Calculate Junction to Ambient Thermal Resistance of a Schottky Diode

Thread Starter

Tony9168

Joined Mar 6, 2016
32
Hi Guys,

I unable to calculate the temperature rise (Temperature rise = Power dissipation * Junction to Ambient Thermal Resistance) of a schottky diode as the Junction to Ambient Thermal Resistance is not specified in the datasheet.

I have read some article online that the Junction to Ambient Thermal Resistance is able to be calculated by << (Junction Temperature - Ambient Temperature) / Maximum power dissipation>>.

Question:
  1. The mentioned junction temperature for determination of junction to ambient thermal resistance is referring to (a) or (b)? From my understanding, it should be (b).
  • (a) The maximum junction temperature of the schottky diode OR
  • (b) The junction temperature when the schottky diode disspiating maximum power.
2. Could anyone guide and show how to compute the junction to ambient thermal resistance for the schottky diode as shown in attachment?

3. I have heard from somebody claims that if the application of the schottky diode is without heatsink, the junction to case thermal resistance in the datasheet is equal to its junction to ambient thermal resistance. Is it true?

Thanks and please guide.

 

Attachments

MrAl

Joined Jun 17, 2014
6,603
Hello,

If you had a heatsink you could calculate the total resistance by adding the junc to case to the case to ambient.
Since you dont have a heatsink, imagine that you had a heatsink that was the same area as the metal on the package.
You could compute the 'heatsink' to ambient based on that.
A rule of thumb is that for one (local) square inch of area the temperature rises by 60 degrees C for every watt.
So based on the flat metal area, calculate the thermal resistance from that which of course is then 60 degrees C per watt divided by area.
The junc to case thermal resistance is 1.5 deg C per watt, so the total is about 62 deg C per watt. However, since the area is probably around 3/4 square inch the case to ambient is more like 80 deg C per watt, so you end up with around 82 deg C per watt.
If the area was only 1/2 square inch, then the case to ambient would be around 120 deg C per watt, with a total of around 122 deg C per watt for everything.
Always test to make sure the package does not overheat regardless what you calculate, and it should be in the intended project enclosure with all the other parts included.

If i rem right for a case size like that i was able to run at 1 watt. That is assuming it is running with an ambient of room temperature. You also have to include the ambient environment temperature to make sure it does not overheat. That was a different part though with the same package size so take that into account also.

You can also look at a flat plate heatsink calculator online which may help.
 
Last edited:

Analog Ground

Joined Apr 24, 2019
169
A tip of the hat to MrAl. For an interpretation of his solution, treat the part without a heat sink as if it was attached to a single fin heat sink with an area of 15mm x 5mm (which is the dimensions of the tab). Then, use an on-line "heat sink calculator" to determine the thermal performance. I suspect you will get an answer close to the one presented by MrAl. A good calculator will also give the option to fold in air flow and other factors.

The answer to your question #3 is "no". It is not true. It would only be true if the part was perfectly attached to an infinitely large heat sink which is the opposite of operating with no heat sink. If the answer was "yes" no one would ever need a heat sink.
 

Thread Starter

Tony9168

Joined Mar 6, 2016
32
A rule of thumb is that for one (local) square inch of area the temperature rises by 60 degrees C for every watt.
Thanks Mr AI for the explanation. Is there any reference material which provide further explanation on the the calculation and do we need to consider the thermal resistance of case to heat sink?
 

MrAl

Joined Jun 17, 2014
6,603
Thanks Mr AI for the explanation. Is there any reference material which provide further explanation on the the calculation and do we need to consider the thermal resistance of case to heat sink?
Hi,

Usually that is small.

There is stuff on the web but it doest seem correct because the temperature rise is too high.
So maybe search the web a little and find something that comes close to that rule of thumb that's all i can say
Otherwise we've got to do a full blown heat equation solution. I meant to do that but never got around to it. If you are interested though, look up the "heat equation".
 

Thread Starter

Tony9168

Joined Mar 6, 2016
32
Hi,

Usually that is small.

There is stuff on the web but it doest seem correct because the temperature rise is too high.
So maybe search the web a little and find something that comes close to that rule of thumb that's all i can say
Otherwise we've got to do a full blown heat equation solution. I meant to do that but never got around to it. If you are interested though, look up the "heat equation".
Thanks Mr AI. You have provide me a great guidance on this issue.
 

MrAl

Joined Jun 17, 2014
6,603
Is the thickness of the heatsink does matter on its thermal resistance?
Hi,

I think it would have to be abnormally thick to see much added effect there and here is why.
The thermal conduction of aluminum is much better than the thermal conduction of air so it would be sort of like having maybe a 20 Ohm resistor and then connecting a second resistor in series which is only 1 Ohm. The 1 Ohm resistor adds little to the mix so it can usually be ignored. If the metal was real thick then it may matter but we'd have to do some calculations or look that up, but usually the metal part is considered to conduct very well and so does not need to be a consideration. The main consideration is the exposed area of the heat sink and sometimes only one of the two main surfaces is exposed to the free air.
 

Berzerker

Joined Jul 29, 2018
550
I'm still trying to figure this out myself. Not real great at it still but I did find some tutorials on it and wrote this down from it.
Don't know if this will help but it comes from my notes

Jc+Cs+Sa = Ja. junc-case + Case-sink + sink-amb = junc-amb temp
(Ja x Pd) + Ja = Tj. junc-amb x power dissipation + temp amb = (temp) @ junction
Volts x current = Pd "power dissipation"
(Tj - Ta) / Pd = Ja. "junc amb temp"
Brzrkr
 
Last edited:

MrAl

Joined Jun 17, 2014
6,603
I'm still trying to figure this out myself. Not real great at it still but I did find some tutorials on it and wrote this down from it.
Don't know if this will help but it comes from my notes

Jc+Cs+Sa = Ja. junc-case + Case-sink + sink-amb = junc-amb temp
(Ja x Pd) + Ja = Tj. junc-amb x power dissipation + temp amb = (temp) @ junction
Volts x current = Pd "power dissipation"
(Tj - Ta) / Pd = Ja. "junc amb temp"
Brzrkr

Hi,

It is really all about thermal resistance. The thermal resistance works in a manner similar to the resistance in a circuit.

For example:
In a circuit if we have 10 ohms and we apply 2 amps, the voltage come out to 20 volts.
With a heat sink if we have 10 degrees C per watt and we apply 2 watts, the temperature rise is 20 degrees C.

As with electrical resistance, thermal resistance can work in parallel or in series. The most typical is in series so if we had a power source connected to a device that had a thermal resistance of 2 deg C per watt and a heat sink 10 deg C per watt, the total is:
2+10=12 deg C per watt.

This is rare, but if we could somehow use two of those heat sinks mounted properly the total thermal resistance would be:
10/2=5 deg C per watt. This is more rare however.
 

Thread Starter

Tony9168

Joined Mar 6, 2016
32
I'm still trying to figure this out myself. Not real great at it still but I did find some tutorials on it and wrote this down from it.
Don't know if this will help but it comes from my notes

Jc+Cs+Sa = Ja. junc-case + Case-sink + sink-amb = junc-amb temp
(Ja x Pd) + Ja = Tj. junc-amb x power dissipation + temp amb = (temp) @ junction
Volts x current = Pd "power dissipation"
(Tj - Ta) / Pd = Ja. "junc amb temp"
Brzrkr
Hi Berzerker,

Ya, i saw this formulae as well.

I have try to determine the junction to ambient thermal resistance by the following methods.

(a), (b) and (c) is using the formulae.
(d) is using the rule of thumb recommended by Mr AI
(e) is using the test result done by my senior previously.

** (a) & (c) is impossible as the calculated result is lower than the junction to case thermal resistance whereas (b), (d) and (e) might be correct. But the result greatly differ from each other... Not sure need to refer to which one.

The datasheet is available at the attachment. Please advise if you able to figure out the junction to ambient thermal resistance from the information available in the datasheet.

*****************************************************************************************************************************************************************

(a) Refer to data sheet, Tj = 150C; Maximum Power Dissipation = Repetitive Peak Surge Reverse Power = 660W; Ta not mentioned, Let Ta = Tc = 25 C.
>> Junction to ambient thermal resistance = (150 - 25) / 660 = 0.19 deg C / Watt.
>> Personally think that this is not the correct answer as it is lower than the junction to case thermal resistance, 1.50 deg C / Watt.​

(b) Refer to datasheet, Tj = 150C; Maximum Power Dissipation = Average Rectified Forward Current * Forward Voltage = 25 * 0.58 = 14.5W; Ta not mentioned, Let Ta = Tc = 25 C
>> Junction to ambient thermal resistance = (150 - 25) / 14.5 = 8.62 deg C / Watt
>> Possible.
(c) Refer to datasheet, Tj = 150C; Maximum Power Dissipation = Peak Surge Forward Current * Forward Voltage = 300 * 0.58 = 174W; Ta not mentioned, Let Ta = Tc = 25 C
>> Junction to ambient thermal resistance = (150 - 25) / 174 = 0.7 deg C / Watt.
>> Personally think that not the correct answer as it is lower than the junction to case thermal resistance, 1.50 deg C / Watt.
(d) Based on the rule of thumb that suggested by Mr AI, consider 1 square inch of area, the temperature rise by 60C / Watt.
>> The area of the flat heat sink in my case is 0.47 square inch, estimated case to ambient thermal resistance = 60/0.47 = 127.6 C/Watt.
>> Possible.
(e) Based on the temperature rise test that done by my senior previously, when the diode is input with 4.0A, the temperature rise is 40C.
>> Since Temperature Rise = Power Dissipation * Junction to ambient thermal resistance,
>> the expected junction to ambient thermal resistance = 40 / (4.0A * 0.58V) = 40 / 2.32 = 17.2 C / Watt.

 

Attachments

Thread Starter

Tony9168

Joined Mar 6, 2016
32
The Rthja - thermal resistance between Junction to Ambient will be around 37°C/W and 1.7°C/W for Rthjc - thermal resistance between Junction to Case.
For this ITo-3 package.

And the "electrical analogy"

View attachment 179944

http://sound.whsites.net/heatsinks.htm
Hi Jony130, may i know how you get the junction to ambient thermal resistance of 37C/W? Is it a standard value for ITo-3 package or by calculation? Could you provide me some guidance?
 

Berzerker

Joined Jul 29, 2018
550
Tony9168 said:
** (a) & (c) is impossible as the calculated result is lower than the junction to case thermal resistance whereas (b), (d) and (e) might be correct. But the result greatly differ from each other... Not sure need to refer to which one.
Yeah I ran into the same problem:eek:. It would seem that if you use one formula and it gives you an answer, you should be able to use another an get the same answer but it doesn't always come out that way (At least for me). As I said I'm still going over these formulas myself to understand "WHY" it doesn't..... Might be we're doing something wrong and it just hasn't slapped us in the face and have that "Oh Crap' moment.
If I understood it more I would do the calculations for you and show you how I got each one of the answers but I'm not going to mess you up with incorrect figures.
Had help from several of the great people from here myself! Don't worry someone will come along and explain it so hopefully you and I will understand it better.
Brzrkr
 

WBahn

Joined Mar 31, 2012
24,850
Hi Guys,

I unable to calculate the temperature rise (Temperature rise = Power dissipation * Junction to Ambient Thermal Resistance) of a schottky diode as the Junction to Ambient Thermal Resistance is not specified in the datasheet.

I have read some article online that the Junction to Ambient Thermal Resistance is able to be calculated by << (Junction Temperature - Ambient Temperature) / Maximum power dissipation>>.

Question:
  1. The mentioned junction temperature for determination of junction to ambient thermal resistance is referring to (a) or (b)? From my understanding, it should be (b).
  • (a) The maximum junction temperature of the schottky diode OR
  • (b) The junction temperature when the schottky diode disspiating maximum power.
2. Could anyone guide and show how to compute the junction to ambient thermal resistance for the schottky diode as shown in attachment?

3. I have heard from somebody claims that if the application of the schottky diode is without heatsink, the junction to case thermal resistance in the datasheet is equal to its junction to ambient thermal resistance. Is it true?

Thanks and please guide.

Many datasheets only give the junction-to-case thermal resistance because that can be highly influenced by the particular die within the package and how it is mounted. But the thermal resistance from the case to ambient is something that is a function of the package only (allowing for the possibilities that packages of the same designation from different manufacturers will be a bit different).

So what you want to do is find the datasheet or reference table that gives the information for the package itself.

Here's an example:

https://www.analog.com/media/en/package-pcb-resources/package/thermal-table.pdf

It shows that the case-to-ambient for the TO-3P (a.k.a., TO-247) is 45 °C/W.

But note that this is not quite an ITO-3P -- this part has pin 2 as common to the case. It makes a difference. In the couple minutes I spent searching I didn't find a table that includes the ITO-3P specifically.
 

Berzerker

Joined Jul 29, 2018
550
@WBahn and others
You guys have to stop explaining in PHD terms. For those of us who are trying to learn your reply just went over our heads. Just like you explain 10X10 to a second grader that never thought of counting ten fingers ten times it messes us up. Just give an example of a calculation of a make believe thermal resistance or anything else and show us how it works with the math in tow would be great. Not all the members here have PHD's in electronics.
Brzrkr
 

Thread Starter

Tony9168

Joined Mar 6, 2016
32
Hi All,

I think that the case to ambient thermal resistance will only affected by:
i) Its surrounding condition (eg. airflow)
ii) The surface area of the parts that dissipate heat.

It will not be affected by the case materials.

Am i right?
 

kubeek

Joined Sep 20, 2005
5,621
Junction to case will be affected by the case materials, but IC plastic is basically allways the same. Metal vs plastic case would be somewhat different, but if they have the same surface area the result should be close.
 
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