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Control car horns in PWM : electronic and thermal design questions

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jerome-eev:
"Patatas de Goma" is a sound installation where I want control voltage of car horns and car light bulbs in PWM :
http://jeromeabel.net/en/art/patatas-de-goma

The past experiences was pretty difficult, because  the car horns have an inductive charge and it takes 12A and peak may be at 20 or 30A. The current electronic system is not reliable at all. It's a lot of intensity.
According to this very good tutorial (EEVblog #105 – Electronics Thermal Heatsink Design), I need to design the heat dissipation very carefully.

The method described in this tutorial is here : http://jeromeabel.net/files/projets/patatasdegoma/EEV105_ThermalDesign.png
The current schematics are here : http://jeromeabel.net/files/projets/patatasdegoma/schematics/
The datasheets are here : http://jeromeabel.net/files/projets/patatasdegoma/materials_datasheets/.

POWER TO DISSIPATE BY THE MOSFETS
If the MOSfet let pass all the voltage 12V and the horn take 20A : P = RDS(on) * I^2 = 0.041 * 20 * 20 = 16,4W
But if the MOSfet let pass 1V or 0.001V : P = (12 - 0.001) * 20A = 240 W !

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MATERIALS DATA SHEETS :
- MOSFET : FDP61N20, 200V N-Channel MOSFET, 61A, 200V, RDS(on) = 0.041Ohms @VGS = 10 V
Rtheta JC Thermal Resistance, Junction-to-Case : 0.3 °C/W
Rtheta JA Thermal Resistance, Junction-to-Ambient : 62.5 °C/W
Tj, Tstg (Operating and Storage Temperature Range) : -55 to +150 °C

- HEATSINK : ML33G
Thermal Resistance 8.50 Ohms
Rth = 10 °C/W

- FAN : kde1206phv2a 12 Vcc - 60 x 60 x 15 mm
Rate speed : 3800 RPM
FPM = RPM x Circumference
Circumference = 3.14 x 6cm = 18.84cm; 18.84cm x 0.033 = 0.62172 feet
FPM = 3800 x 0.62172 = 2363
According to the datasheet if FPM = 2356, the surface to ambient is 2°C/W

- Insulating Kit - VELLEMAN MOUNTING KIT FOR TO220

- PELTIER MODULE CP2 TEC1-12706
(Ceramic Plate Thermoelectric Modules) (plaque a effet peltier)
Intensity : 6 A
Power : 51,4 W
Dimensions : 40 x 40 x 4,7 mm

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THERMAL DESIGN #1 - NO DESIGN ;), WITHOUT ANY HEAT COOLER DEVICES :
In the MOSfet datasheet, Rtheta JA Thermal Resistance, Junction-to-Ambient : 62.5 °C/W
I need to dissipate 240W => 62.5 x 240 = 15 000 °C !!! Am I all right ?
The MOSfet Tj, Tstg (Operating and Storage Temperature Range) is -55 to +150 °C. So It will crash, isn't it ?

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THERMAL DESIGN #2 - WITH A HEATSINK :

Goal : decrease the Rtheta JC
If I attach the MOSfet to a heatsink, the MOSfet Rtheta JC Junction-to-Case thermal resistance is 0.3 °C/W.
I need to dissipate 240W => 0.3 x 240 = 72°C. It is lesser than the temperature range maximum 150 °C, so it could work. Isn't it ? Or I have to sum all other thermal resistances Rtheta HA and Rtheta CH ?
Rtheta JC = 0.3 °C/W

According to the first thermal curve (rise above ambient / heat dissipated watts), the maximum value is 70°C/8Watts = 8.75 °C/W
It is close to the 8.5 thermal resistance of the datasheet. Is it the same ?
If we add 1/3 to be closer to the reality :
Rtheta HA = 8.75 + (8.73/3) = 11.7 °C/W

Let's say I have not a Silicone/Mica component between MOSfet and Heatsink, the common number seems to be :
Rtheta CH = 1.1 °C/W

Total = Rtheta JC + Rtheta HA + Rtheta CH = 0.3 + 11.7 + 1.1 =~ 13 °C/W
If I need 240W   : 240 x 13 + 18° (ambient temperature) = 3138 °C !!
If I need 144W   : 144 x 13 + 18° (ambient temperature) = 1890 °C !!

Hum, Am I all right ?
It is better than 15 000 °C, but I need less heat.

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THERMAL DESIGN #3 - WITH A HEATSINK, A FAN :

Goal : decrease the Rtheta HA
The fan cools the thermal resistance of the heatsink (Rtheta HA).
The first attempt is to see the second graph of the heatsink (Air velocity feet per minute / Surface to ambient °C/W )
The FPM of my fan is :
FPM = 3.14(Pi) x 6cm(Diameter of the fan) x 0.033(cm to feet conversion) x 3800 RPM = 2363
In the graph, after 800 FPM the thermal resistance is stick at 2 °C/W.
So, is it all right to say that the new Rtheta HA is 2 °C/W instead of previously 11.7 °C/W ?

New total = Rtheta JC + Rtheta HA + Rtheta CH = 0.3 + 2 + 1.1 =~ 3.4 °C/W

If I need 240W   : 240 x 3.4 + 18° (ambient temperature) = 834 °C !!
If I need 144W   : 144 x 3.4 + 18° (ambient temperature) = 507 °C !!

Whaaa, still hot.
It is better but I need less heat.

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THERMAL DESIGN #4 - WITH A HEATSINK, A FAN, INSULATING KIT :

Goal : decrease the Rtheta CH
If I add an insulating kit, how calculate the improvement ?

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THERMAL DESIGN #5 - WITH A HEATSINK, A FAN, INSULATING KIT, FEW MOSFETS :

An advice sent to me was to put few MOSfets in parallel to spread the power (not the heat).
How calculate the improvement ? Which thermal resistance will decrease ?

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THERMAL DESIGN #6 - WITH A HEATSINK, A FAN, INSULATING KIT, FEW MOSFETS, PELTIER :

Goal : decrease the Rtheta HA
How calculate the improvement ?

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Jerome

ciccio:
Sorry for not being of any help, but I really appreciated the video.
Thanks for sharing.
Best regards

eevblogfan:
hey

I am "newbie "  so I cant let you pro help here ,

I didn't read anything . but I red few lines , and based on them I can tell you that perhaps , you missed something , the power ratings you gave weren't correct , IE : to power something on . you should provide the whole power into it and "let" the something "limit" the amount of current he need in order to work (I simplified it allot , perhaps I did some mistakes ,) . when you say that you need to provide the 240W , you meant that the resistive load will draw approx 240W , but in order to power it on , you need to switch the power on , by closing the circuit within a conductive "element" (IE . low resistance and the amount of power loss will be low , )

PWM is something that I am not experience enough with , but I know that the efficiency is much higher then a linear solution . because the switching element (usually MOSFET's ) are almost achieve they're lowest Rds on (IE . very low power dissipation ) , so assuming you want to activate some load on 240W , at 50% duty cycle . ignore tons of factors , you'll look for about 120W loss . 90% 24W loss , etcetera etcetera ..

so 500/800/15000C is insane , based on the datasheet you supplied , the temp rise with 24W is 31.2 C above ambient ( heatsink C/W =1 , MOSFET C/W = 0.3 ) ,

and as I said , I am not PWM expert , therefore I cant provide any answer about many factor's here ,

BTW , tank's for the video :P

muvideo:
Hello,
you said the mosfet are driven with pwm signal,
a pair of questions:
what is the pwm frequency?
I understand for lights but is what is the reason of using pwm for horns?
I imagine that horns are self-oscillating inductive things, dont't know
how they behave with a pwm signal to feed them.

A pair of considerations: using pwm you are switching mosfets on and off,
or at least this is what you want to do.
So for the dissipation of the mosfets you must take in accoount the switching
energy loss, that depends strongly on the frequency of pwm, circuit layout
and the behaviour of the loads: horns and lights.
For example I think that you want the recirculation diodes near the mosfets
drains, also the mosfet drivers' strenght must be enough for driving the gates
capacitance (in general terms, I havent' checked the particular components you
have chosen).

The Thermoelectric cooler (Peltier module):
I dont't understand how you want to use it.
Put simply it has two sides: the cold one and the hot one, usually the hot
one "puts out" more than twice the thermal power taken from the cold side,
the power taken from the cold side plus the electrical power it takes.
Also it must function at less than 70°C, usually this means a way bigger
heatsink on the hot side than the heatsink needed by the mosfets alone.

I think you should make these steps:
decide what frequency the pwm signals has to be,
try to design for the least stress on the components and
choose the righ ones, and last design the heat management.

Fabio.

edit:
jerome reading eevblogfan i noted the way you calculated the dissipation:

--- Quote from: jerome-eev on March 12, 2012, 06:58:48 pm ---POWER TO DISSIPATE BY THE MOSFETS
If the MOSfet let pass all the voltage 12V and the horn take 20A : P = RDS(on) * I^2 = 0.041 * 20 * 20 = 16,4W
But if the MOSfet let pass 1V or 0.001V : P = (12 - 0.001) * 20A = 240 W !

--- End quote ---

With pwm you have this process:
-the mosfet turns on:
it is a closed switch with it's Rdson resistance, so it dissipates the power you calculated correctly above: P=RdsON*i^2
-the mosfet turns off and ideally the current stops flowing and it dissipates no current.
- the mosfet turns on again etc...
Ideally you have that for duty on % time the dissipation is the one given above and for the rest time the dissipaiton
is null.
In the real world the loads are non resistive, there are inductor like horns but also parasite inductors like cables that
store and release energy during turn on and off of the mosfets, this makes the mosfets switching a complex thing, and
the energy stored in these inductances has to go or dissipate somewhere.

T4P:
I have a 50amps MOSFET PWM switcher somewhere in my room , i have the schematic on paper , i will scan it for you tomorrow !  ::)