For the network of Fig. 2.188 determine the range of Vi that will maintain VL at 8 V and n...

Calculate the load current.

$$ \begin{aligned} I_{L} &=\frac{V_{L}}{R_{L}} \\ &=\frac{V_{Z}}{R_{L}} \end{aligned} $$

Where,

ZENER voltage is \(V_{Z}=8 \mathrm{~V}\)

Load resistance is \(R_{L}=0.22 \mathrm{k} \Omega\)

Substitute \(8 \mathrm{~V}\) for \(V_{Z}\) and \(0.22 \mathrm{k} \Omega\) for \(R_{L}\) to obtain \(I_{L}\).

$$ \begin{aligned} I_{L} &=\frac{8}{0.22 \times 10^{3}} \\ &=36.36 \times 10^{-3} \\ &=36.36 \mathrm{~mA} \end{aligned} $$

Calculate the maximum ZENER current \(I_{Z_{\max }}\).

\(I_{Z_{\max }}=\frac{P_{Z_{\max }}}{V_{Z}}\)

Here,

Maximum ZENER power \(P_{Z_{\text {max }}}\) is \(400 \mathrm{~mW}\)

ZENER voltage \(V_{Z}\) is \(8 \mathrm{~V}\)

Substitute \(8 \mathrm{~V}\) for \(V_{Z}\) and \(400 \mathrm{~mW}\) for \(P_{z_{\operatorname{mex}}}\) to obtain \(I_{Z_{\max }}\).

\(\begin{aligned} I_{Z_{\text {eax }}} &=\frac{400 \mathrm{~mW}}{8 \mathrm{~V}} \\ &=50 \mathrm{~mA} \end{aligned}\)

Calculate the current through the source resistance \(I_{R_{s}}\).

$$ \begin{aligned} I_{R_{S}} &=I_{Z_{\text {max }}}+I_{L} \\ &=50 \mathrm{~mA}+36.36 \mathrm{~mA} \\ &=86.36 \mathrm{~mA} \end{aligned} $$

Calculate the maximum input voltage.

\(V_{i_{\max }}=I_{R_{s}} R_{S}+V_{Z}\)

Substitute \(8 \mathrm{~V}\) for \(V_{Z}, 91 \Omega\) for \(R_{S}\) and \(86.36 \mathrm{~mA}\) for \(I_{R_{s}}\) to obtain \(V_{i_{\text {eq }}}\).

$$ \begin{aligned} V_{i_{\max }} &=I_{R_{5}} R_{S}+V_{Z} \\ &=\left(86.36 \times 10^{-3}\right)(91)+8 \\ &=7.86+8 \\ &=15.86 \mathrm{~V} \end{aligned} $$

Therefore, the range of input voltage varies between \(11.3 \mathrm{~V}\) and \(15.86 \mathrm{~V}\)