So I'm still very new to electronics but I would like to make a 100 LED Lamp for fun. I got my hands on a few 18650 batteries yesterday so I decided to do this project. I thought it through but I am not sure I have it right, here is my idea: I take two 18650 batteries, they both have 4,2V and output 5,6A. If I put them in parallel I have 4,2V and 11,2A, right? I measured my LEDs, they draw about 100mA each, that means I shouldn't have any problems powering 100 of them with the two batteries in parallel. Now my idea was to take 10 LEDs and put them in sequence and have these 10 sequences in parallel, would that work? Also I need a way to charge the Batteries later on and some sort of voltage cutoff so the batteries don't break. But I will worry about those things once I understood the basics. Thanks in advance :)

  • \$\begingroup\$ 10 LEDs in series are likely to have a 15-20V total Vf, far exceeding your battery pack voltage. You'll need to either obtain a boost regulator, or set up a different arrangement of LEDs. \$\endgroup\$ – ζ-- Apr 28 '18 at 22:29
  • \$\begingroup\$ @hexafraction ok thanks, how did you know they have 15-20V total? Sorry for the noob question but I am still very new and trying to understand this as good as possible ^^ \$\endgroup\$ – Tanonic Apr 28 '18 at 22:34
  • \$\begingroup\$ I happen to remember the rule of thumb that LEDs typically have a voltage drop of approximately 1.5-3V, depending on their color and exact model. With 10 in series, I multiply that value by ten. I realize now that I said 15-20V in my prior comment, which was a typo; it's actually more along the lines of 15-30V for 10 LEDs. The datasheet/specs for your LEDs will indicate their exact forward voltage and current. \$\endgroup\$ – ζ-- Apr 28 '18 at 22:41
  • \$\begingroup\$ @hexafraction I just checked and the voltage is actually documented to be 2,8-3,6V which doesn't make it easier for me. Maybe I will build a smaller model first just to test, with 20 LED's or so. I could put 3 batteries in sequence having 12,6V 5,6A and have sequences of 4 in 5 parallel rows. 4*3,6 is 14,4V but maybe it would work cutting it a bit close. \$\endgroup\$ – Tanonic Apr 28 '18 at 22:56
  • \$\begingroup\$ Where are your datasheets? This is false info two 18650 batteries, they both have 4,2V and output 5,6A more likely max current with charge peak voltage of 4.2V and will only last for 15 minutes at that rate. if you are lucky it dioesn't overheat with 1~3 Watts of self heat. \$\endgroup\$ – Tony Stewart Sunnyskyguy EE75 Apr 29 '18 at 2:21

In comment, you have identified the LED forward voltage as 2.8 - 3.5 volts, which identifies them as white LEDs.

Now, at 100 mA, each LED will dissipate .28 to .35 watts (current times voltage) and let's be conservative and go for .35. Your two batteries will put out 4.2 volts at 11.2 amps (and yes, you got that right). So the power available is, again, current times volts or 47 watts. 47 watts divided by .35 watts/LED says that, in principle, you can drive 134 LEDs.

So what you can do is connect each LED to a resistor of about 10 ohms, which will give you something on the order of (4.2 - 3.5)/10, or 70 mA. Put all 100 LED/resistor pairs in parallel, and you should be in business.

Or maybe not, since if the LEDs are actually running at 2.8 volts, the current will be (4.2 - 2.8)/10, or 140 mA, for a total of 14 amps. So a better starting point would be 15 ohm resistors.

But first, you need to think real hard about cooling. Your LEDs plus resistors will be dissipating something like 40 watts. In the worst case, the LEDs alone will produce 35 watts, and you MUST find a way to cool them. For what you're doing, a professional would be hard-pressed to do the job, and just having at it as you seem to want to do is almost guaranteed to end in heartache. You'll need a metal mounting plate for the LEDs, and very aggressive cooling. A finned heatsink may or may not do the job, and water cooling may be necessary.

Your comment that "Maybe I will build a smaller model first just to test" is the most sensible thing you've said so far.

  • \$\begingroup\$ Low powered (100 mA) LED do not need thermal management if not placed to densely. For example the Bridgelux EB Gen 2 strips put 112 LEDs on a 22"x 1" strip. When driven at 28 Watts (20V x 1.4 amp) they do not need a heatsink on an FR-4 PCB. PS the 4.2V is not correct. 11.2 Amp is not likely correct. 3.6V nominal, and likely no more than 3 AH. Those charlatan vendors lie. Resistors are a bad idea for a battery powered light source. They will dim as battery voltage drops. A current regulator is a reasonable requirement when the batteries are not going to last very long. \$\endgroup\$ – Misunderstood Apr 29 '18 at 1:02
  • \$\begingroup\$ @Misunderstood - I'm afraid your advice about current regulators is off the mark. Do you really think that someone who is "still very new to electronics " is going to build or buy 100 current regulators? Give a little consideration to the skills possessed by the poster. \$\endgroup\$ – WhatRoughBeast Apr 29 '18 at 3:22
  • \$\begingroup\$ 100 regulators??? My answer uses 1. If I were to use resistors I would first need to know the actual Vf. If Vf is 3V, I would use a 4.99Ω 1% for 100 mA. If you look at the discharge curve (see my NCR18650 link) the mid point of the 1C discharge curve is ≈3.5V. An 18650 is not a 4.2V battery, it's 3.6V nominal. A 5Ω would dissipate an average 300 mW for LED and resistor at ≈86% efficiency. A 10Ω would yield 50mA with the same efficiency. A regulator will provide a consistent (brightness) current across the entire discharge curve and will save the labor of 100 soldering resistors. \$\endgroup\$ – Misunderstood Apr 30 '18 at 18:09
  • \$\begingroup\$ You should delete your comment "Your comment that "Maybe I will build a smaller model first just to test" is the most sensible thing you've said so far." It is condescending, rude, and most of all WRONG! You should also rework your numbers as again YOU are wrong. 15Ω is NOT a good starting point. Almost all your "calculations" are wrong. Thermal management is not a huge issue at 100mA. You should use the discharge curve for the target voltage not the 4.2v charge voltage. \$\endgroup\$ – Misunderstood May 1 '18 at 5:42
  • \$\begingroup\$ @Misunderstood - I'm sorry you didn't read the OP closely. The poster is almost entirely inexperienced. He does need to start small. And "Thermal management is not a huge issue at 100mA" is true enough within this context, but 100 times 100 mA is 10 amps/30 watts, and at 30 watts thermal management IS a huge issue, particularly if he wants to make a compact unit. \$\endgroup\$ – WhatRoughBeast May 2 '18 at 20:53

First off a single 18650 battery can handle between 4 amp and 10 amp. For example the Samsung 3,350mAh INR19850 has a specified max continuous discharge rate of 8 amp.
Any 18650 will be able to handle a 2C (twice the rated capacity) discharge rate.
2C is always a safe continuous discharge rate for most Li-ion batteries. You should measure your capacity if it is from a no name vendor.

You need not worry about heat if you space the LEDs correctly. For example a Bridgelux EB series Gen 2 strip of 56 LEDs on a 11" x 1" board needs no heatsink when run at the max 24 watt. The 48 watt 22" x 1" does not need a heatsink either.

With spacing like below, thermal management will NOT be an issue.

enter image description here

You said your LEDs use 100 mA. That's technically an incorrect way of looking at LEDs. LEDs have a max current just as batteries have a max current. They all work at lower currents. The trick is to find the optimum operating parameters.

When selecting an LED the efficacy (lumens per watt) is the most important parameter. Cost per lumen may be a close second if efficiency is not important. In a battery powered lamp, efficiency is very important and will affect brightness and battery life.

The inexpensive Chinese LEDs are using technology where the patents have expired. They are very inefficient because the technology is 20+ year behind. A reasonable cost for a very efficient 100mA white LED is 10-15¢ in small quantities (<1000).

A Bridgelux EB-Series Gen 2 strip is hard to beat at $4 for a 1200 lm, 56 LED strip or $7 for a 2400 lm, 112 LED strip. This works out to a very efficient 175-180 lm/W and only 7¢ per LED including PCB, 4 power connectors, and assembly. And no heatsink required. No assembly required. Wiring is very simple when using 20 ga solid wire known as "Bell Wire" which is pushed into the connectors on the strip. The drawback is you need 20V to power them. A efficient boost driver need at least a 9V input or 3 18650s in series. This way 3 name brand batteries will get you 2 plus hours of 1200 lm per charge.

If you wire 5 strings of 20 LEDs in parallel you will need about 60V.

Using a $12 Mean Well LDH-45A-500 you can boost 9V up to between 12 and 86V at 500 mA giving you 100 mA with enough overhead to handle the max 3.6V.

I suspect you found some no name batteries. LiIon are not 4.2V. The max charge voltage is 4.2 but they quickly drop to 3.6V. The charlatan vendors use 4.2V and lie, exaggerating the mAh. When buying off Amazon or E-bay I would only buy Panasonic e.g. NCR18650PF, Samsung e.g. INR19850 or e.g. ICR19850.

For 100 LEDs at 3V and 100mA you will need 30 Watts.

You have to use the LiIon cutoff voltage as the design parameter. At a discharge rate of 1C the cutoff voltage is 3.2V.
You need 3.33 Amp at 9V (plus 10% for LED driver efficiency) so 3.6 Amp
A realistic capacity for a 18650 is 2500 to 3000 mAh. This will give you over 2 hours (≈150 minutes) for each set of 3 batteries in series.

I would use two $4, 11", Bridgelux EB-Series Gen 2 strips with 56 LEDs or one 22" strip with 112 LEDs. The LDH-45A driver will power the up to three 11" in series. Avoid wiring strips in parallel which will cause current imbalance. Each 11" strip will give you bout 1200 lumens (80W incandescent equivalent) at 14 Watts.

If you have more time than money or like the idea of assembling 100 resistors, you can go the resistor route.

An 18650 cell works very well with white LEDs. The nominal voltage is 3.6v and 4.2 max. A Li-ion battery has a relatively flat discharge curve from 3.6v down to 3.2. Most white LEDs normally operate at 3V or lower. A 6 amp it will from 4.2 to 4.0v in seconds and down to 3.6 in about 10 minutes. Then down to 3.5v after30 minutes.

A 3v LED at 100 mA with a 5Ω resistor will dissipate 350mW (300 LED, 50 resistor) when powered with a 3.5v source.

As you can see in this Panasonic NCR18650 discharge curve the sweet spot is 3.5V. So 100 LEDs would run about 35 Watts. At 3.5V this would require 10 Amp. So two 18650 batteries in parallel will easily provide 10 Amp.

You may want to consider dropping the current to extend battery life.

Li-ion 18650 discharge voltage curve

To select the desired current and resistor use a online resistor calculator. It is important you know the actual Vf. To measure, light one up with a 5-10Ω resistor and measure across the LED leads with a voltmeter.

For an online calculator I like: Hobby Hour LED Series Resistor Calculator

First enter the sweet spot voltage for the battery e.g 3.5v
If you desire 100mA use that.
Enter the LED's measured Vf, e.g. 3v
Try both 5% and 1% and check the bottom line labeled efficiency.

enter image description here

Do not be too excited if your Vf is like 3.4V which gives you an efficiency of 97.14%. First off, it will not work well below 3.4V when the discharge cutoff is 3.2v. At 3.4V it will be old inefficient technology. A higher voltage reduces your lm/w efficacy.

Be sure not to allow the batteries to discharge below 3V opr battery life will suffer. To extend the lifespan of Li-ion batteries charge them to 80% of full charge and discharge to 20%. That applies to your cell phone as well.

A great source for information on batteries is batteryuniversity.com


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