Timeline for Driving an LED from a uC pin with the Anode above uC's Vcc rail
Current License: CC BY-SA 3.0
13 events
| when toggle format | what | by | license | comment | |
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| Sep 25, 2014 at 18:11 | comment | added | Zeph | The answers are interesting - at this point there is a good answer explaining why the high impedance approach would be safer, and another good answer explaining why the active high would be safer. I think I understand both lines of reasoning, but am not sure which is more accurate. The subtleties explored in each answer are exactly what I was looking for, because they are at the edge of my understanding. Now I'm hoping those with more experience will be able to sort out a single best approach. | |
| Sep 25, 2014 at 18:07 | history | edited | Zeph | CC BY-SA 3.0 |
correct resistor value
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| Sep 25, 2014 at 12:53 | answer | added | Evad | timeline score: 1 | |
| Sep 25, 2014 at 3:19 | answer | added | Spehro 'speff' Pefhany | timeline score: 3 | |
| Sep 25, 2014 at 2:00 | answer | added | mixed_signal | timeline score: 1 | |
| Sep 24, 2014 at 21:23 | comment | added | Zeph | @KyranF The 1K2 resistor's purpose is to limit current when the LED is on. When the LED is off (pin high impedance) the current path through the pin would include the Vfwd of the LED (2-3V), the resistor, and the clamping diode. The resistor's effect is negligible, most of the effective resistance comes from there being less than Vfwd between 4.5V Vin and the 3.3v Vcc rail. | |
| Sep 24, 2014 at 21:12 | comment | added | Zeph | @Kyranf I used a red LED as an example, but a blue or green LED with Vfwd closer to 3V gets a bit tighter in terms of margins when powered from 3.3v out of the Arduino Pro Mini's (only about 0.3 V for the current limiting resistor). That's one reason I'd like to explore powering the LEDs directly from the 4.5v Vin. | |
| Sep 24, 2014 at 19:24 | comment | added | Passerby | Agreed. unless the regulator is some ooddly weak thing, the only thing you are doing is shifting the power\heat loss from the regulator to the led resistors. While still having the 40/100/240 pin/port/total mA current limits. Sinking current to a low pin, has the same issues as sourcing current out of the pin. In this case, positive Voltage Droop. (er, lift?) | |
| Sep 24, 2014 at 19:15 | comment | added | gbulmer | @KyranF - If the pin is 'tristated', i.e. an input, then the LED should be below it's threshold, and be a very large resistance; it will be a "VERY low power over-voltage", so it won't be "constant over-voltage condition with only a 1.2k resistor". Having said that, I wouldn't do this unless I needed to fix something already broken, and this was my 'best option'. | |
| Sep 24, 2014 at 18:41 | answer | added | Olin Lathrop | timeline score: 3 | |
| Sep 24, 2014 at 18:25 | comment | added | KyranF | Remember that every pin/port of the ATMEGA has CMOS style clamping diodes to VCC and GND to deal with VERY low power over-voltage/negative voltage issues, but if you have a constant over-voltage condition with only a 1.2k resistor I think it will bust the CMOS diode clamps over time and break your entire port | |
| Sep 24, 2014 at 18:21 | comment | added | KyranF | I -think- that going high impedance or active low should be safe enough. But at these current levels why not just source current from the microcontroller's VCC rail as an output rather than a current sink/active low how it is now? | |
| Sep 24, 2014 at 17:58 | history | asked | Zeph | CC BY-SA 3.0 |