Question 7.14: For the op amp circuit in Fig. 7.55 (a), find vo for t > ...

This problem can be solved in two ways:
METHOD 1 Consider the circuit in Fig. 7.55(a). Let us derive the appropriate differential equation using nodal analysis. If v_{1} is the voltage at node 1, at that node, KCL gives

\frac{0-v_{1}}{R_{1}}=C \frac{d v}{d t}    (7.14.1)
Since nodes 2 and 3 must be at the same potential, the potential at node 2 is zero. Thus, v_{1}-0=v \text { or } v_{1}=v and Eq. (7.14.1) becomes

\frac{d v}{d t}+\frac{v}{C R_{1}}=0    (7.14.2)
This is similar to Eq. (7.4b) so that the solution is obtained the same way as in Section 7.2, i.e.,

\frac{d v}{d t}+\frac{v}{R C}=0     (7.4b)

v(t)=V_{0} e^{-t / \tau}, \quad \tau=R_{1} C    (7.14.3)
where V_{0} is the initial voltage across the capacitor. But v(0) = 3 = V_{0} and \tau=20 \times 10^{3} \times 5 \times 10^{-6}=0.1. Hence,

v(t)=3 e^{-10 t}    (7.14.4)
Applying KCL at node 2 gives

or

v_{o}=-R_{f} C \frac{d v}{d t}        (7.14.5)

Now we can find V_{0} as

v_{o}=-80 \times 10^{3} \times 5 \times 10^{-6}\left(-30 e^{-10 t}\right)=12 e^{-10 t} V,      t > 0

METHOD 2 Let us now apply the short-cut method from Eq. (7.53). We need to find v_{o}\left(0^{+}\right), v_{o}(\infty), and τ. Since v\left(0^{+}\right)=v\left(0^{-}\right)=3 V, we apply KCL at node 2 in the circuit of Fig. 7.55(b) to obtain

v(t)=v(\infty)+[v(0)-v(\infty)] e^{-t / \tau}  (7.53)

or v_{o}\left(0^{+}\right)=12 V. Since the circuit is source free, v(∞) = 0 V. To find τ, we need the equivalent resistance R_{ eq } across the capacitor terminals. If we remove the capacitor and replace it by a 1-A current source, we have the circuit shown in Fig. 7.55(c). Applying KVL to the input loop yields

20,000(1) − v = 0   ⇒    v = 20 kV
Then

and \tau=R_{ eq } C=0.1. Thus,

v_{o}(t)=v_{o}(\infty)+\left[v_{o}(0)-v_{o}(\infty)\right] e^{-t / \tau}
=0+(12-0) e^{-10 t}=12 e^{-10 t} V,            t > 0
as before.