Measuring Current on LED's with a Multimeter

[cron.7]

Oh boy,

so I bought the 1 Ohm resistor and what did I got? 0,03 V across the resistor....

this would mean: I = U/R so I = 0,03 V / 1 Ohm = 0,03 A, but I had 2 LEDs on this one so 0,015 A... so either the software does not allow to draw the full amount of power (@dany_boy) which I doubt or something other is wrong here. It's not the multimeter (it can read 1,5 and 5 V, but maybe it's to bad for this little voltage) and I would have probably noticed if it was my PSU. So LEDs maybe? I don't know...

This was supposed to be easier...

Well lets see how this one goes. 

And maybe I just should go out and buy wires that can carry the amount of current with the official measurements (60 mA). Any thoughts? Would be some longer some shorter wires, but all for max 1 m of LEDs.

 

Cheers Guys

[mariushm]

In case you didn't do it, set your multimeter on the lowest range (I think 2v in your case). That should give you the most decimals after 0, should be at least two decimals for the meter in that picture.

 

If the leds are the kind with brains ( the kind where you set the brightness of each red green and blue by sending data to them), then all you need to be concerned about is making sure they get the minimum voltage they need.

Some versions need at least around 4.5v to work otherwise they become flaky (the chip inside can't function properly). other versions need at least around 3.2v .. 3.5v, (which is the minimum voltage needed to light up the blue led plus maybe 0.2-0.4v)

 

In one meter of cable, assuming you're going to have one led every 3cm or so, you're looking at around 35 leds ... let's round it down to 25 leds, because math works easier this way.

 

If you exagerate a bit and go with 20 mA per led, then your maximum current will be 20 mA x 25 = 500mA which just happens to be the maximum a usb port is supposed to be able to provide (but in reality most can do up to 1A or even slightly more)

 

So what wire should you use... I basically strip  a chunk of ethernet cable and get 8 wires inside, each with different color insulations ... but you can use other kinds of wires.

If your interest is to make sure the leds get a minimum voltage, then you can re-use this basic formula to measure the voltage drop on a cable of a various thickness.

In computers, most cables are AWG style , where AWG is an of American standard which defines the resistance per mile (or whatever americans use) of the cable and other properties... here's a AWG chart: https://en.wikipedia.org/wiki/American_wire_gauge#Tables_of_AWG_wire_sizes

 

In that chart, AWG18 is the thickness of most power supply cables and AWG28 or AWG30 is the thickness of those wires in IDE cables (if you're old enough to remember them)

 

I cut out from Wikipedia the thicknesses most commonly used :

 

 

So let's say you go with AWG24 (the thickness of wires inside an ethernet cable and also on most USB cables). You can look up in the picture above and you can see that the wire has a wire resistance of 84.22 mOhm per meter, but we can round it up to 85 to make the math easier.

So let's say you want 2 meters of cable between you light bar and your power supply, in which case you're going to have 4 meters of cable between power supply and leds (because current has to go to the leds and then back to the psu), so your four meters of wire act as a resistance equal to

R = 4 meters x 85mOhm = 340 mOhm

and basically your circuit is like this :  power supply -> 0.34 ohm resistor - > leds

 

So now again we can use the formula V = I x R and we know the resistance of the wire is 0.34 ohm and the current will be up to 500mA or 0.5A) and therefore we know we may lose up to  V = 0.5v x 0.34 ohm = 0.17v on the cable.

Therefore, with 2 meters of AWG24 wire (pair of wires) and a 5v power supply (usb, phone charger) you know your leds will see at least 4.83v and they'll work.

If you're going to use more leds and therefore more current , redo the math depending on how much length of wire you're going to have and how much current all the leds will use and what's the minimum voltage your leds will be happy with.

 

If your leds will work just fine even with 4v then you don't have to worry much about the thickness of the wires.

 

 

 

 

 

 

[cron.7]

1 hour ago, mariushm said:

In case you didn't do it, set your multimeter on the lowest range (I think 2v in your case). That should give you the most decimals after 0, should be at least two decimals for the meter in that picture.

 

If the leds are the kind with brains ( the kind where you set the brightness of each red green and blue by sending data to them), then all you need to be concerned about is making sure they get the minimum voltage they need.

Some versions need at least around 4.5v to work otherwise they become flaky (the chip inside can't function properly). other versions need at least around 3.2v .. 3.5v, (which is the minimum voltage needed to light up the blue led plus maybe 0.2-0.4v)

 

In one meter of cable, assuming you're going to have one led every 3cm or so, you're looking at around 35 leds ... let's round it down to 25 leds, because math works easier this way.

 

If you exagerate a bit and go with 20 mA per led, then your maximum current will be 20 mA x 25 = 500mA which just happens to be the maximum a usb port is supposed to be able to provide (but in reality most can do up to 1A or even slightly more)

 

So what wire should you use... I basically strip  a chunk of ethernet cable and get 8 wires inside, each with different color insulations ... but you can use other kinds of wires.

If your interest is to make sure the leds get a minimum voltage, then you can re-use this basic formula to measure the voltage drop on a cable of a various thickness.

In computers, most cables are AWG style , where AWG is an of American standard which defines the resistance per mile (or whatever americans use) of the cable and other properties... here's a AWG chart: https://en.wikipedia.org/wiki/American_wire_gauge#Tables_of_AWG_wire_sizes

 

In that chart, AWG18 is the thickness of most power supply cables and AWG28 or AWG30 is the thickness of those wires in IDE cables (if you're old enough to remember them)

 

I cut out from Wikipedia the thicknesses most commonly used :

 

 

So let's say you go with AWG24 (the thickness of wires inside an ethernet cable and also on most USB cables). You can look up in the picture above and you can see that the wire has a wire resistance of 84.22 mOhm per meter, but we can round it up to 85 to make the math easier.

So let's say you want 2 meters of cable between you light bar and your power supply, in which case you're going to have 4 meters of cable between power supply and leds (because current has to go to the leds and then back to the psu), so your four meters of wire act as a resistance equal to

R = 4 meters x 85mOhm = 340 mOhm

and basically your circuit is like this :  power supply -> 0.34 ohm resistor - > leds

 

So now again we can use the formula V = I x R and we know the resistance of the wire is 0.34 ohm and the current will be up to 500mA or 0.5A) and therefore we know we may lose up to  V = 0.5v x 0.34 ohm = 0.17v on the cable.

Therefore, with 2 meters of AWG24 wire (pair of wires) and a 5v power supply (usb, phone charger) you know your leds will see at least 4.83v and they'll work.

If you're going to use more leds and therefore more current , redo the math depending on how much length of wire you're going to have and how much current all the leds will use and what's the minimum voltage your leds will be happy with.

 

If your leds will work just fine even with 4v then you don't have to worry much about the thickness of the wires.

 

 

 

 

 

 

So, my LEDs work just fine at around 5 V (measured when it did this measurment). The LEDs are WS2812b LEDs, which means they are digital as you said. I have 60 LEDs / m and I will run 3,3 m in parallel for every 1m or so to provide a stable enough Voltage. So 3,6 A per 1 m, which will be collectively connected to the PUS.  I've already looked at this table and I probably will need 18 AWG or better for some parts. Power will be directly drawn from the PSUs Molex 5 V and the signal comes directly from the Arduino.

 

My real concern here is the amount of current the wires have to handle (with 18 AWG i should be fine for the 1 m parallel connection as it is supposed to draw only 3,6 A and the 18 AWG should do around that, right? ) and the amount it has to draw from my PSU (the 3,3 m strip that is my main one, should only be drawing 12 A) if it is on full white. It probably wont, but I'd like to know how much headroom I have).  I'd like to confirm those stated numbers of around 60 mA per LED to be safe. 

And the resistance implied by the wires is not that huge. There should be (if I take 18 AWG) around 0,075 V of loss per Meter if I run them how I at the moment think I will. so this should not be a problem.

[mariushm]

Correct. No problem if you go with AWG18 wires.

 

One minor observation I would have: modern power supplies aren't really capable of providing a lot of current on 5v, because they're "optimized" for 12v which is used by all power hungry items in a pc (video card, cpu, mechanical hard disk motors, fans ..  in order of power consumption)

 

Most modern power supplies have a budget of 100-120 watts from which they can produce 3.3v up to some amount of current, and 5v up to some amount of current.

For example, if you look at the label of a power supply you may see 3.3v at 20A and 5v at 20A , but in total maximum power of 120 watts - that means if the pc uses 10A on 3.3v (33 watts), then it can only use 120w-33w = 87 watts to create 5v, so even though the label says it can do 20A , in reality it will top out at 87w / 5v = 17.4A

 

In modern computers, 3.3v is used very little, let's say maybe 3-5A so around 10-15w which means usually power supplies have no problems doing 20A or whatever value is there on 5v, but keep in mind the motherboard (usb ports, chipsets, onboard audio and network, electronics on mechanical drives and SSDs) these all use some amount of 5v. If you reserve let's say 5-8A on 5v for the computer parts, you may be left with quite a small budget for your leds.

 

There are tiny DC-DC converters you can buy and which convert 12v to 5v with very high efficiency, here's this one with 92% efficiency and maximum 3A of current : http://www.ebay.com/itm/DC-DC-Buck-Step-Down-Converter-Regulator-Power-Supply-Module-3-3v-5v-9v-12v-3A-/122050107357

 

Yeah they're from China so it may take a while to reach you, and you have to select the one which outputs 5v from the drop down list. Really, they're adjustable, they just don't have a potentiometer - if you really want to you could always change the value of a resistor on the board to adjust the output voltage.

 

There's also online distributors of electronics like Farnell which should ship across Europe and may even have a version of their store in your language (you linked to Conrad which is think is German, that's why I mention it).

Here's a link to DC-DC converters which can convert 12v (or some other voltage to 5v or other voltage smaller than input voltage) : very long link typical for farnell store

For example, this converter takes 8v..14v and outputs 0.75v .. 5.5v at up to 10A : http://uk.farnell.com/murata-power-solutions/okx-t-10-d12n-c/psu-pol-sip-10a-12-vin-neg-on/dp/1705262

Super easy to use ,  on/off pin (leave unconnected for power on, or connect to input voltage to be off) , trim pin (connect a resistor between it and a ground to set output voltage, it explains how in datasheet on the page) , input voltage, ground and output voltage .. piece of cake.

 

[cron.7]

Yeah, some ohter componets occupy the 3,3 and 5 V rail, but as I not want to doo all LEDs on white, beacsue I could just turn my light on that, I don't think I would run into a limit from the PSU. The current PSUs I have on my watchlist (because mine is not that quiet and I want to go more quiet with my system) are about 20 A max on 5 V so technically there could be a limitation, but I want to look further, as there are PSUs with 25 A max on 5 V which would fit my purpose much better.

I also did thought of DC/C Converters, but having the power drawn directly from the PUS would be better, and I wouldn't need to buy an extra converter. The upside to this would be that I could have more than enough power to drive my LEDs.

But for now I'll stick with the PSUs 5 V.

So your opinion on this measurment craze is to buy the wires that fit best for the stated amount of current that each LED consumes at full white brightness.