Hello everyone,

I tried to search online how to calculate the power dissipation on the triac to select the suitable heat sink.
The problem that faced me that I didn't understand how to select the heat sink part itself also.
In other words,
This website in my county sells aluminum heat sinks but i don't know which one is suitable to withstand a 3A current.
http://ram-e-shop.com/oscmax/catalog/index.php?cPath=146

BTA06 Datasheet:
http://www.st.com/web/en/resource/technical/document/datasheet/CD00000670.pdf

Thank you in advance.

 

I could do all kinds of wonderful math and it still won't get you where you want to go because that selling site does not provide the thermal resistance of their products. I can say you have less than 3 watts to get rid of, so think about the size of a 6 watt resistor and a heat sink that size should work.

Personally, I would just bolt that to about 70 mm by 70 mm piece of 6mm aluminum.

 

I could do all kinds of wonderful math and it still won't get you where you want to go because that selling site does not provide the thermal resistance of their products. I can say you have less than 3 watts to get rid of, so think about the size of a 6 watt resistor and a heat sink that size should work.

Personally, I would just bolt that to about 70 mm by 70 mm piece of 6mm aluminum.

yes unfortunately they don't so I thought maybe someone would have faced that problem before. Thank you for your fast responce.

Any other suggestions would be helpful.

 

Well...here's a page that has some thermal specifications. If you can do the math for the Centigrades per Watt and try to find something that matches what you can actually buy, that's a way.

http://www.mouser.com/Thermal-Management/Heat-Sinks/_/N-5gg0/

The problem is that it's time consuming and it's still a bit of a guess, even after your best efforts. That's why I just go to the shed and grab any old piece of aluminum that looks big enough and bolt it on.

 

looks like i should buy two or three different types and try them if i didn't know the exact one. thank you for you help

 

Can you do the math?

 

Can you do the math?

Actually I can't so that's why I put the link of the IC.
I really don't know which data do I need.

Can you illustrate the math and put an example of a heat sink which have a detailed specification for it as a live example.
Thank you for your efforts. I really appreciate.

 

read this then see if you can do the math...
http://sound.westhost.com/heatsinks.htm
Thermal resistances are stated in the BTA datasheet.

Generic examples..
(junction to ambient) 60degC/W junction to ambient simply means that if you are dissipating 1W with no heatsink that the junction temps will be 60deg C (over ambient) and if dissipating 2W it will be 2 x 60 = 120 deg C over ambient,etc...
(junction to case) 2.7degC/W and assuming you pick a heatsink with a thermal resistance of 2.3degC/W means that if you are dissipating 1W the junction temps will be 5 deg C (over ambient).. (ignoring other thermal resistances like thermal compound,etc..)

any decent heatsink will show its thermal resistance.. If it doesn't avoid it or you will need to do physical testing to get its rating.

 

Ok...Early in my morning. 50 ma into the gate at 1.3 volts max is 65 mW.
Figure 3 = 2.85 Watts at 3 amps.
Worst case thermal resistance is 2.7 C/W
Add 1 for attachment junction.
Operating limit is 125C
Ambient is assumed to be 25C

How many Rs of thermal resistance is the maximum?
100C rise/2.915W = 34.3C/W

How many have we already used up?
3.7 C/W

What's left?
30.6 C/W

That's the heat sink.

Aaand then...you can change the 125C limit to 100C so the internal chip will only get hot enough to boil water under operating conditions and that will make it last longer.

75 C rise/2.915W = 25.73 C/W
25.73C/W - 3.7C/W = 22 C/W
And thats the heat sink.

 

You can get an estimate of the thermal resistance of the heat sink by using this formula:
Thermal resistance = 50 divided by the square root of the surface area in cm.
So if you have a 2cm x 2cm piece of aluminum you have 8 cm of surface area (both sides). so 50/2.8=17.8 C/W.

 

There has to be some thickness to that radiator, but I haven't run the numbers in my long history of fiddling with math and electronics. I just assume the heat sink manufacturer included thickness in his final specification, or I make sure I have 4 mm to 6 mm of thickness just because I'm copying what I see in the retail stores.

 

Wow I really gained a lot of information. Looks like i need to read more to understand this topic more.
Thank you very much for your help. I'll try to implement the circuit and inform you soon with the results.