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Can I use a multimeter for measure voltage drop in LEDs? Answered

I have a really cheap multimeter, but it has a "diode" position in the main switch. As the LEDs are diodes, could I measure the voltage drop for use in resistor calculations? My guess is "yes", but I'am very new to electronics... Thanks a lot.



11 years ago

The cheaper the measuring instruments the more inaccurate the measured results. The lower the voltage involved, the more precice the measurement should be, neccesatating more expensive measuring devices. The diode is not strictly a resistive device. It has a capacitive reactance value and should not be plugged into any calculation as strictly a resistive value. As a rule of thumb all transistor junctions fall into two categories, including LEDs. They will be silicon or germanium. To gate on a junction fully, it takes 0.6vdc or 0.4vdc respectively. Using these voltages in any calculation should get you where you want to be. And if I am remembering correctly there is always a current limiting resistor in the LED circuit.

LEDs are not transistors, and the junction is neither silicon nor germanium, and the voltage drop is always higher than 0.6V. In most cases, if you're going to use a resistor for current limiting, you can just use "typical" Vf data from a datasheet describing similar technology; about 1.7V for GaAlAs RED, 2.2V for green, 3.5V for GaN (blue, cyan, white) However, even a cheap meter should be able to measure the actual Vf. The "diode" setting usually just tells you whether a diode is good or not, I don't think they tend to report the voltage drop. (ok, I tried it on my cheap ($3) HF DVM; the number reported with a silicon diode is suspiciously close to a likely Vf for such a diode in mV ("663", IIRC); but the same setting decided that sample red and white LEDs were simply "bad", even though it made them glow a bit.) Instead, set up a circuit that lights the LED with "about" the target current, and actually measure the voltage between the two LED leads. You can do this at a "safe" current (say, with a 1kOhm current limiting resistor and a 9V battery), but Vf is somewhat dependent on current, so it'll be better if you get close to the target current. Since we have a source resistance of about 1k here, even an old mechaniccal meter should give reasonable results, and the modern cheap DVM shouldn't have any problem. Accuracy is limited to the accuracy of the meter, of course...

Ok, I'll try this. Thanks a lot!