1
\$\begingroup\$

I am asked this for homework:enter image description here

I am trying to do nodal analysis for part A of the question but am running into problems with number of unknowns versus number equations. Assuming I chose my nodes correctly I get:

$$V_a = V_1 + \frac{V_a}{R_2} + \frac{V_a-V_b}{R_F} + \frac{V_a-V_x}{R_1} $$ $$V_b = V_2 + \frac{V_b-V_x}{R_3} + \frac{V_b-V_a}{R_F} + \frac{V_b}{R_4} $$

I don't believe my node equations are correct but I believe that there is 3 unknowns (Va, Vb, and Vx | V1, V2, all resistors are known(?) ) and only two equations. I tried super position as well treating each stage as an inverting and noninverting with sources on and off.

Any help or guidance on how to approach this problem so that I can reduce it down into the given form would be appreciated .

Note: Vref does not equal ground.

\$\endgroup\$
3
  • 3
    \$\begingroup\$ Hint: If negative feedback is working as it should, you will end up with V_a = v_1 and V_b = v_2. \$\endgroup\$
    – The Photon
    Commented Sep 18, 2013 at 20:35
  • 1
    \$\begingroup\$ Your node equations cannot be correct as they add voltage and currents together. A node equations is a KCL equation which involves only currents. \$\endgroup\$ Commented Sep 18, 2013 at 21:47
  • \$\begingroup\$ Yes, pay attention to your units. Volts cannot equal Volts + Amps. \$\endgroup\$
    – dext0rb
    Commented Sep 18, 2013 at 22:12

1 Answer 1

3
\$\begingroup\$

For op-amp problems, assuming negative feedback is present, the inverting and non-inverting inputs (ideally) have the same voltage.

Thus, for example, for the first op-amp, both inputs have a voltage of \$v_1\$. So, how to proceed?

The correct node equation for node A is:

\$\dfrac{v_1 - V_{REF}}{R_2} + \dfrac{v_1 - v_x}{R_1} + \dfrac{v_1 - v_2}{R_F} = 0\$

Can you take it from here?

\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.