I'm trying to supply power to strings of LED lights for emergency lighting
purposes with a 12v car battery. With the 12v supply I figured I could get 3 sets going at the same time.
The power supply that comes with them says:
INPUT: 120V-60HZ,0.06A – OUTPUT: 3.5 =-=-= 0.45A
For using DC only with a battery, I figured that a simple voltage divider would work. So in trying to determine the best value for the resistors in the divider I tried taking a resistance reading on the string of LEDs. I am unable to get a reading and at the 200ohms range setting, the meter is actually lighting the LEDs.
Here's what else I know:
The actual voltage coming out of the power adapter is 4.3v without the LEDs connected and only 3.2v with them connected. The resistance of the DC side of the adapter is 430 ohms.
So, am I right in assuming the following circuit would work?
simulate this circuit – Schematic created using CircuitLab
Best Answer
Your reading of 430 ohms does not reflect the current limit of the supply, although I don't know exactly what it does show.
Assuming 3.2 volts per LED at 0.45 amps, if your battery really is a 12 volt battery at that current (which I doubt),
simulate this circuit – Schematic created using CircuitLab will work - maybe. The 3 LEDs are in series, so they all have the same current. Assuming 3.2 volts for the LEDs, and 12 volts for the battery, this leaves $$V_R = 12 - 9.6 = 2.4 \text{ volts}$$ Then, in order to get a current of 0.45 amps, $$R = \frac{V}{i} = \frac{2.4}{0.45} = 5.33 \text{ ohms}$$
The reason I say "maybe" is that batteries don't usually have exactly the stated voltage, and it varies with both load current and state of charge. I'd suggest starting out with a 10 ohm resistor and get a feel for the circuit.
Oh yes, and you'll need a decent power rating on the dropping resistor. The power dissipated will be the voltage times the current, and I leave that for you to calculate. But it will certainly be more than 1/4 watt.