piclist 2001\11\09\061724a >
Thread: PIC overheating a 7805 !
picon face BY : Russell McMahon email (remove spam text)

Tobie said -

Hi, I've got this system up and running where a series of modules containing
16F628 (running @ 20MHz) and individually fed by LM7805 (TO-220 package) are
all run from the same 24VDC line (long run) coming from a switching supply.
Each module also has an LM7806 for driving an R/C Servo motor. All my
modules work fine except that the 7805s are running rather hot even though
the R/C motors are pulling much more current from the 7806 than the little
PIC is from the 7805. The 7806's temperature is ok though. At first I
thought that it would only be necessary to heat-sink the motor's regulator
but know I think I might have to do the same with the PIC's. I can almost
burn my finger from the 7805 ...

I understand that 24V isn't the optimal voltage to run the 7805 from if I
intended to draw a lot of current from it but National's datasheet says that
it can take up to 35V and considering the very low current consumption I
don't understand why it is overheating. I suspected the length of my supply
line but running a module even a few minutes on a short length gives the
same result. Besides, the voltage drop is barely perceptible and I have
pretty good wire gauge. Could my choice of caps for the regulators be
causing the problem ? I have one 470uF 35V electrolytic before the 2
regulators and a 0.1uF ceramic cap after each of them.


It doesn't SOUND like it makes sense.
IF the 6v supply IS drawing more current from the same input then it DOESN'T
make sense.

- As others note, ensure the regulator is not oscillating. A scope helps a
lot for this.
Make sure caps are per the databook.

- A 7805 / LM340 in a TO220 package is good for about 1 watt (preferably
less) without a heatsink.
At 24 volts in, 5v out this gives Current = 1 watt / (24-5) = ~50 mA.
A PIC by itself will draw much less than that.

- Check that the 5 v supply is not drawing more than you think.

- A series resistor will take the load off the regulator.
eg at 100 mA and 15v drop a 15/.1 = 150 ohm resistor will dissipate V^2/R =
225/150 = 1.5 watts and allow the regulator to dissipate
(24-15-5) * 0.1 = 0.4 watt.
If you place a 150 ohm resistor in series and the regulator can't support
the 5v output then it is drawing rather more than 100 mA.
Note that you need a larger decoupling capacitor at the regulator input if
you use the series input resistor.
Using a 5 watt wire wound here will move most of the dissipation out of the
If the 24v supply varies to below 24v you will have to design the resistor
for the lowest input voltage.

Rseries = (Vin_minimum - 5v - Vheadroom) / (Imaximum)
V headroom is the minimum voltage you want across the regulator. For a 7805
2.0v is safe.

How's the Autonomous Boat coming?


           Russell McMahon

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See also: www.piclist.com/techref/power.htm?key=7805
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