Determine vo for each network of Fig. 2.177 for the input shown.FIG. 2.177

As per Fig. \(2.176\) (a) for the negative pulse of input voltage \(v_{i}\) the output voltage \(v_{o}\) when the

diode is open is given as:

\(v_{o}=0 \mathrm{~V}\)

Hence, for the negative pulse the output voltage \(v_{o}\) is \(0 \mathrm{~V}\).

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) for the positive pulse of input voltage \(v_{i}\) the expression for output voltage \(v_{o}\) is,

$$ v_{o}=V-V_{D}+E $$

Here,

Voltage \(V\) is \(12 \mathrm{~V}\)

Silicon Diode voltage \(V_{D}\) is \(0.7 \mathrm{~V}\)

Battery supply \(E\) is \(4 \mathrm{~V}\)

Substitute the corresponding values to obtain \(v_{o}\).

$$ \begin{aligned} v_{o} &=V-V_{D}+E \\ &=12 \mathrm{~V}-0.7 \mathrm{~V}+4 \mathrm{~V} \\ &=15.3 \mathrm{~V} \end{aligned} $$

Hence, for the positive pulse the output voltage \(v_{o}\) is \(15.3 \mathrm{~V}\).

As per Fig. \(2.176\) (b) for the negative pulse of input voltage \(v_{i}\) the output voltage \(v_{o}\) when the

diode is open is given as:

\(v_{o}=0 \mathrm{~V}\)

Hence, for the negative pulse the output voltage \(v_{o}\) is \(0 \mathrm{~V}\).

The output voltage waveform is given as:

Figure 2