V_{ Z }=4.7 V , P_{ Z (\max )}=400 mW , i_{ L (\min )}=5 mA , i_{ L (\max )}=50 mA , V_{ S (\min )}=20 V , \text { and } V_{ S (\max )}=30 V . Using Eq. (4.37), we have
v_{ Z }=v_{ ZO }+R_{ Z } i_{ Z } (4.37)
V_{ ZO }=V_{ Z }-R_{ Z } I_{ ZT }=4.7-19 \times 20 mA =4.32 V
Also, i_{Z(\max )}=\frac{P_{Z(\max )}}{V_{Z}}=\frac{400 mW }{4.7 V }=85.1 mA \text { (from specifications) }
(a) Since the minimum value of the zener current is not specified, we can assume for all practical purposes that
i_{ Z ( min )}=0.1 \times i_{ Z (\max )}=0.1 \times 85.1 mA =8.51 mA
From Eq. (4.44) and Fig. 4.26(b), we can find the value of R_{ s } :
R_{ s }=\frac{V_{ S ( min )}-\left(V_{ ZO }+R_{ Z } i_{ Z ( min )}\right)}{i_{ Z ( min )}+i_{ L (\max )}} (4.44)
R_{ s }=\frac{V_{ S ( min )}-V_{ ZO }-R_{ Z } i_{ Z ( min )}}{i_{ Z (\min )}+i_{ L (\max )}}=\frac{20 V -4.32 V -19 \Omega \times 8.51 mA }{8.51 mA +50 mA }=265 \Omega
From Eq. (4.45), we can find
R_{ s }=\frac{V_{ S ( max )}-\left(V_{ ZO }+R_{ Z } i_{ Z ( min )}\right)}{i_{ Z (\max )}+i_{ L (\min )}} (4.45)
R_{ s }\left(i_{ Z (\max )}+i_{ L ( min )}\right)=V_{ S (\max )}-V_{ ZO }-R_{ Z } i_{ Z (\max )}
which can be solved to find the actual value of the maximum zener current i_{Z(\max )} :
i_{ Z (\max )}=\frac{V_{ S (\max )}-V_{ ZO }-R_{ s } i_{ L ( min )}}{R_{ s }+R_{ Z }}=\frac{30 V -4.32 V -265 \Omega \times 5 mA }{(265+19) \Omega}=85.76 mA
The power rating P_{ R } \text { of } R_{ s } is
\begin{aligned}P_{ R } & \approx\left(i_{ Z (\max )}+i_{ L ( min )}\right)\left(V_{ S (\max )}-V_{ Z }\right) \\&=(85.76 mA +5 mA )(30 V -4.7 V )=2.3 W\end{aligned}
The worst-case power rating will be
P_{ R (\max )}=\frac{V_{ S (\max )}^{2}}{R_{ s }}=\frac{30^{2}}{265}=3.4 W
From i_{L(\min )}=5 mA \text { and } i_{ L (\max )}=50 mA , we find that the corresponding maximum and minimum values of the load resistance are
\begin{aligned}&R_{ L (\max )}=\frac{V_{ Z }}{i_{ L (\min )}}=\frac{4.7 V }{5 mA }=940 \Omega \\&R_{ L (\min )}=\frac{V_{ Z }}{i_{ L (\max )}}=\frac{4.7 V }{50 mA }=94 \Omega\end{aligned}
The zener voltage regulator for the PSpice simulation is shown in Fig. 4.27. The zener diode is normally modeled by setting the diode parameter B V \cong V_{ Z } in the PSpice model.
(b) The PSpice plot of the output voltage v_{ O } against supply voltage v_{ S } is shown in Fig. 4.28. The zener action begins at an output voltage of v_{ O }=4.74 V , which is close to the expected value of 4.7 V.
