Op-amps how could this possibly be a comparator?

[Femme_physics]

I'm asked which type of op-amps these are

http://imageshack.us/photo/my-images/707/opamps.jpg/I think it's clear the SECOND one is a comparator, but the FIRST one is an inverter IMO. It has a feedback loop. If it has a feedback loop, it can't possibly be a comparator! Am I right or what?

The solution manual my teacher made, had them both written as "comparators"

 

[NascentOxygen]

I think it's clear the SECOND one is a comparator, but the FIRST one is an inverter IMO. It has a feedback loop. If it has a feedback loop, it can't possibly be a comparator! Am I right

No, sorry, you're wrong.

Look closer, there is no negative feedback, no feedback of any description, in fact.

There are inverting comparators, and there are non-inverting comparators.

 

[Femme_physics]

where R1 is connected- that's not feedback?

 

[NascentOxygen]

where R1 is connected- that's not feedback?

R1 just goes to earth. Feedback requires some of the output to feed back to the input (to the (-) input for negative feedback).

(If you ever see positive feedback (i.e., back to the + input) then that makes a comparator.)

 

[Femme_physics]

R1 just goes to earth. Feedback requires some of the output to feed back to the input (to the (-) input for negative feedback).

Duly noted, thanks for this explanation :-)

 

[Femme_physics]

I assume that whenver there is a LED after the comparator, only if V+ > V- , then the LED will light up. Is that a correct assumption to make? Also in my specific case?

 

[NascentOxygen]

I assume that whenver there is a LED after the comparator, only if V+ > V- , then the LED will light up. Is that a correct assumption to make? Also in my specific case?

The output that lights the LED is dependent on where the LED's other pin is connected.
If the LED is connected between output and ground then it will light up when the output goes high.

In your case, the LED is connected between OP-AMP output and +Vcc, meaning that when the output is high, there will be not much voltage across the LED.

So ...

 

[Femme_physics]

That clears it :) Thank you