Timeline for Does LED brightness change with voltage?
Current License: CC BY-SA 3.0
10 events
| when toggle format | what | by | license | comment | |
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| Sep 6, 2016 at 12:32 | comment | added | Scott Seidman | Yes- but reader's should know that this is a rough approximation. $V_b$ is generally not much bigger than $V_f$ -- more often, it's bigger by a factor of 2-5, as $V_f$ is about 2V, depending on color. If you get an LED with a resistor built in, designed to be used with 5V, you may not get near enough to $V_F$ to light two in series. | |
| Sep 6, 2016 at 11:31 | history | edited | Ariser | CC BY-SA 3.0 |
contradiction removed.
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| Sep 6, 2016 at 11:29 | comment | added | KalleMP | Your answer contains a bit too much generalisation in the text if one is working with combined voltages close to the LED forward voltage. a 3V LED + resistor combination will not work well at 1.5V or 6V but a 15V LED + resistor combination will work from 5V to 15V in the manner you hope. I gave you an upvote for "(nearly)". | |
| Sep 6, 2016 at 10:57 | history | edited | Ariser | CC BY-SA 3.0 |
formatting corrected
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| Sep 6, 2016 at 10:47 | comment | added | Ariser | @ScottSeidman: thanks for the correction. Better now? | |
| Sep 6, 2016 at 10:46 | comment | added | Ariser | @pjc50 That's why I wrote \$V_b >> V_f\$. I didn't want to extend it to n LEDs then :) | |
| Sep 6, 2016 at 10:45 | history | edited | Ariser | CC BY-SA 3.0 |
corrected wrong statement. Thanks to @ScottSeidman
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| Sep 6, 2016 at 10:38 | comment | added | Scott Seidman | This is incorrect, until you change "reduced" to "approximated". | |
| Sep 6, 2016 at 10:38 | comment | added | pjc50 | This is true, but relies on Vb > 2Vf, i.e. most of the power being spent in the resistors in the normal use case. | |
| Sep 6, 2016 at 10:26 | history | answered | Ariser | CC BY-SA 3.0 |