Let's analyze and solve the problem step by step. --- ## **Given Data and Assumptions** - The Zener diode maintains a constant Zener voltage \( V_Z \). - \( V_{in} \) varies from 10V to 15V. - \( R_{load} \) varies from 125Ω to 250Ω. - The Zener voltage \( V_Z \) is 5.V. - \( I_{Z} \) (Zener current) must be greater than 20mA for regulation. - The maximum power dissipation in R: \( P_{R} = I_{R}^2 R \). - The supply voltage \( V_{in} \) can go down to 10V and up to 15V. - \( R_{load} \) can be from 125Ω to 250Ω. --- ## **a) Find the maximum value of R such that \( I_{Z} > 20\text{mA} \)** ### Step 1: Determine the Zener current \( I_Z \) The total current \( I_R \) splits into the Zener diode current \( I_Z \) and load current \( I_{load} \): \[ I_R = I_Z + I_{load} \] Where: \[ I_{load} = \frac{V_{out}}{R_{load}} = \frac{V_Z}{R_{load}} \] ### Step 2: Find \( R_{max} \) for the minimum \( I_Z \) To maximize \( R \), minimize the Zener current \( I_Z \). The minimum \( I_Z \) should be 20mA. At the **worst-case scenario** for maximum \( R \), the supply voltage is at its **minimum** (i.e., 10V), and the load resistance is at its maximum (250Ω), because: - Lower \( V_{in} \) reduces \( I_R \). - Larger \( R_{load} \) reduces \( I_{load} \). ### Step 3: Calculate \( I_{load} \) \[ I_{load} = \frac{V_Z}{R_{load}} = \frac{5V}{250\Omega} = .02A = 20mA \] ### Step 4: Find \( I_R \) The supply voltage \( V_{in} = 10V \), and the Zener diode voltage \( V_Z = 5V \), so: \[ V_{R} = V_{in} - V_Z = 10V - 5V = 5V \] Total resistor current: \[ I_R = \frac{V_{R}}{R} = \frac{5V}{R} \] Since: \[ I_R = I_Z + I_{load} \] and we want \( I_Z \geq 20mA \), with \( I_{load} = 20mA \): \[ I_R = I_Z + I_{load} = 20mA + 20mA = 40mA \] ### Step 5: Calculate \( R_{max} \) \[ R_{max} = \frac{V_{R}}{I_R} = \frac{5V}{40mA} = \frac{5V}{.04A} = 125 \Omega \] --- ## **Answer for part (a):** \[ \boxed{ R_{max} = 125\, \Omega } \] --- ## **b) Find the minimum power dissipation in R** Power dissipation: \[ P_{R} = I_R^2 R \] Using the **maximum \( R \)** from part (a) (125Ω), and the **minimum \( V_{in} \)** (10V): \[ I_R = \frac{V_{in} - V_Z}{R} = \frac{10V - 5V}{125\Omega} = \frac{5V}{125\Omega} = .04A = 40mA \] Power dissipated: \[ P_{R} = (.04A)^2 \times 125\Omega = .0016A^2 \times 125\Omega = .2\, \text{W} \] --- ## **Answer for part (b):** \[ \boxed{ P_{R} = .2\, \text{W} } \] --- ## **c) Find the minimum and maximum power dissipation in the Zener diode** ### Step 1: Calculate \( I_Z \) - For **minimum \( I_Z \)**: \( I_Z = 20\, \text{mA} \) (given). - For **maximum \( I_Z \)**: when \( V_{in} \) is at maximum (15V) and load at minimum (125Ω): \[ V_{R} = V_{in} - V_Z = 15V - 5V = 10V \] Current through resistor: \[ I_R = \frac{10V}{R} \] At \( R = 125\, \Omega \): \[ I_R = \frac{10V}{125\Omega} = .08A = 80mA \] The load current: \[ I_{load} = \frac{V_Z}{R_{load}} = \frac{5V}{125\Omega} = 40mA \] Thus, the Zener current: \[ I_Z = I_R - I_{load} = 80mA - 40mA = 40mA \] ### Step 2: Power dissipation in Zener diode \[ P_Z = V_Z \times I_Z \] - **Minimum \( P_Z \)** (at \( I_Z = 20\, \text{mA} \)): \[ P_{Z, min} = 5V \times 20mA = 5V \times .02A = .1\, \text{W} \] - **Maximum \( P_Z \)** (at \( I_Z = 40\, \text{mA} \)): \[ P_{Z, max} = 5V \times 40mA = .2\, \text{W} \] --- ## **Final answers:** | **Part** | **Result** | |------------|--------------| | **a)** | \( R_{max} = 125\, \Omega \) | | **b)** | \( P_{R} = .2\, \text{W} \) | | **c)** | \( P_{Z, min} = .1\, \text{W} \), \( P_{Z, max} = .2\, \text{W} \) | Let me know if you'd like further clarification!

## Solution to the Voltage Regulator Problem ### **Given Data** - Zener voltage, \( V_Z = 5.0\, V \) - Input voltage range, \( V_{in} = 10\, V \) to \( 15\, V \) - Load resistance range, \( R_{load} = 125\, \Omega \) to \( 2500\, \Omega \) - Minimum Zener current, \( I_Z > 20\, mA \) --- ## **a) Maximum Value of R to Maintain \( I_Z > 20\, mA \)** ### Step 1: Calculate Load Current \( I_{load} \) \[ I_{load} = \frac{V_Z}{R_{load}} \] #### For Maximum Load Resistance \( R_{load} = 2500\, \Omega \): \[ I_{load} = \frac{5\, V}{2500\, \Omega} = 0.002\, A = 2\, mA \] ### Step 2: Total Current \( I_R \) \[ I_R = I_Z + I_{load} \] To ensure \( I_Z \geq 20\, mA \): \[ I_R = 20\, mA + 2\, mA = 22\, mA = 0.022\, A \] ### Step 3: Calculate Maximum Resistance \( R_{max} \) Using the minimum input voltage \( V_{in} = 10\, V \): \[ V_R = V_{in} - V_Z = 10\, V - 5\, V = 5\, V \] Thus, \[ R_{max} = \frac{V_R}{I_R} = \frac{5\, V}{0.022\, A} \approx 227.27\, \Omega \] ### **Final Answer for Part (a)** \[ \boxed{R_{max} \approx 227.27\, \Omega} \] --- ## **b) Minimum and Maximum Power Dissipation in R** ### Power Dissipation Formula \[ P_R = I_R^2 R \] ### For Minimum Power Dissipation Using \( R = 125\, \Omega \) and \( I_R = 0.022\, A \): \[ P_{R, min} = (0.022\, A)^2 \times 125\, \Omega = 0.000484 \times 125 \approx 0.0605\, W \] ### For Maximum Power Dissipation Using \( R = 227.27\, \Omega \) and \( I_R = 0.022\, A \): \[ P_{R, max} = (0.022\, A)^2 \times 227.27\, \Omega = 0.000484 \times 227.27 \approx 0.1105\, W \] ### **Final Answers for Part (b)** \[ \boxed{P_{R, min} \approx 0.0605\, W}, \quad \boxed{P_{R, max} \approx 0.1105\, W} \] --- ## **c) Minimum and Maximum Power Dissipation in the Zener Diode** ### Step 1: Calculate Zener Current \( I_Z \) - For **minimum \( I_Z = 20\, mA \)**: \[ P_{Z, min} = V_Z \times I_Z = 5\, V \times 0.020\, A = 0.1\, W \] - For **maximum \( I_Z \)** when \( V_{in} = 15\, V \) and \( R_{load} = 125\, \Omega \): #### Step 1a: Calculate \( I_{load} \) \[ I_{load} = \frac{5\, V}{125\, \Omega} = 0.04\, A = 40\, mA \] #### Step 1b: Calculate Total Current \( I_R \) \[ I_R = \frac{10\, V}{R} \text{ (use \( R = 125\, \Omega \))} \] \[ I_R = 0.08\, A = 80\, mA \] #### Step 1c: Calculate Zener Current \( I_Z \) \[ I_Z = I_R - I_{load} = 80\, mA - 40\, mA = 40\, mA \] #### Step 1d: Calculate Power Dissipation in Zener Diode \[ P_{Z, max} = V_Z \times I_Z = 5\, V \times 0.040\, A = 0.2\, W \] ### **Final Answers for Part (c)** \[ \boxed{P_{Z, min} = 0.1\, W}, \quad \boxed{P_{Z, max} = 0.2\, W} \] --- ## **d) Maximum Power Dissipation in R with Reverse Voltage** ### Step 1: Calculate Total Current with Negative Input If \( V_{in} = -10\, V \): \[ V_R = V_{in} - V_Z = -10\, V - 5\, V = -15\, V \] ### Step 2: Calculate Current \( I_R \) Using \( R = 125\, \Omega \): \[ I_R = \frac{-15\, V}{125\, \Omega} = -0.12\, A = 120\, mA \] ### Step 3: Calculate Power Dissipation in R \[ P_R = I_R^2 R = (0.12\, A)^2 \times 125\, \Omega = 0.0144 \times 125 = 1.8\, W \] ### **Final Answer for Part (d)** \[ \boxed{P_R \approx 1.8\, W} \] --- This concludes the step-by-step solution for all parts of the voltage regulator problem.