Let's analyze and solve the problem step-by-step. --- # **Given Data and Assumptions** - The circuit uses a **Zener diode** as a voltage regulator. - **V\(_Z\)** (Zener voltage): 5. V - **V\(_in\)** varies from 10 V to 15 V - **R\(_{\text{load}}\)** (load resistor): unknown, but we will assume a typical value or analyze generally - **I\(_L\)** (load current): > 20 mA - **R** (series resistor): unknown, to be calculated - **I\(_Z\)** (Zener current): must be > 20 mA - **V\(_\text{Z}\)** is maintained at 5 V - **In** varies from 10 V to 15 V - **R\(_\text{load}\)** varies from 125 Ω to 250 Ω - The goal is to find: - **Maximum R** - **Minimum/Maximum power dissipation in R** - **Minimum/Maximum power dissipation in Zener diode** - **Maximum power dissipation when hooked backward** --- # **Step 1: Determine the series resistor \( R \)** ### **a. Basic relationship** \[ V_{in} = I_{R} R + V_{Z} \] \[ I_{R} = I_{Z} + I_{load} \] Given: - \( V_{Z} = 5\,V \) - \( V_{in} \) varies from 10 V to 15 V - \( I_{load} = \frac{V_{Z}}{R_{load}} \) --- ### **b. Find \( R \) for the minimum input voltage** At the **minimum input voltage (10 V)**, the Zener diode just maintains regulation with the minimum current: \[ I_{Z,\text{min}} = 20\,mA \] The load current varies with load resistance: \[ I_{load} = \frac{V_{Z}}{R_{load}} \] - For \( R_{load} = 125\,\Omega \): \[ I_{load} = \frac{5\,V}{125\,\Omega} = .04\,A = 40\,mA \] - For \( R_{load} = 250\,\Omega \): \[ I_{load} = \frac{5\,V}{250\,\Omega} = .02\,A = 20\,mA \] --- ### **c. Calculate R for the worst case (minimum \( V_{in} \), maximum load current)** The **worst case** for the resistor occurs when: - \( V_{in} = 10\,V \) - \( R_{load} = 125\,\Omega \) (maximum load current) Total current through R: \[ I_{R} = I_{Z} + I_{load} \geq 20\,mA + 40\,mA = 60\,mA \] But since \( I_{Z} \) should be at least 20 mA, and the load current is 40 mA, total: \[ I_{R} = 60\,mA \] Now, apply the voltage: \[ V_{in} = I_{R} R + V_{Z} \] \[ 10\,V = .06\,A \times R + 5\,V \] \[ .06\,A \times R = 5\,V \] \[ R = \frac{5\,V}{.06\,A} \approx 83.33\,\Omega \] --- ### **d. Calculate R for the worst case (maximum \( V_{in} \), minimum load current)** - \( V_{in} = 15\,V \) - \( R_{load} = 250\,\Omega \): \[ I_{load} = \frac{5\,V}{250\,\Omega} = 20\,mA \] Total current: \[ I_{R} = I_{Z} + I_{load} \geq 20\,mA + 20\,mA = 40\,mA \] Voltage: \[ 15\,V = .04\,A \times R + 5\,V \] \[ .04\,A \times R = 10\,V \] \[ R = \frac{10\,V}{.04\,A} = 250\,\Omega \] --- # **Answer:** - To satisfy all conditions, choose \( R \) between **83.33 Ω** and **250 Ω**. --- # **Step 2: Power Dissipation Calculations** --- ## **a. Power dissipated in R** - **Maximum power** occurs at the highest current: \[ P_{R,\text{max}} = I_{R,\text{max}}^{2} \times R \] From previous: - At \( V_{in} = 10\,V \), \( R = 83.33\,\Omega \), \( I_{R} = .06\,A \): \[ P_{R} = (.06\,A)^2 \times 83.33\,\Omega \approx .0036 \times 83.33 \approx .3\,W \] - At \( V_{in} = 15\,V \), \( R = 250\,\Omega \), \( I_{R} = .04\,A \): \[ P_{R} = (.04)^2 \times 250 = .0016 \times 250 = .4\,W \] **Maximum power in R: approximately .4 W.** --- ## **b. Power in the Zener diode** - The **Zener current** \( I_{Z} \) varies: ### **At minimum \( V_{in} \) and maximum load:** \[ I_{Z} = I_{R} - I_{load} = 60\,mA - 40\,mA = 20\,mA \] ### **At maximum \( V_{in} \) and minimum load:** \[ I_{Z} = 40\,mA - 20\,mA = 20\,mA \] So, **\( I_{Z} \)** remains **at least 20 mA** in all cases, satisfying the regulation requirement. - Power dissipation: \[ P_{Z} = V_{Z} \times I_{Z} \] \[ P_{Z} = 5\,V \times 20\,mA = .1\,W \] **Maximum power in Zener diode:** approximately .1 W. --- ## **c. Power dissipation in reverse hookup** - When hooked backwards, the Zener diode acts as a regular diode with a forward voltage drop (~.7 V). - The maximum current could be: \[ I_{forward} = \frac{V_{in} - V_{F}}{R} \] - For \( V_{in} = 15\,V \), \( R = 83.33\,\Omega \): \[ I_{forward} = \frac{15\,V - .7\,V}{83.33\,\Omega} \approx \frac{14.3\,V}{83.33\,\Omega} \approx .171\,A \] - Power: \[ P_{forward} = V_{F} \times I_{forward} \approx .7\,V \times .171\,A \approx .12\,W \] **Maximum power dissipation: approximately .12 W.** --- # **Summary of Results** | Parameter | Value/Range | |--------------|--------------| | **Series Resistance \( R \)** | 83.33 Ω to 250 Ω | | **Max power in R** | ~.4 W | | **Max power in Zener diode** | ~.1 W | | **Max power in reverse hookup** | ~.12 W | --- Feel free to ask for further clarification or specific detailed calculations!

Here’s a step-by-step solution using \( R = 83.33 \, \Omega \) only. --- # **Given Data and Assumptions** - **Zener Voltage \( V_Z \)**: 5 V - **Input Voltage \( V_{in} \)**: 10 V to 15 V - **Series Resistor \( R \)**: 83.33 Ω - **Zener Current \( I_Z \)**: must be > 20 mA - **Load Resistor \( R_{load} \)**: varies from 125 Ω to 250 Ω --- # **Step 1: Calculate Load Current \( I_{load} \)** ### **a. Determine Load Current for Different Resistor Values** 1. **For \( R_{load} = 125 \, \Omega \)**: \[ I_{load} = \frac{V_Z}{R_{load}} = \frac{5\,V}{125\,\Omega} = 0.04\,A = 40\,mA \] 2. **For \( R_{load} = 250 \, \Omega \)**: \[ I_{load} = \frac{V_Z}{R_{load}} = \frac{5\,V}{250\,\Omega} = 0.02\,A = 20\,mA \] --- # **Step 2: Calculate Total Current \( I_R \)** ### **a. Find Total Current for Minimum \( V_{in} (10 \, V) \)** Using \( V_{in} = 10\,V \): \[ I_R = I_Z + I_{load} \] 1. **For \( R_{load} = 125 \, \Omega \)**: \[ I_R = 20\,mA + 40\,mA = 60\,mA \] 2. **For \( R_{load} = 250 \, \Omega \)**: \[ I_R = 20\,mA + 20\,mA = 40\,mA \] --- # **Step 3: Verify Voltage Across Resistor \( R \)** ### **a. Voltage Equation** Using the voltage equation: \[ V_{in} = I_R R + V_Z \] 1. **For \( R_{load} = 125 \, \Omega \)**: \[ 10\,V = (0.06\,A)(83.33\,\Omega) + 5\,V \] \[ 10\,V = 5\,V + 5\,V \quad \text{(valid)} \] 2. **For \( R_{load} = 250 \, \Omega \)**: \[ 10\,V = (0.04\,A)(83.33\,\Omega) + 5\,V \] \[ 10\,V = 5\,V + 3.33\,V \quad \text{(valid)} \] --- # **Step 4: Power Dissipation in Resistor \( R \)** ### **a. Calculate Power Dissipation** Using the formula: \[ P_R = I_R^2 R \] 1. **For \( R_{load} = 125 \, \Omega \)**: \[ P_R = (0.06\,A)^2 \cdot 83.33\,\Omega = 0.0036 \cdot 83.33 \approx 0.3\,W \] 2. **For \( R_{load} = 250 \, \Omega \)**: \[ P_R = (0.04\,A)^2 \cdot 83.33\,\Omega = 0.0016 \cdot 83.33 \approx 0.133\,W \] ### **b. Maximum Power Dissipation** - **Maximum Power in Resistor R** is approximately **0.3 W** when \( R_{load} = 125 \, \Omega \). --- # **Step 5: Power Dissipation in the Zener Diode** ### **a. Calculate Zener Current** From previous calculations, the Zener current when \( R_{load} = 125 \, \Omega \) and \( I_R = 60 \, mA \): \[ I_Z = I_R - I_{load} = 60\,mA - 40\,mA = 20\,mA \] ### **b. Power Dissipation in Zener Diode** Using the formula: \[ P_Z = V_Z \cdot I_Z \] \[ P_Z = 5\,V \cdot 0.02\,A = 0.1\,W \] ### **c. Maximum Power in Zener Diode** - **Maximum Power in Zener Diode** is approximately **0.1 W**. --- # **Summary of Results for \( R = 83.33 \, \Omega \)** | Parameter | Value | |--------------------------------|----------------| | **Series Resistance \( R \)** | 83.33 Ω | | **Max power in Resistor R** | ~0.3 W | | **Max power in Zener diode** | ~0.1 W | --- Feel free to ask if you have any further questions or need additional calculations!