Measuring Current Draw from LED ring (WS2812B)

Hi everyone,

This is confusing me. I have read that Adafruit's neopixels (WS2812B) can draw up to 60mA per pixel (LED) when displaying white at full brightness. (20mA for each colour - RGB).

I am using ICStation's WS2812B LED ring which has 16 LEDs on the ring.

I am measuring the voltage drop on a 1.2ohm resistor in series with the 5V pin of the LED ring, while lighting up an increasing number of LEDs (white at full brightness).

Here are my results:

1 LED = 74 mV 2 LEDs = 120 mV 16 LEDs = 234 mV ??

I would have expected that the current draw (and voltage drop) would have been much higher than that reading when ALL leds are lit up ?? What am I doing wrong ?

I am powering the LED ring using an external 5V 4A power supply.

Regards Scott

they do draw ~60mA each

2 LEDS = 120mV / 1.2 ohms = (drum roll) 120mA 1 LEDS = 0.074V / 1.2 ohms = 61.7 mA

Now, keep in mind that @5V Vin, a 1.2 ohm resistor will reduce the overall current and hence the overall Vf of your LEDS which will draw less current. The chart proves the relationships between VIRP

Thanks rinkrides… I had no problem with the first two results… it was the last one that I was confused about.

The chart proves the relationships between VIRP

The chart proves nothing. It might define relationships but it does not prove them.

The problem with your reading is that the current is being switched in order to control the brightness and so is AC. Therefore you are not getting an accurate reading of this high speed AC voltage on your meter.

16x 60mA is almost 1Amp. That will drop 1.2volt across the 1.2ohm current sense resistor. That current can not be reached, because you're underpowering the LEDs (3.8volt).

The fact that ICStation shows them connected to the 5volt pin of an Uno shows that they now little about electronics. When powered from USB, the polyfuse will trip at full brightness, and/or the Arduino will reset. When powered from the onboard regulator it will surely reset. Two of those rings are even shown. They should be powered from an external 5volt source. Leo..

Thanks Grumpy Mike... so how can I accurately measure the current draw? Do I just assume that the current draw is going to be 16 x 60 mA ?

Wawa - yes - I was powering using an external power supply 5V 4A - when doing these measurements.

Regards Scott

so how can I accurately measure the current draw?

Use a proper AC millivoltmeter. They are expensive. Alternately measure the voltage on an oscilloscope.

Do I just assume that the current draw is going to be 16 x 60 mA ?

Yes. At least for all on maximum and white.

Ok Thanks Grumpy Mike - that is good to know.

I am saving up to buy an oscilloscope - so will test it out when I get one. Is there any chance I could damage an oscilloscope with this kind of measurement ? Any words of caution ??

Is there any chance I could damage an oscilloscope with this kind of measurement ?

Not the scope no.

However where is the resistor? One of the scope leads is connected to mains ground. If the resistor in in the positive lead and the supply is not floating then you could short out the supply through the scope.

The 1.2 ohm resistor is on a breadboard sitting between the Positive terminal of the Power supply (5V 4A Plug-pack connected to wall), and the 5V pin of the LED ring.

ScottC: ... the Power supply (5V 4A Plug-pack connected to wall), ...

Two pin plug packs are by definition, floating.

Not so however, if they have three pins, but that is very uncommon for less than the size of laptop power supplies.

I'd first save up to get a proper multimeter with 10A current range...

The other way to measure this with a scope, that has two channels or more, is to connect a channel to either side of the resistor. Then switch it to differential mode and the display will be the difference between the two sides of the resistor.

This is plug pack power supply that I am using - yes it is a two pin.

http://www.altronics.com.au/p/m8911a-powertran-5v-dc-4a-fixed-2.1mm-tip-appliance-plugpack/

MarkT:
I’d first save up to get a proper multimeter with 10A current range…

Hi Mark T:

You are right - I will probably need to get one eventually.

Would it be able to measure the current draw of these LED rings ?

While my multimeter would not measure high voltages… what else would it do to warrant its purchase that the oscilloscope and my present multimeter could not do ?

When a pixel is programmed to full on (255) then its current can be deduced with a shunt resistor and a voltmeter. When it's programmed otherwise (0-254), something dimmer, the current going through it, when there is current going through it, is no less.

Thanks Runaway Pancake,

I don’t think I can really comment - as it has already been established that my method of using a 1.2 ohm resistor and my multimeter is not suitable for this task.

However, just in case anyone wanted to know what values I got while doing some of these tests - please read below:

While illuminating a SINGLE PIXEL (or LED) on the LED strip while keeping the rest off (brightness set to 0) - here are my initial results.

RGB (255, 0, 0) : Red : 32 mA
RGB ( 0, 255, 0) : Green: 32 mA
RGB ( 0, 0, 255) : Blue: 32 mA

RGB (255, 255, 255) : White: 61 mA

So then I decided to just compare a Blue pixel vs a White pixel, and adjusted the Brightness:
The table and graph can be seen (attached).

As you can see - my results are not consistent with your statement, nor do the individual colours add up to the current draw from displaying white (RGB all on)… so I think until I can get my hands on a suitable / accurate measuring device, I am going to have to put this one on hold.

LED Table.jpg

LED Chart.jpg

I didn't say that a 1.2ohm resistor was appropriate. The greater the current through the shunt, the greater the voltage across the shunt - voltage that won't be across the WS2812, skewing things. Get a milliammeter or use less shunt resistance.

Actually, you mean an ammeter. Milliammeters have a high shunt resistance already.

That’s the difficulty with common multimeters, no 2 A range (but the 10 A range is usually sufficient).

Paul__B: Actually, you mean an ammeter. Milliammeters have a high shunt resistance already.

That's the difficulty with common multimeters, no 2 A range (but the 10 A range is usually sufficient).

You know, you're a pain. Actually... I posted what I meant and I meant what I posted.