Hello,

I have the following operational amplifier:

My task is to calculate the expression that yields the quotient \(~~\frac{u_0}{i_s}\)

Hm, i don't really know how to start, any advice?

 

Consider the features of an ideal opamp.

1. Infinite open loop gain.
2. Infinite input impedance on both input terminals
3. Output impedance is 0 ohms.

Also consider that an opamp's mission is to drives its output to whatever voltage it takes to make the two input terminals equal in voltage. In this opamp configuration, the positive terminal is at 0v. That means that the opamp is going to drive its output to a voltage that will force its negative terminal to be 0V.


hgmjr

 

\(\frac{U_0}{I_s}=R_3(\frac{R_1}{R_2}+\frac{R_1}{R_3}-1)\)
Can someone check me?

 

\(\frac{U_0}{I_s}=R_3(\frac{R_1}{R_2}+\frac{R_1}{R_3}-1)\)
Can someone check me?

That's not correct.

 

Think about it as "What current will flow through R3?"

Since R1 and R2 are voltage dividers across the current source, and R3 is the tap to the output.....

 

Ok. So. There is no current in R3? Then U0=R2(Is)?

 

Ok. So. There is no current in R3? Then U0=R2(Is)?

No one said there is no current in R3. How did you conclude that?

 

No one said there is no current in R3. How did you conclude that?

I am just stumbling in the dark. When I solved it the first time, I thought R2 and R3 would act as current divider, that was shut down. Now I am out of ideas.

 

Here's a tip:
You can convert the T-pad network to a PI-pad network using the Y-Δ transformation:
http://en.wikipedia.org/wiki/Wye-delta

Or you could just write out your node equations



Is = ( Va - Vb) /R1 (1)

Is = Vb/R2 + (Vb - Vout)/R3 (2)

Va = 0V ( thanks to op amp "action" )

 

I am just stumbling in the dark. When I solved it the first time, I thought R2 and R3 would act as current divider, that was shut down. Now I am out of ideas.

Are you the same person as regexp?

 

Greetings Shteii01,

Like RonH, I too am confused. Your replies are phrased in such a manner that it sounds like you are the originator of this thread.

Can you please confirm that you "are" or "are not" regexp?

hgmjr

 

This is not my thread. I am trying to solve it because I am not very good at op-amp circuits, I thought it would be good practice for me. Note that I only posted my answers, I still want regex do his own homework.

 

Here's a tip:
You can convert the T-pad network to a PI-pad network using the Y-Δ transformation:
http://en.wikipedia.org/wiki/Wye-delta

Or you could just write out your node equations



Is = ( Va - Vb) /R1 (1)

Is = Vb/R2 + (Vb - Vout)/R3 (2)

Va = 0V ( thanks to op amp "action" )

I see what you mean. I think I can do node equations on the original circuit.

Ok, I redid the problem using node voltage method on the original circuit. I got following: \(\frac{U_0}{I_s}=-R_3(\frac{R_1}{R_2}+\frac{R_1}{R_3}+1)\)

 

Your equation look good for me

 

Yeah, that's what I got too.

 

The minus sign looks wrong. Everything else looks right.

 

I obtained the same answer as shteii01 using Millman's Theorem.

hgmjr

 

The minus sign looks wrong. Everything else looks right.

The output voltage is bound to be negative, due to the direction of the current and the feedback.

 

Hmm, suit your self, but there are valid ways to work this out so that you do not end up with negaitve ohms as your answer.

 

The minus sign looks wrong. Everything else looks right.

Put your money where your mouth is.