Chapter 2, Problem 2.24P

Consider the Zener diode circuit in Figure 2.19 in the text. Assume parameter values of $V_{ZO}\text{=5}\text{.6V}$ (diode voltage when $I_Z\cong 0$ ), $r_z=3\Omega$ , and $R_i=50\Omega$ . Determine $V_L,I_Z,I_L$ , and the power dissipated in the diode for (a) $V_{PS}\text{=10V,}{R}_L=\infty$ ; (b) $V_{PS}\text{=10V,}{R}_L=200\Omega$ ; (c) $V_{PS}\text{=12V,}{R}_L=\infty$ ; and (d) $V_{PS}\text{=12V,}{R}_L=200\Omega$ .

To determine

The value of $V_L,I_L\text{and}{I}_Z$

Answer to Problem 2.24P

  $\begin{array}{l}{I}_Z=83.01\text{mA}\\ I_L=0\\ V_Z=5.84\text{V}\\ P_Z=484.7\text{mW}\end{array}$

Explanation of Solution

Given:

  $\begin{array}{l}{V}_{ZO}=5.6\text{V}\\ r_Z=3\Omega \\ R_i=50\Omega \end{array}$

The given circuit is

  

Calculation:

For $V_{PS}=10\text{V, }{R}_L=\infty$

Apply KVL in loop 1,

  $\begin{array}{l}{I}_I=I_Z=\frac{V_{PS}-V_{ZO}}{R_i+r_z}\\ I_I=I_Z=\frac{10-5.6}{50+3}\\ I_Z=83.01\text{mA}\\ \text{Since }{R}_L=\infty ,\\ I_L=0\end{array}$

Therefore, the load voltage will be

  $\begin{array}{l}{V}_L=V_Z=V_{ZO}+I_Z{r}_z\\ V_Z=5.6+\left(83.01\times {10}^{-3}\right)\times 3\\ V_Z=5.84\text{V}\end{array}$

Thus, the power dissipated will be

  $\begin{array}{l}{P}_Z=V_Z{I}_Z\\ P_Z=5.84\times 83.01\times {10}^{-3}\\ P_Z=484.7\text{mW}\end{array}$

To determine

The value of $V_L,I_L\text{and}{I}_Z$

Answer to Problem 2.24P

  $\begin{array}{l}{I}_Z=55.81\text{mA}\\ V_L=5.767\text{V}\\ I_L=28.85\text{mA}\\ P_Z=321.85\text{mW}\end{array}$

Explanation of Solution

Given:

  $\begin{array}{l}{V}_{ZO}=5.6\text{V}\\ r_Z=3\Omega \\ R_i=50\Omega \end{array}$

The given circuit is

  

Calculation:

For $V_{PS}=10\text{V, }{R}_L=200\Omega$

Apply KCL at the center node,

  $\frac{V_Z-V_{PS}}{R_i}+I_Z+\frac{V_Z}{R_L}=0$

As $V_Z=V_{ZO}+I_Z{r}_Z$

By substituting the value of $V_Z$

  $\begin{array}{l}\frac{\left(V_{ZO}+I_Z{r}_Z\right)-V_{PS}}{R_i}+I_Z+\frac{\left(V_{ZO}+I_Z{r}_Z\right)}{R_L}=0\\ \frac{\left(5.6+3I_Z\right)-10}{50}+I_Z+\frac{\left(5.6+3I_Z\right)}{200}=0\\ 0.112+0.06I_Z-0.2+I_Z+0.028+0.015I_Z=0\\ \left(0.06+1+0.015\right)I_Z=0.06\\ I_Z=\frac{0.06}{1.075}\\ I_Z=55.81\text{mA}\end{array}$

Hence, the load voltage will be

  $\begin{array}{l}{V}_Z=V_{ZO}+I_Z{r}_Z\\ V_Z=5.6+0.167\\ V_Z=V_L\\ V_L=5.767\text{V}\\ \text{Hence,}\\ I_I=\frac{V_{PS}-V_Z}{R_i}\\ I_I=\frac{10-5.767}{50}\\ I_I=84.66\text{mA}\end{array}$

Thus, the load current can be calculated as

  $\begin{array}{l}{I}_L=I_I-I_Z\\ I_L=\left(84.66-55.81\right)\text{mA}\\ I_L=28.85\text{mA}\end{array}$

Thus, the power dissipated will be equal to

  $\begin{array}{l}{P}_Z=V_Z{I}_Z\\ P_Z=5.767\times 55.81\times {10}^{-3}\\ P_Z=321.85\text{mW}\end{array}$

To determine

The value of $V_L,I_L\text{and}{I}_Z$

Answer to Problem 2.24P

  $\begin{array}{l}{I}_Z=120.75\text{mA}\\ I_L=0\\ V_Z=5.96\text{V}\\ P_Z=719.67\text{mW}\end{array}$

Explanation of Solution

Given:

  $\begin{array}{l}{V}_{ZO}=5.6\text{V}\\ r_Z=3\Omega \\ R_i=50\Omega \end{array}$

The given circuit is

  

Calculation:

For $V_{PS}=12\text{V, }{R}_L=\infty$

Apply KVL in loop 1,

  $\begin{array}{l}{I}_I=I_Z=\frac{V_{PS}-V_{ZO}}{R_i+r_z}\\ I_I=I_Z=\frac{12-5.6}{50+3}\\ I_Z=120.75\text{mA}\\ \text{Since }{R}_L=\infty ,\\ I_L=0\end{array}$

Therefore, the load voltage will be

  $\begin{array}{l}{V}_L=V_Z=V_{ZO}+I_Z{r}_z\\ V_Z=5.6+\left(120.75\times {10}^{-3}\right)\times 3\\ V_Z=5.96\text{V}\end{array}$

Thus, the power dissipated will be

  $\begin{array}{l}{P}_Z=V_Z{I}_Z\\ P_Z=5.96\times 120.75\times {10}^{-3}\\ P_Z=719.67\text{mW}\end{array}$

To determine

The value of $V_L,I_L\text{and}{I}_Z$

Answer to Problem 2.24P

  $\begin{array}{l}{I}_Z=93.02\text{mA}\\ I_L=29.4\text{mA}\\ V_L=5.87\text{V}\\ P_Z=546.02\text{mW}\end{array}$

Explanation of Solution

Given:

  $\begin{array}{l}{V}_{ZO}=5.6\text{V}\\ r_Z=3\Omega \\ R_i=50\Omega \end{array}$

The given circuit is

  

Calculation:

For $V_{PS}=12\text{V, }{R}_L=200\Omega$

Apply KCL at the center node,

  $\frac{V_Z-V_{PS}}{R_i}+I_Z+\frac{V_Z}{R_L}=0$

As $V_Z=V_{ZO}+I_Z{r}_Z$

By substituting the value of $V_Z$

  $\begin{array}{l}\frac{\left(V_{ZO}+I_Z{r}_Z\right)-V_{PS}}{R_i}+I_Z+\frac{\left(V_{ZO}+I_Z{r}_Z\right)}{R_L}=0\\ \frac{\left(5.6+3I_Z\right)-12}{50}+I_Z+\frac{\left(5.6+3I_Z\right)}{200}=0\\ 0.112+0.06I_Z-0.24+I_Z+0.028+0.015I_Z=0\\ \left(0.06+1+0.015\right)I_Z=0.1\\ I_Z=\frac{0.1}{1.075}\\ I_Z=93.02\text{mA}\end{array}$

Hence, the load voltage will be

  $\begin{array}{l}{V}_Z=V_{ZO}+I_Z{r}_Z\\ V_Z=5.6+\left(93.02\times {10}^{-3}\times 3\right)\\ V_Z=V_L\\ V_L=5.87\text{V}\\ \text{Hence,}\\ I_I=\frac{V_{PS}-V_Z}{R_i}\\ I_I=\frac{12-5.87}{50}\\ I_I=122.41\text{mA}\end{array}$

Thus, the load current can be calculated as

  $\begin{array}{l}{I}_L=I_I-I_Z\\ I_L=\left(122.41-55.81\right)\text{mA}\\ I_L=29.4\text{mA}\end{array}$

Thus, the power dissipated will be equal to

  $\begin{array}{l}{P}_Z=V_Z{I}_Z\\ P_Z=5.87\times 93.02\times {10}^{-3}\\ P_Z=546.02\text{mW}\end{array}$