# basic potentiometer wiring

theviperman
Member - Posts: 15
Joined: March 25, 2009
Location: Pennsylvania, United States
Posted: April 14, 2009 at 8:39 AM / IP Logged

Hey all,

I'm working on a project with some LED's.  I'm considering using a 1K, 3/4W potentiometer (technically a trimmer) to dial down the amperage in a parallel circuit and eliminate the need for ANY individiual resistors at each LED.  If I can dial the resistance up to about 630 ohms for 10 LED's, then I don't need 10 LED's.

My question is this:  How do I wire up a trimmer to fulfill this purpose?  It has three pins, with pin 2 being the "Wiper," but I'm getting conflicting reports of what to do with pins 1 and 3.  I want to use this on the POWER side of the 12vdc circuit.  I've read that pin 1 goes to positive, pin 2 goes to my load(s), and pin three to ground??  but doesn't that mean that between pin one and 3 I need to drop 12 volts?  That'd make the potentiometer/trimmer work like a load resistor, and at only 3/4W, it'd melt the thing to a puddle in roughly 5 seconds.

Can you guys provide some guidance?  Thanks.

Jeff

Don't mind me...
i am an idiot
Platinum - Posts: 13,466
Joined: September 21, 2006
Location: Louisiana, United States
Posted: April 14, 2009 at 9:30 AM / IP Logged
Wiring 12 volts to pin 1 and ground to pin 3 will not melt the potentiometer.  If the pot is turned to one end, it will provide 12 volts to your LEDs.  They will not like that.  With another resistor added in series with the potentiometer you can prevent this from happening.  What is the voltage rating of the LEDs?
KPierson
Platinum - Posts: 3,526
Joined: April 14, 2005
Location: Ohio, United States
Posted: April 14, 2009 at 10:10 AM / IP Logged

I, personally, would only use pins 1 and 2 and leave 3 unconnected.  This will turn the "trimmer" in to a variable resistor.  For best results find the absolute max amount of current you will ever need and put a resistor of proper value in series with the pot.  That way, if the pot is dialed down to 0 you won't blow the LEDs as the other resistor will limit the current to your absolute max value.

One thing you should consider doing is using a 7805 voltage regulator instead of 12vdc - this will allow the LEDs to maintain their brightness even as battery voltage moves around.

As far as hooking up power to pin 1 and pin 3 you will be turning the pot in to a load resistor - however at 1K ohms you will be well below your 3/4watt rating at 14.4vdc (.21 watts).  The pot will be fine.  Between Pin 1 and Pin 3 you will ALWAYS have the total resistance of the pot (so 1K no matter when the dial is turned).  The wiper is then resistantly proportional to the dial position (and inversing proportional between pins 1 and 2 and pins 2 and 3.  So, say you have 200 ohms between pins 1 and 2 you will always have 800 ohms between 2 and 3.  Another way to say it is the resistance between Pins 1 and 2 plus the resitance between pins 2 and 3 will always equal the total resistance of the pot (the resistance between pins 1 and 3).

Kevin Pierson
dualsport
Silver - Posts: 983
Joined: September 27, 2005
Location: United States
Posted: May 05, 2009 at 12:51 AM / IP Logged
Is your plan to put 10 LEDs in parallel and drive them all through this single pot?
If each of them take 15mA, that'd be 150mA through your pot resistor, and even if the forward voltage of all the LEDs were uniformly 3V, then the resistor needs to dissipate about 9V x 0.15A = 1.35W.
That's probably going to let the smoke out of the pot.
Why not wire strings of LEDs in series instead, and then trim the current with a fixed resistor? You probably could put 4 LEDs in series and add a small resistor for current limiting.
A more trick setup would be to use a current mirror to drive the LEDs, where you can use transistors to maintain a constant drive current to multiple strings, mirroring a control LED like the one on an alarm (flashing or whatever it normally does)

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