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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?

schematic

simulate this circuit – Schematic created using CircuitLab

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  • \$\begingroup\$ I've updated the schematics as I saw a flaw in the connection points of the LEDs. Would any of these work? \$\endgroup\$ – Jesse Dunn Jan 11 '16 at 2:25
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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),

schematic

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.

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  • \$\begingroup\$ Well it's going to be a car battery that will be trickle charging when AC power is on. There will be a relay that turns the LEDs ON when AC power loss occurs. Correct me if I'm wrong, but I doubt there will be much of a drop in voltage only pulling 1.3A. \$\endgroup\$ – Jesse Dunn Jan 11 '16 at 2:30
  • \$\begingroup\$ Also, I'm a little confused as this doesn't appear to be a voltage divider at all. I thought you needed at least two resistors? \$\endgroup\$ – Jesse Dunn Jan 11 '16 at 2:31
  • \$\begingroup\$ Please note that the use of the LED symbol in my diagram is actually representing a string of 90 LEDs in series. \$\endgroup\$ – Jesse Dunn Jan 11 '16 at 2:43
  • \$\begingroup\$ If each "LED" requires 3.5 volts or so, it can't be 90 LEDs in series. 90 white LEDs in parallel is possible, but unlikely. as typical small LEDs would use about 20 mA (0.02 Amp), and 90 of them in parallel would want 1.8 Amps. \$\endgroup\$ – Peter Bennett Jan 11 '16 at 3:26
  • \$\begingroup\$ @JesseDunn - You don't drive LEDs by controlling voltage - that varies with temperature, among other things. You control current instead, as I have shown. \$\endgroup\$ – WhatRoughBeast Jan 11 '16 at 4:51
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No, your circuit definitely won't work.

What you need is a DC-DC converter (also called "switch mode power supply" or similar) with a 12 volt DC input, and ~3.5 volt DC output, good for 1.5 Amps or more. You should connect all the LED lights in parallel (with no extra resistors) to the output of the DC-DC converter.

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