a. Determine VL, IL, IZ, and IR for the network of Fig. 2.186 if RL = 180 Ω.b. Repeat part...

Here the load voltage \(V_{L}=9 \mathrm{~V}\) is less than the ZENER voltage \(V_{Z}=10 \mathrm{~V} .\) So, the diode does not conduct and hence the current through the diode is zero.

$$ I_{Z}=0 \mathrm{~A} $$

Therefore, the value of the ZENER current, \(I_{Z}\) is \(0 \mathrm{~A}\).

Calculate the load current \(I_{L}\).

$$ \begin{aligned} I_{L} &=\frac{V_{i}}{R_{s}+R_{L}} \\ &=\frac{20}{180+220} \\ &=\frac{20}{400} \\ &=50 \mathrm{~mA} \end{aligned} $$

Therefore, the value of the load current \(I_{L}\) is \(50 \mathrm{~mA}\).

Calculate the reverse biased current \(I_{R}\).

$$ \begin{aligned} I_{R} &=I_{L} \\ &=50 \mathrm{~mA} \end{aligned} $$

Therefore, the value of the reverse biased current \(I_{R}\) is \(50 \mathrm{~mA}\).

(b)

Calculate the load voltage, \(V_{L}\).

$$ V_{L}=\frac{R_{L} V_{i}}{R_{L}+R_{S}} $$

Here,

The load resistance \(R_{L}\) is \(470 \Omega\)

The source resistance \(R_{S}\) is \(220 \Omega\)

Voltage \(V_{i}\) is \(20 \mathrm{~V}\)

Substitute \(470 \Omega\) for \(R_{L}, 220 \Omega\) for \(R_{S}\) and \(20 \mathrm{~V}\) for \(V_{i}\) in \(V_{L}\).

$$ \begin{aligned} V_{L} &=\frac{(470)(20)}{470+220} \\ &=\frac{9400}{690} \\ &=13.62 \mathrm{~V} \end{aligned} $$

Here the load voltage, \(V_{L}=13.62 \mathrm{~V}\) is greater than the ZENER voltage, \(V_{Z}=10 \mathrm{~V} . \mathrm{So}\), the diode conducts and the load voltage is equal to the ZENER voltage.

$$ \begin{aligned} V_{L} &=V_{Z} \\ &=10 \mathrm{~V} \end{aligned} $$

Therefore, the load voltage, \(V_{L}\) is \(10 \mathrm{~V}\).

Calculate the current, \(I_{R_{\mathrm{s}}}\).

$$ \begin{aligned} I_{R_{s}} &=\frac{V_{R_{S}}}{R_{S}} \\ &=\frac{V_{i}-V_{L}}{R_{s}} \end{aligned} $$

Here,

Resistance \(R_{S}\) is \(220 \Omega\)

Load voltage, \(V_{L}\) is \(10 \mathrm{~V}\)

Voltage \(V_{i}\) is \(20 \mathrm{~V}\)

Substitute \(220 \Omega\) for \(R_{s}, 20 \mathrm{~V}\) for \(V_{i}\) and \(10 \mathrm{~V}\) for \(V_{L}\) in \(I_{R_{\mathrm{s}}}\).

$$ \begin{aligned} I_{R_{s}} &=\frac{20-10}{220} \\ &=45.45 \mathrm{~mA} \end{aligned} $$

Therefore, the current through the resistor, \(I_{R_{s}}\) is \(45.45 \mathrm{~mA}\).

Calculate the load Current, \(I_{L}\).

$$ \begin{aligned} I_{L} &=\frac{V_{L}}{R_{L}} \\ &=\frac{10}{470} \\ &=21.28 \mathrm{~mA} \end{aligned} $$

Therefore, the load Current, \(I_{L}\) is \(21.28 \mathrm{~mA}\).

Calculate the ZENER diode current, \(I_{Z}\).

$$ \begin{aligned} I_{Z} &=I_{R_{s}}-I_{L} \\ &=45.45 \mathrm{~mA}-21.28 \mathrm{~mA} \\ &=24.17 \mathrm{~mA} \end{aligned} $$

Therefore, the ZENER diode current \(I_{Z}\) is \(24.17 \mathrm{~mA}\).

(c)

Calculate the ZENER diode current, \(I_{Z}\).

$$ \begin{aligned} I_{Z} &=\frac{P_{\max }}{V_{Z}} \\ &=\frac{400 \mathrm{~mW}}{10 \mathrm{~V}} \\ &=40 \mathrm{~mA} \end{aligned} $$

Calculate the minimum load current \(I_{L_{\sin }}\).

$$ \begin{aligned} I_{L_{\min }} &=I_{R_{5}}-I_{Z_{\operatorname{mix}}} \\ &=45.45 \mathrm{~mA}-40 \mathrm{~mA} \\ &=5.45 \mathrm{~mA} \end{aligned} $$

Calculate the load resistance, \(R_{L}\)

$$ \begin{aligned} R_{L} &=\frac{V_{L}}{I_{L_{\min }}} \\ &=\frac{10 \mathrm{~V}}{5.45 \times 10^{-3}} \\ &=1.835 \times 10^{3} \\ &=1.835 \mathrm{k} \Omega \end{aligned} $$

Therefore, the value of the load resistance, \(R_{L}\) is \(1.835 \mathrm{k} \Omega\)

(d)

Calculate the load resistance \(R_{L}\) when the ZENER diode is absent.

$$ \begin{aligned} &V_{L}=\frac{R_{L} V_{i}}{R_{L}+R_{S}} \\ &10=\frac{20 R_{L}}{R_{L}+220} \\ &10 R_{L}+2200=20 R_{L} \\ &10 R_{L}=2200 \\ &R_{L}=220 \Omega \end{aligned} $$

Therefore, the load resistance \(R_{L}\) is \(220 \Omega\).