Measure current without loosing some - Page 1
Forum section: electronics / beginners

author: haizaar
author_rank: Regular Contributor
timestamp: July 19, 2019, 01:16:05 pm
content: Good day guys,I hooked 30led, 5V strip to a arduino, with additional power, and now I try to measure how much current do they draw. Spec says they should draw 60mA on full brightness, i.e. .06*30 = 1.8A total.I hook in a EEVlog BM235 multimeter and I observe significant drop in led brightness, and the current is just 650mAwith DMM:https://photos.app.goo.gl/WF23rcEP3EhWsRc59without DMM:https://photos.app.goo.gl/t1a73J7ycKD1nqs86I then tried to get rid of clip-on crocodiles and use just the DMM test lead. Brightness improved and current moved to 1A, but still not as bright as without DMM.Am I missing something? Is there a way to measure current without such losses?My clipon crocodiles have a resistance of ~0.5ohm each, yet they caused 350mA current drop. Does it make sense?Thank you!

author: Zbig
author_rank: Frequent Contributor
timestamp: July 19, 2019, 01:21:10 pm
content: Get a DC-capable clamp meter like the great little Uni-T UT210E.

author: haizaar
author_rank: Regular Contributor
timestamp: July 19, 2019, 01:53:46 pm
content: Sounds like a plan... but does it mean that standard DMM ammeter is not of any good?Btw, is it worth in your opinion to get a high-sensitivity one (UNI-T UT251A)?

author: homebrew
author_rank: Frequent Contributor
timestamp: July 19, 2019, 02:03:19 pm
content: As 'normal' DMMs use a shunt and measure the voltage drop accross it, you'll always 'lose' that voltage difference. However, in high current mode this drop should be rather small. Thus, two questions come to mind:Are you sure that all your LEDs are in parallel? If so, how is the current regulated? Dropper resistors or proper LED driver(s)?If a dropper resistor would be present on the LED strip you could instead measure its voltage drop. Knowing the resistance, you could calculate the current accurately and without the need to interfere with the circuit.

author: haizaar
author_rank: Regular Contributor
timestamp: July 19, 2019, 02:22:15 pm
content: Quote from: homebrew on July 19, 2019, 02:03:19 pmYeah, I looked at the specs, and on the 6A range I'm measuring, the burden voltage is 0.04V/A. Since my amps are within 1-2A range, should 40-80mV make such a significant difference?The strip is a pixel strip - APA102, so I give it power and arduino drives data and clocks. Power is connected in parallel - I tested +/- wires from both ends and they "beep".The strip PCB has only one thing soldered to it besides the leds themselves:https://photos.app.goo.gl/c1FzXFRwX6CjSM8C6I'm not sure what they are since they are inside the silicone tubing. There are two pairs of those - in the beginning and midway (it's a 1 meter strip with 30 leds).

author: MrAl
author_rank: Super Contributor
timestamp: July 19, 2019, 02:51:32 pm
content: Hi,All DMM's use some sort of series resistance to measure current thus you loose a small voltage in the process. It's usually small but sometimes it is just not small enough.In such cases sometimes a current shunt is used. A current shunt has a very very small voltage drop that does not usually cause a problem.There are also Hall Effect current measurement devices used to monitor current during the whole up time of the equipment.Sometimes you can get away with using your own low value series resistor, like 0.1 ohms. It depends on how much voltage drop your application can handle.Current shunts usually require a meter that can measure down to millivolts. Hall Effect current devices require a low voltage power supply.

author: bdunham7
author_rank: Super Contributor
timestamp: July 19, 2019, 02:54:35 pm
content: Your LED circuit is right near a "knee" in the voltage/current curve. A resistor, as you might expect, will have a constant current to voltage ratio. The LEDs, on the other hand, will likely have no current at 3.0V and too much current at 4.0V. In between there, the current will have a very steep change in current per unit voltage. I'm not sure what your circuit is, but proper practice with LEDs is to have a current limiter or regulator driving them so that the current doesn't change with small voltage differences or differences in LED FOV due to temperature. Either you don't have one and the current is right on the edge of the curve, or you do have one and it is right at its limits voltage wise. Either way, your observation that the circuit is very sensitive to small voltage burdens is correct.

author: tunk
author_rank: Super Contributor
timestamp: July 19, 2019, 03:29:37 pm
content: That looks like 27 ohm resistors and most likely they're in series with the LEDs.If there's one in series with each LED then the resistor drop is 1.6V (27*0.06) andthe LED drop 3.4V (5.0-1.6). Using the crocodile clips you have an extra 1V drop(1ohm*1A), the resistor drop is 0.9V (27ohm*1A/30) and the LEDs see 3.1V.Edit: This is consistent with what bdunham7 says above.

author: Kasper
author_rank: Frequent Contributor
timestamp: July 19, 2019, 10:29:33 pm
content: APA102 is an addressable LED. The resistors are likely on the data lines and the current draw of each LED is set mostly by their built in controllers.The advice you are getting may be suitable for general LEDs but not so much for these ones.40mV - 80mV burden voltage of DMM should not make much difference.Another picture showing where the supply power comes from could help.Running your LED strip supply power through breadboard is not ideal. Breadboards aren't good for 'high current' and they often have loose connections.What is the voltage between Vcc and GND with both DMM probes close to the LED strip?It should be near 5V. If not, what is the voltage and max current of your supply?If you have a 2nd DMM, you could measure voltage across the current sensing DMM to verify its burden voltage.My guess is you are using a phone charger that has max output current 1A which means its voltage decreases when you try drawing more than 1A. That combined with the wires and breadboard means your LEDs are being powered with near their minimum voltage of about 3.3V so the LEDs are barely decently powered. When you connect the DMM and extra cables it pushes it beyond its lower limit and brownout occurs.My other guess would be you are powering it from arduinos 5V output and arduino regulators don't usually have high enough output current ratings for LED strips. Judging from your picture, you are powering the LEDs separately which is good, as long as you have the grounds tied together and be careful about having the LED strip power on any time you put power into its data / clk lines.On a side note, try getting the electrolytic cap closer to the LED strip, between the current sensor DMM and the strip.

author: David Hess
author_rank: Super Contributor
timestamp: July 19, 2019, 10:56:53 pm
content: There are other ways to measure the current with no voltage drop but an easy way may be to measure the change in current on the unregulated side of the 5 volt power supply.

author: haizaar
author_rank: Regular Contributor
timestamp: July 23, 2019, 12:00:27 pm
content: Wow, that's a lot to digest. That's quite educational, but regardless, I think I'll get clamp amp meter1. I butchered the strip and those SMD things are indeed 27ohm resistors soldered in before data lines (not sure why the writing on them says 270)2. For power I'm using USB power supply that says it's capable to provide 2A (tested again two different models just to be sure).In the original test it passed through breadboard powerboard:https://photos.app.goo.gl/R2zEqWCp6rBhZ4SHAReading the replies again, I tried to replicate the setup, with different strip though, since I already used that one. This time I powered the strip directly from the USB charger. Then tried to hook it through breadboard to clamp in the DMM, and even before hooking DMM there was noticeable drop in brightness. I guess Kasper is right and jumper wires / breadboards are not good for passing current.Direct:https://photos.app.goo.gl/C1TZu2VyqMpcDWKP6Through breadboard:https://photos.app.goo.gl/xyLTHYgBfuXZmcvC6(It's less evident on pictures, but for human eyes different was rather visible)Since breadboard/wires produce so much variance/noise, it does not seem rational to continue experiments about measuring current.When powering directly through usb-pcb the voltage readings on the latter were as follows:No-load: 4.98v10 leds: 4.95v20 leds: 4.75v30 leds: 4.60v40 leds: 4.43v45 leds: 4.31v50 leds: 3.54vEach lead can consume up to 60mA on full brightness, so with 2A power supply it should be able to power 30 leds easily.What's interesting is that even with 50 leds and there was no noticeable drop in brightness. Hence, if every led consumes 60mA max, 50leds * 0.06A * 3.54v = 10.6W, which equals to 2A * 5V as stated on USB charger. Meaning the charger does what's written on it. It least some useful outcomeThanks again everyone for your replies!

author: IanB
author_rank: Super Contributor
timestamp: July 23, 2019, 12:20:23 pm
content: One thing missing from all of the replies so far:When measuring current by inserting a multimeter in the circuit, if you use the 10 A range the internal resistance of the shunt should be negligible. If you are seeing a big voltage drop under these circumstances then most likely the extra resistance is in the test leads and test connections.

author: IanB
author_rank: Super Contributor
timestamp: July 23, 2019, 12:24:04 pm
content: Quote from: bdunham7 on July 19, 2019, 02:54:35 pmThere is no "knee" in the voltage/current curve of an LED or other diode. This is a common misapprehension about what the "forward voltage" means.The voltage/current curve is actually a smooth exponential function with the same shape at all moderate currents within the operating range of the diode.

author: ledtester
author_rank: Super Contributor
timestamp: July 23, 2019, 01:25:37 pm
content: Quote from: haizaar on July 23, 2019, 12:00:27 pmThe third digit is a power of ten exponent, so the resistor value is 27 * 10^0= 27.

author: Kasper
author_rank: Frequent Contributor
timestamp: July 23, 2019, 03:37:28 pm
content: Glad to hear you figured it out.Quote from: ledtester on July 23, 2019, 01:25:37 pmYes this is right. I was wrong on the other thread when I guessed 270 ohms.If it said 271 that would be 270 ohms. 273 would be 27,000 ohms.

author: Kasper
author_rank: Frequent Contributor
timestamp: July 23, 2019, 04:09:19 pm
content: Quote from: haizaar on July 23, 2019, 12:00:27 pmYour numbers add up nicely and that is as expected but fire and electrocution can also be expected, more so with cheap phone chargers.It is best to keep within all the ratings even with good equipment. If it says 2A, don't try to draw more than 2A. Try to leave a bit of safety margin on top of that also. And keep in mind all the other current draws; arduino, etc.