I am doing part a in this homework:
Doing nodal analysis of two op amps and came to these two equations:
$$\frac{V_1-V_{ref}}{R_2} + \frac{V_1-V_{x}}{R_1} + \frac{V_1-V_{2}}{R_f} = 0 $$
$$ \frac{V_2-V_{x}}{R_3} + \frac{V_2-V_{1}}{R_f} + \frac{V_2-V_{out}}{R_4} = 0 $$
$$ where: R_3 = R_1 and R_2 = R_4$$
I am trying to simply these two equations down and get it in this form: $$V_{out} = A_d(V_2-V_1)+V_{const}$$
I simplified down to get
$$V_x = R_1(\frac{v_1-V_{ref}}{R_2} + \frac{V_1-V_2}{R_f}) + V_1 $$
$$V_out = R_4(\frac{V_2-V_x}{R_3} + \frac{V_2-V_1}{R_f}) + V_2 $$
So I try to plug in Vx into Vout and get this monster:
$$ V_out = R_4(\frac{V_2-[R_1(\frac{v_1-V_{ref}}{R_2} + \frac{V_1-V_2}{R_f} + V_1)]}{R_3} + \frac{V_2-V_1}{R_f}) + V_2 $$
and then try to simplify using R3=R1 and R2=R4 but get nowhere.
Any hints on how to simply this circuit down to the required form?
Best Answer
What I recommend doing is finding common denominator before plugging in, like this:
Node A
$$\frac{V_1-V_{ref}}{R_2} + \frac{V_1-V_{x}}{R_1} + \frac{V_1-V_{2}}{R_f} = 0 $$ $$ V_1(R_1R_2 + R_1R_f + R_2R_f) - V_2R_1R_2 - V_x(R_fR_2) + V_{ref}R_fR_1 = 0$$ $$V_x(R_fR_2) = V_1(R_1R_2 + R_1R_f + R_2R_f) - V_2R_1R_2 + V_{ref}R_fR_1$$
Node B
$$ \frac{V_2-V_{x}}{R_3} + \frac{V_2-V_{1}}{R_f} + \frac{V_2-V_{out}}{R_4} = 0 $$ $$ V_2(R_3R_4 + R_4R_f + R_3R_f) - V_1R_3R_4 - V_x(R_fR_4) + V_oR_fR_3 = 0$$ $$ V_x(R_fR_4) = V_2(R_3R_4 + R_4R_f + R_3R_f) - V_1R_3R_4 + V_oR_fR_3$$
since $$R_4=R_2 \space and \space R_3=R_1$$ $$= V_x(R_fR_2) = V_2(R_1R_2 + R_2R_f + R_1R_f) - V_1R_1R_2 + V_oR_fR_1$$
Then plug it in: $$V_1(R_1R_2 + R_1R_f + R_2R_f) - V_2R_1R_2 + V_{ref}R_fR_1 = V_2(R_1R_2 + R_2R_f + R_1R_f) - V_1R_1R_2 + V_oR_fR_1$$ $$V_oR_fR_1 = V_1(R_1R_2 + R_1R_f + R_2R_f) + V_1R_1R_2 - V_2(R_1R_2 + R_2R_f + R_1R_f) - V_2R_1R_2 + V_{ref}R_fR_1$$ I'm sure you can figure out the rest.