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

[Femme_physics]

Homework Statement

http://img16.imageshack.us/img16/9341/shtoeta.jpg

Find Vout.

Homework Equations

http://img687.imageshack.us/img687/7393/formulaspage.jpg

The Attempt at a Solution

This is the first time I see Vin connected to both the + and the -!

I scanned my known op-amps list, and I can't see anyone similar.

Can anyone tell me the name of that op-amp first, please?

 

[technician]

I have never seen anything like this ! Are you certain you have the correct diagram.
There are applications where Vout is connected back to the - input (this gives a gain of x1) but this circuit seems to me to show some inconsistencies.
I cannot see how Va can be equal to Vb.
I look forward to see the response of others

 

[Femme_physics]

Oh, that reminds me! I forgot to ask my teacher about what you confirmed was correct! I have a class with him sunday, I will definitely remember.

Yes, I'm looking forward to other replies as well-- even though you seem to have great knowledge at the field yourself.

 

[technician]

The first thing I did was re-draw your circuit... I kept R1, R3 and R4 exactly as they are but I then drew R2 just above R3 connecting to the - input.
The connection from Vout to point b remains as it is.
This just did not 'look right'...
Thank you for your comment
I love op amps... they look very complicated but the principles are straight forward and I like sorting them out.

 

[gneill]

It's a tricky divide by 4 circuit. That is, Vout = Vin/4. To analyze, recognize that the feedback from Vout to b will make a = b = Vout. Then forget about the op-amp and concentrate on the resistor network, putting in a controlled source "Vout" to stand in for the the op-amp:

All resistors are 5KΩ.

 

[technician]

Thanks gneill I tried something like that to make Vout = Va =Vb and got tied in Knots !
It is not really a normal question about op amps... difficult to see this as a practical application.

 

[gneill]

Thanks gneill I tried something like that to make Vout = Va =Vb and got tied in Knots !
It is not really a normal question about op amps... difficult to see this as a practical application.

Yeah, you might say that it's more of an academic exercise

As for its practicality, I suppose one would have to check to see how accurate the division operation is for the given component tolerances as compared to other configurations.

 

[Femme_physics]

I forgot to add that the supply voltages are +-10

Is there a name for this op-amp, gneill?

That is, Vout = Vin/4.

REally? That's it? That's the answer?

 

[technician]

I can think of a name for it (if it doesn't already have one)

 

[gneill]

I forgot to add that the supply voltages are +-10

Is there a name for this op-amp, gneill?

I don't know if there's any specific name, but it certainly falls under the general heading of non-inverting amplifier. I'll admit that the positive feedback loop in addition to the negative feedback loop makes it a tad unusual. And yes, the amplification factor is 1/4. You should see if you can derive it from the hints given.

 

[Femme_physics]

Can you elaborate more please? I can't see the logic of just dividing by 4...is it purely based on KVL and KCL?

 

[gneill]

Can you elaborate more please? I can't see the logic of just dividing by 4...is it purely based on KVL and KCL?

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.

 

[gneill]

Just in case it's not obvious where the "reduced" circuit came from, here it is again but with the op-amp shown in place. Note that all I really did was to make the op-amp into a controlled source.

If you think about it, the op-amp is in a voltage follower configuration with the output fed back directly to its "-" input. So it really is a controlled source: it's output is 1x the voltage across the resistor at its + input.

 

[Femme_physics]

From what I understand, it appears you're treating the circuit as though the Vin is our V+ and the minus ends are the plus leg of the transistor and the ground? Such as:

http://img841.imageshack.us/img841/1319/elez.png

I don't know how much that makes sense?

 

[I like Serena]

Yes, it is like that.

But your Sum Va is not quite right.
You're forgetting the voltage of the controlled power supply, which is the same as Vout.
I'm afraid gneill left that out of his (otherwise neat and insightful) drawing.

 

[gneill]

Yes, it is like that.

But your Sum Va is not quite right.
You're forgetting the voltage of the controlled power supply, which is the same as Vout.
I'm afraid gneill left that out of his (otherwise neat and insightful) drawing.

It's the diamond shape at the bottom left, labelled with the carefully concealed clue "Vout".

 

[I like Serena]

It's the diamond shape at the bottom left, labelled with the carefully concealed clue "Vout".

Sorry!
Yes it is there.

@Fp: so you should include Vout in your equation.

 

[gneill]

@FP: also note that your path V(b) should include the voltage drop across R1 due to I1.

 

[Femme_physics]

Yes, it is like that.

OK, rather surprising! but ok..
:shy:
-------------------------------

@Fp: so you should include Vout in your equation.

So I3R4 = Vout?

@FP: also note that your path V(b) should include the voltage drop across R1 due to I1.

Ok, think I got it.

http://img72.imageshack.us/img72/6636/pitaron.jpg

 

[I like Serena]

OK, rather surprising! but ok..
:shy:

How so?

So I3R4 = Vout?

Yes.

Ok, think I got it.
http://img72.imageshack.us/img72/6636/pitaron.jpg

Looks good!
Don't forget to use Vout=I3R4.

 

[Femme_physics]

Great! Thanks. Will do it now (had to take a lil nap :P )

How so?

Because I'm wondering...can I solve all of these op-amps questions I had with just KLV and KCL?

 

[I like Serena]

Great! Thanks. Will do it now (had to take a lil nap :P )

That's also what I do between a couple of difficult problems. ;)

Because I'm wondering...can I solve all of these op-amps questions I had with just KLV and KCL?

Yes.
I was wondering when you would start doing that. :)

 

[Femme_physics]

Yes.
I was wondering when you would start doing that. :)

Oh! If you could show me how!

We weren't taught that properly.Edit: How can you tell me for sure I3R4 = Vout? What is your basis for that?

 

[Femme_physics]

I don't want to start a new thread, so I will just ask here. Can I use Kirchhoff law to find Vout here?

http://img194.imageshack.us/img194/5046/webcam1322988230.png

 

[I like Serena]

Oh! If you could show me how!

We weren't taught that properly.


Edit: How can you tell me for sure I3R4 = Vout? What is your basis for that?

You just did.

I3R4 = Vout, because Vout is across R4 and the current through it is I3.
This is just Ohm's law.

I don't want to start a new thread, so I will just ask here. Can I use Kirchhoff law to find Vout here?

http://img194.imageshack.us/img194/5046/webcam1322988230.png

Yes you can.

Just apply the following rules:

1. Treat the op-amp as if its inputs + and - are not connected to the circuit.
That is, they draw no current in KCL, and they are not part of any loop in KVL.

2. Set V+=V- and solve using KCL and KVL, unless you find it's impossible, or if Vout would come outside its bounds.

 

[Quinzio]

I don't want to start a new thread, so I will just ask here. Can I use Kirchhoff law to find Vout here?

Yes, sure you can, but it's like using a flame-thrower to cook your meal.
Once you grasp the idea of the op-amps, there's no need of such tools.

 

[Femme_physics]

You just did.

I3R4 = Vout, because Vout is across R4 and the current through it is I3.
This is just Ohm's law.

What is the meaning of Vout though? The last voltage drop in a loop?

Yes you can.

Just apply the following rules:

1. Treat the op-amp as if its inputs + and - are not connected to the circuit.
That is, they draw no current in KCL, and they are not part of any loop in KVL.

2. Set V+=V- and solve using KCL and KVL, unless you find it's impossible, or if Vout would come outside its bounds.

But, well, where are my loops? I can't see any loops. I just see lines!

 

[Femme_physics]

Yes, sure you can, but it's like using a flame-thrower to cook your meal.
Once you grasp the idea of the op-amps, there's no need of such tools.

Perhaps you're right, but I want it to provide me with a basis of understanding.

 

[I like Serena]

What is the meaning of Vout though? The last voltage drop in a loop?

Vout = the output voltage

Each circuit typically has an input and an output.
You put an external signal on the input, like a measured temperature.
And you get for instance an amplified signal on the output, that you can use to control an external electro-magnet.

But, well, where are my loops? I can't see any loops. I just see lines!

Consider any 2 Earth's connected.
Consider Earth and Vin connected through a battery with voltage Vin.
Consider Earth and Vout connected through a battery with voltage Vout.
There are no connections through an op-amp.

 

[Femme_physics]

I had a long reply for that yesterday, but somehow got an error. :-/

So, I'll start rewriting it.

Vout = the output voltage

Each circuit typically has an input and an output.

So, in this case...

http://img195.imageshack.us/img195/3663/thehell.jpg

Consider any 2 Earth's connected.
Consider Earth and Vin connected through a battery with voltage Vin.
Consider Earth and Vout connected through a battery with voltage Vout.
There are no connections through an op-amp.

So, this is what I'm getting out of all that:

http://img818.imageshack.us/img818/7315/gettingout.jpg

Am I following you?