LED Current

[chydes]

Apologies if this is too basic and obvious but I cannot find an answer anywhere.

My grandson is doing an electronics project using LED's and is using 4 white LED's in series. Each LED has: Vled 2.9v, Iled 20mA. He doesn't want to use a battery and I have 3 PSU's: 11v, 700mA; 12v,500mA and 12v 1.5A.

I am not sure which PSU to offer and what resistor (if any) to use. Using Ohm's Law what current should be used in the formula? Is the Iled the 20mA of the LED's in series?; 80mA of the combined LED's?; or the output current of the PSU, e.g. 500mA?.

It is many years since I last had to use Ohm's Law and never with LED's. My understanding has always been that LED's will pull as much current as possible so, intuitively, if a 500MA PSU is used the LED's would pull 500 mA and blow.

My grandson doesn't want to use trial and error to help keep costs down and I should like to be as helpful as possible and maintain my standing as 'the Grandad that alway has the answer'!

[bfinio]

Hi chydes - I am one of the authors here at Science Buddies, I have a video here that explains this in detail and might help: https://youtu.be/EeCh68a1GEg

Something not covered in the video about the power supplies though - the voltage and current ratings mean that the power supplies will provide a fixed voltage and UP TO the rated current. For example, if you just connect a resistor (forget about LEDs for now) to your 11V, 700mA power supply, it will provide 11 volts and you would calculate the current using ohm's law (voltage = current x resistance). If you get an answer that is greater than 700mA, that will exceed the ability of the power supply to provide current. The rating does NOT mean that the power supply will force 700mA through whatever you connect to it.

Another note - be careful with units. You need to make sure they're all consistent when doing ohm's law - volts, amps, and ohms. So in your case you need to convert milliamps to amps first (e.g. 500mA is 0.5A).

Hope that helps, please write back if you have more questions.

-Ben

[chydes]

Thanks for the info, it clears up the issue.

I had always assumed that the current provided a reserve to tap into but "My understanding has always been that LED's will pull as much current as possible" stems from something read and discussed many years ago and assumed it to be correct. From what you say it seems that LED's act in the same way as other equipment.

[bfinio]

To clarify - your understanding isn't that far off. If you connect an LED without a resistor - assuming the voltage supply is greater than the forward voltage drop of the LED - then it will draw too much current and blow out the LED. However, an LED does not follow Ohm's Law. Ohm's Law is linear (if you make a graph of current vs voltage for a resistor, it will be a straight line). The current vs voltage graph for an LED is nonlinear - it's very flat until you reach the threshold voltage, and then the current skyrockets - which is why you need the current-limiting resistor. See these graphs:

https://www.google.com/search?q=LED+cur ... 20&bih=944

The one exception is if you have a voltage supply with a very low ability to supply current, like a coin cell battery. For example, in this project you can connect the LEDs directly to the coin cell batteries with no resistor. These batteries can't provide enough current to blow out the LEDs:

https://www.sciencebuddies.org/stem-act ... D-stickies

[chydes]

Thanks for your patience and support, I think I may have now got it.

Taking one of the specifics mentioned earlier: if my grandson were to use a 12v PSU rated at 500mA with 4 LED's with Vs of 2.9v each and an Iled of 20mA connected in series, he would need to use a resistance of around 400ohms to protect the LED's from burn-out?

[bfinio]

Depends - is he putting the LEDs in series or parallel? Assuming series (which I think you said in your original post):

4 LEDs with a drop of 2.9V each, 4 x 2.9 = 11.6V drop total over the LEDs.
The current is still 20mA because the LEDs are in series.

(general rule here: voltages add in series and are the same in parallel; currents are the same in series and add in parallel)

To calculate the resistor value you use Ohm's Law, V = IR, rearrange to solve for R. The voltage drop over the resistor is the power supply voltage (12V) minus the total LED voltage (11.6V).

R = (12-11.6)/0.02 = 20 ohms

If you assume the LEDs are in parallel instead, then the voltage drop is only 2.9 and you get, but you need 80mA through the resistor (so that current splits and you in turn get 20mA through each LED):

R = (12-2.9)/0.08 = 113.75

You need to be careful about the power rating if you're only putting one resistor in series with all four LEDs in parallel like that, because the voltage drop over the resistor is much higher. Power = current squared times resistance, or P = I^2 * R. Most common resistors are rated for 1/4 watt.

P = 0.08^2 * 113.75 = 0.728, or nearly 3/4 watt, so that's going to burn out the resistor.

Based on your answer of around 400 ohms, I assume you did the calculation to have the 4 LEDs in parallel, each with their own resistor? In that case you only have 20mA through each resistor, and you get

R = (12-2.9)/0.02 = 455 ohms

which is close to your answer. That will work fine, but it's wasting an unnecessary amount of power due to the large voltage drop over the resistors (compared to putting the LEDs in series).

Sorry if that's confusing, but as you can see the answer depends on how you have the LEDs wired, so want to make sure we have that straight.

[chydes]

Thanks again - the daylight has finally dawned!

Your last comment has cleared everything up, so thank you for the time you have given.

Love your YouTube video too!

[bfinio]

Glad we could help! Please don't hesitate to write back if you have more questions.