Electronics > Beginners
Measure current without loosing some
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haizaar:
Wow, that's a lot to digest. That's quite educational, but regardless, I think I'll get clamp amp meter :)

1. 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/R2zEqWCp6rBhZ4SHA

Reading 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/C1TZu2VyqMpcDWKP6
Through 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: 4.98v
10 leds: 4.95v
20 leds: 4.75v
30 leds: 4.60v
40 leds: 4.43v
45 leds: 4.31v
50 leds: 3.54v

Each 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 outcome :)

Thanks again everyone for your replies!
IanB:
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.
IanB:

--- Quote from: bdunham7 on July 19, 2019, 02:54:35 pm ---Your LED circuit is right near a "knee" in the voltage/current curve.
--- End quote ---

There 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.
ledtester:

--- Quote from: haizaar on July 23, 2019, 12:00:27 pm ---
1. 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)

--- End quote ---

The third digit is a power of ten exponent, so the resistor value is 27 * 10^0 = 27.
Kasper:
Glad to hear you figured it out.


--- Quote from: ledtester on July 23, 2019, 01:25:37 pm ---
--- Quote from: haizaar on July 23, 2019, 12:00:27 pm ---
1. 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)

--- End quote ---

The third digit is a power of ten exponent, so the resistor value is 27 * 10^0 = 27.

--- End quote ---

Yes 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.
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