CIRCUIT ANALYSIS: 8 Resistors, 1 IVS, 1 Differential Op-Amp - Find Vo

[VinnyCee]

Homework Statement

The circuit below is a differential amplifier driven by a bridge, find V_0.

http://img476.imageshack.us/img476/8827/chapter5problem48fi2.jpg

Homework Equations

KCL, v = i R, Ideal Op-Amp relationships

The Attempt at a Solution

I added 4 voltage markers.

http://img259.imageshack.us/img259/8314/chapter5problem48part2fs0.jpg

By the Ideal Op-Amp relationships, we know that there is 0 current at both input terminals of the Op-Amp.

KCL at V_2)

\frac{0.005\,-\,V_2}{40000}\,=\,\frac{V_2}{60000}\,+\,\frac{V_2\,-\,V_4}{20000}

18\,V_2\,-\,12\,V_4\,=\,0.003 < ------ Equation 1

KCL at V_4)

\frac{V_2\,-\,V_4}{20000}\,=\,\frac{V_4}{80000}

4\,V_2\,-\,5\,V_4\,=\,0 <----- Equation 2

Now, using equations 1 and 2, I get V_2\,=\,0.003571\,V and V_4\,=\,0.002857\,V. Does that seem right?

By the relationships of an ideal Op-Amp, V_3\,=\,V_4, so V_3\,=\,0.002857\,V.

KCL at V_1)

\frac{0.005\,-\,V_1}{10000}\,=\,\frac{V_1}{30000}\,+\,\frac{V_1\,-\,V_3}{20000}

11\,V_1\,-\,3\,V_3\,=\,0.03 <----- Equation 3

KCL at V_3)

\frac{V-1\,-\,V_3}{20000}\,=\,\frac{V_3\,-\,V_0}{80000}

4\,V_1\,-\,3\,V_3\,+\,V_0\,=\,0 <----- Equation 4

Using equations 3 and 4, I get V_0\,=\,0.004363\,V.

V_0\,=\,4.363\,mV <------ Is that right?

 

[VinnyCee]

Bump - Bump:)

 

[Mindscrape]

Looks good. Quick question though, why don't you use linear algebra instead of messy substitutions?