Chapter 2, Problem D2.25P

Design a voltage regulator circuit such as shown in Figure P2.21 so that $V_L=7.5\text{V}$ . The Zener diode voltage is $V_Z=7.5\text{V}$ at $I_Z=10\text{mA}$ . The incremental diode resistance is $r_z=12\Omega$ . The nominal supply voltage is $V_I=12\text{V}$ and the nominal load resistance is $R_L=1\text{k}\Omega$ . (a) Determine $R_i$ . (b) If $V_I$ varies by ± 10 percent, calculate the source regulation. What is the variation in output voltage? (c) If $R_L$ varies over the range of $1\text{k}\Omega \le R_L\le \infty$ , what is the variation in output voltage? Determine the load regulation.

To determine

The value of the input resistance, R_i.

Answer to Problem D2.25P

The value of resistance $R_i$ is $257\Omega$

Explanation of Solution

Given:

In the circuit,

  $\begin{array}{l}{V}_L=7.5\text{V}\\ \text{Zener diode:}\\ {\text{voltage, V}}_Z=7.5\text{V}\\ {\text{current, I}}_Z=10\text{mA}\end{array}$

Evaluating the $R_i$ :

  $R_i=\frac{V_I-V_Z}{I_I}\mathrm{..........}\left(1\right)$

Evaluating the value of $I_I$ :

  $I_I=I_Z+I_L\mathrm{..............}\left(2\right)$

Evaluating the value of $I_L$ :

  $I_L=\frac{V_L}{R_L}$

Substituting $\text{1kΩ}\text{for}{\text{R}}_{\text{L}}\text{and}\text{7}\text{.5V}\text{for}{\text{V}}_{\text{L}}$ :

  $\begin{array}{c}{\text{I}}_{\text{L}}\text{=}\frac{\text{7}\text{.5}}{\text{1k}}\\ \text{=}\frac{\text{75}}{{\text{1×10}}^{\text{3}}}\\ \text{=7}\text{.5mA}\\ {\text{I}}_{\text{l}}\text{=10+7}\text{.5}\\ \text{=17}\text{.5mA}\end{array}$

Substituting $\text{7}\text{.5mA}\text{for}{\text{I}}_{\text{L}}\text{and}\text{10mA}\text{for}{\text{I}}_{\text{Z}}$ in equation 2:

  $\begin{array}{c}{I}_l=10+7.5\\ =17.5\text{mA}\end{array}$

Substituting $\text{17}\text{.5mA}\text{for}{\text{I}}_{\text{I}}\text{,12V}\text{for}{\text{V}}_{\text{I}}\text{and}\text{7}\text{.5V}\text{for}{\text{V}}_{\text{z}}$ :

  $\begin{array}{c}{R}_i=\frac{12-7.5}{17.5\times {10}^{-3}}\\ =\frac{4.5}{17.5\times {10}^{-3}}\\ =0.257\times {10}^3\Omega \\ =257\Omega \end{array}$

Therefore, the value of resistance $R_i$ is $257\Omega$

To determine

The source regulation and the variation in the output voltage.

Answer to Problem D2.25P

The source regulation is 4.44% and the variation in the output voltage is $7.64\text{V}\le V_L\le 7.75\text{V}$ .

Explanation of Solution

Given:

In the circuit,

  $\begin{array}{l}{V}_L=7.5\text{V}\\ \text{Zener diode:}\\ {\text{voltage, V}}_Z=7.5\text{V}\\ {\text{current, I}}_Z=10\text{mA}\end{array}$

The voltage $V_I$ varies by $\pm 10$ percent.

Evaluating $V_{I,\min }$ :

  $\begin{array}{c}{V}_{I,\min }=12-\frac{12\times 10}{100}\\ =12-12\\ =10.8\text{V}\end{array}$

Evaluating the $V_{I,\max }$ :

  $\begin{array}{c}{V}_{I,\max }=12+\frac{12\times 10}{100}\\ =12+12\\ =13.2\text{V}\end{array}$

For no load condition evaluating the value $V_{L,\max }$ :

  $V_{L,\min }=V_Z+r_z{I}_z\mathrm{..........}\left(3\right)$

Evaluating the value of $I_z$ for $V_{I,\min }$ =10.8V.

  $\begin{array}{c}{I}_Z=\frac{V_{I,\min }-V_Z}{R_i+r_z}\\ =\frac{10.8-7.5}{257+12}\\ =0.01226A\\ =12.26\text{mA}\end{array}$

Substituting $0.01226A$ for $I_Z$ , $\text{7}\text{.5V}\text{for}{\text{V}}_{\text{z}}\text{and}\text{12Ω}\text{for}{\text{r}}_{\text{Z}}$ in equation (3):

  $\begin{array}{c}{V}_{L,\min }=7.5+\left(12\right)\left(0.01226\right)\\ =7.647\text{V}\end{array}$

Evaluating the value of $V_{L,\max }$ :

  $V_{L,\min }=V_Z+r_Z{I}_Z\mathrm{.................}\left(4\right)$

Evaluating the value of ${\text{I}}_{\text{Z}}\text{for}{\text{V}}_{\text{I,max}}\text{=13}\text{.2V}$

  $\begin{array}{c}{I}_Z=\frac{V_{I,\max }-V_Z}{R_i+r_z}\\ =\frac{13.2-7.5}{257+12}\\ =0.0212\text{A}\end{array}$

Substituting $0.0212\text{A}$ for $I_Z$ , $\text{7}\text{.5V}\text{for}{\text{V}}_{\text{z}}\text{and}\text{12Ω}\text{for}{\text{r}}_{\text{Z}}$ in equation (4):

  $\begin{array}{c}{V}_{L,\max }=7.5+\left(12\right)\left(0.0212\right)\\ =7.754\text{V}\end{array}$

Evaluating the source regulation:

  $\begin{array}{c}S.R.=\frac{\Delta v_L}{\Delta v_I}\times 100\%\\ =\frac{v_{L,\max }-v_{L\min }}{v_{I,\max }-v_{I,\min }}\times 100\%\\ =\frac{7.754-7.647}{13.2-10.8}\times 100\%\\ =4.44\%\end{array}$

Therefore, the source regulation is 4.44% and the variation in the output voltage is

  $7.64\text{V}\le V_L\le 7.75\text{V}$

To determine

The variation in the output voltage and the load regulation.

Answer to Problem D2.25P

The load regulation is $1.18\%$ and the variation in the output voltage is

  $7.6V\le V_L\le 7.7V$ .

Explanation of Solution

Given:

In the circuit,

  $\begin{array}{l}{V}_L=7.5\text{V}\\ \text{Zener diode:}\\ {\text{voltage, V}}_Z=7.5\text{V}\\ {\text{current, I}}_Z=10\text{mA}\end{array}$

The voltage $V_I$ varies by $\pm 10$ percent.

The load resistance $R_L$ varies from $1\text{k}\Omega \le R_L\le \infty$

For $I_L=0$ , the value of the load resistance $R_L=\infty$

Evaluating the value of ${\text{I}}_{\text{z}}\text{for}{\text{V}}_{\text{I}}\text{=12V}$

  $\begin{array}{c}{I}_z=\frac{12-7.5}{257+12}\\ =0.01673\text{A}\end{array}$

Evaluating the value of $v_{L,noload}$ :

  $\begin{array}{c}{v}_{L,noload}=V_Z+I_Z{r}_Z\\ =7.5+\left(16.73\times {10}^{-3}\right)12\\ =7.\text{7V}\end{array}$

When, $R_L=1\text{kΩ}$

Applying Kirchhoff s current law to determine the current:

  $\begin{array}{c}{I}_I=I_Z+I_L\\ I_Z=I_I-I_L\\ I_Z=\frac{V_1-\left[V_Z+I_Z{r}_Z\right]}{R_1}-I_L\\ =\frac{12-\left[7.5+I_Z\left(12\right)\right]}{257}-7.5\times {10}^{-3}\\ I_Z=\frac{4.5-12I_Z}{257}-7.5\times {10}^{-3}\\ I_Z=0.0175-0.0467I_Z-7.5\times {10}^{-3}\\ I_Z=\frac{0.01}{1.0467}\\ =9.55\text{mA}\end{array}$

Evaluating the value of $v_{L,fullload}$ :

  $\begin{array}{c}{v}_{L,fullload}=V_Z+I_Z{r}_Z\\ =7.5+\left(9.55\times {10}^{-3}\right)12\\ =7.61\text{V}\end{array}$

Evaluating the load regulation:

  $\begin{array}{c}Load\mathrm{Re}gulation=\frac{v_{L,noload}-v_{L,fullload}}{v_{L,fullload}}\times 100\%\\ =\frac{7.7-7.61}{7.61}\times 100\%\\ =1.18\%\end{array}$

Therefore, the load regulation is $1.18\%$ and the variation in the output voltage is

  $7.6V\le V_L\le 7.7V$ .