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Electronic Circuits-Compute voltage gain for OP-amplifiers

  • Engineering
  • Thread starter beyondlight
  • Start date
  • #1

Homework Statement



We got the following circuit:

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

u_in is input
u_ut is output



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

:'(

help please.
 

Answers and Replies

  • #2
NascentOxygen
Staff Emeritus
Science Advisor
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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
NascentOxygen
Staff Emeritus
Science Advisor
9,244
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First step: determine, in terms of the input, the voltage at the (+) input of the third op-amp.
 
  • #4
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:

[tex]i=\frac{\frac{v_x}{2}-(u_{in} + v_x)}{R}[/tex]

Then the expression for the output u_o is:

[tex]u_o=\frac{v_x}{2} - R*i = {\frac{v_x}{2}-\frac{v_x-2u_{in}-2v_x}{2R}*R}[/tex]

after simplifying we obtain:

[tex]u_o=-v_x-u_in[/tex]

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


Have i done correct?
 
  • #5
NascentOxygen
Staff Emeritus
Science Advisor
9,244
1,072
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:
Which resistor? Which direction?

[tex]i=\frac{\frac{v_x}{2}-(u_{in} + v_x)}{R}[/tex]

Then the expression for the output u_o is:

[tex]u_o=\frac{v_x}{2} - R*i [/tex]
If we use your above expression for i, then the "-" here should be "+"
 
  • #6
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.
 

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