Question 24.9: Sketch vC for the step input shown in Fig. 24.22. Assume tha...
Sketch v_{C} for the step input shown in Fig. 24.22. Assume that the -4 mV has been present for a period of time in excess of five time constants of the network. Then determine when v_{C} = 0 V if thestep changes levels at t = 0 s.
V_{i}=-4 mV \quad V_{f}=10 mV.
\tau=R C=(1 k \Omega)(0.01 \mu F )=10 \mu s.
By Eq. (24.6),
v_{C}=V_{f}+\left(V_{i}-V_{f}\right) e^{-t / R C} (24.6).
v_{C}=V_{f}+\left(V_{i}-V_{f}\right) e^{-t / R C}.
=10 mV +(-4 mV -10 mV ) e^{-t / 10 \mu s }.
and v_{C}=10 mV -14 mV e^{-t / 10 \mu s }.
The waveform appears in Fig. 24.23
\text { Substituting } v_{C}=0 V \text { into the above equation yields }v_{C}=0 V =10 mV -14 mV e^{-t / 10 \mu s }.
and \frac{10 mV }{14 mV }=e^{-t / 10 \mu s }.
or 0.714=e^{- t / 10 \mu s }.
but \log _{e} 0.714=\log _{e}\left(e^{-t / 10 \mu s }\right)=\frac{-t}{10 \mu s }.
\text { and } \quad t=-(10 \mu s ) \log _{e} 0.714=-(10 \mu s )(-0.377)=3.37 \mu s.
as indicated in Fig. 24.23.
