Question 4.13: Design of a zener regulator The parameters of a 6.3-V zener ...

V_{ Z }=6.3 V \text { at } i_{ ZT }=40 mA , i_{ L (\min )}=0 \text { and } i_{ Z (\min )}=1 mA . Using Eq. (4.37), we have

v_{ Z }=v_{ ZO }+R_{ Z } i_{ Z }                                           (4.37)

V_{ ZO }=V_{ Z }-R_{ Z } I_{ ZT }=6.3-2 \times 40 mA =6.22 V

(a) The maximum power dissipation P_{Z(\max )} of a zener diode is

P_{Z(\max )}=i_{Z(\max )} V_{Z}=0.75 W

or                            i_{ Z (\max )}=\frac{P_{ Z ( max )}}{V_{Z}}=\frac{0.75}{6.3}=119 mA

(b) The zener current i_{ Z } becomes maximum when the supply voltage is maximum and the load current is minimum—that is, V_{ S (\max )}=18 V , i_{ L (\min )}=0, \text { and } i_{ Z (\max )}=119 mA . From Eq. (4.45),

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 }=\frac{V_{ S ( max )}-V_{ ZO }-R_{ Z } i_{ Z (\max )}}{i_{ Z (\max )}+i_{ L ( min )}}=\frac{18 V -6.22 V -2 \Omega \times 119 mA }{119 mA +0}=96.96 \Omega

(c) The power rating P_{ R } \text { of } R_{ s } is

\begin{aligned}P_{ R } &=\left(i_{ Z (\max )}+i_{ L (\min )}\right)\left(V_{ S (\max )}-V_{ ZO }-R_{ Z } i_{ Z (\max )}\right) \\&=119 mA \times(18 V -6.22 V -2 \Omega \times 119 mA )=1.373 W\end{aligned}

The worst-case power rating of R_{ s } will occur when the load is shorted. That is,

P_{ R (\max )}=\frac{V_{ S (\max )}^{2}}{R_{ s }}=\frac{18^{2}}{96.99}=3.34 W

(d)   i_{ L } must be maintained at the maximum when V_{ S } is minimum and i_{Z} is minimum—that is, V_{ S (\min )}=12 V and i_{ Z (\min )}=1 mA . From Eq. (4.44), we get

R_{ s }=\frac{V_{ S ( min )}-\left(V_{ ZO }+R_{ Z } i_{ Z ( min )}\right)}{i_{ Z ( min )}+i_{ L (\max )}}                                    (4.44)

\begin{aligned}I_{ L (\max )} &=\frac{V_{ S ( min )}-V_{ ZO }-R_{ Z } i_{ Z ( min )}}{R_{ s }}-i_{ Z ( min )}=\frac{12 V -6.22 V -2 \Omega \times 1 mA }{96.99 \Omega}-1 mA \\&=58.57 mA\end{aligned}