Hence find the range of ${V}_{i}$ that can be applied.

# 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.

electricalexpert
Begginer

Consider Fig. 5.39. We have the following situations.

(a) I = 0 . Clearly

${V}_{o}={V}_{i}-2.\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\left(5.74\right)$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

$\Rightarrow {V}_{i}<-4V\mathrm{.}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\left(5.75\right)$⇒Vi<−4V. (5.75)

Now

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

$=\mathrm{min}\{2,6\}$=min{2, 6}

$=2\mathrm{m}\mathrm{A}\mathrm{.}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\left(5.76\right)$=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

$\Rightarrow {V}_{i}=-12\mathrm{V}\mathrm{.}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\left(5.77\right)$⇒Vi=−12V. (5.77)

We also have

${V}_{o}=-3I-5\u20131$Vo=−3I−5–1

$=\frac{3{V}_{i}-12}{4}\mathrm{.}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\left(5.78\right)$=43Vi−12. (5.78)

To summarize:

${V}_{o}=\{\begin{array}{cc}{V}_{i}-2& \mathrm{f}\mathrm{o}\mathrm{r}{V}_{i}\ge -4\mathrm{V}\end{array}(3{V}_{i}-12)\mathrm{/}4& \mathrm{f}\mathrm{o}\mathrm{r}-12\le {V}_{i}<-4\mathrm{V}\\ \phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\left(5.79\right)$Vo={Vi−2(3Vi−12)/4forVi≥−4Vfor−12≤Vi<−4V (5.79)

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