I was using a post from in "How To" about wiring a garage door opener directly to the battery of the bike. I was taking pictures to post this variation on the forum since my opener used a 3 volt baterry, not a 12 volt, so...

I checked the exact voltage coming off of the openers 3 volt battery (2.22v)




Found a resistor that dropped the automotive (using my car at the time) 12.6v battery to 2.91v (resistor is going to the neg. terminal of the battery)




Took off the battery holder and soldered wires to the opener



used alligator clips to connet the system to the 12.6v car barrery and... SMOKE!

crap, I just fried a $30 garage door opener




. . . . . . What went wrong?

 

The resistor is supposed to go inlineon the positive side

 

Crap! I was wondering about that, but when I was wiring up LED's, a diagram showed the resistor on the neg. side. It worked fine for the LED's.
Thanks for the response petrorr. lesson learned!

Amps wouldn't be a problem, right? I don't want to waste another $30...

 

well, theres actually a lot of things that went wrong here. first of all, resistors dont drop voltage, they drop amps. im not sure why your meter showed 3v but what the resistor did was keep the voltage at 12v but less amps. so yes, the amps would be a problem, because it probably still didnt drop them low enough either.

you will need to make a different circuit since youre garage door opener requires a 3v source instead, using a simple relay. i will post a new how-to in a few minutes to show how i made my garage door opener activate when i flash my high beams.

EDIT: Look HERE.

 

on the halo lights i made for my bike im running 3.1v LEDs with a 220 ohm resistor in line on the positive side to each one of four LEDS....I used a 12v, 0.5A charger to test them before mounting them on the bike and wired them to my parking light with no problems...if I had connected them in series +-+-+-+- then I wouldnt have had to use the resistors cause the LEDs would create their own resistance...Maybe your using the right resistor, just on the wrong lead...Resistors drop the forward voltage comming from your power source to something more tolerable for the application...

 

well a resistor would work if you knew the current draw of the garage door opener when in use so you could use OHM's law to figure it out. I=V/R and then calculated out the correct resistor for your particular circuit.

In your case I probably would have used a variable voltage regulator and build a steady 3v out. More than likely I would have just soldered leads to a momentary switch to have a remote button and just leave the battery in.

 

delete me!

 

Im sure you figured it out, but what went wrong was you had to much current flowing..and that voltage drop on that resistor was just that, the voltage of that resistor, you still had the remaining voltage going through the door opener.. you need a bigger resistor..one that will drop like 8 volts..

 

A resistor does drop voltage when used in series, but it is a ratio of the resistance of the rest of the circuit. In parallel, resistors don't drop voltage, they are used to split current between the parallel lines. It really all depends on the rest of the circuit...

An LED is an interesting circuit design. It is considered a voltage drop in series, but takes close to negligeable current (in comparison to a resistor that would drop the same current). Anyway, it's not necessarily comparable...

Any battery has a theoretical infinite voltage, so you just dumb down the voltage. You need to spec the resistor correctly (look into ohms law) to ensure that you are getting one that is beefy enough to take the power that is running through it (usually specd in watts).

I would start by trying to measure the resistance through the opener (after the switch). As long as you're bypassing the switch on the opener (you need to do this... see why in the next sentence). The only problem there is if you apply constant power to the open switch, the resistance goes to 'infinite' meaning more voltage and therefore current would be applied to the switch in the opener, and it can fry the switch. If you apply power to the other side of the switch, then you're good. Just dont apply constant power there...

You can then spec out a resistor to use in series with the opener circuit (remember its just a ratio, voltage applied*(opener resistace/total series resistance) = voltage across opener). I would use the resistor before the opener, but it shouldn't matter.

There is a 'better' way, by using a voltage regulator, but I think that would be more expensive...

That's really f'n confusing, but I don't know how to explain it much better...

 

I've always seen and used resistors on the positive side to reduce voltage, never seen them on the negative side.