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| Back before basic (voltage or current)? |
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| FriedMule:
Thanks for all your fantastic answers! :-) Just to confirm what I do understand you say abut LED and Voltage and Current: The LED do not use any Current without Voltage and even with, lets say 0.5V the LED still do not use any current, it is first at it's minimum on-Voltage that the LED starts to use Current. To not burning the LED off, it's the famous Ohm's law that tell what resistance that is needed. It is now wary clear and I think I do know about what I need to power a LED. :-) Just one last question, I have seen dimmer's that regulated the LED by voltage, not current. It was something about current was a failior point because the LED heats up and the resistance changes, which again heightens the amount of current, that again heats the LED up even more... a cycle that fast can shorten the LED's life. |
| james_s:
Again ohms law applies. If the resistance in series with the LED is constant, then varying the voltage will cause the current to vary. |
| FriedMule:
--- Quote from: james_s on October 03, 2019, 12:36:49 am ---Again ohms law applies. If the resistance in series with the LED is constant, then varying the voltage will cause the current to vary. --- End quote --- Yes you are completely correct, that's also why I did not understand that about adjusting Voltage instead of Current, to avoid the "evil cycle" |
| Rick Law:
LED has resistance, ohm law applies, and V=IR. Think of it as LED as a resister with resistance that changes based on voltage. LED is a semi conductor, it kind of conducts but do it in a different way unlike a full conductor. With LED, V=IR, but you have to tell me what is V before I can tell you what R would be.... And, in the case of LED, it conducts one way (R=infinite when V<0) and it turns the energy (from voltage dropping across it) into light. The light comes on visibly only when voltage is at a certain level. To imagine such magic "turn on the light voltage", think of it like boiling water. If you have a pot of water that is not at boiling point, water is evaporating whenever temperature is > absolute zero, so, some steam (a few water molecule that turned into gas form) is coming out, but at such low amount that you wont see the steam - you call that evaporation. When the pot is at boiling point temperature (100C at sea-level and standard atmospheric pressure), well, being unevenly heated, the water closer to the flame probably get to boiling first. Every water molecule in that pot that reached boiling point turns into gas. Once you reach that magic "turn on" point, enough water in the pot is turning into gas that you see the steam, and it can blow a whistle or drive a turbine. So, LED being a resister with voltage-dependent resistance, you apply low voltage, very few photons come out and you have no visible light. I=V/R, but since resistance is not constant, you don't know the resistance to calculate current directly. You have to first find out at that specific voltage what R is, then I=V/R (or V=IR, same thing) can be evaluated. As you increase the voltage to near "turn on", more and more photon are coming so you see the light. Again, resistance changed by now since voltage is increased. V=IR applies. At around that range, you can indeed control brightness a bit by adjusting the voltage but that adjustment range is very small so controlling brightness that way is ineffective. It is easier to control how much current goes in by using a current limiting (ballast) resister - well, the LED is at (around) 3.15V, I am applying X volt, so at 100mA, my ballast resister should be Y. etc, etc. Easy to calculate. The LED may "settle" at 3.145 volt or 3.151 volt or whatever. With your 1% ballast resister, you are not getting better than 1% precision with voltage anyhow, so you just evaluate it as 3.15V. When you increase the voltage beyond LED's operational range, well, the thing burns out. Now, had you started with negative voltage, LED being a resister with voltage-dependent resistance, and at negative voltage, the resistance is near infinite so you have nearly no current going through - just a few electrons leaked across. |
| FriedMule:
--- Quote from: Rick Law on October 03, 2019, 04:55:44 am ---LED has resistance, ohm law applies, and V=IR. Think of it as LED as a resister with resistance that changes based on voltage. LED is a semi conductor, it kind of conducts but do it in a different way unlike a full conductor. With LED, V=IR, but you have to tell me what is V before I can tell you what R would be.... And, in the case of LED, it conducts one way (R=infinite when V<0) and it turns the energy (from voltage dropping across it) into light. The light comes on visibly only when voltage is at a certain level. To imagine such magic "turn on the light voltage", think of it like boiling water. If you have a pot of water that is not at boiling point, water is evaporating whenever temperature is > absolute zero, so, some steam (a few water molecule that turned into gas form) is coming out, but at such low amount that you wont see the steam - you call that evaporation. When the pot is at boiling point temperature (100C at sea-level and standard atmospheric pressure), well, being unevenly heated, the water closer to the flame probably get to boiling first. Every water molecule in that pot that reached boiling point turns into gas. Once you reach that magic "turn on" point, enough water in the pot is turning into gas that you see the steam, and it can blow a whistle or drive a turbine. So, LED being a resister with voltage-dependent resistance, you apply low voltage, very few photons come out and you have no visible light. I=V/R, but since resistance is not constant, you don't know the resistance to calculate current directly. You have to first find out at that specific voltage what R is, then I=V/R (or V=IR, same thing) can be evaluated. As you increase the voltage to near "turn on", more and more photon are coming so you see the light. Again, resistance changed by now since voltage is increased. V=IR applies. At around that range, you can indeed control brightness a bit by adjusting the voltage but that adjustment range is very small so controlling brightness that way is ineffective. It is easier to control how much current goes in by using a current limiting (ballast) resister - well, the LED is at (around) 3.15V, I am applying X volt, so at 100mA, my ballast resister should be Y. etc, etc. Easy to calculate. The LED may "settle" at 3.145 volt or 3.151 volt or whatever. With your 1% ballast resister, you are not getting better than 1% precision with voltage anyhow, so you just evaluate it as 3.15V. When you increase the voltage beyond LED's operational range, well, the thing burns out. Now, had you started with negative voltage, LED being a resister with voltage-dependent resistance, and at negative voltage, the resistance is near infinite so you have nearly no current going through - just a few electrons leaked across. --- End quote --- WOW THANKS!!! This is without doubt the best answer I could ever hope fore!! So a LED with 0.5V would use a tiny amount of current that you may be able to measure with a pA meter or something like that and until the "boiling pint not much will be visible, while at the "on pint" the LED starts to need a "lot" of current, enough to se it on a cheap pocket meter:-) |
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