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

VinnyCee

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

The circuit below is a differential amplifier driven by a bridge, find [itex]V_0[/itex].




2. Relevant equations

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


3. The attempt at a solution

I added 4 voltage markers.



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

KCL at [itex]V_2[/itex])

[tex]\frac{0.005\,-\,V_2}{40000}\,=\,\frac{V_2}{60000}\,+\,\frac{V_2\,-\,V_4}{20000}[/tex]

[tex]18\,V_2\,-\,12\,V_4\,=\,0.003[/tex] < ------ Equation 1

KCL at [itex]V_4[/itex])

[tex]\frac{V_2\,-\,V_4}{20000}\,=\,\frac{V_4}{80000}[/tex]

[tex]4\,V_2\,-\,5\,V_4\,=\,0[/tex] <----- Equation 2

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

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

KCL at [itex]V_1[/itex])

[tex]\frac{0.005\,-\,V_1}{10000}\,=\,\frac{V_1}{30000}\,+\,\frac{V_1\,-\,V_3}{20000}[/tex]

[tex]11\,V_1\,-\,3\,V_3\,=\,0.03[/tex] <----- Equation 3

KCL at [itex]V_3[/itex])

[tex]\frac{V-1\,-\,V_3}{20000}\,=\,\frac{V_3\,-\,V_0}{80000}[/tex]

[tex]4\,V_1\,-\,3\,V_3\,+\,V_0\,=\,0[/tex] <----- Equation 4

Using equations 3 and 4, I get [itex]V_0\,=\,0.004363\,V[/itex].

[tex]V_0\,=\,4.363\,mV[/tex] <------ Is that right?

VinnyCee

Bump - Bump:)

Mindscrape

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