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I'm planning on building an LED circuit to place inside a model miniature for wargaming. The circuit would be in parallel as I'd need a really big battery to power up all the LEDs.

I wanted to check my understanding of setting up the resistors so things work properly. I'm going to have different types of LEDs with different voltages but here's my basic question.

Let's say I have 10 blue lights at 3.4V 20mA and a 9V supply. Based on my calculations each lamp would need a resistor of only 28 ohms? My math is

$$\frac{9-3.4}{\frac{10\cdot 20}{1000}} = 28$$

Is that right? The final circuit will have close to 20 LEDs, and it seems like at that point the ohms of the resistors will get much smaller?

Once I figure out all the different types of lights (all at 20mA), I will need to calculate the resistors individually for each as they have different voltage requirements?

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No, your calculation is not right.

Since your 9V source is a voltage source, and all LEDs are in parallel, the number of LEDs can't have any influence on the resistor. One LED can't "know" of its neighbors!

So, it's simply \$\frac{(9-3.4)\,\text{V}}{20\,\text{mA}}=280\,\text{Ω}\$.

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  • \$\begingroup\$ Thanks! My follow-up is, if I have LEDs of different Voltages (but equal mA) then I just need to calculate their individual resistors, correct? \$\endgroup\$ – vashts85 Jan 14 at 16:50
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    \$\begingroup\$ yep, and that's why you need each of them to have its own resistor: even diodes of the same type have slightly different forward voltage, so to avoid a few being brighter than the others (+ thermal runaway, which at some point simply burns the ones that have slightly lower forward voltage), each of them has to have its own resistor. \$\endgroup\$ – Marcus Müller Jan 14 at 16:56
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    \$\begingroup\$ I don't know your 9V battery, so I can't even give an estimate. Also, if you have the capacity of the battery, that question has been asked a bazillion times on this forum. You draw 20·20 mA = 400 mA, and your battery has x mAh capacity. Divide capacity by current, and get the time. Yes, it's that easy. \$\endgroup\$ – Marcus Müller Jan 14 at 17:18
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    \$\begingroup\$ You can improve the efficiency by putting two LEDs in series in each leg, That means you only have 2-3V across each dropper resistor, so there'll be more variation in current with forward drop variations and supply variations, but it'll reduce the power consumed by half. \$\endgroup\$ – Phil G Jan 14 at 17:31
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    \$\begingroup\$ But then again, you could also just replace the 9V battery with something that isn't nearly three times the forward voltage of your diodes. \$\endgroup\$ – Marcus Müller Jan 14 at 17:48
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Using a resistor for each LED would waste most of the battery energy you have, especially if you want to use a 9V cell. I would not recommend it.

A better way to provide a solution is to run as many LEDs as possible in series. This means having a higher voltage, but you get to use the battery energy more efficiently.

You could design a boost/CC provider but this may be too complex for you.

I'd suggest you consider a simple linear constant current device like the AL5809-20. This is a simple two terminal device that sets a constant current through a string of LEDs.

The device comes in various current levels designated by the last two digits of the part number:

enter image description here

To use them you simply connect as you would a resistor, and it sets the current for the string.

The recommended minimum voltage is 2.5V but the device will work down to a minimum of 1.75V.

enter image description here

You would then be able to use a higher voltage such as two or three LiPo cells to provide the power (and rechargeable). For example if you used 3S than you have a minimum of 3*3.6=10.8V and a maximum of 3*4.2=12.6V. Which LED forward voltages in the 2.2 - 3.4V range depending on color you could get at least 2 or as many as 4 LEDs in each string. If you could even mix lower and higher Vf LEDs together, they all still get only the 20mA.

If you want to be really creative you could use a single LiPo cell (3.6 - 4.2V) and a boost DC-DC convertor (there are many on Ebay etc) to get say 20-24V and connect 5 to 8 LEDs in each string.

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  • \$\begingroup\$ Thank you! Is it possible to reach out to privately to ask more questions? I'm really new to this and would appreciate being able to talk to someone 1-on-1! \$\endgroup\$ – vashts85 Jan 14 at 21:48
  • \$\begingroup\$ I suppose I could use this: lighthouseleds.com/20ma-led-current-limiter-driver.html ? \$\endgroup\$ – vashts85 Jan 14 at 22:02
  • \$\begingroup\$ Would I need a power source for each string? \$\endgroup\$ – vashts85 Jan 14 at 22:05
  • \$\begingroup\$ @vashts85 The Lighthouse device appears to be the same part I suggested in the answer. You would need one current limiting device per string of LEDs. \$\endgroup\$ – Jack Creasey Jan 23 at 1:02
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Consider Eneloop rechargeable NiMH AA cells.

Generally you want the minimum cell voltage to match the minimum acceptable current in the LEDs. This means the battery array voltage matches the LED's as close as possible or slightly higher then the loss of power in the selected series R is much less.

schematic

simulate this circuit – Schematic created using CircuitLab

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    \$\begingroup\$ Thanks for the answer but this feels WAAAAAAYYYY above my head...I'm totally new to this \$\endgroup\$ – vashts85 Jan 14 at 19:57
  • \$\begingroup\$ 9V battery wont last very long and answer above with 3V led and 9V battery is only 33% efficient. Keep reading. Lots of related questions \$\endgroup\$ – Tony Stewart Sunnyskyguy EE75 Jan 14 at 19:59
  • \$\begingroup\$ Not all LEDs are the same brightness either. from 100 mcd to 30,000 mcd @ 20mA so get the best and run at 2~ 10mA using almost matched battery voltage + IR=V drop (Ohms Law) \$\endgroup\$ – Tony Stewart Sunnyskyguy EE75 Jan 14 at 20:02

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