I just wired a blue LED on a breadboard with the anode to an Arduino Uno's 3.3V rail and the cathode to a 220 ohm resistor to ground. Then I measured the current consumption using my Fluke 87-V and it was 1463 microamps.

Here's the strange part: I then added another blue LED, exactly the same type, in parallel to the first one. This, too, was using a 220 ohm resistor. I expected the current consumption to double to 2926 microamps or so. Instad, the measured current consumption was 2121 microamps. Can anyone explain this?

Update: Just added a third blue LED, same type, in parallel, 220 ohm resistor, and the current drawn is now 2681 microamps. Wtf?

Update 2: I just switched my Fluke to the mA range and now I get 5.41 mA drawn with 3 LEDs in parallel. With two in parallel on the mA range I get 3.61 mA and with just one I get 2.02 mA. All measurements in DC mode of course. So I assume my Fluke is broken since it gives different readings on the uA and mA modes?

  • \$\begingroup\$ Different \$V_F\$ between the LEDs? Resistors not precisely 220Ω? \$\endgroup\$
    – Majenko
    Oct 25, 2014 at 14:39
  • \$\begingroup\$ Nope, LEDs are from same purchase batch, same color, same type, same Vf, everything. Same with the resistors. They have a gold band, so 5% tolerance. \$\endgroup\$ Oct 25, 2014 at 14:42
  • \$\begingroup\$ 5% is quite a wide range - anything from 209Ω to 231Ω. And purchased at the same time doesn't mean manufactured at the same time on the same machine, so there may well be a difference in \$V_F\$. \$\endgroup\$
    – Majenko
    Oct 25, 2014 at 14:44
  • \$\begingroup\$ This is 3 LEDs off the same resistance, right? That's about as expected. The turn-on voltage of the LED is almost 3.3V and varies with current. So sharing current between LEDs reduces the current in each, and the different resistance of each range on the Fluke makes the difference between turning the LED slightly on or fairly well on. Use a higher voltage or expect big differences between LEDs, and give each LED its own resistor. \$\endgroup\$
    – user16324
    Oct 25, 2014 at 14:59
  • \$\begingroup\$ Brian, no, each LED has its own 220 ohm resistor. I don't understand the part about the Fluke having a different resistance on the uA and mA ranges. It should just measure the current correctly or show OL on all ranges, correct? \$\endgroup\$ Oct 25, 2014 at 15:09

1 Answer 1


There's nothing strange going on here.

With a 3.3V supply, and a blue LED, there is very little voltage across the 220 ohm resistor(s).


simulate this circuit – Schematic created using CircuitLab

Your Fluke meter (when set to measure current) has quite significant internal resistance, which is higher on lower current ranges. This resistance adds to the 220 ohms/n (where n is the number of LEDs), and explains the lower reading on the lower current ranges. There are times when you can ignore the voltage drop of an ammeter (or the loading of a voltmeter) but this is not one of them.

Another effect is that the output port pin also has some internal resistance (which again adds to the parallel resistors) so you expect the current to drop if you're driving the LEDs with a single port pin.

  • \$\begingroup\$ I didn't know ammeters influenced the circuit like that. I just thought they would accurately measure current. So in what situations can I trust the Ammeter to give me a correct reading? In my example, how can I know what current the circuit is really consuming without the Ammeter plugged in? \$\endgroup\$ Oct 25, 2014 at 19:41
  • \$\begingroup\$ My suggestion is to simply measure the voltage across the resistors and calculate the current from Ohm's law. You should be able to see the effect of the port pin output resistance that way. In general you can either look up how much voltage your meter drops and decide whether that will affect the circuit of use a second meter and measure the volts. Admittedly it's a little tricky when you have a nonlinear load like a ~3V LED operated with so little headroom-- experience will come. \$\endgroup\$ Oct 25, 2014 at 20:50
  • \$\begingroup\$ P.S. This problem is what inspired our Antipodal luna^H^H^H friend David Jones to make his adapter box for meters-- ucurrent he calls it IIRC- it has significantly less voltage drop than a regular DMM. \$\endgroup\$ Oct 25, 2014 at 21:02

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.