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OpAmp output voltage

  • Thread starter monahanj09
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Homework Statement



For the following circuit, if Rth = 50kΩ, Ri = 1TΩ, and Ro = 50Ω, calculate the output voltage. Do not neglect Rth, Ri, or Ro.

Circuit is here.


Homework Equations



Possibly KCL and/or KVL, possibly the non-inverting op-amp equation Vo=(1=Rf/Rs)Vi, Ohm's Law V=iR.

The Attempt at a Solution



This is the first OpAmp problem I've done, and I'm really not even too sure how to start it, and my notes haven't been of much help. All I've done so far is solve for the current through Rth using the 2.3V source and Ohm's Law, but I'm not sure that's even relevant. It also looks to me like the current would be the same through Ri since it looks like they're in series, but that sounds wrong to me, for whatever reason. I would greatly appreciate if someone could walk me through how to do this.
 

Answers and Replies

  • #2
rude man
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You need to get definition of the voltage (current?) source.
 
  • #3
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You need to get definition of the voltage (current?) source.
Wouldn't that just be 2.3V, or am I misunderstanding you?
 
  • #4
gneill
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Inside the OpAmp is a dependent voltage source (the amplifying bit of the OpAmp). You need to specify the voltage gain and what exactly its amplifying.
 
  • #5
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Inside the OpAmp is a dependent voltage source (the amplifying bit of the OpAmp). You need to specify the voltage gain and what exactly its amplifying.
So will that just be Ri/Rth? And it will be amplifying Ro? Or am I completely off?
 
  • #6
gneill
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So will that just be Ri/Rth? And it will be amplifying Ro? Or am I completely off?
The gain of the OpAmp will be specified somewhere... if not, you'll have to keep it as a variable, "A" is often used for it, and your final result will for the output voltage will be an expression involving A. A typical value for A for modern OpAmps is in the neighborhood of 106 or 107.

You should check your course materials to see what it is that an OpAmp does. There's plenty of info available on the Web, too.
 
  • #7
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The gain of the OpAmp will be specified somewhere... if not, you'll have to keep it as a variable, "A" is often used for it, and your final result will for the output voltage will be an expression involving A. A typical value for A for modern OpAmps is in the neighborhood of 106 or 107.

You should check your course materials to see what it is that an OpAmp does. There's plenty of info available on the Web, too.
The gain isn't specified anywhere. So I'm just going to solve it as I normally would, and treat A as an undefined constant, basically?

I've been looking around online to try to get an understanding of what I'm doing and what the OpAmp does, and it's slowly making more sense to me, but I'm still pretty confused. I'm hoping to meet with my professor tomorrow to get some help, but I'd like to get a feel for it before then so I don't have to put off all my homework for this class until tomorrow night.
 
  • #8
gneill
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The gain isn't specified anywhere. So I'm just going to solve it as I normally would, and treat A as an undefined constant, basically?
Yes, that would be your only choice if it's not a given value.

I've been looking around online to try to get an understanding of what I'm doing and what the OpAmp does, and it's slowly making more sense to me, but I'm still pretty confused. I'm hoping to meet with my professor tomorrow to get some help, but I'd like to get a feel for it before then so I don't have to put off all my homework for this class until tomorrow night.
Okay. If you want to proceed further at this point you have to find out what it is that the OpAmp amplifies. Hint: it has something to do with the input terminals and why they're labelled the way they are...
 
  • #9
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Okay. If you want to proceed further at this point you have to find out what it is that the OpAmp amplifies. Hint: it has something to do with the input terminals and why they're labelled the way they are...
Well, from what we've learned from class, I'm inclined to say it's the current that runs through Rth. I'm not exactly sure why though.
 
  • #10
gneill
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Well, from what we've learned from class, I'm inclined to say it's the current that runs through Rth. I'm not exactly sure why though.
No, that's not right (although that current does play a role in analyzing OpAmp circtuis). The OpAmp has very high input resistance -- check the given value for Ri. Almost no current will be passed into the OpAmp itself.

What does that leave?
 
  • #11
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No, that's not right (although that current does play a role in analyzing OpAmp circtuis). The OpAmp has very high input resistance -- check the given value for Ri. Almost no current will be passed into the OpAmp itself.

What does that leave?
The input voltage from 2.3V source, and I think the current being fed back from the 12k resistor.
 
  • #12
gneill
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The input voltage from 2.3V source, and I think the current being fed back from the 12k resistor.
Ignore the rest of the circuit for a moment; Consider the OpAmp alone, all by itself as a component. What does an OpAmp do? Why does it have two input terminals labelled + and - ?

Your course notes or text must describe the operating principle behind an OpAmp. It's practically a cornerstone of modern electronic technology!

It really is important to understand what it is that the OpAmp is doing at a basic level. Once you've got that, all the circuits built with OpAmps will be MUCH easier to understand. It will make analyzing circuits made with them much easier (and trying to memorize a half dozen or so different formulas for minor variations in circuit configurations that you can only hope to be able to recognize on an exam is problematical).
 
  • #13
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Ignore the rest of the circuit for a moment; Consider the OpAmp alone, all by itself as a component. What does an OpAmp do? Why does it have two input terminals labelled + and - ?

Your course notes or text must describe the operating principle behind an OpAmp. It's practically a cornerstone of modern electronic technology!

It really is important to understand what it is that the OpAmp is doing at a basic level. Once you've got that, all the circuits built with OpAmps will be MUCH easier to understand. It will make analyzing circuits made with them much easier (and trying to memorize a half dozen or so different formulas for minor variations in circuit configurations that you can only hope to be able to recognize on an exam is problematical).
It just amplifies a voltage, I think. I know the + and - have to do with if it's inverting vs. non-inverting, but I don't really understand the how/why.

Sorry if I'm being difficult, I'm doing my best to understand all this. I've been reading about them in between replies trying to get a feel for it but I'm having a really difficult time grasping the concept for some reason.
 
  • #14
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Okay, so after doing a little more reading I have a better idea of what's going on.

I understand the output voltage is the difference between the input voltage at the non-inverting input (+) and the inverting input (-), multiplied by the gain. In my problem I know the input voltage at the non-inverting input is 2.3V, but how do I tell what it is at the inverting input?
 
  • #15
nsaspook
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It just amplifies a voltage, I think. I know the + and - have to do with if it's inverting vs. non-inverting, but I don't really understand the how/why.

Sorry if I'm being difficult, I'm doing my best to understand all this. I've been reading about them in between replies trying to get a feel for it but I'm having a really difficult time grasping the concept for some reason.
Maybe a visual will help with the concept.
Think of the + and - as the ends of a balance beam and the slider weight as the output of the opamp that moves (via a feedback) to rebalance the beam as the weights change on one side or the other.

mwFMJwNOMoNsYqp4ACK-Utw.jpg
 
  • #16
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Maybe a visual will help with the concept.
Think of the + and - as the ends of a balance beam and the slider weight as the output of the opamp that moves (via a feedback) to rebalance the beam as the weights change on one side or the other.

mwFMJwNOMoNsYqp4ACK-Utw.jpg
So you feed it two separate voltages, and it gives you a voltage that is somewhere in between?
 
  • #17
nsaspook
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So you feed it two separate voltages, and it gives you a voltage that is somewhere in between?
It gives you the voltage that is needed to rebalance (null) the voltage difference on the inputs.
 
  • #18
rude man
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I would just assume that the voltage generator = A(V+ - V-), then let A approach infinity if you want to.
 
  • #19
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Okay, so after doing some research and going back over my notes, I have a fairly good idea of what's going on.

For a non-inverting op amp, as this one is, the gain can be calculated by 1+Rf (a.k.a. the value of the feedback resistor)/Rs (a.k.a. the value of the source resistor). Multiplying that by the input voltage gives the output voltage.

Now the problem I'm running into is what to do with Ri and Ro. I know that RL is my feedback resistor, and RTH is my source resistor, so I should just be able to plug into the aforementioned equation and get out a voltage. Theoretically I think Ro and Ri can be ignored as opens, but the problem specifically says to not ignore any of the resistor values. How do I account for RO and Ri?

I greatly appreciate all the help up to this point, btw. Thanks a lot, guys/gals.
 
  • #20
rude man
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Okay, so after doing some research and going back over my notes, I have a fairly good idea of what's going on.

For a non-inverting op amp, as this one is, the gain can be calculated by 1+Rf (a.k.a. the value of the feedback resistor)/Rs (a.k.a. the value of the source resistor). Multiplying that by the input voltage gives the output voltage.

Now the problem I'm running into is what to do with Ri and Ro. I know that RL is my feedback resistor, and RTH is my source resistor, so I should just be able to plug into the aforementioned equation and get out a voltage. Theoretically I think Ro and Ri can be ignored as opens, but the problem specifically says to not ignore any of the resistor values. How do I account for RO and Ri?

I greatly appreciate all the help up to this point, btw. Thanks a lot, guys/gals.
You are assuming an infinite-gain op amp. V = ∞ x (V+ - V-) where V is the voltage at the left-hand-side of Ro.
OK.

RL is not your feedback resistor. Your feedback resistor is zero ohms (the lower Ri does nothing. (Why? Notice this Ri is just in shunt with RL. Does RL affect the gain?)

The upper resistor you should be able to recognize divides the input voltage (2.3V) via RTH..

Also, do you think Ro impacts anything? What is the expression for VL/V+?
 

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