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'[PIC]:Re: 7805 heating up...'
2002\04\03@160218 by

Hi

The power dissipated at .2A could be as high as 1.68 watts if the engine is
running and charging to 14 volts.  Regulator voltage is 14 - .6 -5 = 8.4
volts. Power = 8.4 X .2 = 1.68 watts.  The thermal resistance of the 7805 in
the TO220 case is 60 degrees per watt (junction to ambient).  So the
junction temperature could be as high as 1.68 X 60 = 100.8 degrees above
ambient.  If you are running at 12.0 volts and 200ma the power would be 1.28
watts, with the junction temperature 76.8 degrees above ambient.

If a small clip on heat sink does not cool things off you should check for
oscillations.

Dave

>Should it get this hot?
>
>Thanks,
>-Neil.

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Ahhhh...enlightening.  Right now, heat sinking is limited to a small
copper pad on the PCB about 0.5" x 0.75".  I actually have the
TO-220 case bolted to the copper pad/PCB with some thermal
heat-sink compound between the two.

I just looked at the 7805 specs, and it tells me that the operating
temp range is 0C to 70C (158 deg F).  So it sounds like I will
have a problem eventually (assuming that the temp is not
affecting anything else nearby).  BTW, max junction temp is
150C (302 deg F).

I don't think I have room for a regular heatsink, but I can find
1" x 1.75" (at least).  Could that be enough to solve the problem?
I would probably use some aluminium for this, unless I can
find a small copper sheet.

Thanks,
-Neil.

{Original Message removed}
> I just looked at the 7805 specs, and it tells me that the operating
> temp range is 0C to 70C (158 deg F).  So it sounds like I will
> have a problem eventually (assuming that the temp is not
> affecting anything else nearby).  BTW, max junction temp is
> 150C (302 deg F).

fortunately the 7805 has automatic shutdown on over temperature.  That will
protect the 7805 but your circuit will be out to lunch during a shutdown.

Is there no way to bolt a piece of sheet aluminum (copper would be even
better) between the 7805 and the pad and bend it up immediately at the sides
of the 7805?  That failing, string a few 1N4001 diodes in series at the
input of the regulator.  They will drop the voltage to the regulator a bit
and dissipate some of the power for it.  You can also use a power resistor
if you are very sure about the max current draw.  Either solution is ugly
though.

> I don't think I have room for a regular heatsink, but I can find
> 1" x 1.75" (at least).  Could that be enough to solve the problem?
> I would probably use some aluminium for this, unless I can
> find a small copper sheet.

It's worth a try.  You may have use combinations of several methods.

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Ok. Battery voltage can range from a min of 12V with engine off (say) to a
max of around 14.5 with alternator running.

Load is 0.2 amp. 7805 wants minimum of 7.5 volts on it's input - say 8 v.

You could use a resistor in series with the battery and the 7805 to help
reduce it's dissipation.

With 12v on battery and 8 volt on i/p to regulator, at .2 amp, you need a
(12-8)/.2 = 20 ohm resistor.

Nearest standard value is 22 ohms - voltage drop will be 4.4 volts rather
than 4. But the battery will not go below 12.4 volts or so, and the 7805
only needs 7.5, so you are fine.

It will dissipate 22*(0.2)**2 = 0.88 watts - I'd use a 2 watt resistor

Regulator will dissipate (8-5)*0.2 = 0.6 watts at this battery voltage

When alternator is running, and battery is at 14.5 volts, the voltage to
the i/p of the regulator will be

14.5 - (22 * 0.2)  = 10.1 volts and will dissipate (10.1-5)*0.2 = 1.02 watts.

Free air dissipation of the 7805 at 25 degrees C ambient is (150-25)/60 = 2
watts or so
(The 60 is the thermal resistance of the junction to air in Degrees C/watt).

At 1..02 watts, the ambient temp could rise to 150 - (1.02*60) = 88.8 deg C.

So stick in a 22 ohm 2 watt resistor and the 7805 will be just fine.

NOTE that these calc assume a CONSTANT 200 milliamp load. If you load is
different, redo the calculations. And if the load is not constant (e.g
ranges between say 150 and 220 milliamps), you need to redo the calcs to
determine the value of the resistor at the MAX load current and MIN battery
voltage. Then redo the dissipation calcs using the MAX battery voltage and

Have fun!

Larry

At 03:34 PM 4/3/2002 -0600, you wrote:
{Quote hidden}

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Will try the heatsink, but it would be nice to be able to calculate (with
reasonable ease :-) how much heat dissipation to expect per sq in of
heatsink.  To make matters worse, this is in a friend's car (out of town),
so I'll probably have to have him ship me the unit.

I was thinking similarly about dropping the voltage a bit before letting
the regulator have it, but I thought of 2 regulators -- first drop to 9V
or 8V, then the 7805.

Only other thought is to move the power-supply remotely.

> fortunately the 7805 has automatic shutdown on over temperature.  That
will
> protect the 7805 but your circuit will be out to lunch during a shutdown.
>
> Is there no way to bolt a piece of sheet aluminum (copper would be even
> better) between the 7805 and the pad and bend it up immediately at the
sides
{Quote hidden}

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> I don't think I have room for a regular heatsink, but I can find

Couldn4t you fix the TO220 case of the 7805 to a metal part of the
vehicle?, beware that it is connected to ground terminal.

Cheers,
Diego.

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Not in its current location -- only metal in the area is the stereo just
below it, and I don't want to mess with that.  Here's what the board
looks like (back view with 7805)...
http://www.avn-tech.com/VBM/images/VBM_board_2.jpg

and here's where it's installed (sitting "upright" on an edge)...
http://www.avn-tech.com/VBM/images/VBM_completed_02.jpg

{Original Message removed}
> I was thinking similarly about dropping the voltage a bit before letting
> the regulator have it, but I thought of 2 regulators -- first drop to 9V
> or 8V, then the 7805.

With a linear regulator, 1.8W is going to be dissipated somewhere.  That's
too much for a single TO-220 without a heatsink.  If sufficient heat sinking
isn't possible, then you need to spread the dissipation over multiple
components.  A 9V regulator will spread the dissipation to two devices.
This approach may be required if the circuit draws large current spikes.  If
the max draw current is well known then a resistor for dropping the voltage
will be cheaper.

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> Not in its current location -- only metal in the area is the stereo just
> below it, and I don't want to mess with that.  Here's what the board
> looks like (back view with 7805)...
>     http://www.avn-tech.com/VBM/images/VBM_board_2.jpg

Geesh, I thought you said board space was tight.  You've got lots of room.
It should be real easy to kludge in a dropping resistor.

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Actually what I meant earlier was that heatsink space was tight (can't go
too high off the board, nor too much in any direction).  After that pic was
taken, I hacked in the 1N4007 for more protection.  My first attempt
(cause its easy for my friend to do) is to add an inline resistor, but I'll
do this on the power-lead itself -- just stick one male and one female
crimp-on bullet connectors onto either end of a resistor.

Cheers,
-Neil.

{Original Message removed}
Looks like the bulk of your power budget might be in lighting up the LED's.
If you're running them off the 5 v rail now, you might save some heat
trouble by driving them off 12v (or a seperate regulator (7808) if worried
about stable light intensity.)  --I'm getting deja vu -is this the same
project discussed a few weeks ago?

-Dal

{Original Message removed}
Yep, the LED's seem to absorb most of the power.  I need 5V
though, cause the pressure sensor requires 5V.  Keeping the
sensor and PIC A/D ref at the same voltage level seemed a
clean, easy way to avoid unnecessary errors sources, and
keep parts-count low.

Not sure if it was the same project -- last major discussion I
started was the automatic bathtub controller, and prior to
that it was the 7-segment LED experiment.  So many ideas,
so little time...

{Original Message removed}
> I don't think I have room for a regular heatsink, but I can find
> 1" x 1.75" (at least).  Could that be enough to solve the problem?
> I would probably use some aluminium for this, unless I can
> find a small copper sheet.

If you have no room on the PCB itself for a heat sink, you will find that
using a threaded aluminium spacer on the bolt securing the regulator down to
the PCB will act as a heat pipe on which you could then mount a piece of
flat aluminium as a "top hat" heat radiator. Not as efficient as having a
heat sink directly on the regulator itself, but may lower the thermal
resistance enough to keep you out of trouble.

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No, actually what I was suggesting is that you could run everything else at
5v like normal, but run the LED's on the 12v rail to reduce the heat load on
your regulator.  If you invert the pic outputs and drive a cheap (pn2222a)
transistor to switch the ground on each LED to be able to run them at
12v --or maybe a ULN2803 transitor array to lower parts count.

The only issue might be slight brightness changes in time using the
unregulated battery source  --might not be much of an issue depending on
your resistor choice --LED's are pretty non-linear (brightness to current).

I think I was remembering someone going through a similar situation here a
little while ago.

{Original Message removed}

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