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

[I like Serena]

True, but we have different valued resistors. Different valued resistors mean different currents.

Not in this case.
KCL says that current in equals current out.
The current can not go anywhere else (since it doesn't go into the op-amp).

It's the same when 2 different resistors are in series.
The current through both of them is the same.

 

[Femme_physics]

KCL says that current in equals current out.

Agreed.

The current can not go anywhere else (since it doesn't go into the op-amp).

Right, they both go to the minus end of Vi. How does that necessarily mean they're equal?

It's the same when 2 different resistors are in series.

When 2 different resistors are in series 2 different currents flow through them. If this case is the same as 2 different resistors in series, I1 should not equal I2.

The current through both of them is the same.

How can you prove it?

Edit: Oh and...

http://img46.imageshack.us/img46/1332/hardar.jpg

Did I get it?

Haven't figured out Av for a comparator, I'll it up as soon as I see I'm right so far.

 

[I like Serena]

Agreed.

Right, they both go to the minus end of Vi. How does that necessarily mean they're equal?

When 2 different resistors are in series 2 different currents flow through them. If this case is the same as 2 different resistors in series, I1 should not equal I2.

How can you prove it?

Huh?

Consider this circuit:
http://img195.imageshack.us/img195/3663/thehell.jpg

This is "in series".
The same current flows through both of the resistors.

 

[Femme_physics]

Huh?

Consider this circuit:
http://img195.imageshack.us/img195/3663/thehell.jpg

This is "in series".
The same current flows through both of the resistors.

I...can't believe I forgot that! Oops, my bad. Thanks. :shy:

PS: I edited the reply above...You're fast!

 

[I like Serena]

Edit: Oh and...

http://img46.imageshack.us/img46/1332/hardar.jpg

Did I get it?

In your 2nd picture you connected the left leg (b) to the ground.
It's not connected to the ground but to a "controlled" voltage source giving off a voltage of Vout.

In other words, your Sum V(b) is not right.
It should include Vout.

 

[I like Serena]

I...can't believe I forgot that! Oops, my bad. Thanks. :shy:

PS: I edited the reply above...You're fast!

I'm not fast... you're too slow!

 

[Femme_physics]

In your 2nd picture you connected the left leg (b) to the ground.
It's not connected to the ground but to a "controlled" voltage source giving off a voltage of Vout.

In other words, your Sum V(b) is not right.
It should include Vout.

You're right. I forgot. I didn't have a laptop at work today since I forgot my power cord...
------------
Also,
Tried to solve this

http://img194.imageshack.us/img194/5046/webcam1322988230.png [/QUOTE]

http://img51.imageshack.us/img51/1160/cannot.jpg

My only problem was trying to solve it using only KCL and KVL...as you can see.

I'm not fast... you're too slow!

True

 

[I like Serena]

Oh, it can be solved using only KCL and KVL, and I recommend doing so if only as a verification.

In your 1st method you made a calculation mistake (which you can find by verifying with the 2nd method).

In your KCL/KVL method you didn't use a proper loop.

 

[Femme_physics]

Oh, it can be solved using only KCL and KVL, and I recommend doing so if only as a verification.

Yes, that's what am trying to do

In your 1st method you made a calculation mistake (which you can find by verifying with the 2nd method).

Oh, yea, Vout = 6. Oops!

In your KCL/KVL method you didn't use a proper loop.

Ok, hmm. What's wrong with my loop?

 

[I like Serena]

Yes, that's what am trying to do
Oh, yea, Vout = 6. Oops!Ok, hmm. What's wrong with my loop?

Let's see, first you go up from Earth to 2 [V].
We're at V- now aren't we?

Then you follow resistor R1 down, which is good...

And then we're at Earth again.
All finished!

... except that you moved on to R2, and then Vout...
But how did you get there?

 

[Femme_physics]

.. except that you moved on to R2, and then Vout...
But how did you get there?

Because there are 2 ways to get to the ground. Yours is easier, admittedly, but I don't see Vout in your equation?

Also...->

http://img196.imageshack.us/img196/4086/tryouttry.jpg

Yes?

 

[I like Serena]

Because there are 2 ways to get to the ground. Yours is easier, admittedly, but I don't see Vout in your equation?

Okay, let's do this again and follow a different path.

First 2 [V] up from Earth to V-.

Then we follow R2 down (instead of R1).

Then we go down with Vout.

... and we're back at the ground!


Now we have 2 loops...
Can you solve them?

 

[I like Serena]

Also...->

http://img196.imageshack.us/img196/4086/tryouttry.jpg

Yes?

Your equations are good!

But... your solution isn't!

If you substitute your solution in the first equation for V(a), you'll see it doesn't fit.

 

[Femme_physics]

Not in this case.
KCL says that current in equals current out.
The current can not go anywhere else (since it doesn't go into the op-amp).

It's the same when 2 different resistors are in series.
The current through both of them is the same.

But...but the current splits! How is this the same as resistors in series?



Also, I guess I made the loops in reverse than what you had planned but I'm just trying to see if I got the right loops:
http://img194.imageshack.us/img194/9471/yeathis.jpg

http://img842.imageshack.us/img842/8396/iriririr.jpg

Your equations are good!

But... your solution isn't!

If you substitute your solution in the first equation for V(a), you'll see it doesn't fit.

I love it how you check up on my numbers, not just on my method. I guess you know I have a tendency to slip

Ok,

I got now,

I1 = 0.6 mA
I2 = 0.3 mA
I3 = 0.3 mA

Vout = I3R4 = 0.0003 x 5000 = 1.5 [V]

 

[I like Serena]

But...but the current splits! How is this the same as resistors in series?

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

Also, I guess I made the loops in reverse than what you had planned but I'm just trying to see if I got the right loops:
http://img194.imageshack.us/img194/9471/yeathis.jpg

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.

http://img842.imageshack.us/img842/8396/iriririr.jpg

These are the right loops!

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

I love it how you check up on my numbers, not just on my method. I guess you know I have a tendency to slip

Ok,

I got now,

I1 = 0.6 mA
I2 = 0.3 mA
I3 = 0.3 mA

Vout = I3R4 = 0.0003 x 5000 = 1.5 [V]

Yeppers! :)

It's a divide-by-4 circuit!

 

[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