Question 7.14: For the op amp circuit in Fig. 7.55(a), find vo for t > 0......

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{dv}{dt}             (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  or  v_{1} = v   and Eq. (7.14.1) becomes

\frac{dv}{dt} + \frac{v}{CR_{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{dv}{dt} + \frac{v}{RC} = 0            (7.4b)

v(t) = V_{0} e^{-t/\tau} ,        \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 × 10^{3} × 5 × 10^{-6} = 0.1   Hence,

v(t) = 3e^{-10t}                    (7.14.4)

Applying KCL at node 2 gives

C \frac{dv}{dt} = \frac{0 – v_{o}}{R_{f}}

or

v_{o} = – R_{f} C\frac{dv}{dt}                  (7.14.5)

Now we can find v_{0}   as

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

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

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

\frac{3}{20,000} + \frac{0 – v_{o}(0^{+})}{80,000} = 0

or v_{o} (0^{+}) = 12 V .  Since the circuit is source free,v(∞)= 0 V To find \tau   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 yield

20,000(1) – v = 0        ⇒       v = 20kV

Then

R_{eq} = \frac{v}{1} = 20 k Ω

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

v_{o} (t) = v_{o} (∞) + [v_{o}(0) – v_{o} (∞)] e^{-t/ \tau}

= 0 + ( 12 – 0 ) e^{-10t} = 12 e^{-10t} V ,          t > 0

as before.