Why can't we light up 4x 3V leds with a 12V power supply?

Question

To calculate the resistor to limit the current, if we use R=P/I² formula, it gives us wrong parameters rather than using V=RxI, so the 12V-12V is 0 and the P formula gives us nothing.

Why is it so?

Answer 1

Indeed, the formula is irrelevant when the sum of the led voltages is equal or nearly equal to that of the supply voltage. In this case a resistor is not necessary but you must be careful to always use in your calculations the smallest voltage in the voltage range of the led. Each type of led has a voltage range (look at their datasheet). For example 2.8V to 3.4V. In this case, you must count 2.8V. To simplify you can use 3V to match 12V with 4 leds in series.

The other thing to know is that the light output can be lower than expected and can vary with time and temperature conditions.

For non commercial, DIY applications it's acceptable.

Answer 2

LEDs are diodes. They allow current through while exhibiting a voltage drop that is mostly considered "fixed" (but isn't exactly).

The simplest way to limit this current through the LED is to calculate how much voltage you have left between it and the fixed voltage power source, and using a resistor (you chose the value according to Ohm's law).

In your case, the first problem is that 3 V LEDs aren't exactly 3 V so you could end up needing more than 12 V for the 3 LEDs; below that they could happen to not light up at all. Your second problem is that, in theory, since you have no voltage left, you can't use a resistor to control the current. Your issue with the formulas shows that.

The simplest solution I'd consider would be to split those LEDs into two parallel branches each powered by 12 V and each with a resistor. That way, you'd be left with 6 V to calculate the appropriate resistor value for the desired current. Just keep in mind that it's not super efficient and you should use at least 1/4 W resistors.

Answer 3

The first element to understand is that the Vf of the LEDs you use is NOT fixed. Look at this graph of Vf to forward current for various LED materials:

While you give no details of your particular LEDs, let's assume from the graph above the Vf characteristics will be similar to the "White" LED shown above. If you placed 4 of these in series and applied 12V, what might happen?

1. The LEDs will not have identical (it's called matched) characteristics, so at 12V the voltage across each LED will only be approximately 3V.
2. If the LEDs are reasonably matched, you might expect approximately 5mA of current through the string. NOTE: There is NO series resistor required to achieve this.
3. If your LEDs are rated at 20mA, then you could expect to be able to apply up to 13.2V WITHOUT any series resistor and still have the LEDs within their ratings (even though having no defined current management is a bad policy).
4. With your 12V supply set down to 10.5V you could still expect to have about 2mA current. Certainly enough current to be visible.

So to sum up, you could have a supply that varies from about 10.5 to 13.2V and have sufficient current flowing to have visible operation, and with NO series resistor. The LEDs wouldn't explode, but the parameters for operation are dependent on the characteristics of the individual LEDs, and it's unlikely these are matched for your LEDs. The end result is that you'd have difficulty in defining the operating (or maximum) current you want flowing in the string.

Designing you solution means you have to change your configuration:

1. With the LED characteristics as shown above you could break into two strings, each with a separate current defining series resistor.
2. If the operating current you want is about 20mA, then for two LEDs you would expect approximately 3.3V per LED Vf.
3. If your supply is 12V, then you will need a series resistor that drops 5.4V at 20mA. Since R=V/I, this gives 270 Ohms, and you'd use 1.4W resistors. Note: These are cocktail napkin calculations and you will get only approximately 20mA ...but good enough for most indicator purposes.

So your final schematic would look like this:

^{simulate this circuit – Schematic created using CircuitLab}

Answer 4

if we use R=P/I² formula, it gives us wrong parameters rather than using V=RxI, so the 12V-12V is 0 and the P formula gives us nothing.

Correct. The resistor value required is 0 ohms and it will dissipate 0 watts. These parameters are not wrong.

Just one problem - current will be determined entirely by the characteristics of the LEDs.

Here's the I/V curve of one example '3 V' LED:-

"But that graph only shows ~4 mA at 3 V. I want to draw 20 mA!". Too bad, you can't.

Or perhaps you can. Due to manufacturing tolerances, each individual LED will have a slightly different voltage drop, so by carefully selecting individual LEDs you might be able to get the total voltage drop you need. In this case the manufacturer sorts their LEDs into 'ranks'. Rank 2 is guaranteed to draw 20 mA at between 3.0 and 3.2 V, but even then the current could vary significantly between individual units. You would have to either sort them even more tightly yourself, or accept considerable current variation.

And that's only at 25 °C. White LEDs typically have a temperature coefficient of –3 to -5 mV/°C, so they draw more current as they get hotter. This LED's datasheet doesn't provide temperature coefficients, so you don't know what will happen at different temperatures.

"Oh hang on." you say, "When I said 12 V, I actually meant a 12 V auto electrical system". OK, now your LEDs have to work from as little as 11 V to as much as 16 V. At the lower end they may barely light, while at the upper end they could burn out due to excessive current draw - and brightness will vary dramatically whenever the supply voltage changes even slightly.

This is why white '12 V' LED modules always use 3 LEDs in series with a resistor rather than 4 LEDs directly. The lower number of LEDs gives plenty of voltage 'headroom' to ensure they can draw sufficient current, and the resistor greatly reduces current variations caused by LED characteristics and supply voltage.

Answer 5

We can light up 4 LEDs of 3 V using a 12 V battery. We can even light up 4 LEDs of the same forward voltage with a 5 V battery! The secret is to put them in parallel with each other and each LED will see a voltage of 5 V, issue solved and magic trick happened.

Thank you ladies and gentlemen, you were a great audience.