Question 18.38: Determine the range of input voltage that maintains the outp...

As V_{o} = 10 V constant and R_{L} = 10 kΩ constant,
we have

I_{L} =\frac{V_{o}}{R_{L}}=\frac{10}{10 × 10^{3}} = 1 mA

When V_{in} = V_{in (max)}. I_{Z} = I_{Z max}
Now I = I_{Z} + I_{L}
Therefore, I_{max} = I_{Zmax} + I_{L} = 24 mA + 1 mA = 2 mA.

\frac{V_{in(max)} – V_{Z}}{R} = 25 mA

Therefore, V_{in(max)} –10 = 1 × 10^{3}(25 × 10^{– 3})
V_{in(max)} = 35 V.
When V_{in} = V_{in(min)}, I_{Z} = I_{Zmin} = 5 mA
Therefore, I_{min} = I_{Zmin} + I_{L} = 5 mA + 1 mA = 6 mA

V_{in(min)} –10 = 1 × 10^{3}(6 × 10^{– 3})
V_{in(min)} = 16 V
Thus range of input voltage is 16 V to 35 V for which the output voltage will be of 10 V.