# Make an LED (20 LEDs) circuit battery efficient

I want to wire up a miniature wargaming figurine to have 20 LEDs. As I've poked around options my concern is the battery life and how often the battery would need to be replaced. Since each gaming session lasts 4-5 hours, it'd be great if I could find a battery that would last multiple gaming sessions (though I think this is impossible...)

I'm VERY NEW to all of this so I would appreciate explanations/solutions that understand I have little knowledge of the terms or techniques (I still don't understand what current is...)

EDIT: Here are more details on what I'm trying to do

I am trying to light up this figure (Knight Castellan).

Locations for the lights would be (and their specs):

• 8 blue for right arm (3.4V, 20mA)
• 2 Red-Blue-Green for the eyes (3.4V, 20mA)
• 10 red/orange lights for left arm and engine (2.2V, 20mA)

So altogether that's 20 LEDs, all at 20mA and none above 3.4V.

The battery would go inside the torso -- I can't find an exact image but the dimensions are roughly 2" wide, 2" tall, 3" deep.

My concern is battery life, as I'd like the batteries to last at least a few game sessions (so let's say 10-15 hours). I don't mind replacing or recharging a battery, but there's no point if I have to do it halfway through a game.

I should add that I don't know how bright these LEDs are (the specs say 8-10,000 mcd for the blue ones), and that I don't know how to dim them, but would prefer them bright since they are usually in a well-lit room.

• Elliot's right and you should look at a switching current controller for them, but... Your power consumption is .02A*20*3.4V=1.36W and if you want you could run the LEDs at .015A*20*3.4V=1.02W. Let's assume you can find an 80% efficient LED driver, that means you actually need 1.36W\0.8=1.7W or 1.02W\0.8=1.275W. We'll multiply that by 6h because your battery capacity will decrease over time, so 1.7W*6h=10.2Wh or 1.275W*6h=7.65Wh batteries are needed depending on whether you find 15mA brightness acceptable (20mA is absolute maximum continuous).
– K H
Jan 14, 2019 at 20:44
• The question becomes "Will a single high capacity 18650 Li-Ion battery do the trick? If you have the space or the option, due to ubiquity and cost, for disposable batteries you should aim to use AA or larger and for rechargeables, Li-Ion 18650s for best cost effect. If it's a 3.7V lithium ion you need either 10.2Wh/3.7V=2756mAh or 7.65Wh/3.7V=2067mAh. The best Li-Ion of 2018 has a capacity of 3500mAh, common ones have 2600mAH, so yes, a single 18650 will do if it will fit. Then you need to find a battery charging board and LED current controller with a suitable input voltage.
– K H
Jan 14, 2019 at 20:53
• From a systems perspective the 18650 Li-ion would be great. But about half the useful charge comes out after the cell voltage has dropped below 3.4V; that means that a simple resistor+LED solution is right out. Unless the OP really wants to learn about electronics they should see if they can find a suitable LED controller that'll do the job. Jan 14, 2019 at 20:59
• By the way, LEDs don't share current well in parallel, so if you put them in parallel, you'll have to either use individual current controllers or waste a lot of power with series resistors. If you put them in series strings of at least 3 or 4 (depends on the acceptable output voltage of your current controller), they will share current acceptably and using a current controller will allow you not to use series resistors. You'd be looking for a boost driver with a 3.4-4.2V input (can be wider but must cover that range) and an output suitable for at least 3S and 140mA
– K H
Jan 14, 2019 at 21:05
• That many LEDs all running at 20mA will be very bright, try looking at lower currents and seeing how bright they are. If 2mA will do, then that's an instant x10 improvement in battery life. Just a thought. Jan 14, 2019 at 21:12

You don't want to run all of these at 20mA. 20 leds in a 6" cube of a model like that knight (It took a while to find the size of these things) at full brightness are not just visible in daylight, they will absolutely illuminate a room. You can power them at a lower current, and they will still be visible in daylight and quite nice in the dark.

Since you power them at a lower current, they will have a lower forward voltage so you can use a smaller battery for longer. 20 leds at 1 or 2 mA will run for days off a pair of AAA batteries. Now you could power them all individually with a resistor each, but that's a bit wasteful and take up space. They can be powered in parallel with a single resistor (Ignore those that scream that's a sin!). The only issue with that is that you want red/orange, so you would have one string for the blue and one for the red.

The last thing is the RGB you want for his eyes. You haven't said how you want those to work. A fix color (4 pin RGB led that you combine multiple resistors to get a rough color)? Or flashing (2 pin led with a built in micro controller that runs through a pattern, just add power). The flashing led would work just the same, normally expecting 20mA at 3.3 volts, so power it at less and it's good.

simulate this circuit – Schematic created using CircuitLab

Example schematic. 3x AAA batteries for size. About 13mA for both the blue and red strings. 10 for the two color changers. About 22 hours with good Alkaline batteries (plus more due to the way the batteries discharging lead to dimmer leds which pull less current, but that depends on how dim you mind them going). You could use NiMH 1.2V batteries but would probably want to use 75Ω resistors instead. You could use 75Ω resistors if you want them brighter but obviously that changes how long they last by a bit.

• It’s an LED that rotates colors on its own. So it transitions from red to green to blue. What you suggest sounds like an amazing and simple solution, but what would I need? Jan 15, 2019 at 5:12
• @vashts85 Ah, a "flashing" rgb led. Normally 2 pins and cycles the colors on it's own. Just hook it up to the voltage it expects, sometimes with a resistor. Everything here is a basic battery + wire + resistor + led setup, so very simple. Jan 15, 2019 at 6:38
• So I should essentially use a really high resistor (like 5000 ohms) and wire everything in parallel (with a resistor per LED?), thus reducing drain to about 20mA total (and brightness a ton as well), and then I'll be able to power with something like a 9V for very long periods of time? Jan 15, 2019 at 19:50
• Try 100 ohms and 2 or 3 AAA. They will last longer than a 9v. Jan 15, 2019 at 19:58
• Can you explain why they’d last longer and why the lower resistor? And would I give each LED a resistor and wore them all in parallel? Jan 15, 2019 at 20:15

The capacity of this battery is not 1000mAh, it is about 750mAh at low currents.

Note that the linked battery only provides 9V for a short period of time. The rated capacity assumes that your device still works if the voltage drops to 5.4V. You will actually get about 700mAh and the voltage will be 7V to 8V. If the LEDs are not in series and you need 400mA you should expect no more than 1.75 hours of operation.

It will be difficult to use 2 LEDs in series if the voltage drops close to 7V, and the brightness will change greatly as the battery discharges. Even if that was OK, you will get about 3.5 hours of operation in this case.

You should be looking for another approach, possibly one that uses a high-efficiency switching regulator as a constant current source.

At 20mA your LEDs will be incredibly bright for such a small figure. There are several solutions to the problems you have:

1. You could use the current sources I showed in the other answer and provide a small PWM to lower the brightness. This would reduce the power consumption considerably. Using a small simple MCU could give you PWM control of perhaps 3-5 strings each at a preset brightness for the individual LED strings.

2. You could use the WS2812 integrated RGB LED to get any color or intensity you want. It may even be possible to reduce the number of LEDs required by making a Fiber optic connection to a small number of RGB LEDS. Plastic Optical Fiber (POF) is dirt cheap (see here) and readily available. Use a small hole in some plastic to position multiple fibers above one of the LEDs.
The WS2812b comes already in LED strips form from many sources, and might be a good starting point for you. This could be a very efficient means of providing your lighting and getting the runtime out to many hours. The controller for setting brightness and color would not actually need to be within your model, you could send the serial data when you turn on the model, and then remove the controller. The same controller could be used with multiple models (there are endless opportunities and variations for creating the control system).

The WS2812 works from 3.5-5.3V so a single LiPo battery without any regulation is all that is needed; this would significantly simplify your wiring.