# Correct way to measure current off a constant current driver?



## wildstar87 (Jan 16, 2008)

I'm trying to make sure the driver is actually delivering the stated current, I've measured off the tailcap, but I don't think that's the proper way. I have also measured in between the driver and the led, which I believe is the correct way, but I just want to get the official method. 

I didn't have much luck finding an "official" method using search.


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## DonShock (Jan 16, 2008)

Between the driver and the LED will give you the actual drive current to the LED.


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## 65535 (Jan 16, 2008)

Most electronics gurus would recommend using a calibrated shunt, but for most non PWM circuits measuring current in series with the circuit should work fine.


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## Oznog (Jan 16, 2008)

It's a bit tricky to measure some supplies on some DC meters. 
An LED converter could be anywhere from a smooth supply with 10% or 15% ripple to a PWM varying from 0 to 200% of the average value, and a wide range of frequencies is possible. You'd hope a DC meter would read the average but I don't think there's a guarantee it'll measure that way at any freq encountered.

Oscilloscopes are a lot more informative as to what the supply is doing.


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## MatajumotorS (Jan 16, 2008)

Oznog said:


> It's a bit tricky to measure some supplies on some DC meters.
> An LED converter could be anywhere from a smooth supply with 10% or 15% ripple to a PWM varying from 0 to 200% of the average value, and a wide range of frequencies is possible. You'd hope a DC meter would read the average but I don't think there's a guarantee it'll measure that way at any freq encountered.
> 
> Oscilloscopes are a lot more informative as to what the supply is doing.


I totally agree. If you have super-duper wide bandwidth true rms etc meter, maby then the measurment will be acurate.


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## monkeyboy (Jan 16, 2008)

+1 on what the others said. The tailcap measurement is the current directly from the batteries whereas the measurement between the driver and LED is the current you are intested in.

Judging from the nature of the first post, I don't think the OP wants to run out and buy himself a Fluke oscilloscope and run the wave analysis software on his PC. You would just have to hope that the DMM will give an accurate enough average value.


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## SemiMan (Jan 16, 2008)

Well there may be a PWM switch mode power supply, the odds are for a flashlight, the dimming is going to be linear w.r.t. current as opposed to PWM the current on and off as linear dimming gives better efficiencies from the LED at low currents. I am not saying it will always be that way, but most of the time it will be.

If the switch mode supply has not capacitor on the output, you could see 30% ripple. However, most circuits are using capacitors on the output of the switch mode supply in flashlights. Hence, the ripple is likely to be <10%. One nice thing is in either case, the primary ripple may be somewhat triangular wave shaped and hence even a cheap multimeter will be somewhat accurate. I would not get too concerned with the accuracy. You are likely going to be close enough and yes you need to measure the current into the LED.

Semiman



Oznog said:


> It's a bit tricky to measure some supplies on some DC meters.
> An LED converter could be anywhere from a smooth supply with 10% or 15% ripple to a PWM varying from 0 to 200% of the average value, and a wide range of frequencies is possible. You'd hope a DC meter would read the average but I don't think there's a guarantee it'll measure that way at any freq encountered.
> 
> Oscilloscopes are a lot more informative as to what the supply is doing.


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## Oznog (Jan 16, 2008)

MatajumotorS said:


> I totally agree. If you have super-duper wide bandwidth true rms etc meter, maby then the measurment will be acurate.



Actually RMS voltage across a shunt would be wrong!

RMS is the square root of the average of the squares. So if I have a PWM waveform at 700mA for 50% of the time and 0mA for the other 50%, the average current is 350mA and average current is what he needs in this case. However, by measuring the shunt voltage by RMS method that would read as 495mA.


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## SemiMan (Jan 17, 2008)

Oznog said:


> Actually RMS voltage across a shunt would be wrong!
> 
> RMS is the square root of the average of the squares. So if I have a PWM waveform at 700mA for 50% of the time and 0mA for the other 50%, the average current is 350mA and average current is what he needs in this case. However, by measuring the shunt voltage by RMS method that would read as 495mA.



No, the RMS value would be exactly the same as the Average value in this case Oznog. The formula you used, RMS = Vpeak/Sqr(2) only applies to a sinusoidal wave. For a rectified square wave, RMS = Vpeak/2 or 350mA.

A true RMS meter measures power in the wave independant of the waveform. In this case that would be the equivalent of 350mA.

Semiman


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## wildstar87 (Jan 17, 2008)

Heh.. I just have a simple DMM from RadioShack so I guess I'll just be measuring in between the LED and the driver. Thanks for the info guys..


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## Oznog (Jan 19, 2008)

SemiMan said:


> No, the RMS value would be exactly the same as the Average value in this case Oznog. The formula you used, RMS = Vpeak/Sqr(2) only applies to a sinusoidal wave. For a rectified square wave, RMS = Vpeak/2 or 350mA.
> 
> A true RMS meter measures power in the wave independant of the waveform. In this case that would be the equivalent of 350mA.
> 
> Semiman




Actually no. Remember, RMS is the average of the squares. So "cheap RMS approximation shortcut" takes the average value of the waveform (could also use peak) and multiply it by the appropriate constant but it's only correct if the signal is a sinusoid.

A *true *RMS meter will take a lot of samples (or use a thermal converter, or various analog methods) and actually get the sq root of the average of the squares and will not be confused by nonsinusoidal waveforms provided they are within the freq response of the meter.

The figures I gave are correct. The _true_ RMS heating power of the waveform (well provided you don't filter out the DC component) will calculate as a shunt voltage that translates to 495mA even though the true average current is 350mA. Remember true RMS and true average voltage (or current) are two different things, and that's assuming they're properly measured for the "true" value not confused by using a sinusoid on a piece of non-true hardware using an RMS shortcut.


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## whitecloud (Jan 19, 2008)

Check this RMS/DC converter out:

http://www.datasheetcatalog.com/datasheets_pdf/L/T/1/0/LT1088.shtml

The circuit layout on the die is concentric, thus optimized for accurately measuring heat.... interesting device.

Not fooled by waveforms with ugly crest factors.


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## SemiMan (Jan 20, 2008)

Oznog said:


> Actually no. Remember, RMS is the average of the squares. So "cheap RMS approximation shortcut" takes the average value of the waveform (could also use peak) and multiply it by the appropriate constant but it's only correct if the signal is a sinusoid.
> 
> A *true *RMS meter will take a lot of samples (or use a thermal converter, or various analog methods) and actually get the sq root of the average of the squares and will not be confused by nonsinusoidal waveforms provided they are within the freq response of the meter.
> 
> The figures I gave are correct. The _true_ RMS heating power of the waveform (well provided you don't filter out the DC component) will calculate as a shunt voltage that translates to 495mA even though the true average current is 350mA. Remember true RMS and true average voltage (or current) are two different things, and that's assuming they're properly measured for the "true" value not confused by using a sinusoid on a piece of non-true hardware using an RMS shortcut.




OZNOG you are right, sorry, I was thinking about what the meter will read. Unfortunately, almost all RMS meters give you the "rectified" RMS value, not the DC RMS value. The meter is likely to read 350mA even if it is a true RMS. There are meters that automatically give you the proper DC RMS. They tend to be the high end models.


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## SemiMan (Jan 20, 2008)

Oznog said:


> The figures I gave are correct. The _true_ RMS heating power of the waveform (well provided you don't filter out the DC component) will calculate as a shunt voltage that translates to 495mA even though the true average current is 350mA. Remember true RMS and true average voltage (or current) are two different things, and that's assuming they're properly measured for the "true" value not confused by using a sinusoid on a piece of non-true hardware using an RMS shortcut.





Oznog's point is completely accurate. Getting back to this, just a basic DC current reading would be the most accurate as this is pretty close to being a true average of the DC value of the signal. Any PWM of the signal is likely going to be well above the cut-off frequency of the filter used to get the DC value.

Wildstar, Sorry if things got confused on that RMS discussion. If you want to get into the technicality of why RMS is not accurate in this case....... RMS is used to determine the "power" in the signal. However that power assumes the load is resistive, i.e. power = voltage squared/resistance or current squared * resistance. However, in an LED circuit, the load is not a resistance, but essentially a fixed voltage. Hence the average value of the current is actually a lot more important than the RMS value of the current.

Semiman

p.s. Would that Wildstar be Wildstar from Starblazers? ... makes me feel old.


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## Oznog (Jan 20, 2008)

SemiMan said:


> However that power assumes the load is resistive, i.e. power = voltage squared/resistance or current squared * resistance. However, in an LED circuit, the load is not a resistance, but essentially a fixed voltage. Hence the average value of the current is actually a lot more important than the RMS value of the current.




Ah but there's one more point here yet. If we were measuring _LED_ voltage then RMS, true RMS, average, even an instantaneous reading is not very meaningful for as far as knowing how hard it's being driven.

The question as I interpreted it was in regards to LED current by reading the voltage on a shunt. In this situation knowing average current is what we need to know about 99% of the time. Note that the LED's nonlinearity doesn't matter at all- instantaneous LED current is accurately reflected in the shunt resistor voltage. So any effective averaging scheme- whether it be a meter's lowpass filter, relying on a mechanical meter's needle inertia, or an external RC filter- are all effective at producing the desired average current value regardless of the nonlinear properties of the load, provided the current ripple is above the cutoff of the averaging filter. It's the attempt to apply an RMS scheme that will produce a less than meaningful measurement.


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