# LM317 Current question

Hello , Can you inform me , while using the 317 as current regulator, the resistor dissipates all the power ? Or I can use a trimmer ? (to make a variable current regulator)

For instance if set R=1 ohm I out = 1.25A ( Wich a trimmer cannot stand )

Thank you

Yes, your "trimmer" would need to be of the multi-watt wirewound variety.

Oh, then I guess there's no special reason to use that for regulation at all since the wirewound resistor alone would do pretty much the same job

I think we need to see a schematic...

If the resistor is dissipating all of the power, the LM317 can't be doing anything and you are not actively "regulating" current... You are simply limiting the current and the actual current will depend on the supplied voltage and the load.

There's not much to see on schematic, on the out pin there's connected the resistor and after is connected the adjust pin.

I have made tens of voltage regulators positive and negative in lots of combinations using 1.25vref regulators but never made a current regulator..

I will try it these days but I see no reason to use a wirewound resistor and the regulator..

Well I have just remember once upon a time I used 317 with 1ohm 1watt to keep standard current for an audio preamp... and that resistor was quite hot and so it seems that a trimmer cannot in no way stand this current. I will probably use a dual gang 500 ohm potentiometer (wired in parallel) to regulate up to 5mA... Testing will solve the questions

I think we can assume that the schematics is the standard one for a current reg. with lm317, with Vout on one side of R and Vadj on the other side of R , then the current is Iout = Vref/R with Vref =1,25V (LM317) .
As Iout goes through R, then yes, R will dissipate RxIxI watts
You need this if you really want to regulate the current (a resistor alone is not enough). If you need different Iout values, then you could use different resistors and a rotative switch .

If the resistor is dissipating all of the power, the LM317 can't be doing anything and you are not actively "regulating" current...

That makes no sense.

The "resistor" must carry the current of the load... and the voltage drop across the "resistor" is what makes the LM317 regulate to keep the voltage "where it's supposed to" based on the impact of the load. So the LM317 is regulating, oh heck yeah it is.

But if the resistor is dissipating all the power, then the LM317 is dissipating zero power, and can't be doing much, can it?

Maybe you are seeing it wrong… It’s carrying the load current… The load current “through” the resistor creates a voltage drop. The regulator tries to maintain that voltage drop as a “constant” based on the internal reference voltage. Load current is carried by the objects in the load PATH… that means the resistor and the regulator.

Notice that the regulator is no longer GND referenced. It is referenced to the load side of the resistor that is used to measure the voltage drop at the desired load "current “setting”.

See the drawing. Rough estimates: 3 OHMS = 400mA constant current output, dissipation: 0.5W

So the power dissipated in the regulator is (Vin - Vout)/Iload. If that drops to zero, or some minimum based on the drop-out voltage, the regulator is not regulating anymore.

What would happen when you remove the "load"?

Also, the regulator can only adjust it's output voltage (to compensate for load) up to (Vin-dropout) or down to 1.25V... so yes, there are limits.

So the power dissipated in the regulator is (Vin - Vout)/Iload.

Power in Watt dissipated on the 317 is

P = V_317 * I

You may calculate it from this equation (see the schematics above):

12V = V_317 + ( I * R1 ) + ( I * R_load )

where: I is the "constant" current, V_317 is the voltage between input and output pin of the 317, the current from Adj is 50uA so you may neglect it.

What would happen when you remove the "load"?

No current will flow, no power dissipation on 317.

I agree with pwillard and thanks for a reference diagram.