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# would like to know where to wire the 10k linear potentiometer to the Lm317 Answered

I'm trying to build a Led Constant Current driver for my led grow plants but I'm stuck on where i would solder the 10k linear potentiomter on the LM317
By the way this is what I'm using

10k linear pot
1kohms fix resister
2.2ohms resister
LM317

can someone please draw in color code where i would solder the 10k Pot wires to the LM317  please really appreciate it thank you

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## Discussions

The circuit diagram you attached shows where the components are attached to one another.

The values of these resistors,including the setting of the variable resistor, determine how the circuit will work, specifically how much current it will allow through the load, the series string of LEDs.

The part confusing to me, is you want to substitute different components, like a 10K ohm pot in place of a 500 ohm pot, and then naively expect the circuit to work the same.

It's not like the circuit depends only on where you put the resistors. The values of the resistors matter too.

Consider again the more simple circuit, with just one small resistor R1, and Iout = (1.25 V)/R1; e.g. (1.25 V)/(2.2. ohm) = 0.568 A. This more simple circuit is not adjustable by turning a pot, but it will work the way you expect it to.

thanks Jason

im trying to do this and i do have the 500ohm pot but wanted to use the 10k pot i tried getting a hold of maker but he seems lost and not found my only question is where do the leads of the pot meter go or how would i wire it please if you know where i will appericate it thanks in advance

by the way he does not use the resisters in parallel as you see in the pic of the breadboard thats what confuses me

LEDS used all 0.7mA:
6 x Red 630nm
3 x Blue 460nm
2 x Deep Red 660nm

for the constant current driver
2 x LM317
2 x 1.8 ohm resistors, 2.2ohm can also be used if you wanna be more safe
2 x pot-meters 500 ohm
2 x 1000ohm resistors

You can replace the 0R5 with the 1.8 or 2.2ohm resistor

For this circuit, a constant current source based on LM317 regulator, the math is easy when there is just one small feedback resistor R1 placed between Vout and Vadj. For that easy case, Iout = (1.25 V)/R1 + Iadj, where Iadj is small, typically 50 microamperes, usually small enough to ignore, so Iout = (1.25 V)/R1

For a circuit with a more complicated feedback network, like the circuit you posted, with three resistors, {R1, R2, R3}, one of which is variable resistor (also called potentiometer), the math becomes more complicated, as I am interpreting it.

Do you want to see the more complicated math?

If you have a multimeter,

https://en.wikipedia.org/wiki/Multimeter

you can use this tool to discover, with certainty, which wires of a potentiometer

https://en.wikipedia.org/wiki/Potentiometer

go where; i.e. which wire is the wiper terminal, which wires connect to the ends of the resistor, and also if the pot has the resistance you are expecting it to have.

And that is how potentiometers work.

However, I am still somewhat confused about what you are expecting a 10K potentiometer to do for your circuit.

Maybe you want to make output current adjustable between a minimum of,

(1.25 V)/(2.2 ohm) = 0.568 A [minimum]

and 11 times that amount,

= ((10+1)/1)* 0.568 A = 6.25 A [maximum]

I dunno. This would make more sense to me if I knew more about the goal of this circuit, like how much current it supposed to regulate.