I need to work out what resistor I need for a series circuit containing 3 LEDs of 2 different types, and a 9v battery. I can do the maths, but I need to check I'm doing it right as I'm trying to learn this from what I can google!

LED1 VF = 3.2V @ IF = 20mA
LED2 VF = 2.2V @ IF = 20mA

Both LEDs have a Forward Current of 20mA at these Forward Voltages.

The circuit will contain 2x LED1 and 1x LED2.

So, using R = (VS - VF) / IF, do I calculate the total VF as 3.2V + 3.2V + 2.2V, therefore VF = 8.6V?

Using the figures I have, I get:

(9 - 8.6) / (20 / 1000) = 20 ohm, so I'd need a 20Ω resistor?

Also, can anyone advise what type of resistor would be best?

  • \$\begingroup\$ Do you really want to run the LEDs at 20 mA? Have you checked the brightness? Some LEDs are annoyingly bright with 20 mA through them. \$\endgroup\$
    – pipe
    Commented Apr 10, 2016 at 22:18
  • \$\begingroup\$ The data sheets on the LEDs I'm looking at suggest that the luminous intensity should be ideal at 20mA (if I'm reading the graphs right), I'm using fairly bright LEDs. \$\endgroup\$
    – Hannah
    Commented Apr 24, 2016 at 15:46

2 Answers 2


Your math is correct. However, I'd recommend against having the total Vf that close to the battery voltage. Basically, it gives you very little margin for voltage variation. As soon as the battery voltage starts drooping, the LEDs will start dimming very quickly and soon go out as the voltages drops below Vf. On the other hand, if you have a very fresh battery with is above 9V, you will greatly overdrive the LEDs.

If you really want to run those LEDs in series, you should probably use two 9V batteries in series and recalculate the resistor for 18V. Otherwise, remove one of the LEDs.

  • \$\begingroup\$ Correct answer, but note that if you add another 9V battery in series, the resistor will be dropping a little over 9V while passing 20mA. That will cause the resistor to get pretty hot and will waste a lot of power. A typical 0.25W leaded resistor should survive (I wouldn't touch it), but smaller packages may burn up. \$\endgroup\$
    – Dan Laks
    Commented Apr 10, 2016 at 21:49
  • \$\begingroup\$ Ehh, even with a brand new 9V battery at a max of 9.4 volts, that's 0.8V over 20Ω, putting the leds at 40mA, but that will quickly drop under load. The open circuit voltage of a 9V battery rarely stays that way for long. Most leds can survive that for a while. The risk with a 9V battery is greatly overestimated. Now, if they were AA cells... \$\endgroup\$
    – Passerby
    Commented Apr 10, 2016 at 22:01
  • \$\begingroup\$ @Passerby probably true. My other point stands though. As soon as the 9V battery drops much below 8.6 volts, the LEDs will go out. \$\endgroup\$
    – DoxyLover
    Commented Apr 11, 2016 at 2:50
  • \$\begingroup\$ Not really. They will just dim till about 7 volts maybe. Depending on the LEDs they will still be quite visible down to a few hundred microamps. \$\endgroup\$
    – Passerby
    Commented Apr 11, 2016 at 3:09

Spend another penny to get another resistor for the different led type. Match the resister for the intended brightness. Putting all in series may lit one type brighter than the other. You really will not like that.

Make two circuit loop with resistor calculated separately.

All leds in series may push one type very close to the full forward voltage. Spectral emission of led depends on its forward current, thus the led operated very near to the peak forward voltage will heat up more thus conducting more current and will shift in color output.


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