voltage regulator

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Lory

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need some help here.
I built a voltage regulator to drop 16 DC V to 8.25 DC V using a LM 317 but I am losing to much power to the heat sink. this is a portable unit.
Any ideas...?
Thanks for any info.
Lory johnson
 
You cant violate the Laws of Thermodynamics. The power dissipated in the regulator is (Vin-Vout)*Iload.
 
It may be hot but should be ok. Get a heatsink good for 5 or 6 degrees C per watt and it should run at 100 C. or so at 1.5 amps.

Just don't touch it. :=)

Post the whole circuit. Maybe we can cheat a little.
 
I was just hoping their might be a more efficient voltage regulator out there. the device I am using draws about 800ma and it seems like nearly the same amount of current is going into the heat sink. I was just hoping for a little more run time.

I will just keep the second battery on the charger and do it that way.

Lory
 
As MikeMI said, the power dissipation in any linear regulator is (Vin - Vout) * Iload. All linear regulators must follow this relationship.

The options are to reduce the voltage drop across the regulator or reduce the current. And use a big heat sink.

The alternative is to use a switching regulator. With the linear regulator, you're dissipating as much power in the regulator as in the load. A switching regulator can improve efficiency to 80 - 90%.
 
Thanks Jon
I looked into a switching regulater and it might be over my head but i have a guru friend that might help me build one.
Thanks
Lory
 
Hi Jon
I looked at #LM2596 and it seemed to output 5 volts only but I need 8.25 volts so I googled ( #LM2596 ) and looked around the net and found (#LM2576/LM2576HV ) which seems to have adjustable output voltage....looks good to me ..?.as long as it comes with instructions. at least it would be good for an experiment.
Lory
 
Excellent. Glad you found the right thing. I've ordered a number of items from Sure Electronics via ebay and have been very happy with them.
 
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