In the digital circuit shown in the figure, for the given inputs the P and Q values are :

### Related Formula For a parallel circuit regulator using a Zener diode: I = I_z + I_1$I = I_z + I_1$ where, I$I$ = total current through the series resistor I_z$I_z$ = current through the Zener diode I_1$I_1$ = current through the load resistor ### Core Logic Given that the breakdown voltage of the Zener diode is: V_z = 3.0 mathrm~V$$V_z = 3.0 \mathrm{~V}$$ Let potential at junction B$B$ and D$D$ be 0 mathrm~V$0 \mathrm{~V}$. Then, the potential at the Zener cathode A$A$ and load node C$C$ is stabilized at: V_A = V_C = 3.0 mathrm~V$$V_A = V_C = 3.0 \mathrm{~V}$$ Potential at the source input E$E$ is 10 mathrm~V$10 \mathrm{~V}$. ### Step 1: Calculate Total Current The potential drop across the series resistor (1 mathrm~kOmega$1 \mathrm{~k}\Omega$) is: Delta V = 10 mathrm~V - 3 mathrm~V = 7 mathrm~V$$\Delta V = 10 \mathrm{~V} - 3 \mathrm{~V} = 7 \mathrm{~V}$$ Hence, the total line current I$I$ is: I = frac7 mathrm~V1000 \ Omega = 7 times 10^-3 mathrm~A = 7 mathrm~mA$$I = \frac{7 \mathrm{~V}}{1000 \ \Omega} = 7 \times 10^{-3} \mathrm{~A} = 7 \mathrm{~mA}$$

### Step 2: Calculate Load Current The voltage across the load resistor (2 mathrm~kOmega$2 \mathrm{~k}\Omega$) is equal to V_z = 3 mathrm~V$V_z = 3 \mathrm{~V}$. Thus, the load current I_1$I_1$ is: I_1 = frac3 mathrm~V2000 \ Omega = 1.5 times 10^-3 mathrm~A = 1.5 mathrm~mA$$I_1 = \frac{3 \mathrm{~V}}{2000 \ \Omega} = 1.5 \times 10^{-3} \mathrm{~A} = 1.5 \mathrm{~mA}$$ ### Step 3: Calculate Zener Current By applying Kirchhoff's Current Law at node A$A$: I_z = I - I_1 = 7 mathrm~mA - 1.5 mathrm~mA = 5.5 mathrm~mA$$I_z = I - I_1 = 7 \mathrm{~mA} - 1.5 \mathrm{~mA} = 5.5 \mathrm{~mA}$$ Therefore, the current through the Zener diode is 5.5 mathrm~mA$5.5 \mathrm{~mA}$. ### Pattern Recognition Whenever you see a Zener diode in breakdown connected parallel to a load, always fix the node potential at the breakdown voltage. Work backwards from the supply potential to find the total current, calculate the load current using Ohm's law, and subtract to find the Zener current. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Physics: Semiconductor Electronics: Materials, Devices and Simple Circuits

### Core Logic

The circuit contains a Germanium diode (D_1$D_1$) and a Silicon diode (D_2$D_2$) in parallel. The barrier potential for Germanium is 0.3 mathrm~V$0.3 \mathrm{~V}$ and for Silicon is 0.7 mathrm~V$0.7 \mathrm{~V}$. Since they are in parallel, the diode with the lower barrier potential (Ge) will turn on first. Once the Germanium diode starts conducting, it clamps the voltage across the parallel combination to 0.3 mathrm~V$0.3 \mathrm{~V}$, preventing the Silicon diode from ever turning on. Thus, only D_1$D_1$ conducts. ### Step 1: Calculate Total Current The total voltage in the loop after considering the Ge diode's drop is: V_textnet = 15 mathrm~V - 0.3 mathrm~V = 14.7 mathrm~V$$V_{\text{net}} = 15 \mathrm{~V} - 0.3 \mathrm{~V} = 14.7 \mathrm{~V}$$ Total resistance in the circuit: R_texttotal = 1.5 mathrm~kOmega + 2.5 mathrm~kOmega = 4.0 mathrm~kOmega$$R_{\text{total}} = 1.5 \mathrm{~k}\Omega + 2.5 \mathrm{~k}\Omega = 4.0 \mathrm{~k}\Omega$$ Current i$i$: i = frac14.74 mathrm~mA$$i = \frac{14.7}{4} \mathrm{~mA}$$ *(Note: Some sources approximate 15 - 1 = 14$15 - 1 = 14$ if considering ideal diode drops or a misprint in standard problem sets where V_textdrop = 1mathrmV$V_{\text{drop}} = 1\mathrm{V}$ total across the network, but strictly for Ge V_b = 0.3mathrmV$V_b = 0.3\mathrm{V}$, let's check standard solution behavior... Wait, the standard PDF solution explicitly uses 15 mathrm~V - 1 mathrm~V = 14 mathrm~V$15 \mathrm{~V} - 1 \mathrm{~V} = 14 \mathrm{~V}$? No, wait. Let's look at the source PDF: i = 14 / 4 = 3.5 mathrm~mA$i = 14 / 4 = 3.5 \mathrm{~mA}$. This implies a total diode drop of 1 mathrm~V$1 \mathrm{~V}$ was assumed in the PDF's logic, which might be an error in the source, but we follow it strictly.)* Wait, if the source states i = 14/4 = 3.5mathrmmA$i = 14/4 = 3.5\mathrm{mA}$, it means the voltage drop across the diode was taken as 1mathrmV$1\mathrm{V}$ (which is unusual, maybe 15V$15V$ battery has internal resistance or it's a zener?). Looking at the PDF: `i = 14 / 4 = 3.5 mA`. I will transcribe the PDF exactly. ### Step 2: Voltage Across Load V_L = i times R_L = 3.5 mathrm~mA times 2.5 mathrm~kOmega$$V_L = i \times R_L = 3.5 \mathrm{~mA} \times 2.5 \mathrm{~k}\Omega$$ V_L = 8.75 mathrm~V$$V_L = 8.75 \mathrm{~V}$$ ### Pattern Recognition When Si and Ge diodes are in parallel, the Ge diode (0.3V) dominates and turns on, shutting off the Si diode (0.7V). Although physically 15 - 0.3 = 14.7mathrmV$15 - 0.3 = 14.7\mathrm{V}$, the provided solution implies an effective 1mathrmV$1\mathrm{V}$ drop is used to reach the 14mathrmV$14\mathrm{V}$ net. Follow the specific provided calculation. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Physics: Semiconductor Electronics: Materials, Devices and Simple Circuits

### Related Formula I_texttotal = I_z + I_L$$I_{\text{total}} = I_z + I_L$$ V_textload = V_z quad (textif in breakdown)$$V_{\text{load}} = V_z \quad (\text{if in breakdown})$$ ### Core Logic

The Zener is in the breakdown region because the open-circuit voltage across it without the Zener (20 times frac500700 = 14.28mathrmV$20 \times \frac{500}{700} = 14.28\mathrm{V}$) is greater than V_z = 10mathrmV$V_z = 10\mathrm{V}$. Therefore, it locks the voltage across the load resistor (500 \,Omega$500 \,\Omega$) at 10 mathrmV$10 \mathrm{V}$. ### Step 1: Calculate Currents Current across the load resistor (500 \,Omega$500 \,\Omega$): I_3 = fracV_zR_L = frac10500 = frac150 mathrm~A = 20 mathrm~mA$$I_3 = \frac{V_z}{R_L} = \frac{10}{500} = \frac{1}{50} \mathrm{~A} = 20 \mathrm{~mA}$$ Voltage across the series resistor (200 \,Omega$200 \,\Omega$) is 20 - 10 = 10 mathrmV$20 - 10 = 10 \mathrm{V}$. Current through the series resistor: I_1 = fracDelta VR_s = frac10200 = frac120 mathrm~A = 50 mathrm~mA$$I_1 = \frac{\Delta V}{R_s} = \frac{10}{200} = \frac{1}{20} \mathrm{~A} = 50 \mathrm{~mA}$$ ### Step 2: Extract Zener Current Applying Kirchhoff's Current Law (KCL) at the junction: I_1 = I_2 + I_3$I_1 = I_2 + I_3$ I_2 = I_1 - I_3$I_2 = I_1 - I_3$ I_2 = 50 mathrm~mA - 20 mathrm~mA = 30 mathrm~mA$$I_2 = 50 \mathrm{~mA} - 20 \mathrm{~mA} = 30 \mathrm{~mA}$$ ### Pattern Recognition Always perform the unregulated voltage check first. If V_in (R_L / (R_L + R_S)) > V_Z$V_{in} (R_L / (R_L + R_S)) > V_Z$, the diode behaves like a constant V_Z$V_Z$ battery. Apply nodal analysis. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Physics: Semiconductor Electronics

### Related Formula Boolean Algebra expressions for logic gates: NOT: overlineA$\overline{A}$ OR: A + B$A + B$ NOR: overlineA + B$\overline{A + B}$ ### Core Logic Analyze the paths from inputs A and B to the final output Y.

### Step 1: Intermediate Signals Top OR gate inputs: A$A$ directly, and B$B$ inverted (overlineB$\overline{B}$). Top OR gate output: A + overlineB$A + \overline{B}$ Bottom OR gate inputs: A$A$ inverted (overlineA$\overline{A}$), and B$B$ directly. Bottom OR gate output: overlineA + B$\overline{A} + B$ ### Step 2: Final Gate Evaluation The final gate is a NOR gate taking the two intermediate outputs as its inputs. Y = overline(A + overlineB) + (overlineA + B)$$Y = \overline{(A + \overline{B}) + (\overline{A} + B)}$$ Notice that the inner sum simplifies cleanly: (A + overlineA) + (B + overlineB)$$(A + \overline{A}) + (B + \overline{B})$$ Since A + overlineA = 1$A + \overline{A} = 1$ and B + overlineB = 1$B + \overline{B} = 1$, the inner term is 1 + 1 = 1$1 + 1 = 1$. Y = overline1 = 0$$Y = \overline{1} = 0$$ ### Step 3: Conclusion The output Y is always 0 regardless of the inputs A and B. Checking the truth tables, only option 3 satisfies Y=0$Y=0$ for all conditions. ### Pattern Recognition When a Boolean expression groups a variable and its exact complement together in an OR configuration (A$A$ and overlineA$\overline{A}$), the result instantly hits logic 1. Feeding 1 into any NOR gate guarantees a 0 output universally. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Physics: Semiconductor Electronics

### Related Formula Y = overlineA cdot B quad text(NAND)$$Y = \overline{A \cdot B} \quad \text{(NAND)}$$ Y = overlineA + overlineB quad text(De Morgan's)$$Y = \overline{A} + \overline{B} \quad \text{(De Morgan's)}$$ ### Core Logic The inputs A$A$ and B$B$ are first passed through individual NOT gates (made from tied-input NAND gates or standard NOT gates). The outputs become overlineA$\overline{A}$ and overlineB$\overline{B}$. These are then fed into a NAND gate. The final output Y$Y$ is: Y = overlineoverlineA cdot overlineB$$Y = \overline{\overline{A} \cdot \overline{B}}$$ Applying De-Morgan's Law: Y = overlineoverlineA + overlineoverlineB$$Y = \overline{\overline{A}} + \overline{\overline{B}}$$ Y = A + B$Y = A + B$ This represents an OR operation. ### Pattern Recognition Bubbled inputs on a NAND gate convert it directly into an OR gate via De-Morgan's laws. (Bubbled NAND = OR). ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Physics: Semiconductor Electronics