# how to measure current ot the led



## aljsk8 (Jan 24, 2007)

how do i measure current going to an led

i have a multimeter and i know where to set the dial etc and its digital so i know how to read it

all i need to know is do i just turn the light on and put the probes on the positive and negitive while its running?

will this damage the led?

is this the correct way?

id like to try this in the next 1/2hr if possable

thanks


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## greenLED (Jan 24, 2007)

You gotta place the DMM leads in series (as if it were part of the wire connecting the LED to the rest of the circuit) with one of the LED's contacts. 

|-----O DMM O------LED------------|
|_____________Batt______________|


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## aljsk8 (Jan 24, 2007)

dmm?

so i take it in normal flashlight terms this would require de-soldering one led leg/ tab/whatever and putting the probes between the leg and the solder bit on the driver?

i was hopeing i could just put it across in paralell while its all still in one peice - i take it this wont work?

thanks


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## Doug S (Jan 24, 2007)

aljsk8 said:


> how do i measure current going to an led
> 
> i have a multimeter and i know where to set the dial etc and its digital so i know how to read it
> 
> ...



Your possible risk is not to the LED. Some, but not all, electronic driver circuits for LEDs can be damaged if operated without a load. As Greenled as said, the meter needs to go in series with the LED, if you even monentarily operate the circuit without the load, example, you meter probes slip off of where you have them connected, your circuit may be toast.


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## aljsk8 (Jan 24, 2007)

so if the led is attached to the driver curcuit and is switched on - is it not under load

this is what i had in mind

ignore the fullstops - they were to get the picture to work

--------
/........./
/........./
/ dmm. /
/........ /-----------(+)---------\
/........./ ................................o (led switched on)
/........./------------(-)--------/
--------

so if the led is on and you hold the probes on the leds is it not under load?

Alex


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## greenLED (Jan 24, 2007)

aljsk8 said:


> dmm?


DMM- fancy-shmancy for multimeter



aljsk8 said:


> ...this would require de-soldering one led leg/ tab/whatever and putting the probes between the leg and the solder bit on the driver?


Exactly



aljsk8 said:


> i was hopeing i could just put it across in paralell while its all still in one peice - i take it this wont work?


It'll give you "a" reading, but not "the" reading you're after.



aljsk8 said:


> so if the led is attached to the driver curcuit and is switched on - is it not under load


Nope. "Under load" means the light is switched on. Some circuits burn up if you apply power and the LED is connected.


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## MrAl (Jan 24, 2007)

Hi there,

The best way to measure current is indirectly. That is, connect a 0.1 ohm resistor
in series with the LED, then turn the unit on, then measure voltage across
the resistor and use ohms law to calculate the current. The 0.1 ohm resistor
should be SOLDERED in place so that it can not ever come loose, because if it
does, as mentioned above the driver circuit may burn out in less than a half of
a second.
One of the problems with using the meter without a resistor is that if you set
the meter on the wrong current range or a lead comes loose the circuit burns
out and you cant use the flashlight until it's repaired.

The formula to use with a 0.1 ohm resistor is this:
I=V/0.1 or
I=V*10

so if you measure 0.035 volts across the resistor then that means there is 350ma
flowing through the LED.

Play it safe and use a resistor, and solder it in place before turning on the light.


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## Illum (Jan 24, 2007)




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## MrAl (Jan 24, 2007)

Hi there,

I would be careful about that last circuit...some drivers can not take a short circuit
on the output.


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## jrmcferren (Jan 24, 2007)

In the last circuit, disconnecting the LED is an unnecessary step, electricity will follow the path of least resistance (in the case of a meter shorting across an led) which is the meter. The AMPs will read higher for a resistor based supply, don't know the results for an electronic supply.


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## abvidledUK (Jan 24, 2007)

Not quite to the LED, unless it's direct drive...

Just replace the tail cap (switched or twisty) with the leads from the multimeter, in DC High Current (10A) mode, to get a direct measurement of the current taken from the battery, through the LED and any electronics.

ie, tail cap off, leads one on neg of battery, one lead to end of torch casing thread.


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## TJZ (Jan 24, 2007)

MrAl, greenLED, Doug S, jrmcferren, abvidledUK, 

I have a question if anyone of you knows the answer.
Sounds like you know a lot on this subject.

When you take a current reading in series how does having
the wrong current range on the meter burn out the circuit?
Is there voltage induced from the meter going to the circuit also
when measuring current in series and if you have the wrong current range it
is even higher and it can blow out the circuit?

And if you have the proper current range on the meter, does this mean
no voltage is induced from the meter going to the circuit?

Thanks, 

Tom


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## abvidledUK (Jan 25, 2007)

Usually the lower current ranges on the multimeters have some sort of low current fuse, to protect the fine wires in the meter from burning out. Fuse rated at less current flow than the fine wires.

High current range usually measures voltage across built in resistor, converting to amps.


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## aljsk8 (Jan 25, 2007)

abvidleduk - im interested in the tailcap off method - this is a lux iii so i dont want to start dismantling it but taking the cap off sounds good

1 question do i turn it on when i do all this? (im guessing yes)

will it give the same results as cutting the led and doing the in series thing?

thanks


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## aljsk8 (Jan 25, 2007)

thanks for all the help every one - its very usefull especially the pics (illum) but i think its confused me more as to what i need!

i think i need to clarify why im doing this as im not sure if it is the LED current i want

i thought the driver current and the led current would be the same - but mabee not?

anyway heres what i want to do...

i have a stainless steel civictor (lux iii) i want to put the ssc p4 led in it

so i want to know what current the driver is at the different modes (med, low, high) so i can guestimate the lumens (if its 350ma on high then i can guess ill be getting around 90 - 115 lumens on this mode) and i also wat to know how hard the ssc will be being pushed - for the issue of heat - as stainless is a poor conductor.

so will this tailcap method work?

also what is dc high 10a mode? my multi meter just has numbers around it like 2, 20, 200 etc but not high i dont think
ill post a photo of what ive got later so people can help me

i didnt know this was so complex! (well its probably not - im just stupid!)

Alex


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## abvidledUK (Jan 25, 2007)

aljsk8 said:


> abvidleduk - im interested in the tailcap off method - this is a lux iii so i dont want to start dismantling it but taking the cap off sounds good
> 
> 1 question do i turn it on when i do all this? (im guessing yes)
> 
> ...



If the switch, or twisty operation is in the tailcap, it will automatically complete the circuit when you put the 2 multimeter leads across the neg battery and the end threads you have now exposed.

If side or front switch or twisty, yes, turn it on !!

It will only show total current taken by torch, so current to led's may be less depending on electronics, if any.


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## abvidledUK (Jan 25, 2007)

.


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## aljsk8 (Jan 25, 2007)

heres the problem - 

i want to learn to mod! - i wont learn anything buying another torch!?

also - my ss civictor is what i want - no1 has a stainless steel 1aa torch with with a cree or ssc in it (apart from the 2 ever made) - so i cant buy one - and if i could im guessing it would cost a bomb!

although my electrical knowledge from a "theory" point of view is crap
"practically" im not to bad - im very good at handywork - so the "taking out the led, putting in the new one" wont be a problem 

to be honest i dont need these numbers i can just buy the thing stick it in and it will work

but id just like a rough idea of "how" it will work before i try it

sorry if this post sounds ranty - your all right im just a bit dumb with all this amps, volts stuff! lol


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## MrAl (Jan 25, 2007)

-------------------------------------------------------------------------

STARTQUOTE=TJZ
MrAl, greenLED, Doug S, jrmcferren, abvidledUK, 

I have a question if anyone of you knows the answer.
Sounds like you know a lot on this subject.

When you take a current reading in series how does having
the wrong current range on the meter burn out the circuit?
Is there voltage induced from the meter going to the circuit also
when measuring current in series and if you have the wrong current range it
is even higher and it can blow out the circuit?

And if you have the proper current range on the meter, does this mean
no voltage is induced from the meter going to the circuit?

Thanks, 

Tom ENDQUOTE

-------------------------------------------------------------------------

Hi there Tom,

The problem with using a meter alone to measure current is that
on some of the ranges it might switch in a rather high value resistor.
This extra resistance causes the driver to put out a very high voltage
which burns the output transistor out.

The safest way is to use a 0.1 ohm resistor and measure the voltage
across it with a digital meter set on low volts dc.

BTW, some driver circuits can put out over 100v with no LED connected,
for a very short time after which the output transistor blows out due to
overvoltage between collector emitter and/or collector base.


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## Doug S (Jan 25, 2007)

TJZ Asks: *
MrAl, greenLED, Doug S, jrmcferren, abvidledUK, 

I have a question if anyone of you knows the answer.
Sounds like you know a lot on this subject.

When you take a current reading in series how does having
the wrong current range on the meter burn out the circuit?
Is there voltage induced from the meter going to the circuit also
when measuring current in series and if you have the wrong current range it
is even higher and it can blow out the circuit?

And if you have the proper current range on the meter, does this mean
no voltage is induced from the meter going to the circuit?

Thanks, 

Tom*

Tom, expanding a bit on MrAl's excellent and entirely correct answer. 
You ask specifically how meter range can be an issue. If you change range in the course of your measurement, be aware that some meters have ''break before make'' contacts which means that the driver will be momentarily unloaded during the range change. In some cases this is all it takes to fry it. 
Regarding your question about induced voltage, in the strictest engineering sense, if measuring pure DC, No. There will be a voltage *drop* through the meter since it inserts a small amount of resistance. The higher the ammeter range selected the lower this resistance. To minimize the perturbation caused by insertion of the meter into the circuit, use the highest range possible consistant with the resolution you need. If you check the specifications section of your meter manual it should give the ''voltage burden'' of the meter at various ammeter ranges. Many meters have a burden of 0.25V at full scale, e.g., if measuring 100mA on the 200mA scale the drop through the meter will be 0.125V. The resistance of the meter leads can add additional drop. Be aware that the DC ammeter function does best measuring DC. If the DC has a high ripple component or is high frequency pulsing DC, the measurements can be erroneous. Under these conditions voltage *is* induced in the meter's internal inductance. It can cause the erroneous readings but will not be a cause of harm to the circuit.


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## abvidledUK (Jan 25, 2007)

Double post


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## aljsk8 (Jan 25, 2007)

ok i dont know what KISS is - but i detect sarcasm?

ill just do the mod

dont want to start a argument
just looking for a bit of help

were all good at different things - id probably have a little laugh if you tried some of my other interests

that what makes life interesting


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## abvidledUK (Jan 25, 2007)

aljsk8 said:


> ill just do the mod



I agree with you there.

I'm a great believer in "Just doing it"

Practical beats theory every time.


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## TJZ (Jan 25, 2007)

Thank you all for your help with my question.
This answers my question 100%.
(This was not "Way too technical" abvidledUK.)

I do have some electronics background in electronics with circuits
and took a basic course at school.
Been building, testing, and fixing lots of stuff for many years.
Also made many circuit boards from scratch.
I know how to use meters and other test equipment in general.

Here is why I asked this question.
I have a flashlight with two RCR123 3.6v batteries.
The flashlight has an input range of 2.5 - 9 volts.
Hot off the charger puts them each at 4.2V for a
total of 8.4 Volts.
I went to take a current reading on the light at the tailcap leads, 
one on negitive of battery, one lead to end of torch casing thread.
I had the meter in series and set on DC High Current (10A) mode.
Well the driver circuit blew out after a bit of taking readings.
I wanted to make sure that the DMM was not the cause of this
and it sounds like it was not. I figured if the DMM induced any voltage
to the light this did it. I was at 8.4 Volts already and
.6 more volts would fry the circuit.

I did have the DMM hooked up properly and the range set at 
DC High Current (10A).
If this wasn't a DMM issue/DMM range issue, I don't know what happened...?
I will use the 0.1 Ohm resistor from now on too measure current on
my flashlights.


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## macforsale (Jan 25, 2007)

*


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## TJZ (Jan 25, 2007)

macforsale, This leads back to the thought that somehow 
the DMM blew the circuit. Voltage spikes would explain it.
I understand the batteries couldn't have surged
so all that's left is the DMM.
Thanks for the helpful information!


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## mellis (Jan 26, 2007)

Can anyone tell me about measuring amperage in a circuit that is driven by a switching regulator. They seem to have ugly weird waveforms and a simple DC measurement is not accurate. I have a true rms meter but that is just supposed to handle the AC component. I'm trying to run a ZXLD1350 and figuring out if it's working right is kinda confusing. I'm using a shunt resistor.


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## Doug S (Jan 26, 2007)

Are you trying to measure Input current or Output current?



mellis said:


> Can anyone tell me about measuring amperage in a circuit that is driven by a switching regulator. They seem to have ugly weird waveforms and a simple DC measurement is not accurate. I have a true rms meter but that is just supposed to handle the AC component. I'm trying to run a ZXLD1350 and figuring out if it's working right is kinda confusing. I'm using a shunt resistor.


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## mellis (Jan 27, 2007)

I'm measuring current through a string of 3 jupiter leds powered by a buck regulator. I've tried to measure on hi and low side of the string.


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## Doug S (Jan 27, 2007)

Assuming that you are using the ZXLD1350 typical application circuit from page one of the datasheet, just measure the voltage across Rs and compute the value of I. The average current, I, to the LED will be I=V/Rs.



mellis said:


> I'm measuring current through a string of 3 jupiter leds powered by a buck regulator. I've tried to measure on hi and low side of the string.


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## mellis (Jan 29, 2007)

Thanks Doug,

I've tried all of the normal methods; Rs, ammeter and voltage across a sensing shunt. Seems like the hi freq hi power waves are messing up the reading of an average. The scope of the wave that drives the LEDs moves smoothly to a higher freq/lower duty cycle as input voltage is increased. The LED's seem to hold the same brightness (this is a buck regulator) but the amps measured go down like 30%. As I understand it, that's typical of an rms error. I'd just like to know what effective amperage I have ?!


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## Doug S (Jan 29, 2007)

*The LED's seem to hold the same brightness (this is a buck regulator) but the amps measured go down like 30%. * 

The is exactly what should happen. The IC you are using drives the LED at constant current, hence constant power. As you raise the input voltage to the driver the input current should go down since the input power is voltage times amperage.


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## macforsale (Jan 29, 2007)

*


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## MexicoDoug (Sep 23, 2012)

Doug S said:


> check the specifications section of your meter manual it should give the ''voltage burden'' of the meter at various ammeter ranges. Many meters have a burden of 0.25V at full scale, e.g., if measuring 100mA on the 200mA scale the drop through the meter will be 0.125V. The resistance of the meter leads can add additional drop.



Hi, first time poster looking at a very old comment from Doug S which got me to register with the site and hope for a reply (fingers crossed).

Doug's comment about the meter voltage burden is someting I'm trying to Google without success. If l looked up a digital panel meter, let's say 50 mA ammeter for the sake of this, with a rating of something like an impedance of 2.5 ohms and a spec that says "resistance = about +/- 15% of of impedance".

My question is: when measuring current with an panel meter that falls within the meter range, which I assume (?) uses no shunt, but rather a direct reading somehow, is say, 0.37 ohms according the spec is actual resistance. How can I use this information to correct for the mA (if necessary in my range)I am measuring. I am confused whether the ammeters (including DMMs) are affecting the measurement they display such that they display a number that is affected by using them as a measurement tool, and how to calculate from the spec I have for the meter an approximate correction if appropriate, ie that I am thinking about this right. Also, how could someone test a meter, maybe check for a voltage drop across it, so we are not stuck with a spec but can just measure the burden ourselves in our situation and feel good about understanding the spilt between current with and without the panel meter stuck in series in the measurement circuit. I admit to being confused since mA is current, but what we can measure is voltage across it for the trial - does the meter increase the mA drain, or what's happening? Hope this is understandable, it reflects my own confused [email protected]
thanks to all
Doug prime


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