Timeline for LED forward voltage is a range, so how do you calculate resistor value?
Current License: CC BY-SA 3.0
12 events
| when toggle format | what | by | license | comment | |
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| Nov 13, 2017 at 19:17 | vote | accept | Aust | ||
| Nov 13, 2017 at 13:43 | comment | added | Agent_L | That's the reason why cheap LED flashlights use unwieldy number of THREE 1.5V batteries. | |
| Nov 13, 2017 at 12:08 | comment | added | Trevor_G | @ChrisH true, and with the old LEDs at 5V it did not matter. But with more and more low overhead 3.3V stuff they sure would be nice now. | |
| Nov 13, 2017 at 10:20 | comment | added | Chris H | I suspect the effects on power consumption and cost reduce the market significantly compared to just using a slightly bigger rersistor and accepting that you're not getting maximum power. Compare that to illumination LEDs, for which there are many constant current ICs | |
| Nov 13, 2017 at 1:55 | comment | added | Trevor_G | @mkeith ya the issue with a diode is the battery variance. | |
| Nov 13, 2017 at 1:18 | comment | added | user57037 | This is the best circuit once you get passed what can be done with one transistor. I bet you could also omit the linear regulator and just use a zener or regular diode forward biased as a reference. | |
| Nov 12, 2017 at 16:35 | history | edited | Trevor_G | CC BY-SA 3.0 |
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| Nov 12, 2017 at 15:56 | history | edited | Trevor_G | CC BY-SA 3.0 |
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| Nov 12, 2017 at 14:05 | history | edited | Trevor_G | CC BY-SA 3.0 |
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| Nov 12, 2017 at 14:03 | comment | added | Trevor_G | @andre agreed, See my comment under Jonk's answer. | |
| Nov 12, 2017 at 5:53 | history | edited | Trevor_G | CC BY-SA 3.0 |
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| Nov 12, 2017 at 5:18 | history | answered | Trevor_G | CC BY-SA 3.0 |