Reducing current into an LM317

[Helsing]

I want to use an LM317 to regulate a DC voltage between 60-70 down to 48v without the need for a heatsink. I say 60-70 because I want to accomodate variations in mains. I am using a 22-0-22 power transformer.

I will be powering 1 to two microphones. 40ma current capability would be sufficient. Indicator Leds are powered from a separate source.

Im thinking I should add a large series resistor between the filter cap and the 317 to limit the current into the regulator. Im thinking V/I=R 70/.040=1750 W=I^2 x R= 2.8watts

1750 ohm 3 watt resistor.

Am I on the right track here?

Helsing

 

[Winston OBoogie]

...

 

[PRR]

> Am I on the right track here?

No.

> DC voltage between 60-70 down to 48v.... 40ma current capability
V/I=R 70/.040=1750 W=I^2 x R= 2.8watts

So at 70Vin, and 40mA out, you have 0.040A*1,750Ω= 70V of drop in the resistor. That leaves 70V-70V= zero Volts at the input to the regulator. The output will also be zero volts. So what is the use?

> without the need for a heatsink.

A heatsink is often a better plan than a resistor. Cost and space of 1" aluminum or a 2.5Watt R isn't that different. Heat is the same either way. But the regulator is smarter, and won't waste voltage, or burn-up.

Say minimum supply is 60V, and minimum regulator dropout is 5V. (Yeah, it may be less in the data, but they get pretty goofy when not fed well.) SO we always want at least 48V+5V= 53V into the regulator. With 60V supply, we can lose 60V-53V=7V in a resistor. At 40mA, that is 7V/0.040A= 245Ω. That's the biggest resistor you can put in there and meet a 60V 40mA spec. With 70V input and a dead-short, output current is 70V/245Ω= 0.285Amps.

 

[Helsing]

Thanks so much for the wisdom. Got a lot out of both posts.

Im using the standard datasheet adjustment procedure for the lm317. Theres a 120 ohm resistor from output to the adjustment leg that I believe sets a minimum load so the thing wont shut down. Isnt that what its there for?

Helsing

 

[Winston OBoogie]

...