### 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 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