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I recently had an experiment in school where I used resistors that were 130k, 140k, 150k, 160k and 170k ohm to dim a red LED. The light intensity was then measured with a Pasco light sensor. The circuit was powered by a 9v battery.

I was wondering, how could I predict how much the light intensity will drop as I change the resistors out?

Thanks in advance!

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    \$\begingroup\$ Making use of your measurements, you could simply interpolate values (for resistors in between the ones you tried) or extrapolate (for resistors in a different range). Without measurements, from theory alone, is a different matter altogether. \$\endgroup\$ Feb 28, 2018 at 18:29

2 Answers 2

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By consulting the LED's datasheet's forward current/voltage and luminous intensity/forward current curves:

enter image description here

Source: https://cdn-shop.adafruit.com/datasheets/WP7113SRD-D.pdf

Now, by calculating the current through the LED:

  • I = ( 9V - V(LED)) / R

You can then look up that current on the luminous intensity chart to get the luminous intensity as a percentage of the value given for 20 mA; on this datasheet it is given as 250 mcd.

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    \$\begingroup\$ Also... depending on the type of LED, colour can change too. Worth a mention. \$\endgroup\$
    – Trevor_G
    Feb 28, 2018 at 18:35
  • \$\begingroup\$ Great point @Trevor_G. \$\endgroup\$
    – Jim
    Feb 28, 2018 at 18:43
  • \$\begingroup\$ For a LED with currents up to 30 mA resistors with 130 to 170 kiloohm are much too high anyway. The behaviour at currents lower than 100 µA can not be found in these diagramms of the datasheet. \$\endgroup\$
    – Uwe
    Feb 28, 2018 at 18:47
  • \$\begingroup\$ Did you notice any intensity changes or a threshold of light with each value? It would have been pretty dim with anything over 100K and bright with 470 ohm and very sensitive to changes about this (9-Vf)/470 = If \$\endgroup\$ Mar 1, 2018 at 1:19
  • \$\begingroup\$ @TonyStewart.EEsince'75 The luminosity dropped by 25lx every 10k. starting out at 1.46lx at 130k and ending at 0.02lx at 170k. \$\endgroup\$
    – Knightyy
    Mar 1, 2018 at 17:59
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The luminous intensity should change linearly with current in that range of currents. The forward voltage will drop with current so you won't exactly have a linear effect with 1/R unless you decrease the voltage to compensate.

Older LEDs suffered from a threshold effect where luminous intensity dropped below linear response at low currents, but modern ones are much more defect-free and don't have that effect, at least above a uA.

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