Electronic Circuits-Compute voltage gain for OP-amplifiers

1. Dec 30, 2012

beyondlight

1. The problem statement, all variables and given/known data

We got the following circuit:

http://tinypic.com/view.php?pic=1zl7xg2&s=6

u_in is input
u_ut is output

3. The attempt at a solution

I assumed that the upper amp has voltage u_in on the positive terminal.
And that the lower amp has voltage 0 V on the positive terminal.

then i calculated u_ut but the ratio then becomes: u_in/u_ut = 0

Because the current from the upper amplifiers output terminal to the output terminal u_ut is

i=u_in/2R

which then gives us

u_ut=u_in -(i*2R) = u_in -u_in

the right answer should be -1

:'(

2. Dec 30, 2012

Staff: Mentor

I would analyze it by saying the voltage on the lower input is vx, and that on the upper input terminal is uin+vx, with vx being the common mode voltage.

Otherwise you are losing all the information associated with the lower pair of R resistances.

3. Dec 30, 2012

Staff: Mentor

First step: determine, in terms of the input, the voltage at the (+) input of the third op-amp.

4. Dec 31, 2012

beyondlight

So this is how i did it:

On the positive input for the 3rd amplifier we have v_x/2.

And i suppose that the voltage on the output terminal of the upper amplifier is v_x+u_in

The current through the resistor then becomes:

$$i=\frac{\frac{v_x}{2}-(u_{in} + v_x)}{R}$$

Then the expression for the output u_o is:

$$u_o=\frac{v_x}{2} - R*i = {\frac{v_x}{2}-\frac{v_x-2u_{in}-2v_x}{2R}*R}$$

after simplifying we obtain:

$$u_o=-v_x-u_in$$

u_o/u_in = -1 (if v_x = 0)

Have i done correct?

5. Jan 1, 2013

Staff: Mentor

If we use your above expression for i, then the "-" here should be "+"

6. Jan 2, 2013

beyondlight

It is for the resistor after the output of the upper-most amplifier. I thought the current goes out of this amplifier and ends up at the output. So then it should be "-"

But i re-calculated this task with my method again, and got a quite different result. That turned out to be correct. I calculated without adding v_x.

The same current: i=(u_in/R)

u_out=0-Ri= -R(u_in/R)=-u_in

The gain then becomes -1.

I think this is sufficient.