How to power LEDs from a battery using a resistor

JoakimFlorence

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I am going to embark on a somewhat controversial explanation here. There is good reason that this will be considered by many to be controversial, but the underlying principles are sound (though perhaps more in a theoretical way than being sound in actual practice).

Now first I will say that if you are powering LEDs from a battery, it is highly highly recommended you use a power supply, one designed for LEDs that will limit the current (constant current driver).

But it is theoretically possible to use a resistor, if careful calculations are made.


The first thing to understand about this type of circuit is that the (relative) power consumption will be entirely determined by the voltage drop of each component. Voltage dropis how much voltage each component consumes.

Since the resistor and LED are in series, the same quantity of current (amp) is going to be flowing through each of them. It will be impossible for one of them to have more current flowering through it than the other.

In a resistor (with a given resistance), the amount of current that is able to flow through it is going to be entirely dependent on the voltage. More voltage, more current. It's a very linear relationship.

For the LED, it's not such a simple linear relationship. An LED does actually behave in a way pretty similar to a resistor, but only over a very narrow voltage range (around its operating voltage ). If you go below this voltage, suddenly the LED will no longer be able to conduct current. If you go above this voltage, suddenly the LED will be able to conduct huge amounts of current and the LED will instantly burn out. It only takes a very tiny fraction of a second of too much power being taken up by an LED for it to burn out.


That is why in this particular discussion we need to point out there's a huge difference between a direct battery supply and a standard DC power supply driver. The DC power supply circuit is going to have voltage spikes. Even if it's rated for, say, 12 volts, that is really its average voltage , it's still going to have voltage spikes of 15 or 16 volts. Those voltage spikes can easily lead to the LED burning out.

For this reason, we are only going to be focusing our discussion here on using the power supply directly from a battery.

The first thing you need to do is to know the exact voltage of the battery. Not even necessarily its official rated voltage, but you should take out an instrument to actually measure it exactly. Let's just assume for this example that the voltage from the battery is 12 volts.

A typical white or blue LED will be rated for about 3.2 to 3.4 volts. (It's lower for red or green LEDs)
I'm just going to go with 3.3 for the sake of simplicity here.
Now if we have a string of 3 LEDs in series, those LEDs will consume a total of 9.9 volts. We should find a resistor to consume the remaining 2.1 volts.

Why do we want to use a resistor to consume the remaining 2.1 volts? Well, if we supply the LEDs with more voltage than their operational voltage range, the LEDs are no longer going to behave like normal resistors and will start passing through more current than they were designed to be able to handle.
If the voltage supplied to the LEDs is exactly within their operational voltage range, they will be self-regulating and be able to pass through the right amount of current that they need.

Because the operational voltage range of the LEDs is pretty narrow, some exact calculations need to be made here.


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Remember, to calculate the voltage drop across a resistor, find the current that will be flowing through a resistor, then multiply the current in amps by resistance in ohms to find the voltage drop in volts.
The current that flows through the resistor will be determined by the voltage it is externally subject to divided by its resistance value in ohms. (That voltage value will of course be the voltage from the battery minus the sum of the rated voltages of the LEDs)
Thus the voltage drop in this equation turns out to be the same exact amount as the voltage it is externally subject to. The resistor will automatically "eat up" as many volts as it is given.



For the sake of erring on the side of caution, you will likely want to try having the voltage across the LEDs at the lower end of their operation voltage range. For example, if their rated operational voltage range is 3.2-3.4v, you might want to keep it towards 3.2, or maybe 3.25.
This is going to cause some complication though in that the LEDs will be less bright than they are designed to be (maybe half as much) and (for blue and white LEDs) there may be some moderate degree of color shift as well, where the color hue of light will be just a little bit more greenish shifted. For most casual applications, this shouldn't matter at all, and you could always compensate for any loss of brightness by using higher wattage LEDs.

The other important thing you need to be aware of is that the voltage output of batteries decreases over time. As the batteries become more and more drained, their voltage output will decrease.

You could perhaps compensate for this by using a variable resistor. (Particularly a variable resistor in series with a standard resistor)
Thus as the battery loses charge over time, you could turn the variable resistor to reduce the resistance a little bit, reducing the voltage drop in the overall circuit to better match.

For a typical car battery, we're talking about a 1 or 1.5 voltage drop over the reasonable course of its life.
But be careful if you get a new battery or a battery that has recently just been charged. In that case that voltage might be 1 or 1.2 volts above normal for a short time, maybe a few days.
If the battery starts getting severely drained, like if you left a car light on in the parking lot, or the alternator starts failing, obviously there's going to be a problem and the voltage range of the drained battery might start falling below what the LEDs in this circuit need. (But presumably we are talking about decorative ornamental lights, so their reliability at all times in critical situations should not matter)

Now there's one huge huge problem here with using a car battery, and that is it's usually hooked up to an alternator (which is basically a little electric generator to keep it charged). That alternator is going to have voltage spikes.
What this may mean is you may likely need a second separate smaller battery, with some sort of switch to allow the alternator to be able to charge it, but not have that battery powering the LEDs at the same time it is being charged. If you want the lights constantly on, you may even require two of these separate batteries, and have them switched between alternately being charged and supplying the LED lights with power.


Now, for discussing efficiency. It's been said that resistors are not the best way to have a power supply for LEDs because they're not efficient and will waste too much energy as heat. That's true, but let's take a look at the exact calculations here.
In our example we had 9.1 volts being taken up by the LEDs compared to 2.1 volts being taken up by the resistor. That's a 17.5% efficiency loss due to the resistor. ( 2.1 divided by 12 ) The typical cheap constant current power supply (which would normally be used to power these LEDs) has an efficiency loss of around 15%. That's not a big difference.

Now, heat dissipation in that resistor may still be a big issue, if you have a lot of wattage going through the LED lights, that resistor can get quite hot. You should use one designed for heat dissipation. (For reference comparison, a soldering iron only uses 10 or 15 watts, and an electric incense burner only uses 35 watts)

Unfortunately, using this type of circuit design there's no way to use the resistor to limit the available current supply to the LEDs to very safe low levels without also becoming very inefficient. That would result in more energy being dissipated and wasted in the resistor than supplied to the actual LEDs. (Something that very likely would not be practical for a number of reasons) So it's a bit of a precarious situation.

There is one last thing I will point out as well. The possibility of using incandescent bulb lights as the resistor in the circuit for the LEDs. Incandescent bulbs are pretty much just like a resistor except the energy is given off as light, not just wasted as heat. So if you needed for there to be a resistor in the circuit, you might think might as well have that component be used to make some free light.
However, there's a big problem with this idea people might not think of. The filament in an incandescent bulb has about 14 times less resistance when it is cold before it is first turned on. In that fraction of a second before the filament fully turns on and glows to full brightness, the LEDs in that circuit (or at least one of them) would have burned out. There is probably some circuit design to get around this, but it would still require a the presence of a resistor in the circuit design and probably make things a little too complicated. There would probably have to be some type of switch that would turn the incandescent light bulb on through a separate circuit for a second, and then instantaneously reroute power through the glowing bulb to the LEDs. There would also likely have to be some sort of automatic shut-off in case the power supply got cut off to the bulb for more than 2 seconds, long enough for the filament to cool off.
Also, in some cases where an incandescent bulb burns out there can be a short circuit for a fraction of a second, before the built-in fuse in the bulb burns out, so there would be that inherent risk to the LEDs as well. I'm just pointing out this could be theoretically possible, not necessarily practical or worth the complicated circuit design.

Now you can get some idea how it is possible to power LEDs using just a resistor, but not necessarily very practical.

And I gave some disclaimer in the opening of the post that this would, in many ways, be more theoretical than actually practical.

It's actually so much more simple to just use an LED driver power supply.
 
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