voltage balancer

[tubejim101]

I have two strings of LEDs in a single circuit.
They have the same number of LEDs in each string.
One string is brighter then the other.
I am guessing this is do to the resistance being different for each string,  making the voltages different.
Is there a way I can automatically balance the voltages, so both strings of LEDs shine the same brightness?


 

[armandas]

You should supply constant current to the LEDs. There are plenty solutions for that, just search for LED drivers.

[EDIT]

Check out the video here: exactly what you need.

[IanB]

I have two strings of LEDs in a single circuit.
They have the same number of LEDs in each string.
One string is brighter then the other.
I am guessing this is do to the resistance being different for each string,  making the voltages different.
Is there a way I can automatically balance the voltages, so both strings of LEDs shine the same brightness?

Do you have any resistances in series with each string? Current regulation is always required for LEDs and resistors are the simplest way of providing a measure of current regulation. If you are connecting your LED strings directly to a regulated voltage supply you are doing it wrong. Remove some LEDs and put a suitable resistor in their place.

If they are power LEDs running at 300 mA or more then resistors are not so suitable. In that case follow the advice of armandas and get a proper current regulated LED supply.

[tubejim101]

Ok.  Here is what I did.

Put a POT in series with the bright string.
Then dialed it till both LED strings looked the same.
Then took an ohm reading on the POT.  Came out to about 97 ohms.
So I jumpered in a 100 ohm resistor.  Looks great.
Now I am just running the circuit with the resistor soldered in, to make sure it does not heat up.
So far so good.
Thanks for all the tips.
 

[IanB]

You still may have a problem if you don't have a resistor on the other string. What is to stop them burning out with over current? (You have not said anything about your power supply, which could be one of the most important factors in the equation.)