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

[Femme_physics]

Yeppers! :)

It's a divide-by-4 circuit!

Yea! But I can't tell it without calculations. I'm not gneill or you!

That's because the current doesn't split.
Any reason you think it does?

Oh, wait, so 2 currents "emerge" from Vx?

Something is really unclear to me. If current only comes from Vin, it appears to experience quantom leaps. I.e. Do you see what I mean? It doesn't go in a loop, but it starts at one point (Vin) and ends up at another point (Vx) without "flowing" to that point.

In addition, you said:

I'm afraid I didn't plan anything. You did the planning.
I just followed your loop and found it's not consistent - you made some sort of a jump.

But you have the right loops here.

I have the right loops, despite the fact I made a jump? That's exactly what I'm talking about! How am I allowed to do those jumps? Kirchhoff laws don't talk about quantum jumps!

These are the right loops!

And I1=-I2 because of KCL.
(Note the minus sign.)

Noted. Unclear why, though.

For starters, mathmatically speaking, how can a plus equal a minus?

-5 is by no mean +5, rather, -5 is -5.

 

[I like Serena]

Yea! But I can't tell it without calculations. I'm not gneill or you!

I can't tell either without calculations (but perhaps gneill can).

Oh, wait, so 2 currents "emerge" from Vx?

Umm... no...
You chose to draw 2 arrows as if the current "emerges".
But it doesn't.
So one of the currents has to be negative.

Something is really unclear to me. If current only comes from Vin, it appears to experience quantom leaps. I.e. Do you see what I mean? It doesn't go in a loop, but it starts at one point (Vin) and ends up at another point (Vx) without "flowing" to that point.

Actually, current starts at Vout, where it is fed by the voltage supply of the op-amp.

Then it flows through R2, where it turns out that the voltage happens to be 2 [V] (has to be due to the nature of the op-amp).

Then it flows through R1, after which it reaches earth.Where's the quantum leap?

In addition, you said:

I have the right loops, despite the fact I made a jump? That's exactly what I'm talking about! How am I allowed to do those jumps? Kirchhoff laws don't talk about quantum jumps!

You did not make a quantum jump this time around.

Noted. Unclear why, though.

For starters, mathmatically speaking, how can a plus equal a minus?

-5 is by no mean +5, rather, -5 is -5.

You're right! A plus does not equal a minus!
Heh... glad we got that straight! Since you drew arrows signifying both currents leave Vx, KCL says:
I1+I2=0

so:
I1=-I2.

 

[gneill]

I can't tell either without calculations (but perhaps gneill can).

Not every time, but I can usually make a pretty good guess! I put it down to too much time spent at the books and not enough at the pub.

 

[gneill]

I just thought I might offer the following thought regarding analyzing these sorts of circuits. It often turns out that Nodal Analysis (KCL) is the most expedient way to handle op-amp circuits. This is because the nature of the op-amp often fixes voltages at certain nodes or eliminates certain currents right off the bat. It simplifies things considerably when some of the "variables" are taken care of like that.

In the case of our mystery divide by four circuit, KCL at node Vx can make short work of things:

 

[Femme_physics]

Umm... no...
You chose to draw 2 arrows as if the current "emerges".
But it doesn't.
So one of the currents has to be negative.

-------

Actually, current starts at Vout, where it is fed by the voltage supply of the op-amp.

Then it flows through R2, where it turns out that the voltage happens to be 2 [V] (has to be due to the nature of the op-amp).

Then it flows through R1, after which it reaches earth.Where's the quantum leap?

Ah, that's clear (the no quantum leaps part) BUT if current flows from Vout to ground...shouldn't Vout be positive? You told me it's OK despite I wrote Vout as negative.

Which means...

http://img33.imageshack.us/img33/3422/vouters.jpg

I just thought I might offer the following thought regarding analyzing these sorts of circuits. It often turns out that Nodal Analysis (KCL) is the most expedient way to handle op-amp circuits. This is because the nature of the op-amp often fixes voltages at certain nodes or eliminates certain currents right off the bat. It simplifies things considerably when some of the "variables" are taken care of like that.

In the case of our mystery divide by four circuit, KCL at node Vx can make short work of things:

Thanks gneill! I will print it.

 

[I like Serena]

Ah, that's clear (the no quantum leaps part) BUT if current flows from Vout to ground...shouldn't Vout be positive? You told me it's OK despite I wrote Vout as negative.

Yes, Vout could be negative.
In that case current would flow from the ground to Vout.

My point was that the current does not start out of thin air from the middle (the way you drew it, suggested that).

Which means...

http://img33.imageshack.us/img33/3422/vouters.jpg

Yep.
But you made a different choice for the direction of the currents I1 and I2 here.

You can choose the direction of your currents separately from the direction of your loop.
(And you should for more complicated applications of KVL and KCL.)
In this picture you did not draw the direction you chose for the currents.
So I can only assume that you intended to draw the currents in the same direction as the loop.

 

[Femme_physics]

Yes, Vout could be negative.
In that case current would flow from the ground to Vout.

Which is how it's supposed to be (according to convention), yes?

My point was that the current does not start out of thin air from the middle (the way you drew it, suggested that).

Oh, of course. That didn't make sense to me either, wanted to clear it out

Yep.
But you made a different choice for the direction of the currents I1 and I2 here.

You can choose the direction of your currents separately from the direction of your loop.
(And you should for more complicated applications of KVL and KCL.)
In this picture you did not draw the direction you chose for the currents.
So I can only assume that you intended to draw the currents in the same direction as the loop.

Yes, your assumption is correct. You will definitely see me use more KVL and KCL in more op-amps exercises like that. Just as a verification method.

 

[I like Serena]

Which is how it's supposed to be (according to convention), yes?

Yes.

Yes, your assumption is correct. You will definitely see me use more KVL and KCL in more op-amps exercises like that. Just as a verification method.

Nice to see that KVL/KCL works too, isn't it?
And it's not as if the calculations are all that hard!

Verification is good!

 

[Femme_physics]

*bumps*

Yes, KVL, KCL, the usual analysis tools. If you can see how the diagram I gave arises (here it is again):

Then you can start from there. Suppose that all resistors are R. Apply nodal analysis at the node labelled Vx. You'll get an equation involving Vin, Vout, and Vx. Make Vx vanish by noting that i3 = Vx/(2R), and that Vout = i3*R. Re-arrange what's left to find Vout in terms of Vin.

Just reviewing some older material.

What made you determine that Vout = VR4?

 

[gneill]

Just reviewing some older material.

What made you determine that Vout = VR4?

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.

 

[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