I am newbie to electronics. Basically I developed a mini circuit just to understand concept of electronic circuit. What I did was I developed a circuit with 3Volt AA battery and 1 LED. The LED lighted up. When I increased the number of LEDs to two, the second LEDs didn't light up.

So, I increase the number of battery so that it became 6Volt. The two LEDs light up fine. Then I added the third one, but only two light up and the third one didn't. Also I noticed that the brightness of the light is decreased when I add the third one.

For your information, I connected the LED in series. My question is, how can I determine the number of LEDs can be used for specific volt? I understand that LED is not a resistor, so why the volt dropped from LED to LED in series and how to determine this?

I did know the V=IR thingy. Thanks.

  • \$\begingroup\$ The data sheet for these LED devices is needed. Some devices that are called LEDs are not pure diodes, but actually integrated circuits which contain a LED (or several) and packaged like a LED. For example, some LEDs contain oscillators and are able to flash without any external circuitry. \$\endgroup\$
    – Kaz
    Jul 2 '13 at 23:35

The \$ V = I \cdot R\$ "thingy", as you call it is Ohm's Law. A very important one.

LEDs cause a pretty constant drop which, like Malife says, depends mainly on the LED's color, and also varies a bit with current. This chart shows you that all visible light LEDs require at least 1.8V. A red LED will drop about 2.2V, so like you saw it can be powered from a 3V battery. Two LEDs in series require at least 4.4V, so it won't work with the 3V battery, but the 6V is OK.

That three LEDs are weird. You say two light up faintly and the third doesn't. A LED's luminosity is determined by the current, and the same current passes through all LEDs so they should all three light evenly. The only explanations I can think of is that the third one may be defective, or it may be an IR LED. Though a LED which fails because of too much current will usually be open, not shorted. Also a shorted LED shouldn't decrease brightness.
LEDs are extremely sensitive to ESD, and that may have caused the dead LED. If you don't have any other ESD protective tools, touch a large metal object before handling your LEDs.

Now there's a big mistake in Malife's schematic, and that's the absence of a resistor. You'll have a difference in voltage between the LEDs and the battery. For the two LEDs that will be about 6V - 4.4V = 1.6V. You have to do something with that, if you connect the three just like that there might flow a very large current which can destroy your LEDs. So you place a resistor which will handle the 1.6V. Since you know Ohm's Law you can calculate the resistor's value if you know that a typical indicator LED needs 20mA:

\$ R = \dfrac{V}{I} = \dfrac{6V - 2 \times 2.2V}{20 mA} = 80 \Omega \$

For the single LED this would be

\$ R = \dfrac{V}{I} = \dfrac{3V - 2.2V}{20 mA} = 40 \Omega \$

enter image description here

It doesn't matter in which order you place the LEDs and resistor.

If you haven't used a resistor in your experiments and the LEDs didn't go up in smoke it's probably because the battery can't supply too much current.

edit (after your comment)
Next to Ohm's Law there are also Kirchhoff's Laws: Kirchhoff's Voltage Law (KVL) and Kirchhoff's Current Law (KCL). KVL says that the sum of all voltages in a loop is zero. In our case, the battery's voltage is equal to the sum of the voltages over LEDS and resistor. (The voltage over a component is often called the voltage drop over it.)
In the schematic above we start with 3V at the top. The LED "drops" 2.2V, so the voltage at the cathode is 0.8V. There's only the resistor left before we arrive at 0V, so that 0.8V is the resistor's voltage drop.
For more than 1 LED start at the battery's positive contact and walk through the loop, subtracting voltages as you pass components, until you arrive at 0V when you return to the battery.

  • 1
    \$\begingroup\$ Some on them did smoke up lol. Thanks for your explanation. I even tried up to 9V. Basically, the conclusion is that I could draw from my noobish experiment is, i require 3V for a white LED to light up in a series. I tried up to 9V and managed to get 3 leds lighted up. I am still new and trying to figure out the best way to fit in resistor. thanks for your guide again. \$\endgroup\$ May 10 '12 at 6:29
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    \$\begingroup\$ @Haikal - Be sure to have enough voltage drop over the resistor (you need a resistor!), otherwise a small change in voltage may cause a big change in current. \$\endgroup\$
    – stevenvh
    May 10 '12 at 6:31
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    \$\begingroup\$ @stevenh Can you explain to me why resistor after the diode, not before? \$\endgroup\$ May 10 '12 at 6:42
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    \$\begingroup\$ @Haikal - I said in my answer that the order of the components doesn't matter. You can place the resistor at the anode's side if you want (it's often done that way). With two LEDs you can place the resistor between the LEDs if you want, though that's rarely done. \$\endgroup\$
    – stevenvh
    May 10 '12 at 6:45
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    \$\begingroup\$ I couldn't ever imagine LEDs sensitive to ESD, I've pumped 8 amps through them for short periods, ran roughly a hundred volts on a few in a chain with a beefy wirewound resistor, and they survived fine. ESD seems like a little thing comparatively. \$\endgroup\$
    – Tom Paris
    May 10 '12 at 7:18

It really depends on the LED you are using. LED's Voltage drop vary from one part number to another and it is related to the LED's color . Regardless, it is always specified in the LED's data sheet. Typical ranges go from 1.8 to 3.3. Based on what you describe seems like you have something lower than 3 volts (since it does turn with 1 battery).

You can see in the following image I have a 3 volt-drop LED, if I power it with a 6V battery I have both LEDs light up. As I measure the voltage between the LEDs I read 3 Volts, as expected.

3V LED Voltage Drop

Hope this helps

  • 1
    \$\begingroup\$ I bought the LEDs off a shop and they don't provide the sheet :(. So if I have 3V and the second LED in series won't light up, probably the LED is greater than 1.5V? (I assume that the first LED consumes 1.5V, so the next won't have enough volt to light up due to first drop). So there is no calculation besides the technical sheet can be used to determine the drop? FYI its white bright LED. \$\endgroup\$ May 10 '12 at 3:35
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    \$\begingroup\$ You can read Sparkfun's tutorial on LEDs to understand a bit more about them (sparkfun.com/quiz/60 ). Most of the time, you should not connect the LED directly to power without putting a resistor in series to limit the current and match the LEDs the characteristic forward current (Unless you have a string of LEDs in series that match the voltage source). \$\endgroup\$
    – Malife
    May 10 '12 at 4:15
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    \$\begingroup\$ "Unless you have a string of LEDs in series that match the voltage source". That's absolutely wrong! You should always have a resistor and enough voltage drop so it can control the current! \$\endgroup\$
    – stevenvh
    May 10 '12 at 4:23
  • \$\begingroup\$ @stevenvh -- Your always rule has exceptions. In general, for indication purposes, I agree with you. But for power LED purposes, a buck-converter or boost-converter are used for efficiency purposes, and the series resistor is not used in those cases because it would be a waste of efficiency, not to mention an unwanted additional heat producer. To see a simple boost converter, please look up "Joule Thief" on Wikipedia to see what I'm talking about. The general rule of thumb is the number of LED's in series must be greater or equal to the number of cells in series. \$\endgroup\$ Apr 8 at 3:41

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