# What resistors are needed on a DIY LED panel if the power source is changed from 9v DC battery to 12v DC transformer?

I have a "star mural" which consists of 6 white LED lights (to power the fiber optic "stars") and 4 red LED lights of the "lazer cannons" of a space ship. (Yes, I'm a Star Wars fan, OK?)

Anyway, the mural is currently powered by a total of three 9 volt DC transistor batteries. Two of the batteries, each power 3 LED lights for the stars and one battery powers the 4 red LED lights. The white LEDs are wired in parallel, so they should be considered as follows: (One 9v DC battery powers three LEDs) x2. And as mentioned before, one 9v battery powers 4 red LEDs in parallel (with each set of 2 Red LEDs wired in series).

I've included photos of the resistors showing them up-close and from a distance, to see them in thier current wirring configuration. (see below)

The best I can ssume the color code to be for the star LEDs is: Brown, Black, Orange, Gold -OR- Brown, Black Brown, Gold.

For the Red LEDs it looks like: Brown, Green, Brown, Gold.

So, If I change from a 9 DC volt battery, to a 12 DC volt (transformed from 110v AC) do I need to change the resistors currently in place?

Sorry, if I am leaving out important information or not describing my situation clearly enough. I just don't want to fry my LEDs because I've over-powered them. Any help that can be offered would be greatly appreciated. Thanks!

## Discussions

7 years ago

Wow! Powered by three 9V batteries? That's funny! I'd be tempted to just take the whole thing apart and re-wire it. But you can sort of see what the intended logic was in designing it, right?

The group of three parallel white LEDs has about 3.6V across it, the battery has about 8V, so the 100 ohm resistor (brown, black, brown) has about 4.4V across it and thus a current of (4.4V)/(100 ohm) = 44 mA, and ideally that gets divided by three, into each white LED for like 14.67 mA in each white LED.

The group of red LEDs also has about 3.6V across it, only there it's two 1.8V drops in a series. Again the voltage drop across the resistor is about 4.4V. That resistor is 150 ohms (brown,green, brown). So the current going through that resistor is(4.4V)/(150 ohm) = 29.33 mA. Ideally that current gets split in half, to give like 14.67 mA through each red LED.

Anyway, if you were to increase the supply voltage from 8V to 12V, then that's increasing the drop across those resistors from roughly 4V to 8V, so you'd want them to be twice as big; i.e. increase the 100 ohm resistors to 200, and the 150 ohm resistor to 300.

But like I was saying at the beginning, I'd just be tempted to rewire the whole dang thing, because I don't like having LEDs in parallel. I like having a resistor in series with each series-string of LEDs. That's just because I don't trust LEDs wired in series to neatly divide and share the current going to them.

Answer 7 years ago

WoW! Thanks Jack. I was almost able to follow everything you wrote. The one part I didn't get was the part about calculating the voltage for the LEDs.

How did you get 3.6V for the group of 3 white LEDs and 3.6V for the 2 red LEDs?

Also, how did the 100 ohm resistor get 4.4V across it from the 8V battery?

The embarassingthis is that I took Physics and obviously didn't pay enough attention in class when it came to electric circuits.

I really appreciate the information and "refresher course." Thanks again.

-Gio

Answer 7 years ago

The trick for calculating what size resistor to use with an LED starts with assumptions about the characteristic forward voltage of the LED.

For the calculations I did above, I

assumedthe forward voltage of a white LED was 3.6 V, and the forward voltage of a red LED was 1.8 V. Also it is just a coincidence that 3.6 is exactly twice 1.8.For some reason I was also assuming the battery voltage was exactly 8.0V, instead of the nominal 9.0 V from the battery with that name. I guess I was assuming this 9V battery was being loaded a little bit, or maybe part-way dead, or maybe I just thought using 8 instead of 9 would make the math easier.

Anyway that's the way these LED resistor calculations usually go. You start with a source voltage, and an LED voltage, both assumed to be constant. Then that tells you the voltage drop across the resistor, V

_{R}. You have in mind a design current, I_{R}that you want to be flowing through the resistor. Then you divide these numbers to find R = V_{R}/I_{R}, and that last part is just Ohm's law.Here are some instructables that show you this same calculation:

https://www.instructables.com/id/Choosing-The-Resistor-To-Use-With-LEDs/

https://www.instructables.com/id/LEDs-for-Beginners/

I should just draw you a picture of what I was saying before, and attach it... well, I drew you picture, but the file uploader seems to be broken today, so for now I guess it's just words.

Answer 7 years ago

Here's that picture I was trying to attach yesterday.

Big version here:

https://www.instructables.com/file/F1HWBU9GVIYNFZA/?size=ORIGINAL

Not sure if sure if it's all that profound, but sometimes a picture helps to explain these things.