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 -8 \Omega \times 53 mA =4.28 V
(a) For R_{ L }=\infty , the zener current is
i_{ Z }=\frac{ V _{ S }-V_{ ZO }}{R_{ Z }+R_{ s }}=\frac{12-4.28}{8+220}=33.86 mA
The output voltage is
v_{ O }=V_{ ZO }+R_{ Z } i_{ Z }=4.28 V +8 \Omega \times 33.86 mA =4.55 V
(b) A change in the supply voltage by \Delta v_{ S }=\pm 2 V will cause a change in the output voltage, which we can find from Eq. (4.38):
\text { Line regulation }=\frac{\Delta v_{ O }}{\Delta v_{ S }}=\frac{R_{ Z }}{R_{ Z }+R_{ s }} (4.38)
\Delta v_{ O (\text { supply })}=\frac{\Delta v_{ S } R_{ Z }}{R_{ Z }+R_{ s }}=\frac{\pm 2 \times 8}{8+220}=\pm 70.18 mV
The nominal value of the load current is i_{ L }=V_{ Z } / R_{ L }=4.7 / 470=10 mA . A change in the load current by \Delta i_{ L }=10 mA will also cause a change in the output voltage, which we can find from Eq. (4.39):
\text { Load regulation }=\frac{\Delta v_{ O }}{\Delta i_{ L }}=-\left(R_{ Z } \| R_{ s }\right) (4.39)
\Delta v_{ O (\text { load })}=-\left(R_{ Z } \| R_{ s }\right) \Delta i_{ L }=-(8 \Omega \| 220 \Omega) \times 10 mA =-77.19 mV
Therefore, the maximum and minimum values of the output voltage can be found from
\begin{aligned}&v_{ O (\max )}=4.55 V +70.18 mV -77.19 mV =4.54 V \\&v_{ O (\min )}=4-70.18 mV -77.19 mV =4.47 V\end{aligned}
(c) The nominal value of the load current is i_{ L }=V_{ Z } / R_{ L }=4.7 / 100=47 mA , which is not possible because the maximum current that can flow through R_{ Z } is only 33.86 mA. Thus, the zener diode will be off, and the output voltage will be the voltage across R_{ L } . That is,
v_{ O }=\frac{R_{ L }}{R_{ L }+R_{ s }} V_{ S }=\frac{100}{100+220} \times 12=3.75 V
(d) For the zener diode to be operated in the breakdown region, allowing only I_{ ZK } to flow, the maximum current that can flow through R_{ L } is given by \text { (assuming } \left.I_{ ZK }=I_{ Z } \text { at } v_{ Z }=V_{ ZO }\right)
i_{ L (\max )}=\frac{V_{ S ( min )}-V_{ ZO }}{R_{ s }}-I_{ ZK } (4.41)
=\frac{(10-4.28) V }{220 \Omega}-1 mA =25 mA
Therefore, the minimum value of R_{ L } that guarantees operation in the breakdown region is given by
R_{ L ( min )} \geq \frac{V_{ ZO }}{i_{ L (\max )}} (4.42)
\geq \frac{4.28 V }{25 mA }=171.2 \Omega
