How to solve for v_o of this weighted summer circuit?

1. Jun 4, 2013

dla

1. The problem statement, all variables and given/known data
I am trying to derive the expression for v_o for this weighted summer circuit. I've attached the circuit diagram.

2. Relevant equations

3. The attempt at a solution

I get that the first op-amp output is just the total current x R_a, and then you take that output and find the current of R_b + current of v_3/R_3 and v_4/R_4 and multply all that by R_c to get the v_o, this is what I got

$\huge v_{o} = \frac{v_{1} R_{a}}{R_{1}}\frac{R_{c}}{R_{b}} +\frac{v_{2} R_{a}}{R_{2}}\frac{R_{c}}{R_{b}} + \frac{v_{3} R_{c}} {R_{3}} +\frac{v_{4} R_{c}} {R_{4}}$

However the answer says that the last two terms are negative, how come?

$\huge v_{o} = \frac{v_{1} R_{a}}{R_{1}}\frac{R_{c}}{R_{b}} +\frac{v_{2} R_{a}}{R_{2}}\frac{R_{c}}{R_{b}} -\frac{v_{3} R_{c}} {R_{3}} -\frac{v_{4} R_{c}} {R_{4}}$

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Last edited: Jun 4, 2013
2. Jun 4, 2013

Staff: Mentor

output = -1x(total current x R_a)

The input goes to each OP_AMP's inverting input

3. Jun 4, 2013

dla

Thank you! I get it.