What is this unique op-amp's name and how do I find Vout?

[Femme_physics]

From the original circuit diagram in the first post of the thread, Vout is tied to the (-) input of the op-amp (the node labeled b). Assuming an ideal op-amp, the (+) input must be at the same potential. Thus node a is also at Vout potential. Between node a and ground is R4. Thus VR4 must be equal to Vout.

Ahhh...I see that now. Thanks for pointing that out...and good catch! I guess you need an electronics whiz vision for that :)

I was originally trying to search an old thread for an info about op-amps I was once provided about the difference between V- and V+

Speaking of ideal op-amps,

In comparators, or order for them to work, we get different value for V+ and V-

In non-comparators op-amps, V- always equal V+?

I remember that answer being written to me in the past but I can't find the thread :(

 

[gneill]

Speaking of ideal op-amps,

In comparators, or order for them to work, we get different value for V+ and V-

In non-comparators op-amps, V- always equal V+?

That's the basic rule of thumb. Usually you can tell when an op-amp is being used as a comparator because you won't find any feedback path going from Vout to the V- input.

 

[Femme_physics]

Perfect. Gneill, you are a world of help.. Thank u very much

 

[gneill]

Perfect. Gneill, you are a world of help.. Thank u very much

Glad to help