LEDs and circuit board is hot

Thread Starter

salai ezhil Mathi

Joined Jan 24, 2017
20
Hello,

I have removed a LED circuit from a LED light battern, cut the board where 4 LEDs were in series. Batten is similar to this: http://www.amazon.in/Eveready-18-Wa...TF8&qid=1492661407&sr=8-9&keywords=led+batten I have connected 4 SMD LEDs to 12 V Lead acid battery(volt meter reads 12.5V). they are very bright and the circuit board is very hot to touch. The voltage across each of them is 3.1 V. I connected a 18ohm resistor. circuit is like this.

battery +ve - resistor 18ohm - LED1 - LED2 - LED3 - LED4 - Battery -ve.

Is this ok or should I put the resistor between the LEDs. Is there any other way, simple way to reduce the hotness of the circuit board apart from heatsink.
 

Wingsy

Joined Dec 18, 2016
86
Hello,

I have removed a LED circuit from a LED light battern, cut the board where 4 LEDs were in series. Batten is similar to this: http://www.amazon.in/Eveready-18-Wa...TF8&qid=1492661407&sr=8-9&keywords=led+batten I have connected 4 SMD LEDs to 12 V Lead acid battery(volt meter reads 12.5V). they are very bright and the circuit board is very hot to touch. The voltage across each of them is 3.1 V. I connected a 18ohm resistor. circuit is like this.

battery +ve - resistor 18ohm - LED1 - LED2 - LED3 - LED4 - Battery -ve.

Is this ok or should I put the resistor between the LEDs. Is there any other way, simple way to reduce the hotness of the circuit board apart from heatsink.
I don't think you have enough headroom for the circuit to be stable. Just a minor change in battery voltage will result in a significant change in LED current. I would split the LEDs into 2 parallel circuits. Connect 2 LEDs in series with a 300 ohm resistor (1/4 watt or more), and connect the other 2 in series with another 300 ohm resistor. Use 300 ohm if you want 20ma, a typical LED operating current, or use 200 ohm if you want them really bright (30ma).
 

Thread Starter

salai ezhil Mathi

Joined Jan 24, 2017
20
I don't think you have enough headroom for the circuit to be stable. Just a minor change in battery voltage will result in a significant change in LED current. I would split the LEDs into 2 parallel circuits. Connect 2 LEDs in series with a 300 ohm resistor (1/4 watt or more), and connect the other 2 in series with another 300 ohm resistor. Use 300 ohm if you want 20ma, a typical LED operating current, or use 200 ohm if you want them really bright (30ma).
small clarification Please. 300 ohm or 30 ohm?

there was one more circuit I saw that works with 12V lead acid battery. there were 3 LEDs connected with a 20 ohm resistor, all in series. what is your opinion on this circuit? will this work ?
 

Wingsy

Joined Dec 18, 2016
86
small clarification Please. 300 ohm or 30 ohm?

there was one more circuit I saw that works with 12V lead acid battery. there were 3 LEDs connected with a 20 ohm resistor, all in series. what is your opinion on this circuit? will this work ?
I need to remember not to offer advice at 2AM. :) You're right, 30 ohms would be more in order, and much higher power than 1/4 watt.

I think your 3-LED / 20 ohm circuit would be a much better solution than what you started with and better than my first response. You may have to do a little experimenting to get just what you want, but I'd go with the 3-LED/20 ohm circuit for starters. That 20-ohm will be dissipating around 1/3 watt so I'd use a 1-watt 20 ohm. You could probably just use your 18-ohm that you tried with 4 LEDs. In this case (18 ohms) your LED current would be about 0.15 amps. I would also expect it to get quite warm - your 3 LEDs will be dissipating about 1.4 watts.
 
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