| Electronics > Beginners |
| How do I make an LED light up using a comparator? |
| << < (5/7) > >> |
| renzoms:
This is what I have. I believe it's correct. |
| renzoms:
Ok, well if you don't mind helping me here is what I think about this comparator. The + terminal is higher than the - terminal for 4 seconds => the output is "high" for 4s == the output is shorted to ground for 4s. So the output being at "high" for 4s in this case is open circuited? (because the LED is ON, then OFF for 4s, then ON after testing.) I've connected the + terminal at 4V at fully charged capacitor, and the - terminal at .1V. |
| alsetalokin4017:
--- Quote from: james_s on October 22, 2019, 12:23:12 am ---Search engines are your friend. https://www.ti.com/lit/gpn/tlc372 Open-drain is the MOSFET equivalent to open collector, this means that the output can only pull in one direction, down toward ground. So you must connect the LED to +5V through a resistor, and connect the free end to the output of the comparator. When the comparator triggers, the output will become ground, prior to that it is floating (open circuit). --- End quote --- Thanks, I missed that. So option 1) in the top schematic won't work after all. But he's got the LED lighting up so he must have that part wired correctly anyhow as open drain. +4v rail > resistor > LED anode -- LED cathode > Pin 1 , right? |
| renzoms:
5V rail > resistor > LED cathode - anode > Pin 1 |
| alsetalokin4017:
--- Quote from: renzoms on October 22, 2019, 12:30:35 am ---This is what I have. I believe it's correct. --- End quote --- Yes, that's right. When the output goes HIGH, no current can flow because the LED+resistor are also at that same voltage (or the output is simply open). So the LED turns on when the output goes LOW. So you have to swap the inverting and noninverting inputs until you get the behaviour you are looking for. There are only two choices! |
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