Electronic – Calculating forward voltage for a string of fairy lights: multimeter shows “1”

ledresistance

I'm trying to provide a USB power source to a string of fairy lights that look like these.

They all appear to run in parallel, connected horizontally along two wires until the last led.

I tried isolating a single led to calculate the forward voltage drop, but my multimeter shows "1" with the diode test, and successfully lights up the LED. Any other range fails to light the LED, and also shows "1".

Am I going about this the right way? I'm still confused about the difference in being able to supply current, and drawing current. I know a USB can provide 100mA, but there's no specification for a battery except for mAh?

How can I figure out an appropriate resistor, or at least calculate the forward voltage drop?

The waterfall analogy doesn't seem to be working for me…

Can I just assume it's 3 volts because it's in parallel?

Best Answer

  • The LEDs in the linked picture are white and probably have a forward voltage drop of about 3 to 3.6 V.
  • Without any specifications we'll assume they can handle 20 mA.
  • On a 5 V supply we need to drop 1.5 to 2 V across the resistor.
  • From Ohm's law we can calculate a suitable resistance: \$ R = \frac {V}{I} = \frac {1.75}{0.02} = 87 \; \Omega \$ (choosing mid-range on the voltage). 80 Ω will be good enough for now.
  • Connect four of your 20 Ω resistors in series with each other and the LED. You would need to disconnect one led from the string for this.
  • Connect up to your 5 V supply and measure the voltage drop across the LED. You could also measure the current through the line. (Be careful not to measure voltage on current range.)
  • Go back to step 1 and recalculate with the new values.

The image below may help.

enter image description here

Figure 1 shows that a green LED at 20 mA will have a forward voltage drop of about 2.2 V. If the supply voltage is 5 V then the resistor has to drop 5–2.2=2.8V. The required value is \$ R=\frac {V}{I}= \frac {2.80}{02}=140\;Ω\$. The nearest standard value of 150 Ω will do fine. Source: LEDnique.

In your case you will use the 'W' curve for white LEDs.

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