Heatsink calculation with limited data

[allanlewis]

Hi,

I am using a modified version of the lighting controller circuit at...
http://www.edn.com/index.asp?layout=article&articleid=CA46649
...and I'm not sure what heatsink I need for the PNP Darlington power
transistor. As the specified model is no longer available, I am using
a TIP125. (An aside: is this a suitable replacement?) The datasheet is
available at...
http://www.onsemi.com/PowerSolutions/product.do?id=TIP125
...but the only information it gives regarding thermal characteristics
is the following:

Thermal resistance, junction-to-case, R(theta)JC = 1.92°C/W
Thermal resistance, junction-to-ambient, R(theta)JA = 62.5°C/W
Operating and storage junction, temperature range, Tj, Tstg:
-65--150°C

With regards to power dissipation, all it says is:

TC = 25°C => PD = 65W derating at 0.52W/°C above 25°C
TA = 25°C => PD = 2.0W derating at 0.016W/°C above 25°C

Given this, how do I calculate the heatsink I need?

(The maximum current, ideally, is 1.67A, so presumably it's worth
working to a maximum of at least 2A, and the maximum voltage across
the transistor will be about 16V.)

 

[John O'Flaherty]

On Apr 1, 9:28 am, allanlewis <allanlewi...@googlemail.com> wrote:

Hi,

I am using a modified version of the lighting controller circuit at...http://www.edn.com/index.asp?layout=article&articleid=CA46649
...and I'm not sure what heatsink I need for the PNP Darlington power
transistor. As the specified model is no longer available, I am using
a TIP125. (An aside: is this a suitable replacement?) The datasheet is
available at...http://www.onsemi.com/PowerSolutions/product.do?id=TIP125
...but the only information it gives regarding thermal characteristics
is the following:

Thermal resistance, junction-to-case, R(theta)JC = 1.92°C/W
Thermal resistance, junction-to-ambient, R(theta)JA = 62.5°C/W
Operating and storage junction, temperature range, Tj, Tstg:
-65--150°C

With regards to power dissipation, all it says is:

TC = 25°C => PD = 65W derating at 0.52W/°C above 25°C
TA = 25°C => PD = 2.0W derating at 0.016W/°C above 25°C

I think the TA curve is for no heat sink, so it's the TCase curve that
applies.

Given this, how do I calculate the heatsink I need?

(The maximum current, ideally, is 1.67A, so presumably it's worth
working to a maximum of at least 2A, and the maximum voltage across
the transistor will be about 16V.)

That's the voltage when turned off. For a switching circuit, you
need the voltage when turned on and the current. You might need to
consider the switching times, too. The circuit is running at 10 kHz,
or 100 us period, and the switching times may be about 1 us each way.
That would be 2% of the time in a much higher power dissipation
area.* Anyway, the data sheet says the saturation voltage (Vce when
turned fully on) is a maximum of 2 V at a collector current of 3 A;
they don't give a figure for 2 A, so a safe figure for the dissipation
would be 6 W. If you derate 65 W to 6 W, you have 59 W / (0.52 C/W),
or a maximum case temperature of (25 C + 113.5 C) = 133.5 C. Then
decide what the maximum ambient is - pretend it's 100 C; then you have
23.5 degrees C available to you to get rid of 6 W through the heat
sink, so its maximum thermal resistance, case to ambient, is 23.5 C /
6 W, or about 3.9 degrees C / W.
Then you need to look at the specs for heat sinks, to see what you
need, and whether you need thermal compound, or forced air, or not, to
get to that thermal resistance.
You can visualize all this as a string of thermal resistors from one
temperature to another, like voltage drops, with the power like a
current flowing through all the resistors.
--
John
(Offering the usual money-back guarantee on all guesses).

* Suppose all the switching time was spent at 16 V and 2 A, or 32 W.
That times 2% is 0.64 W average additional, so maybe it's not so much.
It also seems safe when looking at the safe operating area curve in
the data sheet.

 

[John O'Flaherty]

On Apr 1, 1:21 pm, "John O'Flaherty" <quias...@yahoo.com> wrote:

On Apr 1, 9:28 am, allanlewis <allanlewi...@googlemail.com> wrote:



Hi,

I am using a modified version of the lighting controller circuit at...http://www.edn.com/index.asp?layout=article&articleid=CA46649
...and I'm not sure what heatsink I need for the PNP Darlington power
transistor. As the specified model is no longer available, I am using
a TIP125. (An aside: is this a suitable replacement?) The datasheet is
available at...http://www.onsemi.com/PowerSolutions/product.do?id=TIP125
...but the only information it gives regarding thermal characteristics
is the following:

Thermal resistance, junction-to-case, R(theta)JC = 1.92°C/W
Thermal resistance, junction-to-ambient, R(theta)JA = 62.5°C/W
Operating and storage junction, temperature range, Tj, Tstg:
-65--150°C

With regards to power dissipation, all it says is:

TC = 25°C => PD = 65W derating at 0.52W/°C above 25°C
TA = 25°C => PD = 2.0W derating at 0.016W/°C above 25°C

I think the TA curve is for no heat sink, so it's the TCase curve that
applies.

Given this, how do I calculate the heatsink I need?
(The maximum current, ideally, is 1.67A, so presumably it's worth
working to a maximum of at least 2A, and the maximum voltage across
the transistor will be about 16V.)

That's the voltage when turned off. For a switching circuit, you
need the voltage when turned on and the current. You might need to
consider the switching times, too. The circuit is running at 10 kHz,
or 100 us period, and the switching times may be about 1 us each way.
That would be 2% of the time in a much higher power dissipation
area.* Anyway, the data sheet says the saturation voltage (Vce when
turned fully on) is a maximum of 2 V at a collector current of 3 A;
they don't give a figure for 2 A, so a safe figure for the dissipation
would be 6 W. If you derate 65 W to 6 W, you have 59 W / (0.52 C/W),
or a maximum case temperature of (25 C + 113.5 C) = 133.5 C. Then
decide what the maximum ambient is - pretend it's 100 C; then you have
23.5 degrees C available to you to get rid of 6 W through the heat
sink, so its maximum thermal resistance, case to ambient, is 23.5 C /
6 W, or about 3.9 degrees C / W.
Then you need to look at the specs for heat sinks, to see what you
need, and whether you need thermal compound, or forced air, or not, to
get to that thermal resistance.
You can visualize all this as a string of thermal resistors from one
temperature to another, like voltage drops, with the power like a
current flowing through all the resistors.
--
John
(Offering the usual money-back guarantee on all guesses).

* Suppose all the switching time was spent at 16 V and 2 A, or 32 W.
That times 2% is 0.64 W average additional, so maybe it's not so much.
It also seems safe when looking at the safe operating area curve in
the data sheet.

Sorry for my arithmetic above; if 100 C ambient, and 133.5 C max case
temp, you have 33.5 C to play with; 33.5 / 6 = 5.6 C / W for the heat
sink.
--
John

 

[allanlewis]

On Apr 1, 8:19 pm, "John O'Flaherty" <quias...@yahoo.com> wrote:

On Apr 1, 1:21 pm, "John O'Flaherty" <quias...@yahoo.com> wrote:



On Apr 1, 9:28 am, allanlewis <allanlewi...@googlemail.com> wrote:

Hi,

I am using a modified version of the lighting controller circuit at...http://www.edn.com/index.asp?layout=article&articleid=CA46649
...and I'm not sure what heatsink I need for the PNP Darlington power
transistor. As the specified model is no longer available, I am using
a TIP125. (An aside: is this a suitable replacement?) The datasheet is
available at...http://www.onsemi.com/PowerSolutions/product.do?id=TIP125
...but the only information it gives regarding thermal characteristics
is the following:

Thermal resistance, junction-to-case, R(theta)JC = 1.92°C/W
Thermal resistance, junction-to-ambient, R(theta)JA = 62.5°C/W
Operating and storage junction, temperature range, Tj, Tstg:
-65--150°C

With regards to power dissipation, all it says is:

TC = 25°C => PD = 65W derating at 0.52W/°C above 25°C
TA = 25°C => PD = 2.0W derating at 0.016W/°C above 25°C

I think the TA curve is for no heat sink, so it's the TCase curve that
applies.

Given this, how do I calculate the heatsink I need?
(The maximum current, ideally, is 1.67A, so presumably it's worth
working to a maximum of at least 2A, and the maximum voltage across
the transistor will be about 16V.)

That's the voltage when turned off. For a switching circuit, you
need the voltage when turned on and the current. You might need to
consider the switching times, too. The circuit is running at 10 kHz,
or 100 us period, and the switching times may be about 1 us each way.
That would be 2% of the time in a much higher power dissipation
area.* Anyway, the data sheet says the saturation voltage (Vce when
turned fully on) is a maximum of 2 V at a collector current of 3 A;
they don't give a figure for 2 A, so a safe figure for the dissipation
would be 6 W. If you derate 65 W to 6 W, you have 59 W / (0.52 C/W),
or a maximum case temperature of (25 C + 113.5 C) = 133.5 C. Then
decide what the maximum ambient is - pretend it's 100 C; then you have
23.5 degrees C available to you to get rid of 6 W through the heat
sink, so its maximum thermal resistance, case to ambient, is 23.5 C /
6 W, or about 3.9 degrees C / W.
Then you need to look at the specs for heat sinks, to see what you
need, and whether you need thermal compound, or forced air, or not, to
get to that thermal resistance.
You can visualize all this as a string of thermal resistors from one
temperature to another, like voltage drops, with the power like a
current flowing through all the resistors.
--
John
(Offering the usual money-back guarantee on all guesses).

* Suppose all the switching time was spent at 16 V and 2 A, or 32 W.
That times 2% is 0.64 W average additional, so maybe it's not so much.
It also seems safe when looking at the safe operating area curve in
the data sheet.

Sorry for my arithmetic above; if 100 C ambient, and 133.5 C max case
temp, you have 33.5 C to play with; 33.5 / 6 = 5.6 C / W for the heat
sink.
--
John

John,

That was very helpful - I'm now looking at getting this heatsink...
http://www.aavidthermalloy.com/cgi-bin/stdisp.pl?Pnum=6400bg
...which has a rating of 2.8°C/W.

Many thanks,
Allan.