How do I calculate this? I'm pretty sure I don't have enough voltage to power this up but how much do I need? I have 3 AAA batteries equating to 4.5 V. I have the attached wiring setup. I measure the voltage of the bus to be about 1V. Not enough to power anything.
The LEDs are these:
https://www.amazon.com/CHINLY-Individually-Addressable-Waterproof-waterproof/dp/B01LSF4QDM
Chinly WS2812B LEDs
They are 0.3W each LED.
The Op's "diagram"
Each LED will take at maximum white brightness about 60mA, so for 60 LEDs you require a current of 55 * 0.06 = 3.3 Amps.
You can not get this much current out of an AAA battery by several orders of magnitude. You need a mains power supply or quite a substantial battery power pack.
OP's link.
An 18650 size li-ion battery might be more appropriate. Get one with built-in over-discharge protection and a button top. You can get holders for this type. You will need a special charger of course. Don't buy any 18650 battery claiming more than about 3500mAh capacity - those are fake. Buy a good brand like LG, Sony, Panasonic, Samsung etc.
The WS2812 is specified for 4.5 to 5.5 V operation.
Neither 3 x 1.5 V batteries or an 18650 reliably provide this.
I measure the voltage of the bus to be about 1V.
That sounds bad. What do you mean by "bus"? Where did you measure 1V and how? What else was connected?
Paul__B:
WS2812 is specified for 4.5 to 5.5 V operation.
Neither 3 x 1.5 V batteries or an 18650 reliably provide this
True, it worked fine for me, but that's no guarantee.
PaulRB:
That sounds bad. What do you mean by "bus"? Where did you measure 1V and how? What else was connected?
So in my picture there's an arduino nano. I removed the nano and placed my meter there. That main trunk of 24VDC and 0VDC...isn't that called a bus? Nothing else was connected. What do you mean "sounds bad"? Should I have planned/wired it differently?
PaulRB:
OP's link.An 18650 size li-ion battery might be more appropriate. Get one with built-in over-discharge protection and a button top. You can get holders for this type. You will need a special charger of course. Don't buy any 18650 battery claiming more than about 3500mAh capacity - those are fake. Buy a good brand like LG, Sony, Panasonic, Samsung etc.
Looking at the 18650s they output as much as 30A per cell? and I need only 3.3A. Why is it a "maybe it will work"? How many over these 18650s should I get to make this config work?
That main trunk of 24VDC and 0VDC...isn't that called a bus?
No it is not.
In electronics a bus is an interface capable of connecting to many devices, like I2S or SPI.
. I removed the nano and placed my meter there. That main trunk of 24VDC and 0VDC...isn't that called a bus? Nothing else was conne
Why remove the nano? It is not a normal thing to do. I assume you are measuring voltage.
Looking at the 18650s they output as much as 30A per cell?
But what about the voltage?
Grumpy_Mike:
No it is not.
In electronics a bus is an interface capable of connecting to many devices, like I2S or SPI.
Why remove the nano? It is not a normal thing to do. I assume you are measuring voltage.
But what about the voltage?
I removed the nano because I was afraid of damaging it (Yes, I'm a noob). Yes I should've measured with the nano. But when I was testing it I was testing to make sure all my connections were made properly. I'm certain the nano will turn on.
yes I was measuring voltage. I only measured 1V with 3x AAAs.
Those 18650s are up to 3.7V per cell. How many should I use in this circuit?
Those 18650s are up to 3.7V per cell. How many should I use in this circuit?
That is the problem one is not enough and two is one too many.
There was a song about that:-
I only measured 1V with 3x AAAs.
Yes you will have, in fact I am supprised it was that high.
When you are taking current out of a power supply if you try and take too much then the voltage drops.
This is because there is, in effect, an "imaginary" resistor in series with all power supplies, this is called the supply's output impedance. This is caused by the battery chemistry or the components in a bench supply. The effect, like any real resistor, is to drop voltage the more current is drawn from it. That is how you know your supply has insufficient current capacity. The supply has too high an output impedance.
Imagine using 8 AAA batteries in series ( 12V ) to try and turn over the started motor in a car. You wouldn't expect that to work and you would be right it wouldn't.
Grumpy_Mike:
That is the problem one is not enough and two is one too many.There was a song about that:-
https://www.youtube.com/watch?v=c1nTwfUs7oA
Yes you will have, in fact I am supprised it was that high.When you are taking current out of a power supply if you try and take too much then the voltage drops.
This is because there is, in effect, an "imaginary" resistor in series with all power supplies, this is called the supply's output impedance. This is caused by the battery chemistry or the components in a bench supply. The effect, like any real resistor, is to drop voltage the more current is drawn from it. That is how you know your supply has insufficient current capacity. The supply has too high an output impedance.
Imagine using 8 AAA batteries in series ( 12V ) to try and turn over the started motor in a car. You wouldn't expect that to work and you would be right it wouldn't.
Wow! I took a look at amazon and these batteries are expensive and they are big! Before I invest is there a way I can test out your theory? Bench top power supply?
Before I invest is there a way I can test out your theory?
What theory is that then?
I don't deal in theories, or when I do so I say so. I have been in electronics over 50 years, both working in industry and teaching at a University ( in the UK this position is called a senior lecturer, in the US and other parts of the world this is known as a professor ).
The comment made by Paul__B is nearly but not quite true.
Paul__B:
The WS2812 is specified for 4.5 to 5.5 V operation.Neither 3 x 1.5 V batteries or an 18650 reliably provide this.
Yes the 18650 will not be able to provide 4.5 to 5.5 V and the data sheet characterises the parameters of the WS2812 at these voltages. It gives the absolute lower limit on working voltage at ~3.5V, the ~ meaning "about". So it will work at lower than 4.5V, but note here that the driving signal from say an Arduino should not exceed this. This means it is advisable to drive the Arduino with the same power supply as the LED strip.
This has the problem that an Arduino like the Mega and Uno is not guaranteed to work at 16MHz at a lower voltage than about 3.7V.
Now a battery like the 18650 will typically discharge like this:-
You will see it starts fresh at just over 4.3V, spends most of its time at 3.7V, and then drops off. You must not let the battery discharge below 3V or it will become damaged. Some batteries claim to be "protected" against this.
Bench top power supply?
Yes you can use a bench top power supply to play about with voltages and see how they react. There are two things to watch out for. First the bench top power supply will have an overall lower output impedance than a battery, so it is best to have a large capacitor in the circuit to make it a more realistic test of the battery. And second be aware that LED characteristics change with age, so any experiments you make on say the minimum voltage they will work on will not give you quite the results as they would in say a years time.
Grumpy_Mike:
Yes the 18650 will not be able to provide 4.5 to 5.5 V and the data sheet characterises the parameters of the WS2812 at these voltages. It gives the absolute lower limit on working voltage at ~3.5V, the ~ meaning "about". So it will work at lower than 4.5V, but note here that the driving signal from say an Arduino should not exceed this. This means it is advisable to drive the Arduino with the same power supply as the LED strip.This has the problem that an Arduino like the Mega and Uno is not guaranteed to work at 16MHz at a lower voltage than about 3.7V.
OK! So I got 2 18650 batteries (INR 18650). Putting them in series gives me about 7.4VDC. Can use a voltage regulator or a voltage divider to drop the voltage to about 5VDC? I believe someone calculated that I will be pulling 3.3A. I found this:
Is it that simple to just drop the voltage and make it work? I'd have the volts, and the amps sooo... it should work, right? I'm worried I'm missing something since no one has suggested it. And even you have said that 1 is not enough and 2 is too many but didn't mention a voltage regulator to drop the voltage. What am I missing here?
@Grumpy_Mike: Thanks Grumpy! You've been a great help!
I'm worried I'm missing something since no one has suggested it.
Yes you need to drop the voltage.
No you can’t use a voltage divider it wastes far too much current about ten times the current you want to take from it.
Using a voltage regulator converts the excess power into heat, so a lot of battery power is wasted.
The most efficient thing to use is a buck converter, but at these sort of currents these are also the most expensive option.
Try 1 battery. Officially it's not enough, but I found it worked well for me.
Grumpy_Mike:
Yes you need to drop the voltage.No you can’t use a voltage divider it wastes far too much current about ten times the current you want to take from it.
Using a voltage regulator converts the excess power into heat, so a lot of battery power is wasted.
The most efficient thing to use is a buck converter, but at these sort of currents these are also the most expensive option.
Will this work?
link
PaulRB:
Try 1 battery. Officially it's not enough, but I found it worked well for me.
How much were you powering? If grumpy_mike is right, I'm powering as much as 3.7A of LEDs 
Will this work?
Well according to a simple reading of the specification it will but in practice I have found boards like that rarely give what they claim. The specification is too simplistic and you assume that you can get the maximum current out at the minimum voltage in, which is not the case. I have found cheap modules like this rarely give more than 1A.
traderjoe:
How much were you powering? If grumpy_mike is right, I'm powering as much as 3.7A of LEDs
Far fewer leds than you. But depending on the brightness, colours and patterns you use, your current consumption could be only a fraction of Mike's figure, which is based on all leds being max brightness white, I suspect. For example, a rainbow fade pattern, even at maximum brightness, would probably only be one third of that maximum.
