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

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In summary, Vin is connected to the - input of the amplifier and Vout is connected to the + input. The amplification factor is 1/4.
  • #1
Femme_physics
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  • #2
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
 
  • #3
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.
 
  • #4
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:blushing:
I love op amps... they look very complicated but the principles are straight forward and I like sorting them out.
 
  • #5
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:

attachment.php?attachmentid=41442&stc=1&d=1322857270.jpg


All resistors are 5KΩ.
 

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  • #6
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.
 
  • #7
technician said:
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 :smile:

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.
 
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  • #8
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?
 
  • #9
I can think of a name for it (if it doesn't already have one):devil::devil::devil:
 
  • #10
Femme_physics said:
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.
 
  • #11
Can you elaborate more please? I can't see the logic of just dividing by 4...is it purely based on KVL and KCL?
 
  • #12
Femme_physics said:
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):

attachment.php?attachmentid=41442&stc=1&d=1322857270.jpg


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.
 
  • #13
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.

attachment.php?attachmentid=41445&stc=1&d=1322869245.jpg


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.
 

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  • #14
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  • #15
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.
 
  • #16
I like Serena said:
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.

:confused: It's the diamond shape at the bottom left, labelled with the carefully concealed clue "Vout".
 
  • #17
gneill said:
:confused: 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.
 
  • #18
@FP: also note that your path V(b) should include the voltage drop across R1 due to I1.
 
  • #19
Yes, it is like that.

:eek:
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 [Broken]
 
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  • #20
Femme_physics said:
:eek:
OK, rather surprising! but ok..
:shy:

How so?
Femme_physics said:
So I3R4 = Vout?

Yes.
Femme_physics said:
Ok, think I got it.
http://img72.imageshack.us/img72/6636/pitaron.jpg [Broken]

Looks good!
Don't forget to use Vout=I3R4.
 
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  • #21
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?
 
  • #22
Femme_physics said:
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. ;)



Femme_physics said:
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. :)
 
  • #23
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?
 
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  • #25
Femme_physics said:
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.



Femme_physics said:
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 [Broken]

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.
 
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  • #26
Femme_physics said:
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. :tongue:
Once you grasp the idea of the op-amps, there's no need of such tools.
 
  • #27
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!
 
  • #28
Quinzio said:
Yes, sure you can, but it's like using a flame-thrower to cook your meal. :tongue:
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. :smile:
 
  • #29
Femme_physics said:
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.


Femme_physics said:
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.
 
  • #30
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 [Broken]

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 [Broken]

Am I following you?
 
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  • #31
Femme_physics said:
I had a long reply for that yesterday, but somehow got an error. :-/

So, I'll start rewriting it.

Yeah, I've had that sort of thing happen to me too.
Femme_physics said:

But what does it do?
120px-Puzzled.svg.png

Nothing is going in and nothing is going out.
It can't be a very useful circuit.
Oh, I know! It's for an exercise!
Femme_physics said:
So, this is what I'm getting out of all that:

http://img818.imageshack.us/img818/7315/gettingout.jpg [Broken]

Am I following you?

Yep, yep, yep, and yep.
As for the Vout question... perhaps you should make a KVL loop including Vin and Vout?
Or just apply Ohm's law using the difference in voltages?
 
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  • #32
Oh, I know! It's for an exercise!
Heh.. yeppers ;)

As for the Vout question... perhaps you should make a KVL loop including Vin and Vout?

But how can I apply loop to an OPEN circuit? I need a beginning and I need an end. Hmm. Oh, I get to call the end Vout, eh?

http://img444.imageshack.us/img444/6113/taktak.jpg [Broken]

Or just apply Ohm's law using the difference in voltages?
But I'm not sure how do I know that I1 = I2. My teacher told me that, but all i know is that

Sigma I = 0 ; It -I1 -I2 = 0

But that doesn't tell me I1 = I2
 
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  • #33
Femme_physics said:
But how can I apply loop to an OPEN circuit? I need a beginning and I need an end. Hmm. Oh, I get to call the end Vout, eh?

http://img444.imageshack.us/img444/6113/taktak.jpg [Broken]

That's it, except for the RL that shouldn't be there.
Femme_physics said:
But I'm not sure how do I know that I1 = I2. My teacher told me that, but all i know is that

Sigma I = 0 ; It -I1 -I2 = 0

But that doesn't tell me I1 = I2

Yeah, actually I1 = -I2, since you drew the arrows in opposite directions, but their magnitudes are the same.
The key thing is that the op-amp does not draw current.
 
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  • #34
I'm still back on the original circuit, the one with 100% negative feedback combined with some positive feedback. Without analyzing it, I'm tipping that this arrangement gives it a very high input impedance. In some applications, a buffer circuit that does not load the signal source can be a useful isolating stage.
 
  • #35
Yeah, actually I1 = -I2, since you drew the arrows in opposite directions, but their magnitudes are the same.
The key thing is that the op-amp does not draw current.

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

That's it, except for the RL that shouldn't be there.

Gotcha :smile:

I'm still back on the original circuit, the one with 100% negative feedback combined with some positive feedback. Without analyzing it, I'm tipping that this arrangement gives it a very high input impedance. In some applications, a buffer circuit that does not load the signal source can be a useful isolating stage.

Will reply to that soon and try to solve the original circuit. :) Thanks.
 
<h2>1. What is an op-amp and why is it unique?</h2><p>An op-amp, or operational amplifier, is an electronic component that amplifies the difference between two input voltages. It is unique because it has a high gain, high input impedance, and low output impedance, making it ideal for use in various electronic circuits.</p><h2>2. How do I identify the name of a specific op-amp?</h2><p>The name of an op-amp can usually be found on the top of the component. It is typically a combination of letters and numbers, such as "LM741" or "OPA2277". You can also check the datasheet or manufacturer's website for more information.</p><h2>3. What is Vout and why is it important?</h2><p>Vout, or output voltage, is the voltage that is produced by the op-amp in response to the input voltages. It is important because it determines the behavior of the op-amp in a circuit and can be used to control other components.</p><h2>4. How do I calculate Vout for a specific op-amp?</h2><p>The calculation for Vout depends on the specific op-amp and the circuit it is being used in. You can refer to the datasheet for the op-amp to find the equation for Vout, or you can use an online calculator or simulation tool to determine the output voltage.</p><h2>5. What factors affect Vout for an op-amp?</h2><p>The output voltage of an op-amp can be affected by several factors, including the input voltages, the supply voltage, the gain of the op-amp, and the load connected to the output. It is important to consider these factors when designing a circuit using an op-amp.</p>

1. What is an op-amp and why is it unique?

An op-amp, or operational amplifier, is an electronic component that amplifies the difference between two input voltages. It is unique because it has a high gain, high input impedance, and low output impedance, making it ideal for use in various electronic circuits.

2. How do I identify the name of a specific op-amp?

The name of an op-amp can usually be found on the top of the component. It is typically a combination of letters and numbers, such as "LM741" or "OPA2277". You can also check the datasheet or manufacturer's website for more information.

3. What is Vout and why is it important?

Vout, or output voltage, is the voltage that is produced by the op-amp in response to the input voltages. It is important because it determines the behavior of the op-amp in a circuit and can be used to control other components.

4. How do I calculate Vout for a specific op-amp?

The calculation for Vout depends on the specific op-amp and the circuit it is being used in. You can refer to the datasheet for the op-amp to find the equation for Vout, or you can use an online calculator or simulation tool to determine the output voltage.

5. What factors affect Vout for an op-amp?

The output voltage of an op-amp can be affected by several factors, including the input voltages, the supply voltage, the gain of the op-amp, and the load connected to the output. It is important to consider these factors when designing a circuit using an op-amp.

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