# Operational Amplifier

Discussion in 'Homework Help' started by Hitman6267, Apr 9, 2010.

1. ### Hitman6267 Thread Starter Member

Apr 6, 2010
82
0
Look at bottom of the thread for my latest question

I tried solving it using the equation
Vo/4 +1 + 1/8 =0

But it didn't work. What am i doing wrong ?

Last edited: Apr 12, 2010
2. ### Jony130 AAC Fanatic!

Feb 17, 2009
3,993
1,116
Hmm,
Solve this

$\frac{(Vs-0V)}{R1}-\frac{(0V-V1-Vout)}{R2}=0$

3. ### Hitman6267 Thread Starter Member

Apr 6, 2010
82
0
That's the same equation I posted. Its answer is Vo = -4.5. The web-form where I submit my answer says it's wrong.

Last edited: Apr 9, 2010
4. ### Jony130 AAC Fanatic!

Feb 17, 2009
3,993
1,116
If we solve this
$\frac{(Vs-0V)}{R1}-\frac{(0V-V1-Vout)}{R2}=0$
then we get:

Vout = - V1 - R2/R1*Vs = -4V - 0.5V = - 4.5V

5. ### Hitman6267 Thread Starter Member

Apr 6, 2010
82
0
Yes exactly like I said in the post above. And it turns out it's right. The system is accepting it as an answer now. Thank you.

6. ### hgmjr Moderator

Jan 28, 2005
9,030
214
Write Millman's Theorem equation for voltage at opamp minus input terminal by inspection:

$V_- =\Large \frac{\frac{V_S}{R_1}+\frac{V_o+V_1}{R_2}}{\frac{1}{R_1}+\frac{1}{R_2}}$

Ideal Opamp Equation:

${V_-=V_+=0}$

Simplify first equation above:

$0=\frac{R_2V_s+R_1(V_o+V_1)}{R_1+R_2}$

$0={R_2V_s+R_1(V_o+V_1)}$

$0={R_2V_s+R_1V_o+R_1V_1}$

$-R_1V_o={R_2V_s+R_1V_1}$

$V_o=-\frac{R_2V_s+R_1V_1}{R_1}$

$V_o=-\frac{R_2V_s}{R_1}-\frac{R_1V_1}{R_1}$

$V_o=-\frac{R_2V_s}{R_1}-V_1$

Plug in values:

$V_o=-\frac{4(1)}{8}-4$

$V_o=-0.5-4$

$V_o=-4.5V$

Just thought someone might be interested in seeing how Millman's Theorem could be used to solve this problem.

hgmjr

7. ### Hitman6267 Thread Starter Member

Apr 6, 2010
82
0

Let the output of the first op amp be Vo and the output of the second Vo'

I calculated Vo from the equation 4/9 + vo/4 = 0
So I got Vo = -1.78

Then I wrote these equations for the second op amp.

(Vo -Vp)/9 + Vp/4 =0
=> Vp =1.42

(Vo-Vn)/4 + (Vo' -Vn)/9 =0

=> Vo' = 8.62

What's wrong here ? My answer is wrong.

8. ### Jony130 AAC Fanatic!

Feb 17, 2009
3,993
1,116
but Vp = R2/(R1+R2) * Vo1 = -0.547V

9. ### The Electrician AAC Fanatic!

Oct 9, 2007
2,301
338
One problem is that (Vo -Vp)/9 + Vp/4 =0 should be (Vp -Vo)/9 + Vp/4 =0

10. ### hgmjr Moderator

Jan 28, 2005
9,030
214
Here is the solution to the second gain stage using Millman's Theorem.

First I form the Millman's Theorem expression for the voltage at the inverting input to the second stage. I am using e(-) to refer to the voltage at the inverting input to the second stage opamp.

$e_{(-)} = \frac{\frac{V_i}{R_2}+\frac{V_o}{R_1}}{\frac{1}{R_1}+\frac{1}{R_2}}$

Second, I form the Millman's Theorem expression for the voltage at the non-inverting input to the second stage. I am using e(+) to refer to the voltage at the non-inverting input to the second stage opamp.

$e_{(+)} = \frac{\frac{V_i}{R_1}}{\frac{1}{R_1}+\frac{1}{R_2}}$

I now set these two expressions equal to each other since that is the case with an ideal opamp.

$\frac{\frac{V_i}{R_2}+\frac{V_o}{R_1}}{\frac{1}{R_1}+\frac{1}{R_2}} = \frac{\frac{V_i}{R_1}}{\frac{1}{R_1}+\frac{1}{R_2}}$

Then I begin the simplification process by first eliminating the denominator on both sides since they are equal.

$\frac{V_i}{R_2}+\frac{V_o}{R_1} = \frac{V_i}{R_1}$

Next I multiply both sides by R1*R2 to get rid of the fractions.

$R_1V_i+R_2V_o = R_2V_i$

$R_2V_o = R_2V_i-R_1V_i$

$R_2V_o = V_i(R_2-R_1)$

$V_o = \frac{V_i(R_2-R_1)}{R_2}$

The final equation above is the expression for the gain of the second stage only of the two stage amplifier. Vi in this expression is the output of the first opamp stage.

hgmjr