Determine vo for each network of Fig. 2.179 for the input shown.FIG. 2.179

For the condition \(v_{i}<4.7 \mathrm{~V}\) the diode is OFF and the output voltage \(v_{o}\) is given as:

$$ \begin{aligned} v_{o} &=v_{i} \\ &=-8 \mathrm{~V} \end{aligned} $$

Hence, the negative peak output voltage \(v_{o}\) is \(-8 \mathrm{~V}\).

The output voltage waveform is given as:

Figure 1

(b)

Refer to the circuit diagram shown in Fig. 2.176 (b) in the textbook.

As per Fig. \(2.176\) (b), the diode is ON when the input voltage \(v_{i} \geq 3.7 \mathrm{~V}\).

So the output voltage \(v_{o}\) is given as:

$$ v_{o}=0.7 \mathrm{~V} $$

Hence, the positive output voltage \(v_{o}\) is \(0.7 \mathrm{~V}\).

As per Fig. \(2.176\) (b), the diode is ON when the input voltage \(v_{i}<3.7 \mathrm{~V}\). The diode current is same as that of the reverse biased current \(I_{D}=I_{R}=0 \mathrm{~mA}\). Now the voltage across the resistor \(2.2 \mathrm{k} \Omega\) is given as:

$$ \begin{aligned} V_{2.2 \mathrm{k} \Omega} &=I R \\ &=(0 \mathrm{~mA}) R \\ &=0 \mathrm{~V} \end{aligned} $$

Now the output voltage \(v_{o}\) is given as:

$$ v_{o}=v_{i}-3 \mathrm{~V} $$

At the input voltage \(v_{i}=0 \mathrm{~V}\) the output voltage \(v_{o}\) is given as:

$$ v_{o}=-3 \mathrm{~V} $$

Hence, the negative output voltage \(v_{o}\) is \(-3 \mathrm{~V}\).

Now when the input voltage \(v_{i}=-8 \mathrm{~V}\) the output voltage \(v_{o}\) is given as:

$$ \begin{aligned} v_{o} &=-8 \mathrm{~V}-3 \mathrm{~V} \\ &=-11 \mathrm{~V} \end{aligned} $$

Hence, the negative output voltage \(v_{o}\) is \(-11 \mathrm{~V}\).

Figure 4