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In the images below I have the same schematic in multisim, one snapshot for each state of the comparator. I've got the VCC pin connected to 12VDC and the ground and VEE pins to ground. There is a distinct difference between the two states, but 15.8 mV for a "high" doesn't seem right. Shouldn't it be VCC? What am I missing?

Note: I just realized that one of the voltage labels is over the pin, that's pin 8 that's covered.

enter image description here

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  • \$\begingroup\$ Man, you're prolific :p And ... no? I don't know what that is. \$\endgroup\$ – Daniel B. May 29 '14 at 2:11
  • \$\begingroup\$ Quick googling says they can be push-pull or open drain ... datasheet says open drain. What does that mean? \$\endgroup\$ – Daniel B. May 29 '14 at 2:13
  • \$\begingroup\$ ti.com/product/lm311-n There's the datasheet. \$\endgroup\$ – Daniel B. May 29 '14 at 2:13
  • \$\begingroup\$ Aaaaand en.wikipedia.org/wiki/Comparator#Output_type. So I need a pull up resistor ... whatever that means. \$\endgroup\$ – Daniel B. May 29 '14 at 2:16
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The LM311 has an open-collector output; when the output is high it's actually high-impedance. If you want a high voltage then you'll need to use a pullup on the output.

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  • \$\begingroup\$ Well, that's obviously my issue. Thanks for the bonk on the noggin' I guess I need to go do some more learning. And probably source a different op amp, I need the output to push some current and I don't think an open drain can? \$\endgroup\$ – Daniel B. May 29 '14 at 2:19
  • \$\begingroup\$ To clarify ... basically with the open collector, there's high impedance on the output when the + is higher than the - input, so the pull up resistor is connected effectively to an open providing a signal of VCC, then when the - is greater than the +, it shorts to ground so it's down to VEE? That would also mean that when it's at a logic high, there's (virtually)no current through the output, it's just a reference, right? \$\endgroup\$ – Daniel B. May 29 '14 at 2:26
  • \$\begingroup\$ That's accurate for the most part. The output won't be exactly Vcc or Vee due to impedances, but they should easily meet CMOS input thresholds. \$\endgroup\$ – Ignacio Vazquez-Abrams May 29 '14 at 2:31
  • \$\begingroup\$ So ... if I put a pull-up and pull-down each connected to the output, I could create a voltage divider, and if I put my relay (that's what this is for) in series with the pull-down, I should be able to actuate it, right? (that would be where R3 is in the schematic) \$\endgroup\$ – Daniel B. May 29 '14 at 2:33
  • \$\begingroup\$ Putting both a pullup and a pulldown will just drain power needlessly. Use a transistor if you need to switch a load. \$\endgroup\$ – Ignacio Vazquez-Abrams May 29 '14 at 2:39

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