Opamp transfer function problem

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Discussion Overview

The discussion revolves around a homework problem related to the transfer function of an operational amplifier (op-amp) circuit. Participants explore various methods for analyzing the circuit, including nodal analysis and superposition, while addressing the complexities introduced by multiple inputs and feedback components.

Discussion Character

  • Homework-related
  • Technical explanation
  • Debate/contested

Main Points Raised

  • One participant proposes to prove the equation vo=(2R2/R1)[1+R2/Rv][vI2-vI1] involving a variable resistor Rv.
  • Another participant questions the assumption that voltage at node 2 is zero, stating that the current from node 3 to node 2 is unknown.
  • It is suggested that without knowing both vo and vI2, the voltage at node 3 cannot be determined.
  • Some participants advocate for using superposition and nodal analysis to simplify the problem, while acknowledging the complexity of the resulting equations.
  • Terminology and definitions related to operational amplifiers and their circuits are emphasized as crucial for understanding the problem.
  • One participant shares their approach to solving op-amp problems, emphasizing the importance of ensuring that the inputs to the op-amp are equal for the circuit to function correctly.
  • Corrections to earlier posts are noted, with one participant acknowledging a mistake in their KCL equations and providing revised equations for analysis.

Areas of Agreement / Disagreement

Participants express differing views on the assumptions regarding node voltages and the methods for analyzing the circuit. There is no consensus on the best approach to take, and multiple competing views remain throughout the discussion.

Contextual Notes

Some participants highlight the limitations of their assumptions and the need for precise definitions in the context of operational amplifier circuits. The discussion includes unresolved mathematical steps and dependencies on specific circuit configurations.

kenok1216
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Homework Statement


1.PNG

to prove vo=(2R2/R1)[1+R2/Rv][vI2-vI1] and Rv is variable resistor

Homework Equations

The Attempt at a Solution


2.PNG

upper: 2 is 0V? and then Rv,2R2 form a t-network?
bottom:3 is 0V? if not how can i calculate?[/B]
 
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kenok1216 said:
upper: 2 is 0V? and then Rv,2R2 form a t-network?
No, you don't know what the current from (3) to (2) might be. So you cannot say that V(2) is zero.
bottom:3 is 0V? if not how can i calculate?
No, you cannot know V(3) without knowing both vo and vI2.

You'll need to analyze the circuit from first principles, perhaps using nodal analysis and the properties of an ideal op-amp.
 
I think, it is a good approach to use superposition - as proposed by you (setting one of both input signals to zero).
Nevertheless, the remaining circuits are still pretty involved and - as suggested by gneill - you should use nodal analyses for both solutions.
 
If you read this post earlier be aware it had a mistake
which is now corrected, in orange ... old jim

They teach us to solve opamp problems by resolving Zfeedback/Zinput
but when you've got two inputs like above, it's not apparent where to start.

Terminology is important - be picayune and exact with your definitions. I usually quote Lavoisier at this point but will spare you that boring lecture today.

Old JIm says:
An "Operational Amplifier" is just a piece of hardware like a LM741, usually a DC coupled amplifier with high gain
An "Operational Amplifier Circuit " is an electronic circuit , usually having an operational amplifier at its heart,
wired up so that it has a transfer function that's some mathematical operation .like add, multiply, divide, or perform some transcendental function
so we say it "operates" in a mathematical sense not just a "does it work" sense..
That's being nitpicky. But it helps me think straight.

Here are two ground rules i use for op-amp problems.
1. An "operational amplifier circuit " operates er, works by forcing the two inputs to its "operational amplifier" equal
2. It is the duty of the circuit designer to surround the "operational amplifier" with a circuit that let's it do that - if he fails , the circuit will neither "work" nor "operate" .

So let's draw your circuit and label some nodes for convenience
Let's call the voltages at nodes where Rv joins the R2's Vu(pper) and Vl(ower)
and voltages at opamp input nodes Vi(nvertng) and Vn(oninverting)
Komeck1.jpg


What do we know? What remains unknown ?
V1 and V2 are inputs, so they aren't unknowns
All resistor values are known, if only by their identifying labels

Well,
if the circuit "operates" then Vi = Vni
meaning Vi and Vni are unknown but equal,
so they count as only one unknown Let's call it Vx

Vu and Vl are unknown, so is Vo

So i see four unknowns. Vu, Vl, Vo and Vx.

Can we write four equations to solve for those four unknowns ?

KCL says we can sum currents at nodes Vu and Vl
EDIT , to fix a mistake on my drawing (Thanks Tom G ! )
node Vu: (Vx - Vu)/R2 + (Vi -Vo)/Rv - (Vu - Vo)/R2 = 0
node Vu: (Vx - Vu)/R2 + (Vl-Vu)/Rv - (Vu - Vo)/R2 = 0
node Vl: (Vx - Vl)/R2 - (Vl - Vo)/Rv - Vl/R2 = 0
node Vl: (Vx - Vl)/R2 - (Vl - Vu)/Rv - Vl/R2 = 0

Since input current to the operational amplifier at the heart of this operational amplifier circuit is zero,
Again using KCL , summing currents at nodes Vi and Vni gives
node Vi: (V1 - Vx)/R1 - (Vx - Vu)/R2 = 0
node Vni: (V2 - Vx)/R1 - (Vx - Vl)/R2 = 0

I count four equations and four unknowns... should be an algebra problem from there...
You're looking for Vo as a function of V1, V2 and the three resistor values R1 R2 and Rv.

Now - my algebra is rusty and i get absolutely frustrated trying to do it on a keyboard
so i'd do it longhand several times and see if i get same answer.

That is my "Poor Man's approach" to Opamp circuits:
Set voltage at amplifier inputs equal and solve using basic circuit analysis.

(in my case "Poor Man" meaning one like me who was endowed with a meager portion of mathematical ability)

Let me know if those four equations aren't adequate. Re-Derive them yourself.

Remember that key point - opamp must be allowed to force its inputs equal else the circuit can't "operate" .

i hope this helps you get off the ground and fly through your course. I remember when the light clicked on for me.

old jim

EDIT Tom G found my mistake , which i corrected above
...
thanks Tom

old jim
 
Last edited:
Post #4 had a mistake in the KCL equations, now fixed..
Thanks for your interestold jim
 
jim hardy said:
They teach us to solve opamp problems by resolving Zfeedback/Zinput
but when you've got two inputs like above, it's not apparent where to start.

Terminology is important - be picayune and

exact with your definitions. I usually quote Lavoisier at this point but will spare you that boring lecture today.Old JIm says:
An "Operational Amplifier" is just a piece of hardware like a LM741, usually a DC coupled amplifier with high gain
An "Operational Amplifier Circuit " is an electronic circuit , usually having an operational amplifier at its heart,
wired up so that it has a transfer function that's some mathematical operation .like add, multiply, divide, or perform some transcendental function
so we say it "operates" in a mathematical sense not just a "does it work" sense..
That's being nitpicky. But it helps me think straight.

Here are two ground rules i use for op-amp problems.
1. An "operational amplifier circuit " operates er, works by forcing the two inputs to its "operational amplifier" equal
2. It is the duty of the circuit designer to surround the "operational amplifier" with a circuit that let's it do that - if he fails , the circuit will neither "work" nor "operate" .

So let's draw your circuit and label some nodes for convenience
https://www.physicsforums.com/attachments/99799

What do we know? What remains unknown ?
V1 and V2 are inputs, so they aren't unknowns
All resistor values are known, if only by their identifying labels

Well,
if the circuit "operates" then Vi = Vni
meaning Vi and Vni are unknown but equal,
so they count as only one unknown Let's call it Vx
Vu and Vl are unknown, so is Vo

So i see four unknowns. Vu, Vl, Vo and Vx.

Can we write four equations to solve for those four unknowns ?

KCL says we can sum currents at nodes Vu and Vl
node Vu: (Vx - Vu)/R2 + (Vi -Vo)/Rv - (Vu - Vo)/R2 = 0
node Vl: (Vx - Vl)/R2 - (Vl - Vo)/Rv - Vl/R2 = 0

Since input current to the operational amplifier at the heart of this operational amplifier circuit is zero,
Again using KCL , summing currents at nodes Vi and Vni gives
node Vi: (V1 - Vx)/R1 - (Vx - Vu)/R2 = 0
node Vni: (V2 - Vx)/R1 - (Vx - Vl)/R2 = 0

I count four equations and four unknowns... should be an algebra problem from there...
You're looking for Vo as a function of V1, V2 and the three resistor values R1 R2 and Rv.

Now - my algebra is rusty and i get absolutely frustrated trying to do it on a keyboard
so i'd do it longhand several times and see if i get same answer.

That is my "Poor Man's approach" to Opamp circuits:(in my case "Poor Man" meaning one like me who was endowed with a meager portion of mathematical ability)

Let me know if those four equations aren't adequate. Re-Derive them yourself.

Remember that key point - opamp must be allowed to force its inputs equal else the circuit can't "operate" .

i hope this helps you get off the ground and fly through your course. I remember when the light clicked on for me.

old jim
gneill said:
No, you don't know what the current from (3) to (2) might be. So you cannot say that V(2) is zero.

No, you cannot know V(3) without knowing both vo and vI2.

You'll need to analyze the circuit from first principles, perhaps using nodal analysis and the properties of an ideal op-amp.
LvW said:
I think, it is a good approach to use superposition - as proposed by you (setting one of both input signals to zero).
Nevertheless, the remaining circuits are still pretty involved and - as suggested by gneill - you should use nodal analyses for both solutions.
jim hardy said:
Post #4 had a mistake in the KCL equations, now fixed..
Thanks for your interestold jim
thank for you teach! i will try to prove it!
 
kenok1216 said:
thank for you teach! i will try to prove it!

If you have severe problems, give notice.
I can describe another (perhaps simpler?) method for deriving the transfer function.
 
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