Determine Vo and ID for the networks of Fig. 2.158.FIG. 2.158

Determine the diode current \(I_{D}\) when the diode is forward biased.

$$ \begin{aligned} I_{D} &=\frac{22 \mathrm{~V}-0.7 \mathrm{~V}}{2.2 \mathrm{k} \Omega+2.2 \mathrm{k} \Omega} \\ &=4.84 \mathrm{~mA} \end{aligned} $$

Hence, the diode current \(I_{D}\) is \(4.84 \mathrm{~mA}\)

Determine the output voltage \(V_{o}\)

$$ \begin{aligned} V_{o} &=I_{D}(2.2 \mathrm{k} \Omega) \\ &=(4.84 \mathrm{~mA})(2.2 \mathrm{k} \Omega) \\ &=10.65 \mathrm{~V} \end{aligned} $$

Hence, the output voltage \(V_{o}\) is \(10.65 \mathrm{~V}\).

(b)

Refer to Figure \(2.158(\mathrm{~b})\) from the text book.

Cathode of the diode is getting \(-20 \mathrm{~V}\) and anode of the diode is getting \(20 \mathrm{~V} .\) So, diode is in forward condition.

Apply Kirchhoff's voltage law.

$$ \begin{aligned} I_{D} &=\frac{20 \mathrm{~V}+20 \mathrm{~V}-0.7 \mathrm{~V}}{6.8 \mathrm{k} \Omega} \\ &=5.78 \mathrm{~mA} \end{aligned} $$

Hence, the diode current \(I_{D}\) is \(5.78 \mathrm{~mA}\)

Determine the output voltage by applying Kirchhoff's Voltage Law.

$$ \begin{aligned} &V_{o}-0.7 \mathrm{~V}+20 \mathrm{~V}=0 \\ &V_{o}=-19.3 \mathrm{~V} \end{aligned} $$

Hence, the output voltage \(V_{o}\) is \(-19.3 \mathrm{~V}\).