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### Topic: Temperature vs Watts (Read 6978 times)previous topic - next topic

#### Jassper

##### Jan 30, 2009, 04:46 amLast Edit: Jan 30, 2009, 04:47 am by Jassper Reason: 1
Ok, I have searched and read a lot of info on the temperature / watt relationship from electrical components and I am probably more confused than I was before. Why can't the data sheets simply state "Hay Idiot, if the current draw reaches X, you need a heat sink".

Anyways, I know you can't really convert watts into temperature in degrees, but I am wondering how hot a 0.5watt is. The reason is, I have a MC7805B 5v, 1 amp regulator that is getting warmer than I think it should be. the total draw is only 0.093 amps on it's input. So if I am figuring this right, the power dissipated is 5*0.093=0.465 watts.

So how hot is a half of watt?

And should a 1 amp 5v regulator be hot to the touch from a >100ma draw? Maybe a faulty regulator?

Adding a heat sink obviously cooled it down a lot but I have had higher amp draws on 5v regulators before and never had them get this hot.

#### retrolefty

#1
##### Jan 30, 2009, 05:33 amLast Edit: Jan 30, 2009, 05:38 am by retrolefty Reason: 1
"So if I am figuring this right, the power dissipated is 5*0.093=0.465 watts."

The actual power dissipation calculation for a linear voltage regulator is the voltage drop across the regulator times the current. So if the input voltage is 15 vdc then the power loss would be 1 watt (10 X .1). So one way to lower the heat is to keep the input voltage as low as possible, but no lower then the drop out limit specification of the device, 8 volts would be nice.

As far as temperature Vs power dissipation, it is pretty complex but can be expressed mathematically. First the max temp spec is for the silicon chip inside the package, not the external package temperature. A heat sinks effectiveness is based on several things, it's size, how good the heat conduction is between the regulator package and the heat sink, the ambient temperature, etc.

A data sheet for the regulator should cover the topic if you are really interested in the math. Of all semi-conductors, linear voltage regulators are the one we usually don't worry too much about because they usually have internal protection circuitry that turns off the current flow if the temperature or current draw goes over specification.

So other then lowering the input voltage if possible, I wouldn't worry to much about it. If the heat really bothers you, or you are planning to use battery power then consider using a switching voltage regulator, they are much more effecient and therefore produce less heat for the same current draw.

Lefty

#### Grumpy_Mike

#2
##### Jan 30, 2009, 12:14 pm
Quote
Why can't the data sheets simply state "Hay Idiot,

Because data sheets are not written for idiots the are written for grown ups. The are not aimed at beginners or hobbyist but at engineers. We are just privileged to be allow to peek into their world.

Quote
So how hot is a half of watt?

It's not.

It depends on what space the watt is squeezed into.

see :-
http://www.thebox.myzen.co.uk/Tutorial/Power.html

#### Jassper

#3
##### Jan 30, 2009, 12:49 pmLast Edit: Jan 30, 2009, 01:50 pm by Jassper Reason: 1
Quote

Because data sheets are not written for idiots the are written for grown ups. The are not aimed at beginners or hobbyist but at engineers. We are just privileged to be allow to peek into their world.

Um, yes I know that. I was being slightly sarcastic. I'm not as dumb as I may appear. I do know that watts and temp are 2 different things and how physically hot something gets depends on many factors. I was hoping for a general rule of thumb here but I guess there isn't one. I just don't think the v-reg should be getting that hot, but thanks for the info.

@retrolefty

#### Anachrocomputer

#4
##### Jan 30, 2009, 01:35 pm
If you look at the specs for a heatsink, it will normally specify degrees C per Watt.  That tells you how many degrees Celcius it will heat up for each Watt dissipated.  Then you can work out how hot your chip will get.  Have a look at the very smallest sizes, and that will give you an estimate of what will happen without a heatsink.

Sometimes, power ICs have a similar rating for non-heatsink use.  But more often, the rule is "always use a heatsink".

#### jluciani

#5
##### Jan 30, 2009, 04:26 pm
To determine the temperature rise you need to calculate the junction
to ambient thermal resistance (Tja) ---

Tja = Tjc + Tcs + Tsa   all the dimensions are DegC/W

Tjc thermal resistance between the IC junction and IC case.  This is
from the IC datasheet.

Tcs thermal resistance between the IC case and your heatsink. This
should also be in the IC datasheet. Usually you place a thermal
compound between the IC case and the heat sink to decrease the
thermal resistance.  The decrease occurs because the thermal
compound fills the air gaps between the two surfaces and conducts
heat better than the air.

Tsa thermal resistance between the heatsink and ambient air. This is
in the heatsink datasheet. Most heatsink datasheets specify Tsa
with and without airflow. You can determine if you need a fan by
reviewing this number.

For a single heatsource the junction temperature would be

Tj = Tja * P + Tamb  (DegC)

where P is the power dissipated in the IC and Ta is the ambient
temperature.

(* jcl *)
www: http://www.wiblocks.com

#### Jassper

#6
##### Jan 30, 2009, 06:01 pm
jluciani
Thanks for that info, I searched for hours last night on the web and never came across anything that put it into simply terms like you did. Thanks

#### jluciani

#7
##### Jan 30, 2009, 06:20 pm
You're welcome.

Since I've answered this question a few times on different lists I decided to
save the answer and re-post as needed

(* jcl *)
www: http://www.wiblocks.com

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