I found an op amp circuit that is designed to be a derivative controller (D-controller)


simulate this circuit – Schematic created using CircuitLab

and the transfer function $$ \frac{V_o}{V_i} = -s\cdot \frac{1}{4}RC $$ is given.

I tried to analyse the circuit with Kirchhoff's laws but the ground potential here is really confusing. How do I solve it?

These are the equations I have so far (assuming the reference potential of \$V_{i}\$ and \$V_o\$ is \$\phi\$ not ground):

$$ I_1 = \frac{V_i}{R + \frac{4}{sC}}$$ $$ \phi = I_1 \cdot \frac{1}{2}R + \frac{I_2}{sC}$$ $$ \frac{I_2}{sC} = I_3 \cdot \frac{1}{2}R + V_o + \phi$$ $$I_1 = I_2 + I_3$$


1 Answer 1


The dc gain for this circuit is zero. So the transfer function is ac only. The reference voltage, \$\phi\$ is ac 0V. So proceed from there.

You can include \$\phi\$ in the analysis. It will cancel out eventually, but it is a distraction. Set it to zero then carry on.

Apply Thevenin to the feedback voltage divider from \$V_o\$ such that \$Z_f=\frac{R}{2}+Z_{Th}\$.

Your method will work with \$\phi=0\$. You want\$Z_f\$ and \$Z_i\$. Treat \$I_2+I_3=I_f\$ then \$Z_f=\frac{V_o}{I_f}\$.


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