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1w LED (multiple; 6-7) fade in and fade out ? Answered

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Heya fellas! Is there any chance that someone could help me out with this task? Almost nailed it for the weaker ones (5 mm) but for some reason i'm losing voltage really hard when i'm connecting them in parallel (12V supply) and it seems that some voltage is lost before that as well because when i compare it to the one simply connected directly to the 3.7v power supply and one on the board - the first one is much brighter than the second one.
And if I try to add 1w one after 5mm - it just won't light up and almost nulifies all the brightness of the 5mm one before it on the board.

My English and explanations might be pretty bad because i'm completely new to this and have a reeeeeally limited time before the event i'm planning to create this thing for (cosplay event to be precise).

So I think that i've got less time now for anything else than a simple direct connection of 6-7 1W LED to 3.7V supply (to make it real bright for every piece of my lantern) but i really want to make that fade in and fade out feature.

I saw some really helpful things at this place before and I really do hope that someone will help me out and get lots of + to their karma and will be mentioned when asked (about LED scheme and help) for sure.

Scheme i've been working with is attached to this post. And if you'll need some extra info - i'll be more than happy to provide it ( i just don't know what is required to be honest)

Thank you for reading!
Best regards,

1 Replies

jespozBest Answer (author)2017-09-20


It's hard to answer if I don't have the component's specs, but I'll try to give you some advice.

1. You didn't mention if you connected resistors when connecting the LEDs in parallel, so probably you're extracting too much juice ( current) from your power supply, and it responds lowering the voltage. If you're connecting 7 LEDs in parallel, the current will be at least 30mA x 7 = 210mA, assuming that the operation voltage of the LED is 3.7V, i. e., the same voltage of the power supply. For example, if the LED as a forward voltage (from datasheet) of 3V, a current of 20mA and a power supply delivers 3.7V, the resistor will be: (3.7 - 3) / 0.02 = 35ohm, in series with the LED.
2. In your circuit, the resistor is too high for a 1W LED. If we assume the current of 350mA at 3.2V, the saturation voltage of the transistor (voltage drop) of 0.2V, and a power supply of 12V, we have for the resistor: (12 - 3.2 - 0.2) / 0.35 = 25ohm for the LED at full power. That's why the 5mm LED looks brighter than the 1W LED. But this is not all!. Is impractical to use a 25ohm resistor for this, because of the power dissipation. In a resistor, the power can be calculated as I^2 x R, so 0.35^2 x 25 = 3W, which is too much for a resistor. Typical resistors only dissipate 0.25W, but there are models for more power.
3. The correct way to use a high power LED is connecting a "driver circuit" which limits the current and avoid blowing the LED. Take a look at https://circuitdigest.com/electronic-circuits/1w-led-driver-circuit-diagram for a simple example. The circuit in the link should be placed at the 12V point on your schematic and without the 470ohm resistor.
I hope it'll help you.

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