25 on: January 15, 2025, 04:36:12 am »
Ah, so it's the *location* of the resistor which was wrong, but not the idea of using them. Good shout, I changed the circuit.

I also noticed it talks about it helping to keep both transistors on the same heatsink. I assume this is to keep them as close as possible to the same temperature, as temperature changes conductivity and so would result in unequal loading.

I know that semiconductors of the same spec are never identical, it's a known problem with computers not every CPU of the same model and batch can handle the same degree of abuse. So I suppose keeping them thermally isolated would exaggerate this and cause them to go even more unequal.

I still think using 2x is worth it, for nothing else than that now there is 2x the surface area between the power transistor(s) and the heatsink.
 

Offline Picuino

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Re: Calculating heat of a TO-220 package
« Reply #26 on: January 15, 2025, 09:39:55 am »
The equivalent thermal circuit is as shown in the figure.
The power generated is equivalent to a constant current generator (current = power). The voltages of the circuit are equivalent to the temperature (voltage = temperature) and the thermal resistances are equivalent to the electrical resistances of the circuit.


Tj = Junction Temperature (Highest)
Tc = Case Temperature
Ta = Ambient Temperature (Lowest)

Source: https://toshiba.semicon-storage.com/eu/semiconductor/knowledge/faq/common/what-is-a-radiation-equivalent-circuit.html
« Last Edit: January 15, 2025, 12:58:09 pm by Picuino »
 

Offline PGPG

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Re: Calculating heat of a TO-220 package
« Reply #27 on: January 15, 2025, 11:44:05 am »
Ah, so it's the *location* of the resistor which was wrong,

I'm surprised you didn't get it yourself alone.

but not the idea of using them.

If for you 'let us use resistors in collectors' and 'let us use resistors in emitters' are almost the same idea than it tells me that you have to learn a lot how transistors work (not internally but from external (circuit) point of view). Open some power transistor datasheets (best would be from different manufacturers) to find if you understand each value and each figure there.

Do you know that Vbe has negative temperature coefficient and it is the main reason resistors are needed to equalize the current sharing between transistors?
Without resistors (in emitters) if one transistor will got little hotter its Vbe will be smaller making this transistor to get higher part of current than the other. This will make it more hot and then farther increasing the current in this one. End effect - practically only one of them will be conducting whole current.

I can't imagine how, according to your idea resistors in collector were supposed to equalize current between transistors.

In lab power supplies that were using linear regulators there were relays selecting different voltages from trafo to keep dissipated in regulator power in limits.
When I was doing my regulator (I think around 1972..1975) I had only an aluminum block (10x10x1cm) I could use as radiator. Without fins it is a very poor radiator. Because of this I decided to be able to supply its input with any AC voltage in 1V steps (and from it is my solution with 6+6+6+6+6+1+1+1+1+1 (V) trafo winding.
 
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Offline special_KTopic starter

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Re: Calculating heat of a TO-220 package
« Reply #28 on: January 15, 2025, 04:02:10 pm »
With transistors I understand enough to know that I understand nothing. My thought was as simple as "power goes across here, throw an impediment in"

Relays switching between different transformer taps is a good idea. I'm thinking how to do it now, probably measure voltage potentiometer, feed that to comparator which drives 2N7000 to turn relay coil on? Perhaps keep it simple, a 12V winding and 24V winding. It's similar to something I already added to my TV for extra inputs.

For now I'm going to add fine/coarse voltage adjustment, then go to work for the evening.
 

Offline BillyO

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Re: Calculating heat of a TO-220 package
« Reply #29 on: January 15, 2025, 04:35:19 pm »
No matter how many transistors yo use you still have to get rid of 100W of heat.  The most you can run a TO220 package to is about 70W regardless of the size of heatsink.  A TO247 will go to about 100W and a TO3 to about 150W.  But in any case, you still need to get rid of 100W and still keep temperatures down.

The heatsink below measures about 120mmX75mmX35mm and will do a decent job of getting rid of 100W of heat with sufficient fan air flow.

It is rated about 1.5 °C/W free air (fins vertical).  With a decent fan blowing air from the back I have seen .35 °C/W.
« Last Edit: January 15, 2025, 04:44:02 pm by BillyO »
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Offline special_KTopic starter

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Re: Calculating heat of a TO-220 package
« Reply #30 on: January 15, 2025, 05:44:32 pm »
Already understood. The point of multiple transistors is not to magically need to remove less energy overall - but have less concentration of that energy in one place to start with. If the surface area of the contact between transistor is 1 cm2 (Made up number), then doubling is it 2 cm2 - we get much better transfer of heat from one to the other.

Heatsinks are actually something I have much practical experience in, because I'm a computer guy. There is a LOT going on with heatsinks for computers, it is an entire industry.



These things are not (publicly) given ratings in C/W. Instead they talk about the maximum thermal power of the CPU it's to cool, and how it's expected to perform. So this one is meant to keep a CPU creating 125W of heat within a sensible range, while the fan noise not more than 26.8dB(A). A 95W CPU on it has a TJ of 38.5C above ambient. It's a very VERY good heatsink.

Surface area of fins is key. So more than 50 fins, but each thin as a razor blade. Aluminum alone doesn't conduct heat fast enough through the whole structure, so instead there are heatpipes. Looks like copper bar, but is actually filled with coolant which continuously boils and condensates, dripping back down through grooves on the inner wall.

It's disappointing that packages other than CPUs don't seem to have these kind of highly developed heatsinks. I suppose there's just no demand. But it doesn't bother me too much, there's a good supply of used PC heatsinks for practically nothing. Drill and tap the base, screw transistor on, job done.
« Last Edit: January 15, 2025, 11:36:59 pm by special_K »
 

Offline Doctorandus_P

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Re: Calculating heat of a TO-220 package
« Reply #31 on: January 15, 2025, 05:57:13 pm »
Power limit for a TO220 package is around 45W. So for 100W you'd have to put some in parallel, or you have to use a more capable package.
 

Offline BillyO

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Re: Calculating heat of a TO-220 package
« Reply #32 on: January 15, 2025, 06:13:46 pm »
Power limit for a TO220 package is around 45W. So for 100W you'd have to put some in parallel, or you have to use a more capable package.
A lot depends on allowable junction temperature and other factors.  The 70W I was referring to was with the heatsink shown (I've got a lot of them), forced air cooling and a device with a maximum junction temperature of 175 °C (CDS19534KCS).  It's happy at 70W with a temperature of about 60 °C at the case, meaning a junction temperature of about 150 °C which is a fine situation for steady state use.

With a close to infinite heatsink (copper block with water cooling) you could do much better.
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Offline MrAl

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Re: Calculating heat of a TO-220 package
« Reply #33 on: January 15, 2025, 07:15:39 pm »
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.

"How do I connect the linkages on my 440 six pack?"

"buy modern car with fuel injection"

You're certainly very opinionated, but that wasn't actually an answer.

What the heck is that supposed to mean.

I am suggesting that you give up on the linear regulator and look into a buck regulator.  It's going to be the same thing except it does the voltage conversion more efficiently and thus less heat.  It's a power supply that wastes a lot less power.

For example, with a buck TO220 package you can run 1 amp output with ANY input voltage, and either not need a heatsink or a very small one.  That means 20v input and 5v output and still a very small heatsink.
With a linear regulator, with 20v in and 5v out at 1 amp you'd need a heatsink that can handle 15 watts which is much larger.

There is another advantage to using a buck instead of a linear.
With a linear, if your input source is only capable of 20v at 1 amp, then the most you can get out of the linear is (ideally) 20v at 1 amp, but even at 10v you only get 1 amp, and at 5v you only get 1 amp, and at 1v you only get 1 amp output.
With a buck, if you input 20v at 1 amp you can get 2 amps out at 10v, 4 amps out at 5v, and theoretically 20 amps at 1v output.  This is often an attractive feature too.

The choice is always yours you can use whatever you want.  I am just offering some suggestions.


This is a common way of doing things.   How's that for an 'opinion' ?

 

Offline PGPG

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Re: Calculating heat of a TO-220 package
« Reply #34 on: January 15, 2025, 08:06:02 pm »
I am suggesting that you give up on the linear regulator and look into a buck regulator.

As OP has a problem with understanding transistor working will he be able to understand buck regulator enough to solve problems he could encounter during running it?

I would suggest OP just to reduce his expectations for linear regulator to for example 1A (or even 0.5A) and 20V and do it just to learn how it works with assumption that much more powerful buck regulator he will construct in future after making some experiments with buck converters to understand them.

If you at the moment don't know what for you will be using 100W supply than probably you just don't need it.

Lab supply (I have at work) for 99% of time is set to 12V and 0.2A and devices I supply typically consume something about 30mA.
Sometimes I change its voltage to 5V or 3V3. Its max current is 1.2A (I've just checked, as I didn't remembered it). I don't remember setting it to this value, ever.
 

Offline special_KTopic starter

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Re: Calculating heat of a TO-220 package
« Reply #35 on: January 15, 2025, 10:35:42 pm »

What the heck is that supposed to mean.

It's analogy.

If you saw someone talking about assembling an engine for his 1960s car as a hobby, it would be very rude to interrupt his conversation by telling him how it's better to just drive a new car with a modern engine in it. He's an alive person in 2025, he knows that, he does not need you to "helpfully" enlighten him. Mr 440 Magnum wants to fit triple carburetors to his vintage car regardless of how much more practical a new Honda is.

You like switching supplies very much. Ok. It is not relevant to the conversation you interrupted though.
 

Offline special_KTopic starter

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Re: Calculating heat of a TO-220 package
« Reply #36 on: January 15, 2025, 10:41:07 pm »

I would suggest OP just to reduce his expectations for linear regulator to for example 1A (or even 0.5A) and 20V and do it just to learn how it works with assumption that much more powerful buck regulator he will construct in future after making some experiments with buck converters to understand them.

If you at the moment don't know what for you will be using 100W supply than probably you just don't need it.

Nah. I'm gonna build a big one just because I want to.

"100W" is not a goal, it is a round number I picked all the way in the first post when I asked how to calculate temperature of a TO-220. I could've also said 10W, or 1kW.

It's like those very classical maths questions that teachers would pose in school. "If the train leaves the station at 12:00 and is travelling at 35mph..." There is no train in reality. They aren't sending your maths exam paper to network rail.
 

Offline MrAl

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Re: Calculating heat of a TO-220 package
« Reply #37 on: January 16, 2025, 06:05:01 am »

What the heck is that supposed to mean.

It's analogy.

If you saw someone talking about assembling an engine for his 1960s car as a hobby, it would be very rude to interrupt his conversation by telling him how it's better to just drive a new car with a modern engine in it. He's an alive person in 2025, he knows that, he does not need you to "helpfully" enlighten him. Mr 440 Magnum wants to fit triple carburetors to his vintage car regardless of how much more practical a new Honda is.

You like switching supplies very much. Ok. It is not relevant to the conversation you interrupted though.

It's interesting that you went through that much just to explain how you prefer one type of supply over another.
You could have just said something like, "I want to work with this supply anyway".
You would rather call it "rude".
If you want "rude", I can show you rude, just say the word.
This is what you get sometimes from trying to help some people like you.
Have fun with your outdated power supply :)


 
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Offline Xena E

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Re: Calculating heat of a TO-220 package
« Reply #38 on: January 16, 2025, 06:51:59 am »

...If you want rude, just say the word...


Someone call for me! Tourettes girl at your service...

Seriously, OP, people do tend to make alternative practical suggestions.

However, if you want to build a linear power supply, I respect your choice. The raw AC could be supplied by "tap changing" on the supply transformer to limit the dissipation in the circuits pass elements. Or use a switching pre regulator... lots of options to avoid the inconvenience and cost of a lot of difficult thermal design problems.

Best regards.
Xena.
 
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Offline Siwastaja

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Re: Calculating heat of a TO-220 package
« Reply #39 on: January 16, 2025, 07:12:09 am »
Already understood. The point of multiple transistors is not to magically need to remove less energy overall - but have less concentration of that energy in one place to start with. If the surface area of the contact between transistor is 1 cm2 (Made up number), then doubling is it 2 cm2 - we get much better transfer of heat from one to the other.

Exactly - in other words, RthJ-C (and Rth case-to-heatsink, including thermal interface materials) values get paralleled. For example, if RthJ-C for a single transistor is 0.8 K/W and thermal interface to heatsink is 0.7 K/W, that is 1.5 K/W total. Use two in parallel and it effectively gets down to 0.75 K/W. This isn't even taking account another effect: how the heatsink Rth (mounting point to ambient) is calculated. If it's calculated for wide coupling area, then a single small transistor behaves worse than the marketing value of the heatsink. Then again, if the heatsink is realistically characterized with small point source of heat, then spreading the heat over multiple transistors even on the same heatsink makes the heatsink perform better than rated. This is because thermal transfer from the mount point to the fins is not infinite, there will be temperature gradient within the heatsink as well.

Maybe in 1970's semiconductors were expensive and aluminum cheap, but nowadays silicon is cheap enough that using more of it makes sense. Given that you choose a constant maximum die temperature (e.g., 110 degC), reducing thermal differential between the dies and the heatsinks mean you can now run the fins hotter, i.e., use a smaller and cheaper heatsink. This is the advantage of paralleling transistors.
« Last Edit: January 16, 2025, 07:14:29 am by Siwastaja »
 
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Offline dietert1

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Re: Calculating heat of a TO-220 package
« Reply #40 on: January 16, 2025, 10:07:48 am »
Let's correct some of the numbers given above. Last week i did some tests with modern TO-220 mosfets. They are specified with a junction to case thermal resistance of about 0.5 K/W and an allowable chip temperature of 175 °C.
If we have an ideal heatsink at an ambient temperature of 25 °C, the temperature margin of 150 K translates into 300 W of power dissipated inside the chip.
The CPU cooler shown above has a thermal resistance of about 38 K / 125 W, that is about 0.3 K/W, similar to the other heatsink shown above, if used with fan. We have to add this to the 0.5 K/W of the TO-220 package, resulting in a total thermal resistance of 0.8 K/W. In this configuration the part can handle about 150 °C / 0.8 K/W = 187,5 W.
If the part needs electrical isolation, thermal resistance will rise once more. TO-220 case temperature will be above heatsink temperature. If the circuit needs to work at higher ambient temperatures as well, the temperature margin will reduce even further.
In the end, i think 100 W per TO-220 is a realistic number that can be obtained with a standard heatsink of some size, including electrical isolation and with a fan. High power equipment dissipating above KW power usually doesn't have hundreds of transistors but special heatsink solutions with water cooling.

Regards, Dieter

Edit: The 0.3 K/W number for the CPU cooler includes junction to case thermal resistance of the CPU. A better number for that cooler may be 0.2 K/W or so.
« Last Edit: January 16, 2025, 10:18:45 am by dietert1 »
 
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Offline BillyO

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Re: Calculating heat of a TO-220 package
« Reply #41 on: January 16, 2025, 01:58:20 pm »
They are specified with a junction to case thermal resistance of about 0.5 K/W and an allowable chip temperature of 175 °C.
Interesting.  Do you have a part number for this device?  I could probably use some of them.
Bill  (Currently a Siglent fanboy)
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Offline dietert1

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Re: Calculating heat of a TO-220 package
« Reply #42 on: January 16, 2025, 06:59:23 pm »
Yes, i was testing IRFB3207 and IRF1405 mosfets at continuous power up to 54 W (13.58 V at 4 A) although i got a warning that those mosfets might be unstable under linear operation.
Anyway their junction to case thermal resistance is specified at 0.45 K/W max. They specify an additional case to heatsink thermal resistance that depends on mounting details like i mentioned above.

Regards, Dieter
« Last Edit: January 16, 2025, 09:47:48 pm by dietert1 »
 

Offline Hr_Satch

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Re: Calculating heat of a TO-220 package
« Reply #43 on: January 16, 2025, 08:18:33 pm »
Hi, I dont know about the Maths for Heatsinking, just don't do it :

a general rule: if you still can touch it by hand, the electronics will survive ;-?
I think that is somewhere above +50°C for the human skin to be touchable.
-- --
Two other concerns that might help.
TO-220 housing is good for space saving.
However for real power applications, nothing beats a good TO-3 housings or even more modern stuff, . . .
Modern power semiconductors, do provide commonly in galvanic isolation aswell, so that is a benefit even better.
And often TO-3 are better off to be build straigth onto the heatsink.
And isolate the heatsink from the chassis to be bolted on, with some isolating bushes.
-- --
Second, I really don't get it: just adapt the input voltage for any given output voltage towards the needs,
and to create the least possible voltage drop, hence lowest possible series regulator dissipation  :-DD

I too have plenty of transformers laying around, but also almost never the ideal parameters, regarding power-
output voltages ratios, etc . . .

I just recently came to the simplest idea, of Re-winding each available transformers secondary winding(s)
-- --
And for those who still believe in free energy: there is no such things !
You have to respect the actual transformers power ratings: for secondary aswell as primary winding(s).
a 250VA remains a 250VA.
And from that equation come the voltage and current possibilities  :phew:

So let us state a 24V secondary for let us say 10Amps.
I needed a 12Vac . . . the secondary windings are always on the outside, and are easy to turn back off, . . .
So I halvated its secondary windings, to get around 16Vac out unloaded.

BUT this never implicates that I can just expect to get at once the double of current ratings  |O
Even one can think always respect the power ratings, especially for the primary that remains the same.

So the 250VA rating in my example, also counts for the primary AC~input.
Hence at 240Vac the max.prim.current remains around 1Amps.

One could think, oh halving the secondary voltage, can give the double of current,
but I am afraid that will not work.
You will need to apply for thicker wires in the secondary, etc . . .













 

Offline special_KTopic starter

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Re: Calculating heat of a TO-220 package
« Reply #44 on: January 16, 2025, 10:51:37 pm »
I only have toroid transformers as anything else is impossible to find at even half way reasonable price.

Rather than unwind a toroid, I'd probably wrap a few turns of wire around it and measure it's output to get an idea of the turns per volt. Then wind on top.

Would probably be noisy with magnetostriction I guess? And only safe when stepping down.
 

Offline special_KTopic starter

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Re: Calculating heat of a TO-220 package
« Reply #45 on: January 17, 2025, 01:35:12 am »
Evening all. New experiment: automatically switching between 12V and 24V transformer output

Before this point I changed the voltage control to coarse/fine, with two pots

Now coarse is a stereo pot. It's second half is feeding a signal to a comparator. the other input of the comparator also has a pot, to adjust when it triggers.

Comparator toggles relay, disconnecting centre tap and connecting to the end of the winding.

This little bit of control is on an isolated 5V circuit, simply because I already have a small 5v transformer. I figure it'd also be good later to use for a circuit breaker that disconnects main PSU output when there is too much current.

Probably there's mistakes in it, it's 1:30am and I had an annoying day at work
 

Offline Hr_Satch

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Re: Calculating heat of a TO-220 package
« Reply #46 on: January 18, 2025, 05:28:15 pm »
Hi, Special-K, this circuit could work -if you'd understand transistors- sorry,
but the 0.1 ohms don't belong in the Collector lines !
These could balance in the Emitter lines, because of no BE-junction is ever identical.
Sorry, I don't have the time to do electronics coarses, but that's the way it should be: in the E-lines.

Furthermore: NPN's are not ? stronger? than PNP ? ? ? ?
That is mumbo-jumbo language. It are just different polarisations.

And Like N-mosfets are also more common, than their P-variants.
Why they are cheaper and more frequent is because of the Manufacturing process difficulties,
and we commonly work with negative lines in the common ground or return paths.
If you would like to use an NPN instead of that 2N2905 PNP, than that's a whome different ballgame,
regarding circuit desin.

You should dig into the differences between P -or N semiconducting materials,
as seen from the Majority and/or Minority carrier charges: ie. electroncs -vs- gaps, . . .
Also search which ones are more Temperature sensitive ;-)

Moreover, the excessive heat dissip remains the same for a given power,
with this schematic you just devide it over more components, . . .

If possible just adapt the applied input voltage magnitude
 
Why not put two or more LM338's in parallel, balanced against each others difference in output ;-?
You can find datasheat applications enough about this.
and again the balancing can be done in series with their outputs, but this is again a waste of energy.
Better do it with an OpAmp.
 

Offline PGPG

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Re: Calculating heat of a TO-220 package
« Reply #47 on: January 18, 2025, 06:15:29 pm »
but the 0.1 ohms don't belong in the Collector lines !

This was already explained not only once.

Furthermore: NPN's are not ? stronger? than PNP ? ? ? ?
That is mumbo-jumbo language. It are just different polarisations.

You read 'too fast'.
I didn't siad 'are stronger' but: "also NPNs were 'stronger' than PNPs in the time when this schematic was created".
So not 'are' but 'were' and word 'stronger' used with apostrophes.

Saying that I'm speaking mumbo-jumbo language is just rude.

I was speaking of times when silicon NPN were available for higher currents than PNP (those time available only germanium).
I see nothing wrong in naming higher current silicon NPN being stronger than smaller current germanium PNP.

How do you explain than in most DCDC step-down ICs NMOS is used as key even it needs to get bootstrapped voltage to drive it (PMOS wouldn't need it) other than that it is easier to get stronger NMOS than PMOS in these ICs? Stronger in this case means smaller RDSon.
 

Offline special_KTopic starter

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Re: Calculating heat of a TO-220 package
« Reply #48 on: January 18, 2025, 11:54:38 pm »
Hi, Special-K, this circuit could work -if you'd understand transistors- sorry,
but the 0.1 ohms don't belong in the Collector lines !
These could balance in the Emitter lines, because of no BE-junction is ever identical.
Sorry, I don't have the time to do electronics coarses, but that's the way it should be: in the E-lines.

This was fixed a few days ago.

Moreover, the excessive heat dissip remains the same for a given power,
with this schematic you just devide it over more components, . . .

If possible just adapt the applied input voltage magnitude

That is in the last version of the schematic which I posted as an attachment. Two things come to mind:

Firstly, its impossible now to find a new multi tapped transformer for regular sale. If you want a transformer you must buy a toroid with 2x identical secondaries. So this limits what I can do. I do not want to use 2nd hand transformer because I could be waiting months or years to find one by pure chance.

(or I could wind one myself, but that is a long project too. I do not know where to buy a new transformer core, for example)

Secondly I already implemented it - 12v and 24v in the newest drawing, which after I published it very quickly became 15v and 30v to take into account voltage drops and ripple. But in the drawing it won't actually work, because I forgot the results of the voltage setting potentiometer are not absolute, but depend on regulator input voltage. If built this way, voltage control would misbehave severely.

Now I'm thinking about creating a more elaborate method to control the regulator and to switch between transformer taps.
 
Why not put two or more LM338's in parallel, balanced against each others difference in output ;-?
You can find datasheat applications enough about this.
and again the balancing can be done in series with their outputs, but this is again a waste of energy.
Better do it with an OpAmp.

LM338 now only comes in TO-220, which as we discussed in this thread, is bad at handling high power dissipation. There used to be ones in TO-3 but they stopped making it a long time ago, and TO-3 seems like huge pain in the ass anyway.

I saw the multiple regulators controlled by op-amp in the datasheet. I will make it in LTSpice and experiment to see how it works.
 

Offline special_KTopic starter

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Re: Calculating heat of a TO-220 package
« Reply #49 on: January 19, 2025, 05:01:09 am »
Hmm

No luck simulator the mentioned circuit where an op-amp controls multiple copies of the regulator.

LTSpice lacks models for any op-amp I've ever heard of. Googling, I found a PSpice model, and many contradictory instructions to convert it which don't work. I drew the circuit in kicad using LM358, but the built in spice there refuses to run the simulation due to arcane nonsenses like "Error: Not enough tokens in line 25" or whatever that means.

Spice is such old crusty junk. But, looking and thinking about the circuit, I understand the principle of it. The op-amp is set up a lot like a comparator, it's monitoring the difference between the I-in of the master & servant regulators and controlling their adj pins until it matches. Clever.
 

Offline BillyO

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Re: Calculating heat of a TO-220 package
« Reply #50 on: January 19, 2025, 05:16:03 am »
LTSpice lacks models for any op-amp I've ever heard of.

For something like this a generic "op-amp" model will do fine.  For most things, actually.  Just stick in the default "op-amp" model and you'll be fine.
Bill  (Currently a Siglent fanboy)
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Offline special_KTopic starter

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Re: Calculating heat of a TO-220 package
« Reply #51 on: January 19, 2025, 06:21:58 am »
LTSpice lacks models for any op-amp I've ever heard of.

For something like this a generic "op-amp" model will do fine.  For most things, actually.  Just stick in the default "op-amp" model and you'll be fine.

EDIT: turns out that for this component, and only for this component, you have to write the magic spell ".inc opamp.sub" on the drawing sheet. I wonder about the mind of the man who made this program.

Now it works, and I get to see all kinds of interesting behaviors when pushed too hard
« Last Edit: January 19, 2025, 06:54:50 am by special_K »
 

Offline Vovk_Z

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Re: Calculating heat of a TO-220 package
« Reply #52 on: January 19, 2025, 09:41:35 am »

You need a large passive heatsink to dissipate 70-100 W but much smaller and compact with acive cooling. Something a size of AM2 cooler should do the job. I used small 50-60 C thermal relay to turn on a fan.

(I used XL4016 preregulator for my main laboratory bench power supply (with LM338? don't remember exact partnumber 5-Ampere IC regulator). )
 


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