What will the function of the output voltage be?

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My work, and am I right?:

$$\text{V}_{\text{out}}(t)=\frac{\text{R}_9\text{V}_{\text{in}}(t)}{\text{R}_{10}}+\frac{1}{\text{R}_{11}\text{C}_1}\int\text{V}_{\text{in}}(t)\space\text{d}t+\text{R}_{12}\text{C}_2\cdot\frac{\partial\text{V}_{\text{in}}(t)}{\partial t}$$

  • \$\begingroup\$ It's a PID block - what more to say? \$\endgroup\$
    – Andy aka
    Feb 3, 2016 at 13:51
  • \$\begingroup\$ Yes it is, but is my output voltage function right? \$\endgroup\$ Feb 3, 2016 at 13:52
  • 1
    \$\begingroup\$ how have you accounted for the gain due to op amps 1 and 2? \$\endgroup\$ Feb 3, 2016 at 13:53
  • \$\begingroup\$ No the gain isn't one, I named it number 1 \$\endgroup\$ Feb 3, 2016 at 13:54
  • \$\begingroup\$ The gain is 0.75 by the look of it. \$\endgroup\$
    – Andy aka
    Feb 3, 2016 at 13:56

1 Answer 1


Ok, so op amps 3, 4 and 5 form a PID block. Considering their respective output voltages as Vp, Vi and Vd:

$$ V_p = -\frac{R_9}{R_{10}}V_{in} $$

$$ V_i = -\frac{1}{R_{11}C_{1}}\int{V_{in}dt} $$

$$ V_d = -R_{12}C_2\frac{dV_{in}}{dt} $$

Op amp 2 and resistors R3-8 are used to sum these voltages. Using superposition, one can solve this block's output (Vx). Let's say Vi and Vd are 0; R5, R7 and R8 are then in parallel and the voltage on the op amp '+' will be Vp/4.

$$ V_{x(V_p)} = \left(\frac{R_3}{R_4} + 1\right)\frac{V_p}{4} = 0.75V_p$$

Therefore, with superposition, Vx is 0.75(Vp + Vi + Vd). The last op amp is just a regular negative gain circuit, and Vout = -Vx. Final answer:

$$ V_{out} = 0.75\left(\frac{R_9}{R_{10}}V_{in} + \frac{1}{R_{11}C_{1}}\int{V_{in}dt} + R_{12}C_2\frac{dV_{in}}{dt}\right)$$

  • \$\begingroup\$ Why the 0.75? Because Vi and Vd aren't 0 \$\endgroup\$ Feb 3, 2016 at 14:57
  • \$\begingroup\$ Superposition is an analysis tool for linear circuits. It takes the basic property of a linear operator L: L(a+b+c) = L(a) + L(b) + L(c). Considering the circut "simmetry" between different inputs, when one solves for L(a) (special case when b and c are 0), there will be a similar answer for L(b) and L(c), and the final result will be the sum of such answers. \$\endgroup\$ Feb 3, 2016 at 15:09
  • \$\begingroup\$ Oh yes I understand, but is there a way to make it 1, I've to change the value of the resistors but how? \$\endgroup\$ Feb 3, 2016 at 15:10
  • \$\begingroup\$ (1 + R3/R4) should result in 4, not 3. R3 could be 30k, for example. \$\endgroup\$ Feb 3, 2016 at 15:12
  • \$\begingroup\$ So, if I make R3 30k than it will be the ouput function that I supposed in my question?! \$\endgroup\$ Feb 3, 2016 at 15:17

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