The following diagram shows a Zener diode as a voltage regulator. The Zener diode is rated at V_Z = 5V$V_{Z} = 5V$ and the desired current in load is 5 mA. The unregulated voltage source can supply up to 25V. Considering the Zener diode can withstand four times of the load current, the value of resistor R_S$R_{S}$ (shown in circuit) should be ____ Omega$\Omega$.

### Related Formula De Morgan's Laws: overlineA + overlineB = overlineA cdot B$$\overline{A} + \overline{B} = \overline{A \cdot B}$$ overlineoverlineA = A$$\overline{\overline{A}} = A$$ ### Core Logic Analyzing the circuit diagram:

- Top branch has a NOT gate on A, so P = overlineA$P = \overline{A}$ - Bottom branch has a NOT gate on B, so Q = overlineB$Q = \overline{B}$ - They enter a NOR gate, giving output R = overlineoverlineA + overlineB$R = \overline{\overline{A} + \overline{B}}$ - Finally, R passes through a NOT gate to give S = overlineR$S = \overline{R}$ ### Step 1: Boolean Simplification R = overlineoverlineA + overlineB = overlineoverlineAB = AB$$R = \overline{\overline{A} + \overline{B}} = \overline{\overline{AB}} = AB$$ S = overlineR = overlineAB$$S = \overline{R} = \overline{AB}$$ The expression overlineAB$\overline{AB}$ is the Boolean expression for a NAND gate. ### Pattern Recognition Two NOTs feeding into a NOR equals an AND gate (AB$AB$). Adding a final NOT gate turns the AND into a NAND. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Physics: Semiconductor Electronics

### Related Formula textNOR Gate: overlineA+B, quad textNAND Gate: overlineA cdot B$$\text{NOR Gate: } \overline{A+B}, \quad \text{NAND Gate: } \overline{A \cdot B}$$ ### Core Logic

LED will glow in forward biasing when point P is at higher potential (1) and point Q is at lower potential (0). Testing option (4) [1101]: - Inputs A=1, B=1 through NOR and NAND gates yield P = 1. - Inputs C=1, D=0 through NOR gate yields Q = 0. - Forward bias established, LED glows. ### Pattern Recognition Sees: Logic gates combination with LED forward biasing condition. Shortcut: Check forward bias requirement (P=1, Q=0) for each option combination. Check: Option (4) satisfies the condition. ✓ ### Chapter Mix Class 12 Physics: Semiconductor Electronics

### Related Formula Y = overline(overlineA cdot B) cdot (C + D)$$Y = \overline{(\overline{A \cdot B}) \cdot (C + D)}$$ Applying De Morgan's Law: Y = (A cdot B) + (overlineC + D)$$Y = (A \cdot B) + (\overline{C + D})$$ ### Core Logic Analyzing the expression step by step: 1. Upper branch: A$A$ and B$B$ pass through NAND gate rightarrow overlineA cdot B$\rightarrow \overline{A \cdot B}$. 2. Lower branch: C$C$ and D$D$ pass through OR gate rightarrow (C + D)$\rightarrow (C + D)$. 3. Combined in AND gate followed by NOT gate (NAND equivalent output stage): Y = overline(overlineA cdot B) cdot (C + D) = (A cdot B) + (overlineC + D)$$Y = \overline{(\overline{A \cdot B}) \cdot (C + D)} = (A \cdot B) + (\overline{C + D})$$ Testing conditions for Option (2): - For A=1, B=1, C=0, D=1 implies (1 cdot 1) + (overline0 + 1) = 1 + 0 = 1$A=1, B=1, C=0, D=1 \implies (1 \cdot 1) + (\overline{0 + 1}) = 1 + 0 = 1$. - For A=0, B=0, C=1, D=1 implies (0 cdot 0) + (overline1 + 1) = 0 + 0 = 0$A=0, B=0, C=1, D=1 \implies (0 \cdot 0) + (\overline{1 + 1}) = 0 + 0 = 0$. - For A=1, B=0, C=1, D=0 implies (1 cdot 0) + (overline1 + 0) = 0 + 0 = 0$A=1, B=0, C=1, D=0 \implies (1 \cdot 0) + (\overline{1 + 0}) = 0 + 0 = 0$. - For A=1, B=1, C=1, D=1 implies (1 cdot 1) + (overline1 + 1) = 1 + 0 = 1$A=1, B=1, C=1, D=1 \implies (1 \cdot 1) + (\overline{1 + 1}) = 1 + 0 = 1$.

### Step 1: Final Conclusion Option (2) presents the correct truth table matching the evaluated output values. ### Pattern Recognition Boolean reduction: overlinebarX cdot Y = X + barY$\overline{\bar{X} \cdot Y} = X + \bar{Y}$ where X = A cdot B$X = A \cdot B$ and Y = C+D$Y = C+D$. Hence Y = (A cdot B) + (overlineC+D)$Y = (A \cdot B) + (\overline{C+D})$. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Physics: Semiconductor Electronics

### Related Formula Boolean Algebra rules: A + AB = A$A + AB = A$ A cdot A = A$A \cdot A = A$ NAND Gate: Z = overlineX cdot Y$Z = \overline{X \cdot Y}$ ### Core Logic Gate 1 (Bottom): This is an OR gate with inputs n$n$ and m$m$. Output = n + m$n + m$ Gate 2 (Top): This is a NAND gate. Its inputs are direct n$n$ and the output of the OR gate (n + m$n + m$). Output z = overlinen cdot (n + m)$z = \overline{n \cdot (n + m)}$ ### Step 1: Simplify the Boolean Expression First, expand the inner term: n cdot (n + m) = (n cdot n) + (n cdot m)$$n \cdot (n + m) = (n \cdot n) + (n \cdot m)$$ Since n cdot n = n$n \cdot n = n$, this becomes: n + n cdot m = n(1 + m) = n cdot 1 = n$$n + n \cdot m = n(1 + m) = n \cdot 1 = n$$ So the overall output is: z = overlinen cdot (n + m) = overlinen$$z = \overline{n \cdot (n + m)} = \overline{n}$$ ### Step 2: Construct the Truth Table Since z = overlinen$z = \overline{n}$, the output z$z$ only depends on n$n$ and is its exact inverse. When n=0, m=0 implies z=1$n=0, m=0 \implies z=1$ When n=0, m=1 implies z=1$n=0, m=1 \implies z=1$ When n=1, m=1 implies z=0$n=1, m=1 \implies z=0$ When n=1, m=0 implies z=0$n=1, m=0 \implies z=0$ ### Pattern Recognition Always simplify the Boolean expression using standard absorption laws (like A(A+B) = A$A(A+B) = A$) before manually evaluating every row. It saves significant time. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Physics: Semiconductor Electronics