Question on the LM317

[Coyoteboy]

I'm using one of these set as a constant current regulator. The
datasheet says 2.2A max, internally limited by thermal shutdown.

If I try to pass more than that, despite it being very well heatsinked,
will the component shut down completely or will it just refuse to pass
more current? If i shuts down completely it will mess up my timing
issues, and I'll use a resistor. If not I'll not use the res to grab
every last ounce of power it can pass.

Cheers
J

 

[JeffM]

The datasheet says 2.2A max, internally limited by thermal shutdown.

If I try to pass more than that
J (Coyoteboy)

Max means MAX.
Add a pass element.
http://groups-beta.google.com/group/sci.electronics.design/msg/304aaaf0f957db07?&fwc=1

 

[John Musselman]

J,

Depending on your circuit and load profile, you should be able to find a way
to reduce the power dissipation in the part, especially good heatsinking and
putting a power resistor at the input of the regulator, reducing the voltage
into the regulator and the power dissipation in the regulator. Makes the
part run cooler.

BUT...

2.2A is the maximum allowed. If you draw more than that the operation of
the part is "not guaranteed". I'll guess that you can get a little more but
not much. Drawing too much will cause the part to fail, not from heat but
from too much current concentrated in one location in the silicon.

If you need to draw more, use a part that can handle more current.
Especially, look into a switching regulator.

John Musselman

 

[Landon]

As was previously said, use something that can handle the load, maybe
there is a way two paralelle two of them together.

-Landon

 

[loedown]

Use a current pass transistor, that will increase maximum current
throughput, although, not through the LM317

Paul

 

[John Musselman]

Read the data sheet. If you try to draw more than the part can
handle, the device will _limit_ the current through itself, precisely
for the purpose of keeping itself from failing.

I stand corrected. Sorry for my hasty reply. The part has thermal current
limiting AND short circuit current limiting. So to answer the original
question, if you try to draw too much current OR if the part gets too hot
from whatever current you are drawing, it will current limit.

By the way, the data sheet has an example of paralleling multiple parts.
But I would suggest a switching regulator if your circuit can use it.

John Musselman

 

[Coyoteboy]

I'm already paralleling them I was hoping to grab about 2A out of
each one - they should cope i *think*, with a lot of cooling (which
they have, fan cooled heatsink). I'll give them a shot, if they go pop
I'll replace them lol.

My main confusion was the difference between current limiting and
thermal shutdown for power limiting i guess.

Cheers
James

 

[John Musselman]

James,

Since it's a constant current circuit, it should be simple. Take a look at
Figure 4 in the datasheet. (I'm looking at the Texas Instrument datasheet).
Just one resistor to set the current. Multiply that circuit by the number
of sections you need. Make sure you have a big enough resistor (1.2V times
2A = 2.4W, better use a 5W)!

Don't use the other example of parallel parts (Fig 13), it's for constant
voltage output.

John

 

[Coyoteboy]

Cheers John, will follow that diagram and bought 10w resistors as i
couldnt get 5's to save my life - should be safe

Cheers
James