Timeline for Is a comparator the correct part to choose?
Current License: CC BY-SA 3.0
6 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jan 17, 2017 at 9:48 | comment | added | Bimpelrekkie | In practice these inverters have a very narrow window where the output voltage is not properly determined. From the datasheet it looks like that window is 0.4 V (1.2 V - 0.8 V) but these are the guaranteed values. In practice most of these inverters will decide (1 or 0) right in the middle so 1.0 V when Vsupply = 2 V. So at 3.3 V supply I would expect the typical "decision point" to be at 3.3 /2 = 1.65 V so well above your 1.2 V. If you want to do mass production with this circuit then you will need proper data from the manufacturer at a 3.3 V supply voltage. | |
| Jan 17, 2017 at 8:55 | comment | added | Florian Castellane | You could use a diode in series to drop the voltage. Otherwise, yes, a comparator will work, although it is a more complex solution. | |
| Jan 17, 2017 at 8:47 | comment | added | John | @Florian Castellane In that case I'm not sure if I can use the inverter solution. I know I wrote 0.8v above but during special moon phases the voltage might even be 1.1v but should still be considered as low. What to do in that case? Comparator? | |
| Jan 17, 2017 at 8:38 | comment | added | Florian Castellane | @John the output depends on the moon phase when the chip was manufactured -- only guaranteed values are 0,8v low and 1,2v high. | |
| Jan 17, 2017 at 8:29 | comment | added | John | Thanks for your suggestion, one additional question to that: What if there is a voltage between 0.8v and 1.2v? In that case the inverter does not change the output? Whats the output for, lets say, 1.1v ? | |
| Jan 17, 2017 at 8:08 | history | answered | Bimpelrekkie | CC BY-SA 3.0 |