to prove vo=(2R2/R1)[1+R2/Rv][vI2-vI1] and Rv is variable resistor
The Attempt at a Solution
upper: 2 is 0V? and then Rv,2R2 form a t-network?
bottom:3 is 0V? if not how can i calculate?[/B]
No, you don't know what the current from (3) to (2) might be. So you cannot say that V(2) is zero.upper: 2 is 0V? and then Rv,2R2 form a t-network?
No, you cannot know V(3) without knowing both vo and vI2.bottom:3 is 0V? if not how can i calculate?
(in my case "Poor Man" meaning one like me who was endowed with a meager portion of mathematical ability)Set voltage at amplifier inputs equal and solve using basic circuit analysis.
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 "
operateser, 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 lets 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
View attachment 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
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.
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.
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.
thank for you teach! i will try to prove it!!!Post #4 had a mistake in the KCL equations, now fixed..
Thanks for your interest