(a) R_{ D }=50 \Omega, V_{ TD }=0.7 V , R_{ Z }=20 \Omega, V_{ Z }=4.7 V , R_{ s }=1 k \Omega \text {, and } v_{ S }=15 \sin (2000 \pi t) . Using Eq. (4.37), we have
v_{ Z }=v_{ ZO }+R_{ Z } i_{ Z } (4.37)
V_{ ZO }=V_{ Z }-R_{ Z } I_{ ZT }=4.7 V -20 \Omega \times 20 mA =4.3 V
There are four possible intervals, depending on the value of v_{ S } . If 15 \sin 2000 \pi t=V_{ ZO }+V_{ TD }=5 , then
\begin{aligned}2000 \pi t &=\sin ^{-1}\left(\frac{5}{15}\right) \\&=0.34 rad\end{aligned}
Interval 1: This interval is valid for 0 \leq v_{ S } \leq\left(V_{ ZO }+V_{ TD }\right) .
\begin{aligned}&i_{ D }=0 \\&v_{ O }=v_{ s }=15 \sin (2000 \pi t) \quad \text { for } 0 \leq 2000 \pi t \leq 0.34 \text { and }(\pi-0.34) \leq 2000 \pi t \leq \pi\end{aligned}
Interval 2: This interval is valid for v_{ s } \geq\left(V_{ ZO }+V_{ TD }\right) . From Fig. 4.31(c), we can find the instantaneous diode current i_{ D } :
i_{ D }=\frac{v_{ S }}{R_{ s }+R_{ D }+R_{ Z }}-\frac{V_{ ZO }+V_{ TD }}{R_{ s }+R_{ D }+R_{ Z }} (4.48)
=\frac{15 \sin (2000 \pi t)}{1 k \Omega+50 \Omega+20 \Omega}-\frac{(4.3+0.7) V }{1 k \Omega+50 \Omega+20 \Omega}=[14.02 \sin (2000 \pi t)-4.67] mA
The instantaneous output voltage v_{ O } is given by
v_{ O }=V_{ ZO }+V_{ TD }+\left(R_{ D }+R_{ Z }\right) i_{ D } (4.49)
Substituting for i_{ D } , we get
\begin{aligned}v_{ O } &=(4.3+0.7)+(50+20) \times[14.02 \sin (2000 \pi t)-4.67] \times 10^{-3} \\&=4.67+0.981 \sin (2000 \pi t) \quad \text { for } 0.34 \leq 2000 \pi t \leq(\pi-0.34)\end{aligned}
Interval 3: This interval is valid for 0 \geq v_{ S } \geq-\left(V_{ ZO }+V_{ TD }\right) .
\begin{aligned} &i_{ D }=0 \\ &v_{ O }=v_{ S }=-15 \sin (2000 \pi t) \quad \text { for }-0.34 \leq 2000 \pi t \leq 0 \text { and }-\pi \leq 2000 \pi t \leq(-\pi+0.34) \end{aligned}
Interval 4: This interval is valid for v_{ S } \leq-\left(V_{ ZO }+V_{ TD }\right) .
\begin{aligned}&i_{ D }=-[14.02 \sin (2000 \pi t)-4.67] mA \\&v_{ O }=-4.67-0.981 \sin (2000 \pi t) \quad \text { for }(-\pi+0.34) \leq 2000 \pi t \leq-0.34\end{aligned}
The peak diode current i_{ p (\text { diode })} occurs at 2000 \pi t=\pi / 2 . That is,
i_{ p (\text { diode })}=\left[14.02 \sin \left(\frac{\pi}{2}\right)-4.67\right] mA =14.02 mA -4.67 mA =9.35 mA
(b) The symmetrical zener limiter for PSpice simulation is shown in Fig. 4.32. The PSpice plot of instantaneous output voltage v_{ O } is shown in Fig. 4.33, which gives +5.435 V, compared to the expected value of 4.67 + 0.981 = 5.65 V (from the expression of v_{ O } for the interval 2).
