Modifying battery operated lights

[lineman]

Hi DLAers I am In need of some advice . First I want to use some halloween devices that run of batterys the kind when you push a button witches eyes light up that kind of stuff. What i want to know is there any way to modify these type of devices to use with my MR16 . The devices run of 2 or three double A's. I want to cut off the push button and wire to a MR16 channel

[dmaccole]

Jeff:

Relays. Either mechanical or electronic (SSR).

I'm pretty sure one of these would work: You are not allowed to view links. Register or Login

\dmc

[tpctech]

I used a MR16 board and Biglots battery operated LED lights last year.  The Biglot lights are a string of 20 LED and run off 3 AA batteries which is 4.5 volts.  I used a 2 watt resistor to drop the 12v down 4.5 v.  I am not near my work bench right now but I will get you the value a little later today.

KEN

[lineman]

I used a 2 watt resistor to drop the 12v down 4.5 v.  I am not near my work bench right now but I will get you the value

  Ok so all I have to do is solder a resistor on the + side . how do you calculate what resistor to use. I am new at MOD never messed with this stuff before any help would be thankful

[rm357]

The best way, if you have a multimeter is to hook up the lights with the normal batteries and use the meter to measure the current (Amps) that it is drawing. Set the meter for the AMPS scale, then put the meter in the path with the batteries (black lead on the positive end of the battery and red lead touching the wire that would hook to the positive end of the batery. Start with the highest amperage setting on the meter and work your way down until you get a good reading. A single LED is usually satisfied with 10 to 20 milliamps at about 1.5 to 3 volts, I expect your string is a mix of series and parallel strings that will likely draw in the 50 to 100 milliamp range, but I don't know. At any rate, once you know the current, just divide the voltage drop you want by the current and that will be the size resistor you need. Lets let the batterys are run down a little and they are at 4 volts.

If we use 100 ma:
12v - 4.0 = 8 volts that you want to get rid of. Just guessing 8v/100ma = 80 ohms. for the power rating, multiply the voltage and current - 8x100mili = .8 watts.


If we use 50 ma, the resistor becomes 160 ohm and .4 watts. A better option would be to hook three strings in series ...

[rm357]

There is another option as well. The MR16 works by controling the "grounding" - you do not have to use the positive power rail...

Cut the wire that goes to the negative end of your batter pack - conect the negative end of the battery pack to the MR16 input ground and connect the other end of the wire that you just cut to the appropriate channel ground connector (row closest to the transistors) leaving the pos connector unused. The 4.5v battery pack will power the LEDs and the the other battery will power the controller.

In my prior post, the power rating is the minimum - anything equal to or greater is fine, it's the ohms value that is important - too big and the LEDs will be dimmer than they should be, too small they can easily be burned out.

RM