Find (there is no need to draw) versus 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 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 diodeis 2 mA.
Hence find the range of that can be applied.

Consider Fig. 5.39. We have the following situations.
(a) I = 0 . Clearly
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=Vi
⇒I=4Vi−4>0
⇒Vi<−4V. (5.75)
Now
Imax=min{ID,max, IZ,r,max}
=min{2, 6}
=2mA. (5.76)
Since
Imax=2mA
=4−Vi−4
⇒Vi=−12V. (5.77)
We also have
Vo=−3I−5–1
=43Vi−12. (5.78)
To summarize:
Vo={Vi−2(3Vi−12)/4forVi≥−4Vfor−12≤Vi<−4V (5.79)
The range of Vi is −12<Vi<∞ V.