Electrical – will this timer/delay circuit works as intended with almost equal opamp inputs

comparatoroperational-amplifiervoltage divider

I just saw this video about a so called delay circuit using opamp in comparator configuration. The author mentions to use a pot as a voltage divider to feed the (+) input and the node between a series R and C to the (-) input.
however the assembled circuit on the breadboard is missing the connection of the pot to ground (like the schematic I included bellow).
so basically after some amount of time the comparator should see almost equal inputs thus the output is indeterminable. I hooked up the circuit using same values but a TL082 and I got white noise as expected, yet I can't understand does this circuit really outputs a steady low output? the SW1 is meant to restart the timing. does the pot and internal input resistance forms a divider? the author confirms that only two terminals of the pot are used in the comments.
do i miss something else?

highly appreciate your help.

schematic

simulate this circuit – Schematic created using CircuitLab

Best Answer

the author confirms that only two terminals of the pot are used in the comments.

Then he's stupid.

This will not work as I see it working. You need to set a defined voltage at +Vin by using a potential divider and not rely on the giga ohm input impedance of the op-amp's input resistance as some kind of valid value.

Other problems: -

  • The TL082 minimum recommended supply is 10 volts and not 9 volts.
  • Input common mode range is guaranteed to be +/- 11 volts on a +/- 15 volt rail and if your supply rail is (say) +10 volts then the input range is from +4 volts to +6 volts.
  • Output voltage range is guaranteed to be +/- 12 volts on a +/- 15 volt rail hence, on a + 10 volt supply, the output cannot be relied upon outside the range +3 volts to + 7 volts i.e. your LED will likely always be on. Typical limits might cause the LED to switch off but you shouldn't rely on typical limits because you'll get bitten.