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'Heat sinks (OT)'
1998\04\20@122120 by Dave Celsnak

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To heat sink, or not to heat sink, that is the question.

I have a 7805, 220 style case running my PIC board.  I am using a total of
0.700 amps.  Do I need to heat sink the 220 case/ 5v regulator??????
(This is a 3.0 amp device!)

Cya-
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1998\04\20@123620 by John Bellini

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What is the voltage drop across it?

You have know what the regulator is drooping to get the 5Volts.
Example:  If your input to the 5V regulator is 15V, the you are drooping
10V, at 0.7Amps.
Calculate your power this way and then determine whether you need a heat
sink or not.

> {Original Message removed}

1998\04\20@124234 by Thomas Magin

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At 12:08 20.04.1998 -0400, you wrote:
>To heat sink, or not to heat sink, that is the question.
>
>I have a 7805, 220 style case running my PIC board.  I am using a total of
>0.700 amps.  Do I need to heat sink the 220 case/ 5v regulator??????
>(This is a 3.0 amp device!)

Thats a question of the Inputvoltage !

Pv = (Uin - Uout) * Iout. For a TO220 this should not be more then 1W
without heatsink. So you'll probably need one.

So long

Thomas
=8-)

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1998\04\20@124831 by Scott Newell

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>I have a 7805, 220 style case running my PIC board.  I am using a total of
>0.700 amps.  Do I need to heat sink the 220 case/ 5v regulator??????
>(This is a 3.0 amp device!)

You didn't provide enough info--the unregulated supply voltage (under load)
is really important.  I'll do the math assuming the best case scenario.

Ok, according to my datasheet (National), the 7805 needs 7.5Vdc to maintain
line regulation.  That's probably cutting it close, but since you didn't
tell us the loaded unregulated supply voltage, we'll have to guess.  That
makes 2.5V across the device, so it's got to dissipate 1.75W.

Now according to my datasheet (you've got the datasheet, right?), Junction
to ambient thermal resistance is 54 degrees C/watt.  That means the temp
rise will be (54*1.75) above ambient, or about 95 degrees C.  Max junction
temperature (again, this is all in the datasheet) is 150 degrees C, so
you'd be ok if the ambient temperature is always below 55 degrees C.

This is right on the edge, but I wouldn't run a chip this hot without a
good reason.

This is all covered in the book "Art of Electronics" by Horowitz and Hill,
which should be a required document for all companies that design or
produce electronic devices.


newell

1998\04\20@162323 by wwl
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On Mon, 20 Apr 1998 12:08:50 -0400, you wrote:

>To heat sink, or not to heat sink, that is the question.
>
>I have a 7805, 220 style case running my PIC board.  I am using a total of
>0.700 amps.  Do I need to heat sink the 220 case/ 5v regulator??????
>(This is a 3.0 amp device!)
>
>Cya-
>                         6500
>                              7000
>Dave Celsnak                    ___7500
>                               / /     8000     (Shift!)
>                             /-/
>                           / /              RPM
>                         / /
>                       / /
Yes - even if the input is as lows 7V, you;re burning 1.4W which will
get very hot without a sink.
If your input is much more than this, I'd look at using a switching
regulator - much more efficient (input current drops as voltage
increases), and won't need a heatsink at this current level.
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1998\04\21@005523 by tjaart

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Dave Celsnak wrote:

> To heat sink, or not to heat sink, that is the question.
>
> I have a 7805, 220 style case running my PIC board.  I am using a total of
> 0.700 amps.  Do I need to heat sink the 220 case/ 5v regulator??????
> (This is a 3.0 amp device!)

The current rating is the maximum current it can handle. You have to
limit the power dissipated.

As an example, consider the difference between running your regulator
from 20V, and 9V

Power dissipated :
       9V : (9-5)*.7 = 2.8W = Very hot - I'd use a big heat sink.
       20V : (20-5)*.7 = 10.5W = Soldering iron.

Consider using a switch mode supply.

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1998\04\21@111510 by Harold M Hallikainen

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       In general, I'd say yes.  to figure it out, take the input
voltage minus the output voltage, then multiply the difference by the
load current.  This is approximately how much power the regulator will
dissipate.  Then look on the data sheet for the temperature rise in
degrees C per watt with no heat sink.  Multiply this by the power
dissipation figured above to determine the temperature rise.  Add in the
maximum ambient and see how you're doing relative to the rating on the
chip.
       For a more detailed analysis, look at temperature difference as
being similar to voltage difference.  Thermal resistance is measured in
degrees C per watt.  Add up the thermal resistance of the heat sink, the
connection between the heat sink and the case, and between the case and
the junction.  You need to dissipate so many watts, so this is simlar to
forcing a certain current through these resistances in series.  It
results in a voltage (or temperature) drop between the junction and
ambient temperature.    Determine this drop (in degrees C), then add it
to ambient to determine the junction temperature.  Again, see how you're
doing.

Harold



On Mon, 20 Apr 1998 12:08:50 -0400 Dave Celsnak <EraseMEx93celsnakspam_OUTspamTakeThisOuTWMICH.EDU>
writes:
{Quote hidden}

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