Hello fellows. Need to pick your brain again. I bought one of those small led motion detector lights on ebay. It has nine led's. It was powered by 3 AA batteries. (4.5V) .

I mounted this light outside my back door to just light up the entrance of the door at night. Instead of replacing batteries all the time, I hooked the light up to a cell phone power supply I had laying around. Its rated output is 5 VDC @ 800 ma. Its been up about four months and only two of the leds light up now. From doing some reading on here I now realize that I need to hook up a diode and/or a resistor to limit the current, which I assume took these led's out. I'm still confused on how to figure these equations out. What do I have to do to get this light working correctly? (I ordered another light). Thank You for any guidance!

 

Check the cell phone charger and make sure the output is a constant 5 volts. The diode will help but my experience with these inexpensive devices is the LEDs probably would have burned out anyway.
EDIT: Thinking back now I remember one unit that had half the LEDs out. The problem was the LED current was set too high. Replaced all the series current limiting resistor.
SG

 

Thanks but this doesn't tell me what I need to do.

 

As a "standard" Silicon rectifier diode drops about 0.7V across it as current flows on the forward direction, no real calculation is needed. Just add a series diode, the Anode (non stripe end) connected to the 5V + wire, and the Cathode (end with the stripe) goes to the LED driver + connection. Then you will have around the 4.5V (near 4.3V) to be closer to the battery supply volts. The voltage drop is around the same over the current range, varying a bit but for your application, precision is not needed.
A 1N4007 diode, or just about any in the 1N400x series will do. They are 1Amp diodes and the final number just indicates their max reverse voltage rating.
Does that help?

 

As a "standard" Silicon rectifier diode drops about 0.7V across it as current flows on the forward direction, no real calculation is needed. Just add a series diode, the Anode (non stripe end) connected to the 5V + wire, and the Cathode (end with the stripe) goes to the LED driver + connection. Then you will have around the 4.5V (near 4.3V) to be closer to the battery supply volts. The voltage drop is around the same over the current range, varying a bit but for your application, precision is not needed.
A 1N4007 diode, or just about any in the 1N400x series will do. They are 1Amp diodes and the final number just indicates their max reverse voltage rating.
Does that help?

Dendad, that is exactly the kind of information I was looking for! Thanks a million.

 

Once you verify that the charger is verified to be putting out 5 volts, place three each 1N4001 through 1N4007 (whatever you can get) in series to drop the voltage by 2 volts.

Dick, why would you do that? You would then be at 3.0 volts correct? The light would be somewhat dim, no?

 

Thank you. When I spotted the error then saw that dendad had already given the correct answer I deleted my post.

 

And to work out a series resistor for an LED, R = (SupplyV - LEDV)/I
eg, for....
5V supply
15mA LED current
LED forward voltage drop varies with colour and LED type, but we can go with 2.2V for this example.....

R = (5-2.2)/.015
R = 2.8/.015
R = 186.7ohms
so I'd use a 180ohm or 220ohm resistor.

If your multimeter has a diode test, you can use that to measure the forward voltage drop of the diode and the LED.

 

And to work out a series resistor for an LED, R = (SupplyV - LEDV)/I
eg, for....
5V supply
15mA LED current
LED forward voltage drop varies with colour and LED type, but we can go with 2.2V for this example.....

R = (5-2.2)/.015
R = 2.8/.015
R = 186.7ohms
so I'd use a 180ohm or 220ohm resistor.

If your multimeter has a diode test, you can use that to measure the forward voltage drop of the diode and the LED.

Dendad, so your saying I need a 180 or 220 ohm resister in series with the diode that you mentioned before?

 

Dendad, so your saying I need a 180 or 220 ohm resister in series with the diode that you mentioned before?

No. That is just how you work out the resistor needed if you want to run an LED off any voltage. The existing board will already have resistors as needed. So just the diode is needed to drop the 5V to 4.3V

 

Thanks but this doesn't tell me what I need to do.

Yes it did. Check the voltage of the charger. Troubleshooting is done in steps. I wouldn't be surprised if there is only one current limiting resistor in the circuit. Good luck
SG

 

OK, fellows, I think I got it now. Btw, here is a pic of the light. Cheap but pretty bright. Thanks again