Theoretically, on a 12v with a max of 10w power line, how many LED's could you chain in series before the LEDs further down the chain begin to dim? Where the LEDs have a forward voltage of 3.2 - 3.5 and a forward current of 20mA?

I would be using a 470 ohm 1/2w resistor before the first LED.

  • \$\begingroup\$ That's 470 ohm, not 4700 ohm. \$\endgroup\$ – Wayfaring Stranger Aug 19 '14 at 2:40

Each one you add will make the chain dimmer. You could run 3 (using a lower value resistor) in series.

If the forward voltage of all 3 was 3.5V you'd have a current of 1.5/R, if the forward voltage of all 3 was 3.2V you'd have a current of 2.4/R or about 60% more. You do have to be be careful of this if you're running the LEDs close to their maximum capability.

Edit: If you need to run 21, say, you can run 7 parallel strings of 3 LEDs in series, so you need 7 resistors total. Each resistor will be about 1.95V/ILED where ILED is the current you want the LEDs to pass. 470R would give you about 4mA, so you could run as many as 600 LEDs from a 10W supply (but 4mA may not be as bright as you'd like).

  • \$\begingroup\$ I am looking to run around 20 LEDs, so it looks like i am going to have to run in parallel. What would you recommend? \$\endgroup\$ – bizzehdee Aug 18 '14 at 22:48
  • \$\begingroup\$ See my edit above. \$\endgroup\$ – Spehro Pefhany Aug 18 '14 at 22:50
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    \$\begingroup\$ This might be a dumb question, but what would happen if you tried to run 4 leds in this case? Would they simply not turn on at all or just be dimmer? \$\endgroup\$ – Chris James Champeau May 10 '18 at 0:18
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    \$\begingroup\$ @ChrisJamesChampeau It's not a dumb question. The minimum voltage for the desired current is 12.8V and the maximum is 14V so for sure they will be dimmer than desired, even with zero ohms in series, but how dim is difficult to predict and will vary from one sample to the next and with temperature (and with how close the '12V' is to exactly 12V). \$\endgroup\$ – Spehro Pefhany May 10 '18 at 0:33
  1. It's in series, not serial.
  2. Assuming the LEDs are the same, those "further down the chain" aren't going to be any dimmer. Since the LEDs are in series, they all have the same current through them, and will therefore have the same brightness.
  3. To answer the question otherwise, do the math. You have a 12 V supply and each LED might need up to 3.5 V. (12 V)/(3.5 V) = 3.4 LEDs max. Since you can only have whole LEDs, the answer is 3.

To calculate the current limiting resistor value: You know each LED will drop at least 3.2 V, which means 3 of them will drop 9.6 V. That leaves 2.4 V across the resistor. You want the current to be 20 mA, so the optimum resistance for max brightness is (2.4 V)/(20 mA) = 120 Ω.

  • 3
    \$\begingroup\$ Whoever downvoted this, what exactly do you think is wrong, misleading, or badly written? Silent downvotes do a disservice since we don't know what you object to, and the misconception could just as well be on your end. \$\endgroup\$ – Olin Lathrop Aug 19 '14 at 2:03
  • \$\begingroup\$ I UPVOTED this answer. It seems quite good to me. \$\endgroup\$ – Duncan C Jan 14 at 16:20

The answer should be the source voltage divided by the voltage drop of 1 LED. So if your source is 12v, and your LEDs have a voltage drop of 3.5, 12/3.5v, gives you 3 LEDs. If the the forward voltage drop of the LEDs was exactly 3 volts, you'd be able to drive 4 of them, and would not need a current limiting resistor.

The resistance of the current limiting resistor will vary with how many LEDs you're driving in series.

If you have 3 3.3 volt LEDs, you need to figure out the amount of leftover voltage to calculate the current limiting resistor. Thats 12-(3.3*3), or 2.1 volts. Use 2.1 volts in your current calculation. 2.1/.02= 105 ohms. You want the next larger E24 resistor value, or 110 ohms. If you assume your LEDs have a 3.2 volt drop, you'd use (12-(3.2*3))/.02 to calculate your current limiting resistor. That gives 120, which is an exact match. You'd use a 120 ohm resistor.

Edit: If you want to run 20 LEDs, arrange them in groups of 3 series LEDs, each with a 120 ohm resistor, and all the 3 LED chains wired in parallel. Like this:


simulate this circuit – Schematic created using CircuitLab

P.S. A 4.7k resistor would probably prevent even 1 LED from lighting. It would only allow about 1.8 mA to flow through a single LED from a 12 volt supply, and with 3 in series you'd only get about 4.4 microamps through each LED.

  • \$\begingroup\$ Forgive a noob, but why does each "series" have it's own resistor rather than using one resistor on the common ground or supply voltage end? \$\endgroup\$ – Bernhard Hofmann Jan 14 at 12:16
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    \$\begingroup\$ Each parallel path to ground will have a different current flowing through it. You want to control the current through each path separately. If you used one common resistor, you could get more current through one "series" than the others, over-driving those LEDs. \$\endgroup\$ – Duncan C Jan 14 at 13:13
  • \$\begingroup\$ Thank you so much, that really helps explain it well. I can imagine that slight differences in wire/trace quality and length could affect things. Interestingly, I found a website (falstad.com/circuit) that let me simulate the two designs and the resulting current flow is indeed different, even though the circuits there are purely theoretical. \$\endgroup\$ – Bernhard Hofmann Jan 15 at 14:13

You can chain as many as you have voltage for. If your forward voltage is 3.2V thats 3.75 LEDs. Since you obviously can only have whole LEDs, that would 3 plus your resistor. If you are desiring a 20mA drive current though, that resistor is too large for even a single LED. A 470ohm resistor flowing 20mA will see a 9.4V drop, leaving 2.6V for your LEDs...not enough for their forward voltage.

  • \$\begingroup\$ I am looking to run around 20 LEDs, so it looks like i am going to have to run in parallel. What would you recommend? \$\endgroup\$ – bizzehdee Aug 18 '14 at 22:48

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