Measure current to LED? ADDED: Cree XCB3590 image

[Charles Krebs]

I'm working on a couple LED projects and have some questions. The drivers utilize PWM. The output can be varied on some via an internal pot, and/or an external potentiometer.

In one situation I have a 350mA constant current driver with an internal trim pot that can modify output between 75-110% (that should take it down to about 263mA). Without getting into details, I would like to get this down to about 175mA max. With this component I can get this additional reduction via an external potentiometer. My question is... how can I measure the current "fed" to the LED. My understanding is that because of the PWM you can't expect accurate results by simple running through a basic multimeter.

The driver manufacturer provides the following suggestion for doing this:

-Temporarily place a 0.1 ohm, 1% resistor in series with the LED+ output.
-Read the voltage across the 0.1 ohm resistor.
-The voltage, in millivolts X 10, will equal the output current in mA.

Seems simple enough, but my question is what power rating should the resistor have. Does it need to be as high as the LED wattage (which would be about 5W at the 175mA level)

This may be a stupid question but since this driver unit ( http://www.ledsupply.com/content/pdf/le ... tation.pdf ) is itself powered by an external DC source would it be possible to take measurements off of the DC inputs (instead of the PWM output to LED), to determine how much power the driver is using?

_____________________________

In another case I have a Cree XCB3590 COB that can be run at a max of 3.6A with a forward voltage of 39volts (140 watts! ) I don't plan on ever running it near that! But I have a Meanwell HGL-150-48 (AC line input, DC output) that can drive it up to 3.2A. That is still much higher than I want to get involved with, but it can be adjusted downward (PWM) with an external 100K potentiometer. I would like to set it somewhere between 1.4A to 2.4A (~50 to 85W) depending on the cooling method I decide to use. But how can I measure the current through the LED as I vary the resistance?

Also... the Cree has a nice "pad" provided for thermocouple connection to get Tc readings. I have a unit that uses "standard" Type K thermocouples. I plan to attach a thermocouple to this pad to monitor the effectiveness of various cooling methods at different power levels. I have a question on these readings. I have only used the thermocouple measuring with "temporary" contact with the item to be measured. Can these sensors be left permanently attached (via solder or thermal epoxy) Or will the reading go "off" after a while due to being in constant contact (probe wire heating up?)

Keep in mind that I have basic electronic equipment and pretty basic electronics experience. I plan on setting things up so that there will be plenty of "headroom" left, so I am not necessarily after ultra-precise readings, and don't really want to invest in expensive measuring devices if it not necessary.

[elf]

The thermistors in my 3d printer are on continuously and the temperature is around 250C. I would expect thermocouples to work the same. An infrared thermometer is used to verify the temperatures and they are stable.

[rjlittlefield]

my question is what power rating should the resistor have. Does it need to be as high as the LED wattage (which would be about 5W at the 175mA level)

No, power dissipation in the resistor will be quite low. Even at 500 mA = 0.5 A, the voltage across the resistor will be only 0.5 amp * 0.1 ohm = 0.05 volt, so power dissipation only 0.5 amp * 0.05 volt = 0.025 watt. For lesser currents, the power dissipation falls off as current squared.

--Rik

[ChrisR]

For accuracy you could draw your own version of Fgure2, page 2, Output current vs Control Voltage, by measurement with a couple of meters.
It should be within their quoted 5%.
Then you can monitor the control voltage - with some of the devices you're using.

I keep noticing "oscilloscopes" of various degrees of simplicity, down to well below $100. They are (if not TOO simple!!) very handy for seeing what's actually "going on". Usually they're not very high bandwidth devices, but your pulses will probably not be very fast either (except for their edges).

I have 2-3 conventional scopes, so none of these littlun's. I expect a member or two can advise on a recent aquisition.

[mawyatt]

Hi Charles,

Power dissipation in a resistor is simply V*V/R, V*I or I*I*R, all in Root Mean Squared terms. Since you are dealing with pulses the RMS and Average value are not the same. To provide a convenient and accurate means to measure things that doesn't require expensive "True RMS" equipment, will require some knowledge of the waveform. If the LED currents you are interested in are based upon average and not RMS (which is likely), then you just need to measure the average value.

The simplest method to measure current is place a small valued resistor in series with the load and measure the voltage drop across the resistor. The current is V/R, and resistor power dissipation is V*V/R. To get the RMS value we can use the fact that the Average and RMS are related for pulses. Without this information, then proceed with the average value assumption and for power dissipation use a 1.4 multiply factor for the RMS value for power dissipation calculations.

A low pass filter to extract the Average value is really simple. Use a 1K resistor and 0.1uF capacitor like this: Place the 1K is series with your DVM meter + lead and connect to one end of the sensing resistor, the meter - lead goes to the other end of the sensing resistor. Connect the capacitor across the meter + and - terminals, that's all you need to do. Just about any value capacitor above 0.01uF should work, ceramic caps are cheap and work fine. Any value resistor around 1~10K should work fine also, any power level is fine (1/8, 1/4W for example).

For the sense resistor, a 0.1 ohm 1% is going to be expensive and hard to come by. My suggestion is get a a group of say ten 1 ohm resistors and put them in parallel. These can be small, cheap low power ones from eBay. If your DVM has an accurate resistance measurement, measure the DVM lead resistance shorting the leads and recording the reading, then measure each resistor in the bunch. Subtract the lead resistance measurement from each resistor measurement. The total resistance with the 10 resistors in parallel is the recipocle of sum of the recipocle measurements (without lead resistance). After you bundle the resistor together measure the bundle if you can (0.1 ohm is difficult to measure and extract an accurate measurement without a 4 terminal method which is beyond what's needed here). The value you computed from the 10 resistors in parallel is what you use for the sense resistor. It doesn't have to be exactly 0.1 ohms, I'm sure 0.15 or even 0.2 ohms should work fine and give an accurate result.

If you know how much current is being used by the power supply converter you can assess the actual LED load current, however this most likely will vary depending on many things and probably not give you an accurate result.

An option to allow you to get lower current thru the LED than the constant current supply converter can deliver is to shunt the current away from the LED, thus the total supply current is being split between the LED load and shunt. For example, if you want the LED current to be 175ma and the constant current source provides a minimum 263ma, then you need to divert 88ma. If the LED voltage is Vled at 175ma, then the resistor value is Vled/88ma or 113 ohms for a Vled of 10 volts. The resistor dissipation will be I*I*R or 0.88 watts, with the RMS equivalent (I*1.4 times), or 1.7 watts, so use a 2 watt or greater resistor or pot.

BTW since your supply is stated as a "constant current" type, then the 0.1 ohms current sense resistor should not be critical at all. Something like 0.5 ohms might work just fine for the sense resistor at smaller current less than 1 amp.

Anyway, hope this helps.

Best,

Mike

[mawyatt]

For accuracy you could draw your own version of Fgure2, page 2, Output current vs Control Voltage, by measurement with a couple of meters.
It should be within their quoted 5%.
Then you can monitor the control voltage - with some of the devices you're using.

I keep noticing "oscilloscopes" of various degrees of simplicity, down to well below $100. They are (if not TOO simple!!) very handy for seeing what's actually "going on". Usually they're not very high bandwidth devices, but your pulses will probably not be very fast either (except for their edges).

I have 2-3 conventional scopes, so none of these littlun's. I expect a member or two can advise on a recent aquisition.

Chris,

Heres a cheap scope kit, not sure how good though.

http://www.tmart.com/DSO138-DIY-Digital ... 52880.html

Best,

Mike

[ChrisR]

Nor am I, but that's the sort of thing I meant.
Analog Bandwidth 0 - 200KHz
is a tad slow perhaps, I saw one claiming 1MHz I think, for about £/E/$ 70.

I have seen one which simply plugged into the mic socket on a phone, but nobody could find the app to go with it.

[rjlittlefield]

I have 2-3 conventional scopes, so none of these littlun's. I expect a member or two can advise on a recent aquisition.

Picoscopes are good. See https://www.picotech.com/products/oscilloscope . I was moved to get a small one, by reading footopa's posts. (Search the forum for "picoscope".) Actual performance seems to be in line with the specs.

--Rik

[Charles Krebs]

Quick partial follow up.
Working first with the Cree XCB3590 I measured the current two ways. First the voltage across a 0.1 Ohm 1% resistor in series (then multiplied by 10). The second time I simply used the standard basic method of running the circuit directly through the multimeter. For my purposes the readings were essentially identical through the range of 670mA to 3.25A. (By essentially identical I mean for example... when it read 1.001 amp in one method it was 0.998 amp the other method).

As an aside... I have always used passive heat-sinks in the past. Of necessity they get pretty large, and run pretty warm with the higher power LEDs. In this case I mounted the Cree onto a fairly small small copper water-block (1.62 x 1.62 x 0.55 inches. http://www.customthermoelectric.com/Wat ... -CU-01.pdf Got it new at a very good price on Ebay). Since this is being "worked up"in my old darkroom I had access to water and ran a very low tap flow through the block. (Haven't measured flow rate yet but I'll guess somewhere between 0.2 to 0.4 gpm). I was stunned at how effective this was. Running at the full 3.2A (so the LED COB was somewhere between 115 and 120W) the Tc temperature only went up a few degrees above ambient. At first I thought my cheap thermister meter wasn't working properly, so I cautiously used the reliable old "fingertip" thermometer . The copper block was cool to the touch. Even when I touched the ceramic substrate of the LED COB itself, it only seemed a bit warm. Still a little incredulous, so If it does all go up in a puff of smoke I'll let you know!

When/if I get to implementing this the problem might be heat caused by proximity to the intense light produced and not the problem of cooling down the LED itself.

I can now understand why extreme "over-clockers" love water cooling!

[mawyatt]

Charles,

What brand DVM were you using? I have Flukes, both older 77 type and the True RMS 87 type. They developed (and patented many years ago) an incredibly clever method of measuring accurate RMS waveforms, as well as the usual average valued filtered method. Some of the better DVM meters have built-in Low Pass filters for measuring average value for pulses and other non-stationary waveforms.

I'm not surprised your water cooled heat sink is working well, water has one of the lowest thermal impedance and highest thermal masses of anything common, that's why it's used in our cars!!

BTW don't let the smoke out of your LEDs or any electronic component, it's the magic stuff that makes all electronics work!! Don't ask how I I know

Best,

Mike

[Charles Krebs]

My DVM is a fairly inexpensive Chinese "Fluke Wannabe" branded "Thsinde". (The company that owns that name is Shenzhen Lixinde Industry and Trade Co.,Ltd.) They even label the model as an "18B+"
https://www.amazon.com/gp/product/B06ZZ ... UTF8&psc=1

It does state that it is "True RMS" but who knows. It seems nicely made, but comes with very little documentation.

[mawyatt]

Charles,

Wow, that looks like a great deal!! Think I paid $300 for the Fluke 87 (the 77 was about $150) but that was over 30 years ago!! I had no idea these copies (even looks like my 87!) were that good & accurate!! I suspect that most of the old meter's electronics cost has been integrated into a couple chips today, thus the great price.

Fluke's RMS solution was a brilliant adaptation of electronics and thermal properties at the time. RMS is the effective heating value of the waveform, thus the thermal implications. Way back then the usual solution for RMS was to use a analog square circuit, followed by an integrator (mean) then a square root circuit, thus Root of Mean Squared. The analog circuit for the square had a limited dynamic range, which imposed a "crest factor" of less than 10 to 1. Thus, waveforms that had an peak to average ratio of over 10 to 1 weren't accurately computed analog-wise, and narrow pulses were very inaccurate, but sine waves, noise, and high duty cycle pulses were accurate.

Fluke's solution was two identical silicon bipolar transistors with two 50 ohm thin film resistors integrated as close to each emitter base junction as possible, but the transistors were positioned at each end of the chip and somewhat separated. The signal was imposed across one 50 resistor which extracted the true RMS signal value as heat, the base emitter junction of the local transistor responded with the solid state physics based result relating collector current and base emitter junction voltage vs. temperature. The other resistor was driven by a closed loop differential integrator which created a similar heating effect but with a almost stationary DC voltage/current. The differential signal for the integrator was the two transistor base emitter junction voltages, with each transistor biased with an identical current source.

What would happen is the integrator would servo the result so the two resistors had the same heating effect, one controlled the signal the other controlled with an almost DC representation from the integrator. Since the RMS of a DC waveform is just the DC value, the output of the integrator was the True RMS representation of the signal heating content! Thermal coupled negative feedback system, absolutely brilliant!!

To make this work in the real world, Fluke thinned the silicon chip to something like 50 microns or less to reduce the thermal mass, then suspended the chip with 6 wire bonds for electrical connections inside a hermetically sealed, evacuated TO-99 can to thermally isolate the chip since the 6 wire bonds don't provide much of a thermal path to the case. Being thinned, the physical mass of the chip was so small as not to impose a risk of breaking the wire bonds under vibration.

Best,

Mike

[ChrisR]

Brilliant! But I'd probably put my finger on it to see how warm it feels...

[mawyatt]

Brilliant! But I'd probably put my finger on it to see how warm it feels...

Chris,

Always lick your finger first, that way you can pull it back when you hear the sizzling! It seems this responds quicker than my old nerve system which informed me that my finger was burning about 1/2 second later

I remember having Motorola "M" branded on my finger caused by touching a TO5 transistor can to see if it was hot. Yup is was hot, real hot!!

Long ago IBM used thermal imaging of complex computer PCBs to fix defective PCBs in their production line. They simply had a known good PCB fired up and thermal imaged one half the viewing screen and the test PCB fired up on the other half. An unskilled worker simply painted dots on the parts that were of different color (temperature), which were then replaced. No complex test equipment, debugging or skilled labor required.....Brilliant!! If it's too hot or cold it's probably not working right, they got a very good yield from this procedure I recall, and exceptional cost benefit.

My episode, not so Brilliant!!

Best,

Mike

[ChrisR]

I was forgetting, you need to use the back of your finger in case the thing is at mains voltage, or rectified mains up at 300VDC. Your muscles make your hand come away then instead of onto the component - or so I'm told .