Timeline for How can Ohms law be used to calculate the resistor value for an LED when multiple voltage/current pairings give the same resistance?
Current License: CC BY-SA 4.0
14 events
| when toggle format | what | by | license | comment | |
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| Apr 30, 2019 at 21:49 | vote | accept | JShorthouse | ||
| Apr 30, 2019 at 21:42 | comment | added | The Photon | @JShorthouse, edited. | |
| Apr 30, 2019 at 21:42 | history | edited | The Photon | CC BY-SA 4.0 |
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| Apr 30, 2019 at 21:24 | comment | added | The Photon | @JShorthouse, yes that's it. I wish I had thought to express it that way to begin with. | |
| Apr 30, 2019 at 21:23 | comment | added | JShorthouse | Ah ok, I think I get it now. 3V and 20mA aren't two separate 'conditions' that need to be met, but rather a point on a curve and you can't achieve one without the other. So if you're trying to reduce the current down the 20mA then you can just assume that the voltage across the LED will be 3V in your calculations (and hence the voltage across the resistor will be 6V) because that's the only voltage split that can happen at that current. | |
| Apr 30, 2019 at 21:16 | history | edited | The Photon | CC BY-SA 4.0 |
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| Apr 30, 2019 at 21:11 | comment | added | The Photon | In any case, you do know the resistor's IV curve goes through 20 mA @ 3 V because the datasheet told you that, so if you choose your resistor to hit 20 mA at 6 V, you'll be as close as possible (you'd be neglecting part-to-part and thermal variations in the \$V_f\$). If the datasheet told you 20 mA @ 3 V and you wanted to run the LED at 10 mA, you'd need to guess a bit about how much lower the \$V_f\$ would be at 10 mA. Probably a matter of 10's of mV though. | |
| Apr 30, 2019 at 21:08 | comment | added | The Photon | See also the Wiki aritcle on load line analysis. | |
| Apr 30, 2019 at 21:07 | comment | added | The Photon | Yes, if all you know about the LED is "\$V_f=3\ {\rm V}\$", then you essentially have to treat it like a 3 V source (so long as the current is going in the forward direction). If you know the differential resistance of the LED, you could do a more accurate analysis. But practically it would only change the operating forward voltage by a few 10's or maybe 100 mA, because the differential resistance will be much less than 300 ohms. | |
| Apr 30, 2019 at 21:05 | comment | added | JShorthouse | So will the LED always drop 3V regardless? I think my misunderstanding comes from the me thinking that the resistor needs to both limit the current and divide the voltage, is the only purpose of the resistor to limit the current? | |
| Apr 30, 2019 at 21:04 | comment | added | The Photon | If you want to know why, maybe check out this old answer of mine. | |
| Apr 30, 2019 at 21:03 | comment | added | The Photon | A 300 ohm resistor with 6 V across it will give you 20 mA. A 300 oh resistor with 3 V across it will give you 10 mA. If you put a 300 ohm resistor between a 9 V source and a 3 V LED, it will have 6 V across it, not 3 V, so the 3 V result is irrelevant. | |
| Apr 30, 2019 at 21:00 | comment | added | JShorthouse | I completely agree with the second part, but that's the entire basis of my question. Saying "a 300Ω resistor will give you 6V and 0.02A" to me is like saying "If you drove at 60 miles an hour then you must have driven 30 miles in 30 minutes". Why couldn't you have driven 120 miles in 120 minutes? Why can't a 300Ω resistor instead be dropping 3V and letting through 0.01A? | |
| Apr 30, 2019 at 20:54 | history | answered | The Photon | CC BY-SA 4.0 |