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Feb 03, 2013, 10:21 PM
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AnthonysQuad's Avatar

Photocell and leds

Hello i have no electic skills but i managed to make this

I have
Two 3.7 - 4.5 leds blue
Four 3.3 - 3.6 leds white
I placed them inside a led holder and then placed that into a pen tube and cut a hole on the side for the wires to come out.I would like to know how i can hook up a photocell so when i fly my x4 mini quadcopter into the dark the lightswill turn on.
My battery is 3.7 volts.Someone please help.
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Feb 03, 2013, 10:38 PM
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AnthonysQuad's Avatar
Last edited by AnthonysQuad; Feb 03, 2013 at 10:53 PM.
Feb 04, 2013, 08:45 PM
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AnthonysQuad's Avatar
Ok i found this link it has alot on the subject but i just dont know how to read these blue prints.

I just want to solder things together without a board.Help please.
Feb 05, 2013, 09:27 AM
Dave the Rave
dmccormick001's Avatar
Your first problem is that you need a bigger battery to power your LEDs. The numbers you listed, 3.7-4.5 for the blue ones and 3.3-3.6 for the white ones, are the Forward Voltages (Fv) of the LEDs, that is the minimum voltage needed to make the diode (Light Emitting Diode) conduct, which is what causes it to put out light you see. You'll need about a volt more than the Fv of the LED with the largest Fv, so you really need about a 5 volt supply.

Now, second problem is you need some current limiting resistors. Since an LED is a diode, once it conducts, or turns on, it's like a one-way valve for electricity, so if you don't have some kind of resistor in the circuit, the LED will act as a short and the battery will try to dump ALL it's current through the LED, which will burn it and/or your wiring out. Never connect an LED directly to a battery without a current-limiting resistor! Each LED will need a resistor connected to either of it's legs, doesn't matter which one, a 100 ohm should be about right for your LEDs, assuming you use a 5 volt battery.

Now, lastly, I'd lose the photocell. It doesn't do what you think it does. When it gets light, it's resistance goes down, allowing current to flow. But in the dark it will hold the resistance high, which will prevent anything from turning on. You'd have to invert that action, which will require some parts and electronic skills I don't think you have. Just put a servo connector somewhere in your wiring, take an old extention and cut it in half, use the red and black wires, ignore the whire one, and use it to plug/unplug the LEDs when you want to use them.

Hope this helps!
Feb 06, 2013, 12:23 PM
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AnthonysQuad's Avatar
It is able to one led right from the battery but i did burn out a smaller led like that.
I dont know what will happen when i add the rest of the leds.I was planning on wireing all the leds to the battery.I have designed the led holder already so could i add those resistors to the wire instead of the leds so i dont have to reshape the led holder?
Last edited by AnthonysQuad; Feb 06, 2013 at 01:42 PM.
Feb 07, 2013, 09:28 PM
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AnthonysQuad's Avatar
Heres the leds im useing.The battery is 3.7 and foward supply is max 3.6.I didnt hook up any resistors because of the minor diffetence in those numbers.its also wired to the flight control board so im hopeing its not createing a short circut.Is it?
Feb 08, 2013, 09:27 AM
Dave the Rave
dmccormick001's Avatar
The problem isn't with too much voltage, the problem is with too much current. Once the LED conducts, assuming it ever does with a Fv that's so close to the voltage of the battery, it behaves much like any other diode (remember, an LED is a diode, not a bulb) and acts like a one-way valve to the current from the battery. With little internal resistance of it's own, and no current limiting resistor, the battery will be free to dump all it's current through the LED, which will burn it out much too quickly. It may work for a while, but it will eventually burn out, and will drain the battery way too quickly in any event.
Feb 08, 2013, 10:02 AM
srnet's Avatar
Vf for a diode is often quoted as a maximum value at a particular current.

What they mostly don't tell you is what the minimum might be, so the diode could have a Vf at its rated current of maybe 3.2v. The actual current when the Vf is 3.7v could then be a LOT higher.

And the '3.7v' battery is most likely a single Lipo, so its going to be 4.2V when fully charged.
Last edited by srnet; Feb 08, 2013 at 11:53 AM. Reason: More info

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