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

active| newest | oldestthat's my circuit, but there is another version somewhere where i put the maths in to calculate the component values. The maximum output current, is 1.22/r, which with the componentS shown, would be 2.4 amps. The current

thanks rick i know where the wiper is but dont know where to hook up the wiper to the vout or adjust please help me out

ADJ - The wiper is the one with the arrow on it in your diagram.

no jason all i want to see is where do i hook up the wiper to Iout or Iadjust

The circuit diagram you attached shows

wherethe components are attached to one another.The values of these resistors,including the setting of the variable resistor, determine

howthe circuit will work, specificallyhow much currentit 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

whereyou put the resistors. Thevaluesof 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.

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 itplease if you know where i will appericate it thanks in advanceby 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?

http://cornerstonerobotics.org/illustrator.php

all explained in a diagram. all pots are laid out the same - wiper in the middle.

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.