Power Supply for a LM317

YUP !!

Rheilly P

(Dont forget to read books)

Er, I don't know much about electronics, but isn't current the actual current being drawn and not necessarily the current that is available ?

RP would be correct only if the device being powered suddenly drew more than 1.5A.

Since you are only drawing 0.5A I can't see any problem, however depending on what you are testing there might be an inrush that could exceed 1.5A and then the 4A would come into play. A DC motor comes to mind where at start up current for a fraction of a second can be many times the run current.

I would think you could use a current limiting resistor in the 4A line and keep your hands warm this Winter.

I am NOT an expert on this so I suggest you wait for more definitive answers.

Dave

You will have no problem using the 16V wall wart with the LM317. The current rating on the wall wart tells you want it is CAPABLE of. In other words, it tells you the maximum power output of the device. In your case, you'll never need that much. Of course, the LM317 will draw a SMALL amount of extra power, and so you want a power supply that can provide GREATER than 0.5A. So you're in good shape with your supply and your IC.

*) On a related note, it is easy to increase the current carrying ability of the LM317. If you look at the application examples at the end of the datasheet, you'll see how you can add external pass transistors around the LM317. As the LM317 draws more current, it will signal to the external power transistors that they should provide that current. So the LM317 becomes a control circuit driving pass transistors that actually carry most of the current. *) On another related note, if you look at the LM317 schematic, you'll find that internally it uses a Darlington pair as a set of pass transistors that do exactly the same as what I just mentioned. It has a bandgap reference to hold the 1.25V, and that reference drives a buffer which drives the base of the internal pass transistor which delivers all of the power to the output.

--Ted

No, it isn't a problem. The LM317 will limit the current.

Regards, Bob Monsen

Not a problem. The load only takes what it needs - in this case, the "extra" 3.5 amps are sort of "in reserve".

's_law

Do be careful that at low currents, the wall wart's output might want to increase beyond the allowed input V of the regulator; some wall warts have poor regulation.

Have Fun! Rich

--- The 4 amps won't overdrive the LM317, but the difference in voltage between the input and output of the LM317 and the current through the load might.

For instance, with 16V from your wall-wart into the LM317, 5V out of the LM317, and 0.5A into the load, the LM317 will dissipate: P = IE = I * (Ein - Eout) = 0.5A * (16V - 5V) = 5.5 watts

The LM317 in a TO-220 package has a maximum thermal resistance (from junction to ambient, with no heat sink) of 50C per watt and a maximum junction temperature of 150C:

That means that with an ambient temperature of 25C and a maximum junction temperature of 150C , the maximum power which the LM317 can dissipate before it goes into thermal shutdown will be:

50C * watts 50C * 5.5W Pd = ------------- = ------------ = 2.2 watts 150C - 25C 125C With 5.5 watts being dissipated by the LM317, that means you'll need to use a heat sink to keep the junction temperature below 150C.

Which one?

From National's data sheet, the LM317A has a maximum thermal resistance, (Rtjc) from junction to case, of 5C per watt.

Assuming the LM317 is mounted to the heat sink using grease and has a thermal resistance through the interface (Rtcs) of 1C/W, Then we can write:

Tj - Ta Pd = -------------------- Rtjc + Rtcs + Rtsa

and solve for Rtsa:

Tj - Ta Rtsa = --------- - (Rtjc + Rtcs) Pd

150C - 25C = ------------ - ((5C/W) + (1C/W) 5.5W

= 22.7C/W - 6C/W

= 16.72C/W

Which is the highest thermal resistance required.

shows a likely candidate, and one which I've used in the past.

Nice heat sink.

JF

--
That\'s confusing; sorry.

What I meant to say was that that\'s the highest thermal resistance which
can be used.  Lower is better and will keep the part cooler.

JF

No. The amp rating is what you are able to connect to the supply before it damages the supply.

Voltage is the factor when being concerned about the connected device how ever, for safety matters and learning etc..., I have found it cheaper when working with youngsters getting into the field to use a series incandescent lamp rated for the supply or connected device in the event of a direct short. The lamp will illuminate at full brightness to indicate this condition and create a high resistance in the circuit there by protecting the components.. This also impresses the little kiddies

Of course, one could always use a selectable current regulated supply :)

"

--
The current rating of the load, don\'t you mean?

ha, A lot more than you obviously know about me..

I'm sorry you have the wrong impression of me John. I always placed you above most others here when it came to judgment calls how ever, in this case, maybe I am the one that made an error in judgment?

If i'm correct in this assumption, please place me in your ignore list because I don't need to see any disgruntle users here on account of me.

Have a nice day if you can stand your self.

"

:I have a 16 Volt, 4 amp wall I wart to use with a LM317 variable power :supply. The LM317 can only handle 1.5 amps, but the circuits that I :will be testing are rated at less than .5 amps. Will the 4 amp :overdrive the LM317 even though I will be using less than an amp? :Thanks

NO! The current required by the load device is all that will be drawn from any of the preceding power supply units even if those units have very high current output capability.

As an example, a telephone exchange battery has current capability in the thousands of amps range yet the small electronic devices being driven by that supply don't get fried.