Look at bottom of the thread for my latest question
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I tried solving it using the equation
Vo/4 +1 + 1/8 =0

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

 

Hmm,
Solve this

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

 

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.

 

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

 

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.

 

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

 

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.

 

but Vp = R2/(R1+R2) * Vo1 = -0.547V

 

View attachment 18323

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.

One problem is that (Vo -Vp)/9 + Vp/4 =0 should be (Vp -Vo)/9 + Vp/4 =0

 

View attachment 18323

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.

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