Measuring Current on LED's with a Multimeter

[cron.7]

Hey Guys,

I think I'm going insane. I want to measure how much current i need to operate an WS2812b RGB LED at full brightness (on all color channels).

I'm measuring the amount of current that flows through 2 LEDs. I don't really do hobby electronics that much, but doesn't this say, that one LED consumes up to 15mA ? Or  7,5mA? My problem with this value is, that it is supposed to be 60mA per LED. I found on various sites, that the amount of current that is needed for such an LED at full brightness is 60mA or (at 60 LEDs/m) 3,6A per m. On the product page of the LEDs that I've bought it states, that every SMD consumes 0,3W e.g. 60mA.

I have absolutely no clue what I should do. 

It would be very nice, if you could tell me what current is consumed per LED. And at best explain you calc and reasons. I'm stuck with this since February...

 

MEASURMENT PICTURE IMPORTANT TO SEE!!!!!!

Spoiler

 

Thanks for helpin' out!

schmidti

[Enderman]

It is never accurate to use a multimeter like that to measure current because of the internal resistance of the multimeter.

To measure current properly you should buy a clamp meter such as the UT210E which measures current using the Hall effect.

[cron.7]

Hm okay. I may buy it test with it and send it back to amazon, as I have no use to it more than that. But good to know. And yeah the Hall Effect is more precise and has no resistance.  May gonna try

[dany_boy]

45 minutes ago, Enderman said:

It is never accurate to use a multimeter like that to measure current because of the internal resistance of the multimeter.

To measure current properly you should buy a clamp meter such as the UT210E which measures current using the Hall effect.

Are you for real? Or just trolling?

[dany_boy]

28 minutes ago, 5chmidti said:

Hm okay. I may buy it test with it and send it back to amazon, as I have no use to it more than that. But good to know. And yeah the Hall Effect is more precise and has no resistance. Had the Hall Effect last year in school  May gonna try

Keep using the meter that you have, clamp meters are only good for fairly high currents otherwise the precision will screw you over big time.

[Enderman]

12 minutes ago, dany_boy said:

Are you for real? Or just trolling?

If you NEED to measure current with a multimeter you use a shunt.

Putting the multimeter in line with the circuit will never give accurate values.

 

10 minutes ago, dany_boy said:

Keep using the meter that you have, clamp meters are only good for fairly high currents otherwise the precision will screw you over big time.

Clamp meters such as the 210E can measure down to a few milliamps of both AC or DC current.

They are the most accurate way of measuring current.

Who taught you all that incorrect information?

[dany_boy]

Just now, Enderman said:

If you NEED to measure current with a multimeter you use a shunt.

Putting the multimeter in line with the circuit will never give accurate values.

 

Clamp meters such as the 210E can measure down to a few milliamps of both AC or DC current.

They are the most accurate way of measuring current.

Dude, come on man.

Ok Here we go. First of all I apologize if I come across as pedantic or condescending, it is really not my intention.

That being said, please inform yourself before giving ill advice on how to use electronic equipment. Take a look at the uni-t product specification for the clamp mete, particularly at the current capabilities:

DC Current (A)

2A/20A/100A

±（2%+3)

and consider this is a 2000 count meter. that means that if you are gonna measure a current in the order of (3*10^-2) A, your precision is gonna be ±2% (not a big issue), and ±3 counts which considering the order of magnitude of your measurement will equate to about 10% of additional relative error, which is huge! Plus the fact that even after zeroed out, clamp meters tend to drift. This one in particular has been found to drift in the 2A range up to 40 counts after 1 minute with crappy batteries and 10 counts in a minute with decent batteries (source). Again, for big currents, that is really not a big issue, but for the case here it is.

 

Now take a cheap-o multi-meter without a μA range. This is important because the  μA  range has a rather large 400-ish ohm shunt in series. The mA range on the other hand, will have a shunt of a couple of ohms for most meters. Looking back at the error, most crappy meters will have 2000 counts too, with a precision of 2% ± 5 counts (based on experience with them). At a current in the order of (3*10^-2) A, you will get ± 2% (again not a big deal), but now with the 200mA range those 5 counts will only equate to about 1% of added relative error.

 

So for small currents, a crappy meter is somewhat better than a clamp meter. And if @5chmidti does not have a clamp meter already, then that is money he could have spent buying a good multi-meter in the first place.

Cheers!

[cron.7]

5 minutes ago, dany_boy said:

Dude, come on man.

Ok Here we go. First of all I apologize if I come across as pedantic or condescending, it is really not my intention.

That being said, please inform yourself before giving ill advice on how to use electronic equipment. Take a look at the uni-t product specification for the clamp mete, particularly at the current capabilities:

DC Current (A)

2A/20A/100A

±（2%+3)

and consider this is a 2000 count meter. that means that if you are gonna measure a current in the order of (3*10^-2) A, your precision is gonna be ±2% (not a big issue), and ±3 counts which considering the order of magnitude of your measurement will equate to about 10% of additional relative error, which is huge! Plus the fact that even after zeroed out, clamp meters tend to drift. This one in particular has been found to drift in the 2A range up to 40 counts after 1 minute with crappy batteries and 10 counts in a minute with decent batteries (source). Again, for big currents, that is really not a big issue, but for the case here it is.

 

Now take a cheap-o multi-meter without a μA range. This is important because the  μA  range has a rather large 400-ish ohm shunt in series. The mA range on the other hand, will have a shunt of a couple of ohms for most meters. Looking back at the error, most crappy meters will have 2000 counts too, with a precision of 2% ± 5 counts (based on experience with them). At a current in the order of (3*10^-2) A, you will get ± 2% (again not a big deal), but now with the 200mA range those 5 counts will only equate to about 1% of added relative error.

 

So for small currents, a crappy meter is somewhat better than a clamp meter. And if @5chmidti does not have a clamp meter already, then that is money he could have spent buying a good multi-meter in the first place.

Cheers!

Well at this time this multimeter is all I got. I may borrow a multimeter from school next week, as they are more recent than the one I used and they may have a bigger Range of operation for measuring current. Yet any thoughts about my current measurements? I mean with the other multimeter I can do a new test with 30LEDs or 0,5m, but the test I did should indecate somewhat how much current is needed, yet it may not be as accurate due to resistance.

[cron.7]

As for research of power consumption of these types of LEDs, evereything I've seen so far indicates 0,3W or 0,06A, so 3,6A per meter (at 60 LEDs/m). yet the amount of current I've measured comes nowhere near that kind of power. There is a 4x difference as I see it.

[dany_boy]

9 minutes ago, 5chmidti said:

Well at this time this multimeter is all I got. I may borrow a multimeter from school next week, as they are more recent than the one I used and they may have a bigger Range of operation for measuring current. Yet any thoughts about my current measurements? I mean with the other multimeter I can do a new test with 30LEDs or 0,5m, but the test I did should indecate somewhat how much current is needed, yet it may not be as accurate due to resistance.

I just went out to eat, if you give me a couple of hours I can provide you with a detailed explanation and calculations. In a nutshell, at full brightness a single well fed led will consume a little under 60 mA. (Big emphasis on single and well fed)

Cheers!

[cron.7]

Just now, dany_boy said:

I just went out to eat, if you give me a couple of hours I can provide you with a detailed explanation and calculations. In a nutshell, at full brightness a single well fed led will consume a little under 60 mA. 

Cheers!

Thanks man, sure.  That amount of power aligns with all other sources I've read, yet I ask, because my measurments make no sense. better off asking  

[Enderman]

20 minutes ago, dany_boy said:

Dude, come on man.

Ok Here we go. First of all I apologize if I come across as pedantic or condescending, it is really not my intention.

That being said, please inform yourself before giving ill advice on how to use electronic equipment. Take a look at the uni-t product specification for the clamp mete, particularly at the current capabilities:

DC Current (A)

2A/20A/100A

±（2%+3)

and consider this is a 2000 count meter. that means that if you are gonna measure a current in the order of (3*10^-2) A, your precision is gonna be ±2% (not a big issue), and ±3 counts which considering the order of magnitude of your measurement will equate to about 10% of additional relative error, which is huge! Plus the fact that even after zeroed out, clamp meters tend to drift. This one in particular has been found to drift in the 2A range up to 40 counts after 1 minute with crappy batteries and 10 counts in a minute with decent batteries (source). Again, for big currents, that is really not a big issue, but for the case here it is.

 

Now take a cheap-o multi-meter without a μA range. This is important because the  μA  range has a rather large 400-ish ohm shunt in series. The mA range on the other hand, will have a shunt of a couple of ohms for most meters. Looking back at the error, most crappy meters will have 2000 counts too, with a precision of 2% ± 5 counts (based on experience with them). At a current in the order of (3*10^-2) A, you will get ± 2% (again not a big deal), but now with the 200mA range those 5 counts will only equate to about 1% of added relative error.

 

So for small currents, a crappy meter is somewhat better than a clamp meter. And if @5chmidti does not have a clamp meter already, then that is money he could have spent buying a good multi-meter in the first place.

Cheers!

You can clearly see how incorrect the vales are when measuring current with a multimeter here.

This is because the shunt changes the resistance of the circuit, and the current+voltage at each component.

A clamp meter even with just a 2A scale will still be more accurate because it does not affect the circuit at all.

 

I have a strong feeling you have never used a multimeter or clamp meter from the stuff you're making up...

[cron.7]

Oh boy.... this is gonna be interesting

[dany_boy]

9 minutes ago, Enderman said:

You can clearly see how incorrect the vales are when measuring current with a multimeter here.

This is because the shunt changes the resistance of the circuit, and the current+voltage at each component.

A clamp meter even with just a 2A scale will still be more accurate because it does not affect the circuit at all.

 

I have a strong feeling you have never used a multimeter or clamp meter from the stuff you're making up...

<sarcasm> sure I'm just pulling all of this out of my ass to troll people </sarcasm>

Ill be back in a couple of hours, then we'll keen on discussing.

[dany_boy]

Ok Im back (the seafood was absolutely awesome BTW).

2 hours ago, Enderman said:

I have a strong feeling you have never used a multimeter or clamp meter from the stuff you're making up...

HAHAHA it really cracked me up. Almost reminds me of the YouTube comment section. For your peace of mind, here are my meters:

And here is why I say that that the shunt resistor in the mA range of a meter is only a couple of ohms:

I have been doing electronics for the last 7 years of my life, 2 of my uncles (who have masters degrees in electrical telecommunications engineering), as well as my professor of electric circuits (with a PH.D in some digital electronics design subject) have taught me that the resistance in a good meter while measuring current, is not significant to the current measurement value for any given range. If you have another explanation, or you feel like I have been lied to, please explain yourself with actual theory, data, and calculations. I am more than happy to listen and change my mind. But the way I see it, a component with an ESR of 83 ohms is not gonna be hugely affected by 3 ohms more in a very significant way.

 

@5chmidti Here are some measurements that I took with a 30cm length of wire between the ardiuno and the LED strip:

1 LED all white:   3.945v -- 16.35mA

2 LEDs all white: 3.832v -- 31.31mA

5 LEDs all white: 3.702v -- 72mA (had to change the scale)

This is of course a non-ideal scenario, since I'm feeding the arduino + LEDs with a PC USB 1m long cable. But it should give you a rough idea of what to expect. Next week I might have time to collect some more controlled data. Let me know if you want me to do this full on "lab report" along with uncertainties and all that cool stuff. Cheers!

 

[cron.7]

9 hours ago, dany_boy said:

Ok Im back (the seafood was absolutely awesome BTW).

HAHAHA it really cracked me up. Almost reminds me of the YouTube comment section. For your peace of mind, here are my meters:

And here is why I say that that the shunt resistor in the mA range of a meter is only a couple of ohms:

I have been doing electronics for the last 7 years of my life, 2 of my uncles (who have masters degrees in electrical telecommunications engineering), as well as my professor of electric circuits (with a PH.D in some digital electronics design subject) have taught me that the resistance in a good meter while measuring current, is not significant to the current measurement value for any given range. If you have another explanation, or you feel like I have been lied to, please explain yourself with actual theory, data, and calculations. I am more than happy to listen and change my mind. But the way I see it, a component with an ESR of 83 ohms is not gonna be hugely affected by 3 ohms more in a very significant way.

 

@5chmidti Here are some measurements that I took with a 30cm length of wire between the ardiuno and the LED strip:

1 LED all white:   3.945v -- 16.35mA

2 LEDs all white: 3.832v -- 31.31mA

5 LEDs all white: 3.702v -- 72mA (had to change the scale)

This is of course a non-ideal scenario, since I'm feeding the arduino + LEDs with a PC USB 1m long cable. But it should give you a rough idea of what to expect. Next week I might have time to collect some more controlled data. Let me know if you want me to do this full on "lab report" along with uncertainties and all that cool stuff. Cheers!

 

Well, as I intend to feed my LEDs directly from my PSU (5V molex) it would be nice if you could tell me the amount of current you measured from that. And there is no need for a full report, but if you think there is a need for that.... I suggest just measure those three test samples from the 5V molex and if it says something in the neighbourhood of 60mA everythings fine for me, as i know how thick of a wire i need to purchase (running parallel 6*0,5m so i don't have voltage dropouts) which should carry, if 60mA is the amount of current needed, around 2A.

Thanks for helpin out mate

[Enderman]

9 hours ago, dany_boy said:

Ok Im back (the seafood was absolutely awesome BTW).

HAHAHA it really cracked me up. Almost reminds me of the YouTube comment section. For your peace of mind, here are my meters:

And here is why I say that that the shunt resistor in the mA range of a meter is only a couple of ohms:

 

I have been doing electronics for the last 7 years of my life, 2 of my uncles (who have masters degrees in electrical telecommunications engineering), as well as my professor of electric circuits (with a PH.D in some digital electronics design subject) have taught me that the resistance in a good meter while measuring current, is not significant to the current measurement value for any given range. If you have another explanation, or you feel like I have been lied to, please explain yourself with actual theory, data, and calculations. I am more than happy to listen and change my mind. But the way I see it, a component with an ESR of 83 ohms is not gonna be hugely affected by 3 ohms more in a very significant way.

 

@5chmidti Here are some measurements that I took with a 30cm length of wire between the ardiuno and the LED strip:

1 LED all white:   3.945v -- 16.35mA

2 LEDs all white: 3.832v -- 31.31mA

5 LEDs all white: 3.702v -- 72mA (had to change the scale)

This is of course a non-ideal scenario, since I'm feeding the arduino + LEDs with a PC USB 1m long cable. But it should give you a rough idea of what to expect. Next week I might have time to collect some more controlled data. Let me know if you want me to do this full on "lab report" along with uncertainties and all that cool stuff. Cheers!

 

The amount of current an LED uses increases exponentially as the forward voltage increases.

An extremely small difference in circuit resistance, even a few milliohms from a length of wire, will easily change the output of an LED, as well as the current it uses.

This is why you should ALWAYS use a clamp meter when measuring LED current, because it does not affect the circuit at all.

 

Clearly the  last 7 years of your life you haven't been dealing with LEDs, if you want to learn more then go to a forum such as BLF or CPF where people know how LEDs work and how to properly measure their current. Unfortunately it seems like I'm trying to argue with an entitled brick wall, so I'll just stop wasting my time and let you measure your current however you want.

[mariushm]

No,you don't use a clamp meter to measure mA's of current. The clamp's accuracy at such low magnetic fields is too low. Even the physical position of the wire carrying current to the led will affect measurement and the meter itself is maybe only 1-2% accurate on DC range.

 

Multimeters will have a burden voltage, they will affect the reading a bit when placed in series in the circuit to measure the current. As said above, it's because meter will place a resistor in series with the circuit and measure the voltage drop across the resistor.. 1v drop on a 1 ohm resistor means 1A of current in circuit, but the circuit also sees a DC voltage 1v lower.

Also,one must be aware of leads resistance and always short the leads and use the REL feature on the meter to remove leads resistance.

 

The internal resistance of a meter can also vary with the range but it's usually small. For example a 4000 count meter may have a 10 ohm resistor for up to 3.999 mA and then switch to 1 ohm for 4..399.9 mA

Anyway, unless you know your multimeter welll, it's safer to just NOT use resistance mode for such low measurements.

 

Just find a 0.01 ohm shunt resistor or even a 0.1 ohm resistor, place it in series with the power supply and then use the multimeter to measure the voltage drop across the resistor.

You have the plain boring formula V = I x R (voltage = current x resistance)

 

So if you measure 0.2v drop on the 0.1 ohm resistor then : 0.2v = I x 0.1 => I = 0.2/0.1 = 2A

 

Here's an example circuit here (Falstad circuit simuator) : http://tinyurl.com/ydh9vute

You have a 10 ohm load and a 10 mOhm shunt resistor and 12v power supply. The 10 ohm resistor eats 1.2A and there's a 12mV drop on the 10mOhm resistor so the load will see only 11.88v instead of 12v but it's good enough.  Put the mouse over the resistors and you should see at the bottom some values including voltage drop.

 

At < 100mA currents, with a 0.1 ohm resistor you'll have mV levels of voltage drop which any cheap meter will still be able to measure just fine (you have <0.5% on DC voltage on most $20+ meters) and which really won't affect the led.

The led's performance really won't change so much with just a few mV of forward voltage change.

 

Also note that you'd have to place the shunt resistor close to the led and measure the input voltage at the shunt and the voltage drop over the shunt... even a trace on a circuit board can have some resistance which could affect the accurate measurement.

A 1m usb cable will also have some resistance and some voltage drop over it, which will vary with the current... you won't have 5v at the shunt resistor, that's why you also have to measure the voltage.

 

-

 

The forward voltage of a pure led will change with HEAT. As the led heats up, the forward voltage will drop slightly and more current will flow through it... there's a potential for cascade failure (led heats too much, current increases, more energy flows, more heat is produced, led dies shorted so suddenly the voltage available for other leds in series is higher therefore the current may also increase.

For these reasons, led driver chips are preferred because they monitor the current flowing through the series of leds and limit the current constantly, there isn't just a "peak current limit" like what you'd do with a plain resistor.

 

However, those addressable leds aren't pure leds, those have a tiny chip which uses PWM to turn on and off individual leds on the die to achieve some brighness level and some mix of colors, and the chip itself does some sort of basic current monitoring (let's say with 5-10% accuracy) aiming to keep the leds from killing themselves on the die. They're sort of state machines, not even true "microcontrollers" , very basic :

 

 

Edited August 20, 2017 by mariushm
said 0.01 ohm twice, meant 0.01 or 0.1 ohm

[dany_boy]

8 hours ago, Enderman said:

The amount of current an LED uses increases exponentially as the forward voltage increases.

An extremely small difference in circuit resistance, even a few milliohms from a length of wire, will easily change the output of an LED, as well as the current it uses.

This is why you should ALWAYS use a clamp meter when measuring LED current, because it does not affect the circuit at all.

 

Clearly the  last 7 years of your life you haven't been dealing with LEDs, if you want to learn more then go to a forum such as BLF or CPF where people know how LEDs work and how to properly measure their current. Unfortunately it seems like I'm trying to argue with an entitled brick wall, so I'll just stop wasting my time and let you measure your current however you want.

Hot damn! salt much? I feel like we both need to calm down and watch this video instead of turning the forums into youtube comment section:

That being said, I still stand by the point I said, specially given the lack of concrete evidence that you have provided. Either way, on Monday I'll borrow some equipment from my Uni labs and perform measurements with a remote sense feature to eliminate the small resistance in the wires. I will use a clamp meter too if I can and we can them compare concrete results. 

Cheers!

[bob345]

As someone who has designed integrated current measurement and voltage measurement circuits professionally, a current shunt is the ONLY way to accurately measure relatively low currents. The change in resistance makes next to no difference in reality, and even if you need that extra accuracy the data sheet for the current shunt you are using will give you a lookup table you can use to compensate for that change in resistance.

[cron.7]

So the probably best way to do this is to use a resistor (shunt/measurement), because we only need to measure the voltage drop across this resistor in order to calculate the amount of current that flows through it. So there is no resistor from the Multimeter that potentially affects the measurement. 

Now some short questions about that:

• "shunt" necessary? or does a regular resistor be enough (I know shunt is validated for measuring)

should I go (searched a bit, I found those very quickly) for

• 0,1 Ohm (predicted voltage drop: 0,006 V)
• 9 Ohm (predicted voltage drop: 0,54 V)

Would a huge voltage drop lead to more current being drawn from the psu, so the amount of power that is necessary will add up? By using the 9 Ohm resistor this would potentially add up to additionally +6,6 mA. Which would be still close enough to the predicted amount of Current (if it adds up to 66 mA)

 

Cheers 5chmidti

[mariushm]

A current shunt is just a term used for resistors specially designed for this purpose, measuring current,

They're basically sightly fancier resistors.that can handle variations in temperature better, as in their value doesn't change much

as the resistor warms up or cools down (from current flowing through it)

 

For example, cheap 0.1 ohm resistors may be 1% accurate out of the box, so if you randomly pick resistors from a box you'll get 0.1 ohm +/- 1% and then the resistor may be let's say 0.1003 ohm at 25c (room temperature), at let's say 60c in circuit after current flowing through it for an hour or so, its value may drift down to 0.0997 ohm

So if you design something that must be really accurate and precise, such high drift with temperature is not desired.

A current shunt will drift less with temperature change and generally will be more accurate out of the box, let's say +/- 0.1% 

 

Anyway... you don't need super precise measurements and your current values are not so high as to heat up resistors and make them change their value significantly

You can use plain resistors, just try to get 1% precision or better, and ideally thin or think metal film type (they're usually blue or dark gray with stripes of color), there's a slightly cheaper carbon kind which will also work just fine but these thin film resistors are just overall better.

 

You should aim for as little resistance as possible, my advice is under 1 ohm.

Your aim should be to create as little voltage drop as possible, but still big enough at low currents that your multimeter will be able to measure it.

For example, if you have a 2000 count multimeter (cheap < 10$ meters usually are, 2000 count means they usually change range at 200mV, 2v, 20v, 200v), you may want to measure (10mV) 0.001v at 1mA and 50mV (0.005v) at 5mA because the multimeter may not be capable of finer readings.  If that's the case V=IxR so 0.005v = 0.005A xR => R = 0.005/0.005= 1 ohm resistor

With a better multimeter you may use common values like 0.47 ohm or 0.22 ohm or 0.1 ohm

 

You can produce various values by placing two or more resistors of same value in parallel in circuit (two resistors in parallel = half the resistance) so for example you can have two 1 ohm resistors in parallel to have a 0.5 ohm value

 

You can always calculate the current with the formula I told you V=IxR, so R can be anything ... 0.1 ohm and 1ohm are just easier, less math in your brain

 

You should be able to find 1ohm resistors and even smaller in local stores. If you don't look at electronic parts distributors like Digikey, Mouser , Newark / Farnell, TME.eu (for EU people mostly)

Most of the above will charge around 5-10$ for shipping (one time fee, no matter how many items you buy), so obviously it sucks if you only order ONE resistor but you could add other things you may need.

Here's digikey's resistor selection ...tens of thousand of them (hundreds of thousands of them) : https://www.digikey.com/short/31rcw8  (through hole, as in with leads)  and   https://www.digikey.com/short/31rcdq  (surface mount)

 

[cron.7]

8 hours ago, mariushm said:

A current shunt is just a term used for resistors specially designed for this purpose, measuring current,

They're basically sightly fancier resistors.that can handle variations in temperature better, as in their value doesn't change much

as the resistor warms up or cools down (from current flowing through it)

 

For example, cheap 0.1 ohm resistors may be 1% accurate out of the box, so if you randomly pick resistors from a box you'll get 0.1 ohm +/- 1% and then the resistor may be let's say 0.1003 ohm at 25c (room temperature), at let's say 60c in circuit after current flowing through it for an hour or so, its value may drift down to 0.0997 ohm

So if you design something that must be really accurate and precise, such high drift with temperature is not desired.

A current shunt will drift less with temperature change and generally will be more accurate out of the box, let's say +/- 0.1% 

 

Anyway... you don't need super precise measurements and your current values are not so high as to heat up resistors and make them change their value significantly

You can use plain resistors, just try to get 1% precision or better, and ideally thin or think metal film type (they're usually blue or dark gray with stripes of color), there's a slightly cheaper carbon kind which will also work just fine but these thin film resistors are just overall better.

 

You should aim for as little resistance as possible, my advice is under 1 ohm.

Your aim should be to create as little voltage drop as possible, but still big enough at low currents that your multimeter will be able to measure it.

For example, if you have a 2000 count multimeter (cheap < 10$ meters usually are, 2000 count means they usually change range at 200mV, 2v, 20v, 200v), you may want to measure (10mV) 0.001v at 1mA and 50mV (0.005v) at 5mA because the multimeter may not be capable of finer readings.  If that's the case V=IxR so 0.005v = 0.005A xR => R = 0.005/0.005= 1 ohm resistor

With a better multimeter you may use common values like 0.47 ohm or 0.22 ohm or 0.1 ohm

 

You can produce various values by placing two or more resistors of same value in parallel in circuit (two resistors in parallel = half the resistance) so for example you can have two 1 ohm resistors in parallel to have a 0.5 ohm value

 

You can always calculate the current with the formula I told you V=IxR, so R can be anything ... 0.1 ohm and 1ohm are just easier, less math in your brain

 

You should be able to find 1ohm resistors and even smaller in local stores. If you don't look at electronic parts distributors like Digikey, Mouser , Newark / Farnell, TME.eu (for EU people mostly)

Most of the above will charge around 5-10$ for shipping (one time fee, no matter how many items you buy), so obviously it sucks if you only order ONE resistor but you could add other things you may need.

Here's digikey's resistor selection ...tens of thousand of them (hundreds of thousands of them) : https://www.digikey.com/short/31rcw8  (through hole, as in with leads)  and   https://www.digikey.com/short/31rcdq  (surface mount)

 

So my multimeter is the one featured in the picture from the OP, so that is my measurment range. So mV won't be a problem. But I think I should go with 1 Ohm since measuring 6 mV is on the edge of my measurment range (potentiall inaccurate) and using 1 Ohm would result in a voltage drop of 60 mV which would be somewhat more accurate. I'll look into those stores you mentioned and look if my lokalstore has some too.

[cron.7]

23 hours ago, mariushm said:

A current shunt is just a term used for resistors specially designed for this purpose, measuring current,

They're basically sightly fancier resistors.that can handle variations in temperature better, as in their value doesn't change much

as the resistor warms up or cools down (from current flowing through it)

 

For example, cheap 0.1 ohm resistors may be 1% accurate out of the box, so if you randomly pick resistors from a box you'll get 0.1 ohm +/- 1% and then the resistor may be let's say 0.1003 ohm at 25c (room temperature), at let's say 60c in circuit after current flowing through it for an hour or so, its value may drift down to 0.0997 ohm

So if you design something that must be really accurate and precise, such high drift with temperature is not desired.

A current shunt will drift less with temperature change and generally will be more accurate out of the box, let's say +/- 0.1% 

 

Anyway... you don't need super precise measurements and your current values are not so high as to heat up resistors and make them change their value significantly

You can use plain resistors, just try to get 1% precision or better, and ideally thin or think metal film type (they're usually blue or dark gray with stripes of color), there's a slightly cheaper carbon kind which will also work just fine but these thin film resistors are just overall better.

 

You should aim for as little resistance as possible, my advice is under 1 ohm.

Your aim should be to create as little voltage drop as possible, but still big enough at low currents that your multimeter will be able to measure it.

For example, if you have a 2000 count multimeter (cheap < 10$ meters usually are, 2000 count means they usually change range at 200mV, 2v, 20v, 200v), you may want to measure (10mV) 0.001v at 1mA and 50mV (0.005v) at 5mA because the multimeter may not be capable of finer readings.  If that's the case V=IxR so 0.005v = 0.005A xR => R = 0.005/0.005= 1 ohm resistor

With a better multimeter you may use common values like 0.47 ohm or 0.22 ohm or 0.1 ohm

 

You can produce various values by placing two or more resistors of same value in parallel in circuit (two resistors in parallel = half the resistance) so for example you can have two 1 ohm resistors in parallel to have a 0.5 ohm value

 

You can always calculate the current with the formula I told you V=IxR, so R can be anything ... 0.1 ohm and 1ohm are just easier, less math in your brain

 

You should be able to find 1ohm resistors and even smaller in local stores. If you don't look at electronic parts distributors like Digikey, Mouser , Newark / Farnell, TME.eu (for EU people mostly)

Most of the above will charge around 5-10$ for shipping (one time fee, no matter how many items you buy), so obviously it sucks if you only order ONE resistor but you could add other things you may need.

Here's digikey's resistor selection ...tens of thousand of them (hundreds of thousands of them) : https://www.digikey.com/short/31rcw8  (through hole, as in with leads)  and   https://www.digikey.com/short/31rcdq  (surface mount)

 

So my local store got something. Yes, it is not a shunt resistor, but should be suitable, right? Its a metal layer(/film?) resistor with 1 Ohm, 1W, +-1% and 50 ppm for 0,18€ : https://www.conrad.de/de/metallschicht-widerstand-1-axial-bedrahtet-0414-1-w-mfr1145-1-st-419206.html . I could buy it on Thursday and try to test it as soon as possible.

If you think this resistor is good enough for this amount of power let me know. As long as there is not more than around +-3 mA (or volt) of a false reading I'm totally okay with that.

And because this is 1 Ohm the measurment will be pretty straight forward. If I get around 60 mV of voltage drop I have all I need to know (I'll do one or two tests more with 2 and 3 leds, but it should be confirmed by then). If those 60 mA are confirmed it's over with this measurement craze...

[mariushm]

Yes, sure, it's perfectly fine.

Even better that it's rated for 1w, as it won't heat up much with low currents, and 50ppm is a good value.

Power dissipated = Current²xResistance so for example, even at 100mA (0.1A) , you're only going to produce 0.1x0.1x1= 0.01w which will barely warm up the resistor