Solve OpAmp Symbolic Gain: KVL/KCL

[Learnphysics]

Homework Statement

What is the gain of this Op-AMP. Express symbolically (Eg. in terms of R1, R2..)

http://img195.imageshack.us/img195/3389/opampproblem.png



Uploaded with ImageShack.us

Homework Equations

KVL/KCL

The Attempt at a Solution

i1=i2

Vx (the voltage at the middle node)

i2 = Vi/R1 = i1

i3 = Vx/R2

Vx = Vi + i2*R4

KCL at middle node:
i2 + i3 = i4.

Vo = Vx + i4(R3)

substituting things in, in place of the currents.

Vo = Vi + i2(R4) + i4(R3)

Vo = Vi + (Vi/R1)(R4) + i4(R3)

i4 = i2+i3.

i4 = Vi/R1 + (Vi + (Vi/R1)*R4)/R2

Vo = Vi + (Vi/R1)(R4) + (Vi/R1 + (Vi + (Vi/R1)*R4)/R2) (R3)

Vo/Vin = (1/R1) + (R3/R1) + [R3 + (R4*R3)/R1]/R2

But the answer is apparently
((R1+R4)*(R2+R3)+R2*R3)/(R1*R2)

is there some Flaw in my reasoning? or have i just made a calcualtion/algebra error?

 

[The Electrician]

You haven't identified I1, I2, I3 and I4. Looking over your equations, apparently I1 is the current in R1, I2 is the current in R2, etc. When you're asking for help, you should provide a schematic with the currents and voltages identified. Otherwise, those who would help you may see that you are asking them to make assumptions, and they may just not bother.

Your first problem is that you are assuming that I1 = I2. For this to be true, R1+R4 would have to be equal to R2. Looking at the resistor network, we see that since no current can leave the network through the minus input to the opamp, the current leaving the network through R1 and R2 must equal the current entering through R3.

I haven't considered what else you have done. Fix your error and try again.

 

[Learnphysics]

I'm sorry.

I'v indicated the currents I'v assume on this diagram.

http://img824.imageshack.us/img824/3389/opampproblem.png

Uploaded with ImageShack.us

 

[The Electrician]

OK. For the current directions you've indicated, your first error is a sign error. This is very common for people just learning circuit analysis, and you must be very careful with your signs.

You have I1=I2, but for the current directions indicated, I1 = -I2, because according to KCL, I1 + I2 = 0. You can't have two currents leaving a node and no currents entering.

Further on, you have Vo = Vx + i4(R3).

This should be Vo = Vx - i4(R3).

I'm assuming you're using "conventional" current where the current is assumed to flow from + to - through a resistor, which is the customary assumption.

Re-calculate your problem, with due care in regard to the signs.

 

[Learnphysics]

OK. For the current directions you've indicated, your first error is a sign error. This is very common for people just learning circuit analysis, and you must be very careful with your signs.

You have I1=I2, but for the current directions indicated, I1 = -I2, because according to KCL, I1 + I2 = 0. You can't have two currents leaving a node and no currents entering.

Further on, you have Vo = Vx + i4(R3).

This should be Vo = Vx - i4(R3).

I'm assuming you're using "conventional" current where the current is assumed to flow from + to - through a resistor, which is the customary assumption.

Re-calculate your problem, with due care in regard to the signs.

i1= -i2

-i2 = (Vi/R1)

-i3 = Vx/R2 <= Vx being the voltage at the middle node

-i2 -i3 = i4

Vx = Vi -i2(R4)

Vo = Vx - i4(R3)


Vx = Vi +(Vi/R1)(R4)

i4= (Vi/R1) + (Vx/R2)

Vo = Vi +(Vi/R1)(R4) -[(Vi/R1) + (Vx/R2)]*R3

Am i on the right track so far?

 

[The Electrician]

For your fourth equation you have: -i2 -i3 = i4

KCL says that the sum of all the currents entering a node is zero.

That means that I2 + I3 + (-I4) = 0, or I2 + I3 = I4, or -I2 -I3 = -I4

Once more into the breach!

 

[Learnphysics]

For your fourth equation you have: -i2 -i3 = i4

KCL says that the sum of all the currents entering a node is zero.

That means that I2 + I3 + (-I4) = 0, or I2 + I3 = I4, or -I2 -I3 = -I4

Once more into the breach!

Ah... Thanks for helping btw...

Is there any kind of... system or set of practices i can use to make sure these little sign errors don't happen or is it just something that gets better with practice.i1= -i2

i2 = -(Vi/R1)

i3 = -(Vx/R2) <= Vx being the voltage at the middle node

i2 + i3 = i4

Vx = Vi -i2(R4)

Vo = Vx - i4(R3)Vx = Vi + (Vi/R1)(R4)

-i4= (Vi/R1) +(Vx/R2)

Vo = Vi +(Vi/R1)(R4) -[(Vi/R1) + ({Vi + (Vi/R1)(R4})/R2)]*R3

 

[The Electrician]

It gets better with practice.

Have you tried to put your last expression in the form Vo/Vi = ...?

 

[Learnphysics]

I ended up getting

Vo/Vin = [(R1*R2)+(R4*R2)-(R3*R2)-(R1*R3)-(R4*R3)]/(R1*R2)

This looks rather similar to the correct answer, but some of the signs don't match up.

 

[The Electrician]

Look back carefully at how you got:

Vo = Vi +(Vi/R1)(R4) -[(Vi/R1) + ({Vi + (Vi/R1)(R4})/R2)]*R3

I think it should be:

Vo = Vi +(Vi/R1)(R4) +[(Vi/R1) + ({Vi + (Vi/R1)(R4})/R2)]*R3

Does that fix your problem?

 

[Learnphysics]

Look back carefully at how you got:

Vo = Vi +(Vi/R1)(R4) -[(Vi/R1) + ({Vi + (Vi/R1)(R4})/R2)]*R3

I think it should be:

Vo = Vi +(Vi/R1)(R4) +[(Vi/R1) + ({Vi + (Vi/R1)(R4})/R2)]*R3

Does that fix your problem?

After a few re-tries i think i got it looking like that. I suppose i need a little more raw practice dancing with these voltages/resistances.

Thanks for your help!

 

[VinnyCee]

I_1\,=\,\frac{V_i}{R_1}\,=\,-\,I_4

I_2\,=\,\frac{V_x}{R_2}

I_4\,=\,I_2\,+\,I_3\,\longrightarrow\,I_3\,=\,I_4\,-\,I_2\,\longrightarrow\,I_3\,=\,-\,\frac{V_i}{R_1}\,-\,\frac{V_x}{R_2}

[PLAIN]http://img17.imageshack.us/img17/4164/pfhwhelpopampcircuit.jpg

Now we need some voltage equations...

-\,R_1\,I_1\,+\,R_4\,I_4\,+\,R_3\,I_3\,+\,V_0\,=\,0\,\longrightarrow\,V_0\,=\,R_1\,I_1\,-\,R_4\,I_4\,-\,R_3\,I_3\,\longrightarrow\,V_0\,=\,R_1\,\left(\frac{V_i}{R_1}\right)\,-\,R_4\,\left(-\,\frac{V_i}{R_1}\right)\,-\,R_3\,\left(-\,\frac{V_i}{R_1}\,-\,\frac{V_x}{R_2}\right)

V_0\,=\,V_i\,+\,V_i\,\frac{R_4}{R_1}\,+\,V_i\,\frac{R_3}{R_1}\,+\,V_x\,\frac{R_3}{R_2}\,\longrightarrow\,V_0\,=\,V_i\,\left(1\,+\,\frac{R_4}{R_1}\,+\,\frac{R_3}{R_1}\right)\,+\,V_x\,\left(\frac{R_3}{R_2}\right)

Now we need an expression for V_x in terms of V_i...

V_x\,=\,V_i\,-\,R_4\,I_4\,\longrightarrow\,V_x\,=\,V_i\,-\,R_4\,\left(-\,\frac{V_i}{R_1}\right)\,\longrightarrow\,V_x\,=\,V_i\,\left(1\,+\,\frac{R_4}{R_1}\right)

Now we have...

V_0\,=\,V_i\,\left(1\,+\,\frac{R_4}{R_1}\,+\,\frac{R_3}{R_1}\right)\,+\,V_i\,\left(1\,+\,\frac{R_4}{R_1}\right)\,\left(\frac{R_3}{R_2}\right)\,\longrightarrow\,V_0\,=\,V_i\,\left(\frac{R_1\,+\,R_4\,+\,R_3}{R_1}\right)\,+\,V_i\,\left(\frac{R_3\,R_1\,+\,R_3\,R_4}{R_2\,R_1}\right)

V_0\,=\,V_i\,\left(\frac{R_1\,+\,R_4\,+\,R_3}{R_1}\,+\,\frac{R_3\,R_1\,+\,R_3\,R_4}{R_2\,R_1}\right)

\frac{V_0}{V_i}\,=\,\frac{R_1\,R_2\,+\,R_2\,R_3\,+\,R_2\,R_4}{R_1\,R_2}\,+\,\frac{R_1\,R_3\,+\,R_3\,R_4}{R_1\,R_2}

\frac{V_0}{V_i}\,=\,\frac{R_1\,R_2\,+\,R_1\,R_3\,+\,R_2\,R_3\,+\,R_2\,R_4\,+\,R_3\,R_4}{R_1\,R_2}

Does that look right?