hello

this problem has been driving me insane! "prob3" refers to the original hw problem, and "sol3" refers to my attempts at simplifying it... i just don't know how to simplify it enough that it's workable.. that Vs in the middle of the resistors is confusing me. i tried converting the delta's to pi's, but i don't really see where that's going either.... help

oh, btw, i forgot to erase my pencil markings on the problem.. the nodes labeled "Vs" and "-Vs" and the 0A currents going into the op-amp are all my additions to the initial problem.

thanks in advance!

 

hello

this problem has been driving me insane! "prob3" refers to the original hw problem, and "sol3" refers to my attempts at simplifying it... i just don't know how to simplify it enough that it's workable.. that Vs in the middle of the resistors is confusing me. i tried converting the delta's to pi's, but i don't really see where that's going either.... help

oh, btw, i forgot to erase my pencil markings on the problem.. the nodes labeled "Vs" and "-Vs" and the 0A currents going into the op-amp are all my additions to the initial problem.

thanks in advance!

Try redrawing it as below. If you're still stuck, post your attempt at the solution.

 

thanks! i haven't really attempted to solve what you posted, but by glancing at it i noticed you introduced a couple more unknown variables. i'll try to work it out tomorrow, i hope i'll be able to figure out a way around those variables. thanks once more


i just glanced at it once again.. can i add the 12ohms and the 6ohms, since the wire in between them carries 0A? in other words, can i treat them as if they are in series, is this a correct first step?

 

thanks! i haven't really attempted to solve what you posted, but by glancing at it i noticed you introduced a couple more unknown variables. i'll try to work it out tomorrow, i hope i'll be able to figure out a way around those variables. thanks once more

I didn't really think of them as variables, although I suppose they are. I just labeled the nodes as an aid in solving for the output voltage.
Keep in mind that Vn=Vp. This is one of the keys, but you hopefully know this if you are studying op amps.

 

Ahh...the classic bridge.

 

Greetings caramel,

Keep in mind that Vn=Vp.

Were you able to get any further down the solution path using the important hint that ronh provided concerning the voltages at the op-amp's positive and negative input terminals?

hgmjr

 

thanks for the input so far.

yes, i know that Vp=Vn, but now i have 3 unknowns instead of 2. i haven't worked through it, but i will right now and re-post my result.

i'd also like to learn how the initial circuit was simplified to this one.. and what is "bridge?"

 

Here is an introduction to bridge circuits located on this website in the tutorials.

hgmjr

 

thanks for the link i'll check it out asap.

here's my attempt at solving the circuit.. i got stuck, again! i think i need another another KCL and then solve the system of equations, but i don't see where i can apply KCL. after plugging in 1.5Vp for Vs, i end up with one equation 2 unknowns... so what to do now?

sorry i had to zip it because the pdf file was too big

 

Greetings caramel,

I looked at your most recent solution and the one thing that appears to be hanging you up is your tendency to try to confer the value of 0 volts on the input terminals. This is not a valid assumption for this design.

The fact is that there is enough information in the problem for you to clearly determine the equation for the voltage at the positive terminal using simple voltage divider analysis.

The equation for the voltage at the op-amp's positive terminal is a good place to begin to break the problem down into smaller bites.

Can you take a stab at writing that equation and then post it here?

hgmjr

 

hi

the 0's are currents going into the op-amp, sorry forgot to write in my units. i don't really understand what you're asking for. i think i've labeled all my voltage nodes correctly, right?

 

hi

the 0's are currents going into the op-amp, sorry forgot to write in my units. i don't really understand what you're asking for. i think i've labeled all my voltage nodes correctly, right?

Sorry, my bad. You did label them as 0A which I now understand was your way of noting to yourself that there was no current flowing into or out of the op-amp terminals.

What I was referring to was the voltage being supplied to the positive op-amp terminal. You should be able to write the equation for the voltage that appears at the junction of the 6 ohm resistor and the 12 ohm resistor as a function of Vs.

hgmjr

 

you mean to divide the voltage so that the 6ohm gets Vs/3 and the 12ohm gets 2Vs/3 ... i still don't see where that would lead to

 

What that leads to is a critical piece of the puzzle.

What you have calculated is the voltage Vp.

With an expression for Vp all you need to do is write a similar expression for Vn and you will then be able to set them equal to each other and solve for Vo.

hgmjr

 

i'm sorry, i'm still not seeing what you're talking about.. so i found that Vp = 2/3Vs, but i've known that and it didn't help. and the only "similar" expression i can write for Vn would be Vn/3.. how would setting them equal to one another get me Vo? for some reason, i'm not seeing the big picture in this, i don't see where all of this is going at all.

i tried reading that page you gave me the link to about bridge circuits, and i learned new concepts from it, but nothing that would really help with this.

i hate being so lost

 

No problem with being lost. That is part of the learning process.

Notice that the voltage at the negative input is a function of the 3 ohm resistor from Vs, the 4 ohm resistor from Vo, the 3 ohm resistor to ground, Vs and Vo.

This is just a two source and three resistor circuit that you have seen before. Sketch the circuit on a piece of paper and then you can use Kirchoff's Voltage Law to derive the voltage at the negative input terminal of the op-amp.

You can do this.
hgmjr

 

i'm probably completely off, but would Vo=20/27 Volts?

 

Caramel, do you understand the concept of "virtual ground"? You must understand this before you can solve this problem. Actually, in this case we don't really have virtual ground, we have Vn=Vp. The concept is the same.
And you said,

i'd also like to learn how the initial circuit was simplified to this one.

There is no simplification. The redrawn circuit is identical to the original circuit.

 

I am probably repeating what others have already said, but what I would do is find the symbolic expression of V+ and then substitute it back into the KCL equation for the V- node. Then I would ultimately solve for Vout/Vin.

What is your equation that shows Vo = 20/27V ?

This will be a good circuit to put up on my website!

 

knightofsolamnus ... what is your website?