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My son's fish-lava light LEDs burned out. Here is the simplified version of the circuit (some components removed for clarity):

enter image description here

I've spent the last few days researching LED circuits. I realize that the circuit is not an ideal configuration and I see the reason it burned out. Since I'm working with a fixed board, I was looking for a formula to calculate the resistance for the resistor on the left (between the sets of LEDs).

I found a similar question was asked a few months ago but the answer there is "don't do that". Unfortunately, that's what I'd like to do only because it seems easier to replace a few 30c LEDs and resistors each time it burns out than to create a new board. In addition, I'd really like to understand the theory even though the practical application isn't desirable.

How does one go about calculating the resistance for a single resistor protecting LEDs in a mixed series/parallel configuration?

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  • \$\begingroup\$ "components removed for clarity" is usually a big mistake. Where does it end - a blank sheet of paper. Answer not provided to keep this question page as clean and clear as possible. \$\endgroup\$
    – Andy aka
    Commented Oct 30, 2015 at 17:54
  • \$\begingroup\$ The + and - connect through a bridge rectifier to a 12v power source. That's the entire circuit. \$\endgroup\$
    – NRF
    Commented Oct 30, 2015 at 18:06
  • \$\begingroup\$ The single resistor argument is compelling! Bloody cheapskate designers! \$\endgroup\$
    – Andy aka
    Commented Oct 30, 2015 at 18:08
  • \$\begingroup\$ Indeed! I'm considering cutting down a perfboard into a circle and doing it the "right way" but the theory has me hooked. \$\endgroup\$
    – NRF
    Commented Oct 30, 2015 at 18:11
  • \$\begingroup\$ Two series LEDs can share the same resistor so you need one extra resistor but both new resistors have to be double the value of the original resistor. \$\endgroup\$
    – Andy aka
    Commented Oct 30, 2015 at 18:15

2 Answers 2

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If you are connecting a resistor in series with 1 or more LEDs do this:

You will need to know the voltage drop beetwen LEDs (normally around 2V per LED) and the voltage input. The voltage that gonna drop beetwen your resistor is what didn't drop in the LEDs. So:

$$V_{in} = V_r + V_{led}$$

$$V_r = V_{in} - V_{led}$$

Then the current going thought the resistor gonna be the current you want to flow in the LEDs, most of the time I use 10mA, the max the LED can handle is normally around 20mA. That will determine how much bright your LED gonna be.

With your resistor voltage and current you can use ohms law to calculate its resistance.

$$V=R \cdot I$$

One thing to pay attention is the power dessipated in the resistor.

$$P=V_r\cdot I_r$$

If you have parallel LED association I would consider using this to calculate a resistor for each parallel. The problem with using only one resistor is that all the current from each LED will go though this resistor, so it may need to have a better power rate. Even if you get a resistor that has a good power rate you may consider calculating separated LED resistors because power rate won't prevent it from heating up.

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    \$\begingroup\$ You can use mathjax which is very similar to latex directly to format mathematical expressions. No need to create images and add them to the post. I suggested an edit that replaces the images. =) I changed the symbol \$i\$to a big \$I\$ for the current in ohms law to stay consistent with your other formula. \$\endgroup\$
    – Magic Smoke
    Commented Oct 30, 2015 at 22:49
  • \$\begingroup\$ Thanks a lot buddy, I've searched for that a while ago, and those s*** images was the best I've found. Thanks! \$\endgroup\$
    – Pedro Nadolny
    Commented Oct 31, 2015 at 0:10
  • \$\begingroup\$ The math people over at math.SE made this meta post which explains a few things \$\endgroup\$
    – Magic Smoke
    Commented Oct 31, 2015 at 0:19
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    \$\begingroup\$ I'm not in a position where the pre-fabricated board has room to insert a resistor per series of LEDs. Thanks for the tip on getting a resistor with a better power rating, though! \$\endgroup\$
    – NRF
    Commented Oct 31, 2015 at 1:47
  • \$\begingroup\$ @NRF: Sometimes one has to either cut and scrape traces or resort to point-to-point construction. \$\endgroup\$
    – Ignacio Vazquez-Abrams
    Commented Oct 31, 2015 at 1:56
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Since you already read the posts you mentioned, you already know that you can't calculate the resistor because you don't know how the current is divided between parallel diodes/LEDs.

You didn't give details about the board. Would it be possible to enlarge one of the LED terminal holes (if it is a PTH LED) and solder a small resistor in series with each LED?

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  • \$\begingroup\$ Can't calculate the resistance? Could we calculate it given ideal LEDs and worst case LEDs, then guesstimate what resistor to use? Apart from that, that's a good idea to try and put a resistor per LED. Space is a bit tight, but that may be doable. I'll play around with the components... \$\endgroup\$
    – NRF
    Commented Oct 31, 2015 at 1:49

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