Chapter 29, Problem 43GP

(a)

To determine

Range of supply voltages.

Answer to Problem 43GP

Range of supply voltages is: $146\text{ V}\le V\le \text{362 V}$

Explanation of Solution

Given:

Resistance, $R=1.80\text{ k}\Omega =1.80\times {10}^3\Omega$

Maximum current, $I=120\text{ mA}=120\times {10}^{-3}\text{A}$

Reverse voltage, $V_{output}=130\text{ V}$

Load resistance, $R_{load}=15\times {10}^3\Omega$

Supply voltage, $V_{\text{supply}}=200\text{ V}$

Formula used:

Ohm’s law is given by:

  $V=IR$

Calculation:

The minimum supply voltage can be calculated as follows:

  $I_{load}=\frac{V_{output}}{R_{load}}=\frac{130\text{ V}}{15\times {10}^3\Omega }=8.67\times {10}^{-3}\text{A}$

  $\begin{array}{l}{V}_{\min }=V_r+V_{output}=\left(I_{load}R\right)+V_{output}\\ V_{\min }=\left(8.67\times {10}^{-3}\text{A}\right)\left(1.80\times {10}^3\Omega \right)+\left(130\text{ V}\right)\\ V_{\min }=146\text{ V}\end{array}$

The maximum supply voltage can be calculated as follows:

  $\begin{array}{l}{I}_{\max }=\left(120\times {10}^{-3}\text{A}\right)+\left(8.67\times {10}^{-3}\text{A}\right)\\ I_{\max }=128.7\times {10}^{-3}\text{A}\end{array}$

  $\begin{array}{l}{V}_{\max }=V_r+V_{output}=\left(I_{\max }R\right)+V_{output}\\ V_{\max }=\left(128.7\times {10}^{-3}\text{A}\right)\left(1.80\times {10}^3\Omega \right)+\left(130\text{ V}\right)\\ V_{\max }=362\text{ V}\end{array}$

Conclusion:

Hence, the range of supply voltages is: $146\text{ V}\le V\le \text{362 V}$

(b)

To determine

Range of load resistance.

Answer to Problem 43GP

Range of load resistance is $3.34\text{ k}\Omega \le R_{load}\le \infty$

Explanation of Solution

Given:

Resistance, $R=1.80\text{ k}\Omega =1.80\times {10}^3\Omega$

Reverse voltage, $V_{output}=130\text{ V}$

Supply voltage, $V_{\text{supply}}=200\text{ V}$

Calculation:

The source voltage is,

  $\begin{array}{l}{V}_s=200\text{V}-130\text{V}\\ V_s=70\text{V}\end{array}$

Load resistance range can be calculated as follows:

  $I_r=\frac{V_s}{R}=\frac{70\text{V}}{1.80\times {10}^3\Omega }=38.9\text{ mA}$

  $R_{load}=\frac{V_{output}}{I_r}=\frac{130\text{V}}{38.9\text{ mA}}=3.34\text{ k}\Omega$

So, the range is given by:

  $3.34\text{ k}\Omega \le R_{load}\le \infty$

Conclusion:

Range of load resistance is $3.34\text{ k}\Omega \le R_{load}\le \infty$