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'[EE] Heat Dissipation calculations'
2010\02\23@072005
by
Moreira, Luis A

Hi All,
I have an existing assembly with two power diodes in series mounted on
an aluminium heat sink, I am increasing the power dissipated on them and
I am trying to do the calculations, to make sure that the diodes will
still be fine using the current setup.
I started by modelling the assembly:
I will have two branches in parallel of Rjunctioncase in series with
Rcaseheatsink and they will be in series with Rheatsinkambient.
I am just having some difficulty looking at it and need someone to check
if I am thinking correctly. Both diodes will dissipate, in this case,
the same power. The heat will flow from both to the heat sink and then
to ambient.
Hence if I use the formula
Tj (Tcase + Theatsink + Tambient) = Q*(
Rjunctioncase + Rcaseheatsink +
Rheatsinkambient)
Where Q is the sum of power dissipated on both diodes
In this case as the diodes are identical is it correct to say that the
Tj you get out of this calculation is the junction temperature for both
the diodes which is also the Tj each of them?
How would you do it for two diodes dissipating different power? It seems
to me that it would not matter, but at the same time I can not be sure,
especially if this power is pulsed instead of continuous.
Any thoughts welcomed.
Best Regards
Luis
2010\02\23@074046
by
Wouter van Ooijen
> I am just having some difficulty looking at it
without giving a specific answer: translate the heatstuff to
electronics. The thermal resistance is a resistor, a heat sources is a
current source, and a temperature is a voltage.
It will surely matter when the diodes dissipate different amounts. Take
the extreme cases: when all heat is dissipated by one diode, the
junctiontosink resistance of the other one is out of the equation.
When they both dissipate the same (taking your word for it) you can
replace them with one diode which dissipates the sum, and has half the
junctiontosink resistance.

Wouter van Ooijen
 
Van Ooijen Technische Informatica: http://www.voti.nl
consultancy, development, PICmicro products
docent Hogeschool van Utrecht: http://www.voti.nl/hvu
2010\02\23@074822
by
Russell McMahon
An easy "trick" when you have a thermal "Y" structure is to start at the
"bottom" ambient and work up. There are some issues with this simplitic
approach but it will probably give you a good enough result.
NB  the following is far far farmore horrid to read and write than to
understand and to calculate.
It ends up as advanced common sense once you follow it.
E&OE  odds are I've got a subscript etc wrong below somewhere
Here the total power flows from sink to ambient so
Call diodes x & y.
Tsa = Q(x+y) . Rth_sa
so
Ts = Ta + Tsa = Ta + Q(x+y) . Rth_sa
Now you have the common sink temperature.
Each diode or other device can now be individually calculated.
Tjx.s = Qx.Rth_jx.s
Tjx = Ts + Tjx.s = Ts + Qx.Rth_jx.s
or
Tjx = Ts + Qx.Rth_jx.s =
Tjx = Ta + Q(x+y) . Rth_sa + Qx.Rth_jx.s
Similarly
Tjy = Ta + Q(x+y) . Rth_sa + Qy.Rth_jy.s
Words  easier here
Total energy flow through sink to ambient cause sink rise based on sink to
ambient resistance.
Then
Junction to sink rise resulsts from junction energy flow through junction to
sink thermal resistance.
If Junction to sink resistances and energy flows are the same for both
devices then junction temperatures are the same.
In that case you can treat it as one devie with double the energy flow of
one device and half the thernmal junction to sink resistance.
FAR easier to understand than to read :)
Russell McMahon
On 24 February 2010 01:20, Moreira, Luis A <spam_OUTLuis.MoreiraTakeThisOuTccfe.ac.uk> wrote:
{Quote hidden}>
>
> Hi All,
> I have an existing assembly with two power diodes in series mounted on
> an aluminium heat sink, I am increasing the power dissipated on them and
> I am trying to do the calculations, to make sure that the diodes will
> still be fine using the current setup.
> I started by modelling the assembly:
>
> I will have two branches in parallel of Rjunctioncase in series with
> Rcaseheatsink and they will be in series with Rheatsinkambient.
>
> I am just having some difficulty looking at it and need someone to check
> if I am thinking correctly. Both diodes will dissipate, in this case,
> the same power. The heat will flow from both to the heat sink and then
> to ambient.
>
> Hence if I use the formula
>
> Tj (Tcase + Theatsink + Tambient) = Q*(
> Rjunctioncase + Rcaseheatsink +
> Rheatsinkambient)
>
> Where Q is the sum of power dissipated on both diodes
>
>
> In this case as the diodes are identical is it correct to say that the
> Tj you get out of this calculation is the junction temperature for both
> the diodes which is also the Tj each of them?
>
> How would you do it for two diodes dissipating different power? It seems
> to me that it would not matter, but at the same time I can not be sure,
> especially if this power is pulsed instead of continuous.
>
> Any thoughts welcomed.
>
>
> Best Regards
> Luis
>
>
>
> 
2010\02\23@075708
by
Russell McMahon
Pulsing is not too too important to the heat sink if thermal time
constant is long compared to pulse width. Note that thermal tc and
thermal resistance do not have to be at all well correlated BUT often
have some relationship as big fat bits of metal tend to be both heat
stores and have lower thermal resistance.
The Tjs is more susceptible to pulse effects as the junction thermal
mass is small. Most power devices have tables in the data sheet that
related Tj temperature rise w1ith pulse frequency and duty cycle. So
Tjs will probably be pulse affected but Tsa not so much so.
Again, once you get a feel for what is happening it is truly trivial
in concept  BUT not necessarily trivial in practice due to various
practical effects. Wouters suggestion to use ohms law type
calculations and mindsets is entirely appropriate.
Do note that device characteristics may be severely affected by Tj
but manufacturers stupidly usually quote them in the most favorable
and entirely unrealistic conditions. Marketing says to do this 
common sense says not to.
eg MOSFETS usually have Rdson quoted at eg 1% dutycycle at say 10
kHz.. Tj gets a chance to drop back between pulses. Rule of thumb 
MOSFET real world Rdsons are typically about 2:1 greater than data
sheet claims.
Russell McMahon
2010\02\23@104237
by
Moreira, Luis A
Hi Wouter/Russell,
I am not getting the half junction to sink resistance... I can see it
from the electrical model analogy, but when I think about it, it does
not fell right. If I have one diode on the heatsink, then it only needs
to dissipate the heat generated by that diode, but when I have two
diodes on the same heatsink generating double the heat then it will be
harder for the heatsink to dissipate that heat hence Tj will be higher.
Hence in real terms you would say that the junctiontosink resistance
is higher.
Sorry if I am missing something basic, but I am having one of those
cases of using these calculations for years for a single component and
now faced with more than one, things started not making sense.
Best Regards
Luis
{Original Message removed}
2010\02\23@112245
by
Wouter van Ooijen
Moreira, Luis A wrote:
> Hi Wouter/Russell,
>
> I am not getting the half junction to sink resistance...
Maybe because you missed the next part: half the junctiontosink
resistance, but also the sum of the dissipated powers.

Wouter van Ooijen
 
Van Ooijen Technische Informatica: http://www.voti.nl
consultancy, development, PICmicro products
docent Hogeschool van Utrecht: http://www.voti.nl/hvu
2010\02\23@122631
by
Russell McMahon
> I am not getting the half junction to sink resistance... I can see it
> from the electrical model analogy, but when I think about it, it does
> not fell right. If I have one diode on the heatsink, then it only needs
> to dissipate the heat generated by that diode, but when I have two
> diodes on the same heatsink generating double the heat then it will be
> harder for the heatsink to dissipate that heat hence Tj will be higher.
> Hence in real terms you would say that the junctiontosink resistance
> is higher.
You have it correct.
What you are saying is what I tried to show
ie note here fo the sink calculations I have Q(x+y)  the total heat
flux from all sources 9two in this case) flowing through the common
Rth_Sa
I said:
>>
Here the total power flows from sink to ambient so
Call diodes x & y.
Tsa = Q(x+y) . Rth_sa
so
Ts = Ta + Tsa = Ta + Q(x+y) . Rth_sa
Now you have the common sink temperature.
>>
Q(x+y) is the combined heat flux.
Rth_sa is the sink to air resistance shared by both of them.
So, you have it right, "go round power please" . :)
Just keep going from there
R
> Sorry if I am missing something basic, but I am having one of those
> cases of using these calculations for years for a single component and
> now faced with more than one, things started not making sense.
>
> Best Regards
> Luis
>
>
> {Original Message removed}
2010\02\26@062411
by
Moreira, Luis A
Hi Wouter/Russell,
After looking at this for a bit it starts to come together...
I can then say that if one of the diodes is on the heatsink but is not dissipating any power, its Tj will be the same as the calculated heatsink, which will be due to the power dissipated by the other diode?
Related to this subject I would like to look into cooling of the air around the heatsink.
So now I am thinking, what happens if you have this heatsink inside a box? You will heat the air and hence your Tj will start to increase, hence I need cooling, to reduce ambient temperature in the box. How do I calculate this?
I heard a lot about "air volume changes per minute". Also what about water cooling, what is the process to calculate requirements?
Thank you for help.
Best Regards
Luis
{Original Message removed}
2010\02\26@064000
by
Wouter van Ooijen
> I can then say that if one of the diodes is
> on the heatsink but is not dissipating any power,
> its Tj will be the same as the calculated heatsink,
> which will be due to the power dissipated by the other diode?
Of course, except for the probably a small effect of the cooling via the
Rth bodyambient of the inactive diode.
> So now I am thinking, what happens if you have this heatsink inside a
> box? You will heat the air and hence your Tj will start to increase,
> hence I need cooling, to reduce ambient temperature in the box.
> How do I calculate this?
I dunno. Safest thing to do is to provide enough air flow, forced
(ventilator) or natural.
> Also what about water cooling, what is the process to calculate
> requirements?
In most cases the water part will be a low Rth, it simply dislocates the
Rth Xtoair.

Wouter van Ooijen
 
Van Ooijen Technische Informatica: http://www.voti.nl
consultancy, development, PICmicro products
docent Hogeschool van Utrecht: http://www.voti.nl/hvu
2010\02\28@135540
by
Rich
'[EE] Heat Dissipation calculations'
2010\03\05@024204
by
Moreira, Luis A
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