Problem

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

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

FIG. 2.175

#### Step-by-Step Solution

Solution 1

Refer to the circuit diagram in Figure $$2.175$$ in the textbook.

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

Write the expression for output peak voltage $$V_{o_{m d}}$$.

$$V_{o_{\mathrm{md}}}=\frac{R_{\mathrm{nd}}(V)}{R_{\mathrm{nd}}+R} \ldots \ldots \text { (1) }$$

Here,

The net resistance $$R_{\text {ner }}$$ is,

\begin{aligned} R_{\text {mer }} &=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}

Voltage $$V$$ is $$170 \mathrm{~V}$$

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 $$36.04 \mathrm{~V}$$.