How easy is it to fry an LM317?

I'm trying to get an LM317 hooked up to generate 5.0V.

R1 = 1k and R2 = 3k,

Vo = 1.25 * (1 + R2/R1) = 5.0

Except that it doesn't. I'm getting Vo of about .25V less than Vi.

(I've tried Vi from 2.5V to 12V).

Any possibility that I've fried the thing?

 

Subject: How easy is it to fry an LM317?

This kind of thing happens if it's not hooked up correctly (View in fixed font
or M$ Notepad):

.---------.
| |
| o |
| |
|---------|
| LM317 |
| FRONT |
| VIEW |
| |
'---------'
| | |
| | |
| | |
| | |

A O I
D U N
J T

You can fry 'em if you hook them up incorrectly -- seen it many times. The
first time I hooked up an LM317 (back when the earth's crust was still
cooling), I wired it like a 7805, and got similar results. If you've got a
spare, best to substitute -- *after* you check your wiring. But who knows --
it might still work.

By the way, if you're using an R1 greater than the standard 240 ohms, you want
a minimum output load current. The data sheet says you'll probably want at
least 4 mA of load current. I'm not sure if that would cause your problem, but
you might want to check it, just in case.

Good luck
Chris

 

Is that voltage measured with no load? A 317 has a minimum load
current spec, and the 4K load of the divider may not be enough to keep
it down. Usually the divider is much lower in resistance for this
reason.

John

 

Jeff,

You need at least 7.5volts in to get 5volts out. Use a 240 ohm (or 270 ohm)
resistor for R1, and for R2 use a 2K potentiometer (one side and wiper only,
as in variable resistor). Tweak the potentiometer till the output is at
5Volts. Then turn power off, and measure the potentiometer resistance. Use
the closest standard value resistor to the potentiometer resistance (as R2),
or just keep the potentiometer in the circuit.

Quiquid latine dictum sit altum viditur

hth,
Joe