Find (there is no need to draw) V o ​ versus V i  for the circuit in Fig. 5.39. Assume cut-in voltage of the zener to be 1 V and the breakdown voltage to be 5 V. The maximum forward bias current rating ( I Z , f , max ⁡ ) of the zener is 4 mA, and the maximum reverse bias current rating  ( I Z , r , max ⁡ )  is 6 mA. The resistance of the zener when it is conducting is zero. The zener knee current is also zero. The diode has a cut-in voltage of 1 V and zero forward resistance. The peak current rating of the diode ( I D , m a x )  is 2 mA. Hence find the range of V i  that can be applied.

Consider Fig. 5.39. We have the following situations.

(a) I = 0 . Clearly

$V_o=V_i-2. \left(5.74\right)$Vo​=Vi​−2. (5.74)

(b) I > 0. Then D must be ON and Z must be in the breakdown region. The resulting circuit is shown in Fig. 5.40. We have:

$-3I-5-1+2-I=V_i$−3I−5−1+2−I=Vi​

$\Rightarrow I=\frac{V_i-4}{4}>0$⇒I=4Vi​−4​>0

$\Rightarrow V_i<-4V\mathrm{.} \left(5.75\right)$⇒Vi​<−4V. (5.75)

Now

$I_{\mathrm{m}\mathrm{a}\mathrm{x}}=\min \{I_{D,}, I_{Z,r,}\}$Imax​=min{ID,max​, IZ,r,max​}

$=\min \{2, 6\}$=min{2, 6}

$=2\mathrm{m}\mathrm{A}\mathrm{.} \left(5.76\right)$=2mA. (5.76)

Since

$I_{\mathrm{m}\mathrm{a}\mathrm{x}}=2\mathrm{m}\mathrm{A}$Imax​=2mA

$=\frac{-V_i-4}{4}$=4−Vi​−4​

$\Rightarrow V_i=-12\mathrm{V}\mathrm{.} \left(5.77\right)$⇒Vi​=−12V. (5.77)

We also have

$V_o=-3I-5–1$Vo​=−3I−5–1

$=\frac{3V_i-12}{4}\mathrm{.} \left(5.78\right)$=43Vi​−12​. (5.78)

To summarize:

$V_o=\{\begin{array}{cc}{V}_i-2& \mathrm{f}\mathrm{o}\mathrm{r}{V}_i\ge -4\mathrm{V}\end{array}(3V_i-12)\mathrm{/}4& \mathrm{f}\mathrm{o}\mathrm{r}-12\le V_i<-4\mathrm{V}\\ \left(5.79\right)$Vo​={Vi​−2(3Vi​−12)/4​forVi​≥−4Vfor−12≤Vi​<−4V​ (5.79)

The range of $V_i$Vi​ is $-12<V_i<\mathrm{\infty } V\mathrm{.}$−12<Vi​<∞  V.