1) I have 6 4.5V 20mA LEDs and have connected them to a common 9V battery. Doing this I made 3 parallel circuits with 2 LEDs in each.

Now, I see that most pages that describe LED circuitry say that you should always have a resistor in the circuit. In my case, do I need one? Why?

This is a hobby project, but the LEDs are sealed in glue so fixing a blown bulb would be a big hassle. I'd rather have them live for as long as possible.

2) If I happened to have the same circuit as above, but with 3.4V 20mA LEDs instead, is it likely that the LEDs would blow quickly? The LEDs are of a cheap type:


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    \$\begingroup\$ NO Ls are 4.5V - where did you get that figure? See table on page you cited. You MUST use a resistor or constant current source. Note that 9V battery drops to ~=6V when almost flat. \$\endgroup\$ – Russell McMahon Feb 24 '12 at 18:17
  • \$\begingroup\$ The ebay page you linked shows that the white LEDs have a forward voltage of 3.2 - 3.4 V. What makes you think its okay to apply 4.5 V to them? \$\endgroup\$ – The Photon Feb 24 '12 at 18:45
  • \$\begingroup\$ And even if they were rated for 4.5 V, since the current is exponential you would still risk to push a too high current (and now you are doing that). A resistor is recommended in any case. \$\endgroup\$ – clabacchio Feb 24 '12 at 19:12
  • \$\begingroup\$ possible duplicate of What would the current be if I plug an LED into a perfect circuit \$\endgroup\$ – Kellenjb Feb 24 '12 at 19:38
  • \$\begingroup\$ I'm sitting here with a packet of 5V LEDs right in front of me. Plus a bag full of the ones in the ebay-link. Plus a whole range of other ones. That part of the question was related to which LEDs I should choose for my project! \$\endgroup\$ – Pedery Feb 24 '12 at 22:38

Ideally the LEDs should be driven by a constant-current circuit. This will maintain a constant brightness and color as the battery drains, or as the LEDs heat up or cool down.

But the real world isn't ideal, so you can often get away by just using resistors. Yes, you should definitely use them. The resistors are there to limit the current through the LED and keep them from overheating and burning up. A 9V battery has a fairly high internal resistance, so you may be able to get away with two in series and no resistor, but it will be unreliable (changing to a different brand of battery could be enough to blow out the LEDs, etc.)

For the worst case of two white LEDs in series running at 20mA, the lowest forward voltage shown in your link is 3.2V, so you would have (9 - 6.4)/.02 = 130 ohms. The current is low, so a 1/4 watt resistor will be fine. Select the closest value to this you can find. Running at 20mA the LEDs will be pretty bright and this is a benefit: as the battery drains or the LED forward voltage changes, the apparent brightness probably won't change that much. Human vision is more sensitive to dim lights and it's harder to tell that a bright light has changed 10% than a dim light has changed 10%.

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    \$\begingroup\$ Consider that he might have red LEDs, that have a drop of 1.8-2.2 V; you may add also the values for that kind of LED. \$\endgroup\$ – clabacchio Feb 24 '12 at 19:17
  • \$\begingroup\$ Nope, I have three types of LEDs - 3.4V, 4.5V and 5V. I think I'll go for the 3.4V ones since I have so many of them. \$\endgroup\$ – Pedery Feb 24 '12 at 22:42

Yes, you need a series resistor. The reason is that the LED voltage isn't exactly 4.5V, and it also varies with current. Having a series resistor is the only way to be sure you supply the right amount of current. You need some voltage drop across the resistor, however, so putting two LEDs in series is out. You need 4.5V across the resistor for 20mA so

\$R = \dfrac{9V - 4.5V}{20mA} = 220\Omega\$

A 9V battery can't supply very much current, but without series resistor the current may still be too much for the LEDs. I'm currently working with white LEDs with a typical current of 20mA, which have only 25mA as maximum.

edit (re your comment)
In theory you could do without resistor, but the LEDs are never exactly 4.5V and your battery is never exactly 9V. If the battery's voltage decreases to 8.9V your LEDs will light (much) dimmer, or (much) brighter if the LEDs' voltage is 4.4V, since the current will be much higher, and maybe become too high.

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  • \$\begingroup\$ Yes, but my question was related to TWO LEDs in series. That would make your equation R = (9V - 2*4.5V)/20mA = 0Ω. And this is henceforth the core of my question. Can I get away with not using a resistor in this case? \$\endgroup\$ – Pedery Feb 24 '12 at 22:45
  • \$\begingroup\$ @Pedery - Edited my answer. \$\endgroup\$ – stevenvh Feb 25 '12 at 6:53
  • \$\begingroup\$ Ah, I see. Resistors it is then :) \$\endgroup\$ – Pedery Feb 25 '12 at 16:34
  • \$\begingroup\$ @Pedery - Like others have commented 4.5V is very odd for a LED, so it's possible that the resistor is integrated. In that case the LED's voltage is lower and the resistor is calculated to bridge the difference between 4.5V and the LED's actual voltage. The best way to test this is to apply a variable voltage and decrease it from 4.5V. If at 4.3V or 4.2V you still have more or less the same brightness the resistor is integrated. \$\endgroup\$ – stevenvh Feb 25 '12 at 16:44
  • \$\begingroup\$ I see. One of my "odd" LEDs is a blinking one, so it probably has an IC. As for the other ones I don't know, and I don't have a variable power source. Anyway, this is a pet project, so no biggie. I'll go for the common 3.4V LEDs, add a resistor, and problem solved :) \$\endgroup\$ – Pedery Feb 26 '12 at 18:37

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