Sketch vo for the network of Fig. 2.175 and determine the dc voltage available.FIG. 2.175

Substitute \(170 \mathrm{~V}\) for \(V, R=2.2 \mathrm{k} \Omega\) for \(R\), and \(1.1 \mathrm{k} \Omega\) for \(R_{\text {met }}\) in equation (1).

$$ \begin{aligned} V_{o, m d} &=\frac{R_{\mathrm{met}}(V)}{R_{\mathrm{met}}+R} \\ &=\frac{1.1 \mathrm{k} \Omega(170 \mathrm{~V})}{1.1 \mathrm{k} \Omega+2.2 \mathrm{k} \Omega} \\ &=56.67 \mathrm{~V} \end{aligned} $$

As per Fig \(2.175\) for the Negative Pulse for the input voltage \(v_{i}\) the top left diode is in ON state and the bottom left is in the OFF state.

Write the expression for output peak voltage \(V_{o_{m}} .\)

$$ V_{e_{m+}}=\frac{R_{\text {nt }}(V)}{R_{\text {net }}+R} $$

Here,

The net resistance \(R_{\text {wet }}\) is given as:

$$ \begin{aligned} R_{\mathrm{mec}} &=R \| R \\ &=2.2 \mathrm{k} \Omega \| 2.2 \mathrm{k} \Omega \quad[R=2.2 \mathrm{k} \Omega] \\ &=1.1 \mathrm{k} \Omega \end{aligned} $$

Substitute \(170 \mathrm{~V}\) for \(V, R=2.2 \mathrm{k} \Omega\) for \(R\), and \(1.1 \mathrm{k} \Omega\) for \(R_{n e t}\) in equation (1).

$$ \begin{aligned} V_{o_{\mu<<}} &=\frac{R_{\mathrm{net}}(V)}{R_{\mathrm{met}}+R} \\ &=\frac{1.1 \mathrm{k} \Omega(170 \mathrm{~V})}{1.1 \mathrm{k} \Omega+2.2 \mathrm{k} \Omega} \\ &=56.67 \mathrm{~V} \end{aligned} $$

Thus, the peak voltage is \(56.67 \mathrm{~V}\).

Sketch the output voltage as shown in Figure 1.

Determine the \(\mathrm{dc}\)-voltage \(V_{\mathrm{dc}}\).

For a full-wave rectifier the dc-voltage is,

$$ \begin{aligned} V_{d e} &=0.636\left(V_{o_{m d}}\right) \\ &=0.636(56.67 \mathrm{~V}) \\ &=36.04 \mathrm{~V} \end{aligned} $$

Thus, the dc-voltage \(V_{d c}\) is \(36.04 \mathrm{~V}\).