Timeline for Questions about LEDs in a circuit
Current License: CC BY-SA 3.0
11 events
| when toggle format | what | by | license | comment | |
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| Sep 15, 2015 at 11:18 | comment | added | stefandz | Depends what the resistor protecting the LEDs is. Is it's 15 ohms, don't ignore! If it's 150, it's partially worth it, and at 1k then don't bother at all. Also remember CR2032's aren't rated at a very high discharge current, so don't exceed that. | |
| Sep 15, 2015 at 11:14 | comment | added | RJSmith92 | Hi again, just a quick question, don't know if you can remember but here you said that when we are using a resistor we ignore internal resistence due to it being so small. Just looking at the example you gave above of the CR2032 that has 15 ohms of resistance, would you still ignore this amount even if you were using a resistor in the circuit, seems like a lot? If it was a few miliohms I'd undersetand but 15 ohms seems a lot to ignore? Thanks. | |
| Aug 17, 2015 at 0:03 | comment | added | stefandz | Yes, but this is such a small amount as to be negligible compared to temperature effects and device-to-device variation. I wouldn't consider it in my design calculations, I would calculate, build and measure (but only to ensure I was within whatever tolerance I had set myself - probably 10% for LEDs). | |
| Aug 16, 2015 at 23:59 | comment | added | RJSmith92 | Sorry I'm being really, really picky here, just want to confirm something, In the circuit with a 3V battery and 2V LED, if the LED is rated 2V at 20mA, then technically at 66mA the LED will have a voltage drop a minuscule amount above 2V, and the battery will have slightly more voltage available externally (the same amount as the LED is dropping)? Please son't feel you have to write a long answer, you have done more than enough! | |
| Aug 16, 2015 at 23:01 | vote | accept | RJSmith92 | ||
| Aug 16, 2015 at 23:01 | comment | added | RJSmith92 | Brilliant, thanks for the quick response as always, much appreciated! | |
| Aug 16, 2015 at 22:39 | comment | added | stefandz | I'll be honest with you - KVL never gets broken until you're way up in RF territory, as @LorenzoDorati mentions in a comment above. The 3V battery is no longer a 3V battery at this stage - it's a 3V battery with an internal resistance (inside its casing) dropping 1V so that at its output you're only getting 2V. That is to say there's 3V inside, 1V dropped internally and 2V available externally. | |
| Aug 16, 2015 at 22:32 | comment | added | RJSmith92 | I might just to do that :) One last thing that I forgot to ask regarding your example with the 3V battery and 2V LED. The LED will still have voltage drop of around 2V at 66mA, is the 3V battery now only pushing with a voltage of around 2V so KVL is broken? | |
| Aug 16, 2015 at 17:32 | comment | added | stefandz | Yes, but at 120mA you're starting to get beyond the absolute maximum rating of a normal LED (again, this is LED specific) and likely to get into thermal runaway where the voltage drop starts to decrease with temperature increase (counterintuitive, but that is because the LED is a semiconductor device and you are giving the charge carriers higher mobility), you get higher current draw and this runs away until the LED goes pop! You know, given your interest, if I were you I would buy a grab-bag of resistors, a bag of cheap LEDs, a power supply and try out some of these scenarios for yourself :) | |
| Aug 16, 2015 at 17:23 | comment | added | RJSmith92 | Thanks again @stefandz, much appreciated answering all those. Regarding point 3, OK, maybe not a 50% increase but small differences with a small changes in current? Regarding point 2, don't know much about this yet and will read more later, but quickly, in your example with the 3V battery and an LED with a 2V drop at 20mA, the current would be around 66mA and the LED would have slightly higher voltage drop than 2V. If i then used battery with 5V and an internal resistance of 25 Ohms (just made it up), the current would be (5-2)/25 = 120mA and the voltage drop of the LED would be a bit higher? | |
| Aug 16, 2015 at 13:18 | history | answered | stefandz | CC BY-SA 3.0 |