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I want to wire 10 of these LEDs into a circuit with a 9-V battery. The site states that the forward voltage is 3.2-3.8 V at 20 mA current. Should I assume 20 mA is the max amount of current that can flow throw this circuit?

Using V = RI, I calculate that I need a resistor that is 450 Ω (9/0.02). Is this correct? Secondly, if I want to have 10 LEDs here do I need a higher voltage (38 V)?

This is the first LED circuit that I am making, so I'm not familiar with most of the basics.

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    \$\begingroup\$ You may want to look here for future LED math questions: led.linear1.org/led.wiz \$\endgroup\$ – Earlz Dec 11 '10 at 19:24
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If you put all these LEDs in series you'll indeed need 38V, plus a bit for the series resistor. But you can make a circuit with several branches. Put 2 LEDs in series and you'll need 7.6V. So the remaining 9V - 7.6V = 1.4V is the voltage over your resistor. If you want 20mA through your circuit You divide this 1.4V / 20mA = 70 ohm, so a standard 68 ohm will do nicely.
Now that's for 2 LEDs, you could go for more if you had a higher voltage. the way to calculate the resistor is the same. If you want 10 LEDs from 9V, in theory you could put the circuit 5 times in parallel. That would be a total current of 100mA though (5 x 20mA), and that's a bit much for a 9V battery.

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It's been suggested that you could replace the 5 resistors by a single one, and branch from beneath this. In an ideal world that would be true; the resistor value would then be 1.4V / 100mA = 14 ohm. But this isn't an ideal world, and there may be small differences in LED voltages. In that case the branch with the lowest voltage will draw most of the current (100mA!) while the other LEDs will hardly light at all.

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  • \$\begingroup\$ Could I use 5 pairs of 2 LEDs in parallel? Would the resistor for each branch be 90 Ohm? \$\endgroup\$ – cor Dec 11 '10 at 14:16
  • \$\begingroup\$ @cor - assuming 3.6V across each LED and exactly 9V from the battery, 90 ohms will give you 20mA \$\endgroup\$ – JustJeff Dec 11 '10 at 14:25
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    \$\begingroup\$ @cor: yes, unless you're pretty sure that the 9V will not go higher. Also, if the LED voltage would be lower, you'd also have a higher voltage drop over the resistor, resulting in a higher current. Always calculate for worst case. Note that most LEDs can eat more than 20mA, but won't light much brighter above 20mA. Since you mention about 3.5V I assume you're using blue or white LEDs. I don't know what max. current for those are. \$\endgroup\$ – stevenvh Dec 11 '10 at 16:00
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    \$\begingroup\$ @Cor - No, what you'll be measuring will be the voltage across the LED, which will be fairly constant. Think of the LED as a 3.6V voltage source when it's conducting. With one resistor, one LED, and one 9V battery, the voltage across the resistor is 9-3.6=5.4V. With 2 LEDs in series, it's 9-2*3.6=1.8V. \$\endgroup\$ – Kevin Vermeer Dec 11 '10 at 18:14
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    \$\begingroup\$ @cor: try not to think of diodes as resistors though. they behave differently. a resistor has the same resistance no matter what, but a diode's "resistance" changes with circuit conditions. the usual way to treat a diode is to look at the V-I curve in the datasheet, select an operating point that you want (e.g., if you have to have 20mA, the graph will tell you the voltage), and then use operating point's voltage or current as needed in your circuit analysis. \$\endgroup\$ – JustJeff Dec 11 '10 at 18:19
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I recently made a circuit with 10 LEDs which works on 6 V. This site was of great help for me. You just supply parameters you gave us to it and it will produce a schematic of a circuit and will provide information on type of resistors which need to be used and so on.

Also, if you want to use a battery, check out its datasheet for its voltage when discharged. This way, you'll be able to calculate how well circuit works with used batteries.

Also, I agree on Leon Heller that 9 V batteries are a bad choice. the thing is that they start out at 9 V, but for example VARTA No.4922 battery has end voltage between 4.8 and 5.2 volts. It's too low to use 2 LEDs in series. Also capacity is lower than in the 1.5 V batteries.

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A 9V battery is a poor choice. Use a 6V battery with a boost converter; National Semiconductor has some suitable designs using their Simple Switcher chips.

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  • \$\begingroup\$ I'm using a 9v battery as it was convenient - I have some for an arduino board that I am also using. Arduino does give me the option of a 5.5v and 3v output - would it be better to use these? \$\endgroup\$ – cor Dec 11 '10 at 14:11
  • \$\begingroup\$ You could use the 5.5V output for a boost converter, enabling you to put all your LEDs in series. \$\endgroup\$ – Leon Heller Dec 11 '10 at 15:08
  • \$\begingroup\$ Why the downvote? \$\endgroup\$ – AndrejaKo Dec 11 '10 at 20:08
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    \$\begingroup\$ What makes 9V poor choice? Why should he go for lower voltage and then compensate it with some DC/DC converter? \$\endgroup\$ – Lukas Dec 12 '10 at 23:45
  • \$\begingroup\$ @Lukas I'm guessing it's not about the voltage, but about the battery type; Standard (PP3) 9V batteries are no good at medium current levels and up; 100mA is asking a lot from those. \$\endgroup\$ – marcelm Dec 5 '17 at 17:47

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