LM350 regulator problem

[slfowler1]

Hi – I recently built a simple voltage regulator circuit using LM350, straight off the datasheet. (I’m looking to power a piece of equipment requiring 12.6V at 2.6 Amps.) I first tested it with about a 1 Amp load and everything was fine. I then hooked it to an automotive bulb (printed rating 27W). It stayed lit for a few seconds and then the chip’s thermal and/or current protection kicked in. I had the 350 on what I thought was a reasonable heatsink for the job, but to make sure I replaced it with a large 3”x4” heatsink and added a fan too. The bulb stayed on a few more seconds until it dimmed out. The big heatsink wasn’t that hot. The LM350 says it’s “guaranteed” to put out 3 Amps. What am I doing wrong?
- In a perfect world, 27W / 12.6V would be 2.1A. I understand that AC Watts may be derated based on efficiency, but I thought DC Watts equaled V x A? My ammeter shows just over 2 Amps while the bulb is lit.
- Do you think I might just have a bad LM350, or that there is some reason why it’s acting like a regular 1.5A LM317? Or do I really need gi-mongous cooling for such a seemingly modest circuit?
- Sure, I could redo the circuit with a pass transistor to share the load, but there would seem no reason to – the 350 says it can handle the Amps.
I would appreciate any ideas you might have. Thanks!
- Steve

[infinia]

What is the final application?
Bulbs tend to make poor loads due to variable resistance> unless current limit is set high in relation to bulbs cold resistance.

Remeasure all voltages with a 1A load and then again with the bulb. You're either in current limit or not enough V headroom? You must have around 3 V or more for dropout under full load.

The LM350K package has better thermals for heatsinking so you may need to claculate that at full load and worse ambient.

[macboy]

You didn't mention input voltage. This is important, as it will affect the power dissipated by the LM350; the power will be = (Vin - 12.6)*(I).

I would assume that if you ask the poor little device to dissipate more than 20 W, then it will shutdown quite quickly. So with a current of 2.6 A, that means a Vin of no more than 7.7 V higher than Vout, so Vin must be less than about 20 V. And if it was me, I'd want it even lower. Once you account for the heatsink and insulator thermal resistances, and add that to the 3 to 4 degC/W of the device itself (junction-to-case), you probably will have less than 20 W to play with.