3
\$\begingroup\$

Okay, I'm rather a bit confused here now. I am trying my first real circuits, for some LED lighting. I tried making a little circuit last night, matching two LEDs in parallel on 4.5v of battery. However I didn't make any accounting of the amperage. I had no resistors in the circuit and the LEDs quite literally exploded, plastic shrapnel flying around the room!

So today, I looked up the formula to figure out how many ohms of resistance you need, picked up some new LEDs and a couple of resistors to try again. Here is what I figured based on my limited knowledge so far...

My new circuit consists of:

  • 3V, 2xAAA batteries
  • 2xLEDs, red, in parallel; 1.8V, 20mA

I plugged into the formula

R = (Vsource - Vdraw)/A
R = (3 - 1.8)/0.040 ==> R = 30

Vdraw = 1.8 because the LEDs are in parallel, so V stays the same
Amps = 0.040 because the LEDs are in parallel, so A sums up.

This led me to a result of needing 30Ω . So I bought a pack of 15Ω resistor, planning to put them in series in the circuit. Here is the circuit I planned:

schematic

simulate this circuit – Schematic created using CircuitLab

I felt pretty confident in my calculations. When I put it all on the breadboard, I forgot to put the resistors in though before connecting the power source. As I connected the power source, I saw the LEDs light up, and realized I'd forgotten the resistors. I expected another spray of plastic shrapnel. HOWEVER - the LEDs just lit as expected, and there was no explosion. I then put in the resistors, and it did not change the output of the circuit at all. The LEDs still lit, and did not seem any dimmer for the resistors being there.

So apparently, 30Ω is not enough to worry about blowing an LED.
So my question is, how many ohms DO I need to worry about before adding resistors?

\$\endgroup\$
  • 2
    \$\begingroup\$ The failed experiments are at least good experience. You know you are doing something wrong. This type of question has been asked many times before. Try searching through this site and if something is not clear, post a question with what is not clear. \$\endgroup\$ – efox29 Jul 18 '15 at 21:11
  • 4
    \$\begingroup\$ You'll also learn why paralleling LEDs is not a good idea. \$\endgroup\$ – efox29 Jul 18 '15 at 21:13
  • \$\begingroup\$ Indeed! B) It's nice to know when something doesn't work and why; now I'm in the annoying position of having something work and NOT knowing why! LOL \$\endgroup\$ – eidylon Jul 18 '15 at 21:13
  • 1
    \$\begingroup\$ It's the best part I think. If something works, you learned nothing. You only applied what you knew. If something doesn't work, well, now you have an opportunity to learn. \$\endgroup\$ – efox29 Jul 18 '15 at 21:14
  • 3
    \$\begingroup\$ You also need to know that a real voltage source has an internal resistance, limiting the current! \$\endgroup\$ – d3L Jul 18 '15 at 23:00
4
\$\begingroup\$

You said that in the first test, where the LEDs blew up, you used a 4.5 volt battery. That would force MUCH more current through the LEDs than the 3 volts you used in your second test.

The 3 volt test would still be forcing much more current through the LEDs than their maximum rating, and may greatly shorten their life.

What you should do is:

schematic

simulate this circuit – Schematic created using CircuitLab

Since LEDs are not guaranteed to have identical forward voltages, if you connect two LEDs in parallel, with a common series resistor, one LED will draw more current that the other, so will get a little warmer, so its forward voltage will drop slightly, making it draw still more current.... which may cause one LED to be destroyed (not too likely) or one LED to be significantly brighter than the other.

\$\endgroup\$
  • \$\begingroup\$ After efox29's comment above, I went and found this question and was wondering then what the proper way to resistorize the LEDs would be. Your diagram is what I was guessing... cool. So you actually can parallelize diodes, it is just better to give them all their own resistor instead of sharing, yes? Or even with individual resistors is parallelizing still frowned upon? \$\endgroup\$ – eidylon Jul 18 '15 at 22:10
  • \$\begingroup\$ So the resistors also, while not apparently necessary in this particular circuit, would theoretically extend the lives of the LEDs, yes? \$\endgroup\$ – eidylon Jul 18 '15 at 22:13
  • 1
    \$\begingroup\$ A recommended practice for multiple LEDs is as I show above - a resistor in series with each LED. Any number of these resistor/LED combinations can be connected in parallel (subject to power supply limitations). Connecting multiple LEDs in parallel, with a single series resistor is not recommended. If you have a high enough power supply voltage, you can connect several LEDs in series, with a single resistor. \$\endgroup\$ – Peter Bennett Jul 18 '15 at 22:15
  • 2
    \$\begingroup\$ You ALWAYS require a resistor or other current-limiting device in series with an LED. Cheap LED flashlights often depend on the internal resistance of the battery to limit the current, but that's not recommended for most applications. \$\endgroup\$ – Peter Bennett Jul 18 '15 at 22:18
  • \$\begingroup\$ I was wondering about how that worked... my very first project I did with some lighting, I used the LEDs out of Dollar Store flashlights, and there were zero resistors in the circuits... but they still worked. That kinda set me off on some false assumptions as I started out. LOL \$\endgroup\$ – eidylon Jul 18 '15 at 22:20
1
\$\begingroup\$

There might be times when your LEDs survive even if you give them more voltage or current than they are rated for.However,this will most likely damage them.It is best that you stick to their respective specs.I destroyed a few LEDs with a small bang in the past,connecting them directly to a 9 volt battery without adding something in between.I would never expect LEDs like those you used to be fine in my configuration.As the others stated,it's not recommended to put LEDs in parallel with a single resistor,not if they are not the same.I connected two like that in the past and one was bright and the other was completely off,while the goal was to have them both light up.

\$\endgroup\$
1
\$\begingroup\$

A battery has an internal resistance, and in this case that resistance was enough to limit the current to a "safe" value. But that was just luck.

\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.