Electronic – Op-amp analysis with capacitor

circuit analysisdcoperational-amplifier

I have the next exercise:

Circuit

I proceed as follows (assuming ideal op-amp):

$$C = 2 \mu F$$
$$R_{1} = 50k \Omega$$
$$R_{2}=10k \Omega$$
$$v_{2} = 0V$$

For \$t>0\$:

  • Node \$v1\$:

$$i_{s}=C \frac{d(v1-v2)}{dt}+ \frac{v1-v2}{R_{1}}$$
$$ \implies i_{s}=C \frac{dv1}{dt}+\frac{v1}{R_{1}}$$

  • Node \$v2\$:

$$ C \frac{dv1}{dt}+ \frac{v1}{R_{1}} = \frac{v2-v_{o}}{R_{2}}$$
$$ \implies R_{2}C \frac{dv1}{dt}+ \frac{R_{2}}{R_{1}}v1 = -v_{o}$$

Solving for \$v_{o}\$ using last nodal equations:

$$v_{o}(t) = -i_{s}R_{2}=-100mV$$

Based on book answers, \$v_{o}(t)=100(2e^{-10t}-1)mV\$. What I did wrong?

Best Answer

One of the "tricks" I've taught in EE fundamentals class is that placing a network in series with a current source changes the voltage across the current source only.

(This is most easily seen by enclosing the network inside a supernode and applying KCL - the current entering the supernode is the current from the current source and this must equal the current leaving the supernode.)

Thus, the solution for \$v_o(t)\$ is unchanged by replacing the RC network with a wire and then it is clear that this circuit is a current-to-voltage converter with trans-resistance:

$$R_M = -10k \Omega $$

and so,

$$v_o(t) = -i_s(t) \cdot 10k\Omega$$

The answer given by the book is clearly not for the circuit as drawn.

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