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Topic: Calculating balast resistor (Read 242 times) previous topic - next topic

maker-20_

Jan 18, 2021, 04:26 pm Last Edit: Jan 18, 2021, 05:21 pm by maker-20_
Hey, I have 108 LEDs that are connected in parallel with 15 ohm resistor to each LED. I have 5V and 2.1A on it. That is not the right resistors to the LEDs if using 5v and 2.1 A. How can I calculate limiting resistor from battery to the LEDs when I already have resistor on LEDs. Each LED is 3.6V and 15mA.

Thanks!

WattsThat

Initially, you ignore the existing resistor.

(5 - 3.6) / 0.015 = 93 ohms. So you need 93 - 15 ohms in series = 78 ohm. 75 is a standard value. Power I^2R so .015*.015*75=0.016 watts. An 0805 part will be more than sufficient.

Why not replace the existing parts with say 91 ohm parts? How are you going to mount the additional resistors?
Vacuum tube guy in a solid state world

maker-20_

#2
Jan 18, 2021, 04:44 pm Last Edit: Jan 18, 2021, 06:07 pm by maker-20_
Okey! I have 100 Ohms on the latest, been using this tool https://www.amplifiedparts.com/tech-articles/led-parallel-series-calculator

I have some LEDs already mounted and that is the reason I want to reuse the old ones so thats why I want to use limiting resistor.


raschemmel

What's the forward voltage (the voltage across them) of the leds ?

DVDdoug

#4
Jan 18, 2021, 07:23 pm Last Edit: Jan 18, 2021, 07:24 pm by DVDdoug
Bad Design...

The 15 Ohm resistors won't have much effect and because of variations in the LEDs there's no guarantee the the current through all of the LEDs will be exactly the same, so the brightness may vary and there's even a slight chance that you'll blow one or more LEDs...

But here goes...    15mA through 15 Ohms is 0.225V.   Add that to the LED voltage and you have 3.825V across each LED & resistor.

That leaves 1.175V across the unknown resistor.

You'll have 108 x 0.015A = 1.62 Amps through that resistor so we calculate 1.175V / 1.62A = 0.725 Ohms.   (You'll probably have to use a 1-Ohm resistor.)   

1.175V x 1.62A is 1.90W so you'd need a 2 Watt resistor.   (The current, power, and brightness), will be a little lower with a 1-Ohm resistor.)   


maker-20_

#5
Jan 18, 2021, 09:18 pm Last Edit: Jan 18, 2021, 09:19 pm by maker-20_
What's the forward voltage (the voltage across them) of the leds ?
3.6V


Bad Design...

The 15 Ohm resistors won't have much effect and because of variations in the LEDs there's no guarantee the the current through all of the LEDs will be exactly the same, so the brightness may vary and there's even a slight chance that you'll blow one or more LEDs...

But here goes...    15mA through 15 Ohms is 0.225V.   Add that to the LED voltage and you have 3.825V across each LED & resistor.

That leaves 1.175V across the unknown resistor.

You'll have 108 x 0.015A = 1.62 Amps through that resistor so we calculate 1.175V / 1.62A = 0.725 Ohms.   (You'll probably have to use a 1-Ohm resistor.)   

1.175V x 1.62A is 1.90W so you'd need a 2 Watt resistor.   (The current, power, and brightness), will be a little lower with a 1-Ohm resistor.)   


Okey! Thanks for the explination, I have two LED panels that are 54 LEDs on each panel. Can I use one 1Ohm 1W resistor infront of each panel then? I do not have 2W resistor avalible... So I split everything in half, do you understand me?

MarkT

#6
Jan 19, 2021, 02:27 pm Last Edit: Jan 19, 2021, 02:28 pm by MarkT
Are you saying you already have a module with 15 ohm resistors?  If so what voltage is it designed for?
I'd guess 3.7 LiPo battery perhaps?

Then you need a 3.7V buck regulator to step down the 5V to 3.7V for best efficiency.

Or use your scheme and calculate the resistor from the voltage difference (if my guess
is right, 5.0 - 3.7 = 1.3V at 1.62A (108 x 15mA))
1.3 / 1.62 ~= 0.82 ohm (3W or more recommended)
[ I DO NOT respond to personal messages, I WILL delete them unread, use the forum please ]

maker-20_

Are you saying you already have a module with 15 ohm resistors?  If so what voltage is it designed for?
I'd guess 3.7 LiPo battery perhaps?

Then you need a 3.7V buck regulator to step down the 5V to 3.7V for best efficiency.

Or use your scheme and calculate the resistor from the voltage difference (if my guess
is right, 5.0 - 3.7 = 1.3V at 1.62A (108 x 15mA))
1.3 / 1.62 ~= 0.82 ohm (3W or more recommended)
That last is what I was looking for! Thanks, now I know how to calculate that!

Smajdalf

For 54 LEDs (single panel) you need twice the resistance (~1.5 Ohm)! The power dissipation will be half. 1 W may be enough but using two 1 W resistors would be better (I would use two 3R3 resistors in parallel). Of course it depends what value of resistors you have available but I doubt the exact value of current is critical: you can hardly notice difference in brightness at 10 mA or 15 mA. Just make sure the absolute maximum is not exceeded (also note the source may produce more than 5.0000 V).

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