Question:

Please give me the below content same as it is without any changes in rendered equations ### **Given Data** [ V_Z = 5V, \quad I_Z(\text{min}) = 20mA, \quad V_{in} = 10V \text{ to } 15V, \quad R_L = 125\Omega \text{ to } 250\Omega ] We are asked to find: a. Maximum value of ( R ) such that ( I_Z > 20mA ). b. Min/max power dissipation in ( R ). c. Min/max power dissipation in Zener diode. d. Power dissipation in ( R ) if ( V_{in} ) is accidentally reversed (–10 V to –15 V). --- ## ? Step (a): Find maximum ( R ) for ( I_Z > 20mA ) ### KCL at node across Zener: [ I_R = I_Z + I_L ] [ \frac{V_{in} - V_Z}{R} = I_Z + \frac{V_Z}{R_L} ] We must guarantee ( I_Z \ge 20mA ) **even under worst case** conditions: * ( V_{in} ) **lowest** → 10 V * ( R_L ) **lowest** → 125 Ω (draws most load current) Substitute: [ \frac{10 - 5}{R} = 0.02 + \frac{5}{125} ] [ \frac{5}{R} = 0.02 + 0.04 = 0.06 ] [ R = \frac{5}{0.06} = 83.33\ \Omega ] ✅ **Maximum ( R = 83.3\ \Omega)** to maintain ( I_Z > 20mA ). --- ## ? Step (b): Min/max power dissipation in ( R ) Power in ( R ): [ P_R = I_R^2 R = \frac{(V_{in} - V_Z)^2}{R} ] * ( V_{in,\text{min}} = 10V \Rightarrow P_{R,\text{min}} = \frac{(10-5)^2}{83.33} = 0.3W ) * ( V_{in,\text{max}} = 15V \Rightarrow P_{R,\text{max}} = \frac{(15-5)^2}{83.33} = 1.2W ) ✅ **( P_R = 0.3W \text{ to } 1.2W )** --- ## ? Step (c): Min/max power dissipation in Zener diode [ P_Z = V_Z I_Z ] We’ll find ( I_Z ) for extreme conditions. --- ### **Case 1 — Minimum Zener current (already used to find R)** At ( V_{in}=10V, R_L=125Ω ): [ I_Z = \frac{V_{in} - V_Z}{R} - \frac{V_Z}{R_L} ] [ I_Z = \frac{5}{83.33} - \frac{5}{125} = 0.06 - 0.04 = 0.02A = 20mA ] [ P_{Z,\text{min}} = 5 × 0.02 = 0.1W ] --- ### **Case 2 — Maximum Zener current** Occurs when: * ( V_{in} ) **highest** → 15 V * ( R_L ) **highest** → 250 Ω (smallest load current) [ I_Z = \frac{15 - 5}{83.33} - \frac{5}{250} = 0.12 - 0.02 = 0.10A = 100mA ] [ P_{Z,\text{max}} = 5 × 0.10 = 0.5W ] ✅ **( P_Z = 0.1W \text{ to } 0.5W )** --- ## ? Step (d): If ( V_{in} = -10V \text{ to } -15V ) (reversed polarity) In reverse polarity: * The Zener diode becomes **forward-biased**, dropping about ( V_F ≈ 0.7V ). * The resistor ( R ) now drops ( |V_{in}| - 0.7 ). Hence: [ I_R = \frac{|V_{in}| - 0.7}{R} ] [ P_R = I_R^2 R = \frac{(|V_{in}| - 0.7)^2}{R} ] For ( |V_{in}| = 15V ): [ P_{R,\text{max,rev}} = \frac{(15 - 0.7)^2}{83.33} = \frac{14.3^2}{83.33} = 2.46W ] ✅ **Maximum power dissipation in R (reverse connection) = 2.46 W** --- ## ✅ **Final Answers Summary** | Quantity | Expression | Result | | | | ---------------------------- | -------------------- | ----------------- | -------------- | ---------- | | (a) Max ( R ) | ( R = 5 / 0.06 ) | **83.3 Ω** | | | | (b) ( P_R ) range | ( (V_{in}-5)^2 / R ) | **0.3 W – 1.2 W** | | | | (c) ( P_Z ) range | ( 5 × I_Z ) | **0.1 W – 0.5 W** | | | | (d) ( P_{R,\text{max,rev}} ) | ( ( | V_{in} | - 0.7)^2 / R ) | **2.46 W** | ---