I am trying to understand how the circuit shown below works. More specifically, I am trying to understand how the comparator outputs only 0 or 12V. (the circuit is from this paper.

Without the 2.2K resistor connecting +12V to the output, I think you have, for the output, +12V if VB>VREF and -12V if VB>VREF.

I am not sure what's going to happen with the resistor in place.

Naively, I am thinking there is no current from the +12V source to go. So the voltage after the 2.2K resistor should just remain at +12V ? enter image description here

  • \$\begingroup\$ It can go into pin 7 of the comparator... \$\endgroup\$
    – user253751
    Aug 27 '18 at 23:04

If you look at the schematic in the datasheet for the LM306, you will find that pin 1 (not shown in your schematic) is Ground, and the output transistor will pull the output close to Ground. The 2K2 resistor is required to pull the output up to +12 V. Like most comparators, the output is "open collector" - the chip can only pull the output to Ground when low. It just "lets go" of the output pin when it should be high.

The negative power input is only used in the input stage.

  • \$\begingroup\$ electronics-tutorials.ws/opamp/opamp103.gif. I was looking at this and was under the impression that the output is either -12 or 12 volts. I guess that impression was wrong and it's component specific? I looked at the datasheet and it's 0 to 5 volts. With the output pin connected to +12V source via a 2.2k resistor, wouldnt the voltage be fixed at some value? \$\endgroup\$
    – Blackwidow
    Aug 27 '18 at 23:15
  • \$\begingroup\$ As I said, the output of the LM306 (and most other comparators) is Open Collector - the output transistor can only pull the output Low. An external resistor or other device is required to pull the output High, and that pull-up device will determine the output High voltage. In your diagram above, the 2K2 resistor will pull the output up to +12 Volts when the comparator output is High. Note that an analog comparator is not an op-amp, although they usually use the same schematic symbol. \$\endgroup\$ Aug 27 '18 at 23:28
  • \$\begingroup\$ thank you. I had to read up on comparators to understand your answer (new to this stuff) looking at the datasheet, if i was happy with HIGH output voltage being about 5 volts, would i not have needed the 12v pull up resistor connected to the output? \$\endgroup\$
    – Blackwidow
    Aug 28 '18 at 3:31
  • \$\begingroup\$ @Blackwidow: if you want a 5 volt High output level, you would connect the 2K2 pull-up resistor to +5 Volts, rather than to +12 Volts. \$\endgroup\$ Aug 28 '18 at 3:57

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