Choose the correct logic circuit for the given truth table having inputs A and B.

Inputs		Output
A$A$	B$B$	Y$Y$
0	0	0
0	1	0
1	0	1
1	1	1

Related Formula

The Boolean algebraic relationships for standard gates are:

• AND Gate: Y = A · B$Y = A \cdot B$
• NOT Gate: Y = A$Y = \bar{A}$
• OR Gate: Y = A + B$Y = A + B$

Core Logic

Analyzing the connections of the logic circuit shown in the diagram:

• Input A$A$ is passed directly to the top AND gate.
• Input A$A$ is also passed through a NOT gate, giving A$\bar{A}$, which is supplied to the bottom AND gate.
• Input B$B$ is connected directly to both the top and bottom AND gates.

Hence:

• Output of the top AND gate is: A · B$A \cdot B$
• Output of the bottom AND gate is: A · B$\bar{A} \cdot B$

Both outputs are then combined by an OR gate to give the final output Y$Y$:

Step 1: Simplify the Boolean Expression

Using Boolean algebra, factor out B$B$:

Since A + A = 1$A + \bar{A} = 1$:

Y = 1 · B$Y = 1 \cdot B$

Y = B$Y = B$

Step 2: Construct the Truth Table

Because the output Y$Y$ is functionally identical to B$B$, the truth table columns for B$B$ and Y$Y$ must be exactly the same:

• When B = 0$B = 0$, Y = 0$Y = 0$
• When B = 1$B = 1$, Y = 1$Y = 1$

This yields the matching truth table in Option 2.

Pattern Recognition

Observe: Y = A · B + A · B$Y = A \cdot B + \bar{A} \cdot B$. Whenever B$B$ is a common factor to both paths of the AND-OR configuration, it suggests the expression can be simplified directly to B$B$ because the state of A$A$ becomes irrelevant (A + A = 1$A + \bar{A} = 1$).

Chapter Mix

Class 12 Physics: Semiconductors

Core Logic

For a p-n junction diode to be reverse-biased, the p-side must be connected to a lower electrical potential relative to the n-side.

Evaluating option (4): The p-side is at -10 V$-10\text{ V}$ and the n-side is at +2 V$+2\text{ V}$. Since Vₚ < Vₙ$V_p < V_n$, this circuit is explicitly reverse-biased.

Pattern Recognition

Always calculate Vₚ - Vₙ$V_p - V_n$. If Δ V < 0$\Delta V < 0$, it is reverse biasing; if Δ V > 0$\Delta V > 0$, it is forward biasing.

Chapter Mix

Class 12 Physics: Semiconductor Electronics

Related Formula

For a parallel circuit regulator using a Zener diode:

I = Iz + I₁$I = I_z + I_1$

where, I$I$ = total current through the series resistor Iz$I_z$ = current through the Zener diode I₁$I_1$ = current through the load resistor

Core Logic

Given that the breakdown voltage of the Zener diode is:

Let potential at junction B$B$ and D$D$ be 0 ~V$0 \mathrm{~V}$. Then, the potential at the Zener cathode A$A$ and load node C$C$ is stabilized at:

Potential at the source input E$E$ is 10 ~V$10 \mathrm{~V}$.

Step 1: Calculate Total Current

The potential drop across the series resistor (1 ~kΩ$1 \mathrm{~k}\Omega$) is:

Hence, the total line current I$I$ is:

Step 2: Calculate Load Current

The voltage across the load resistor (2 ~kΩ$2 \mathrm{~k}\Omega$) is equal to Vz = 3 ~V$V_z = 3 \mathrm{~V}$. Thus, the load current I₁$I_1$ is:

Step 3: Calculate Zener Current

By applying Kirchhoff's Current Law at node A$A$:

Therefore, the current through the Zener diode is 5.5 ~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.

Chapter Mix

Class 12 Physics: Semiconductor Electronics: Materials, Devices and Simple Circuits

Core Logic

The circuit contains a Germanium diode (D₁$D_1$) and a Silicon diode (D₂$D_2$) in parallel. The barrier potential for Germanium is 0.3 ~V$0.3 \mathrm{~V}$ and for Silicon is 0.7 ~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 ~V$0.3 \mathrm{~V}$, preventing the Silicon diode from ever turning on. Thus, only D₁$D_1$ conducts.

Step 1: Calculate Total Current

The total voltage in the loop after considering the Ge diode's drop is:

Total resistance in the circuit:

Current i$i$:

(Note: Some sources approximate 15 - 1 = 14$15 - 1 = 14$ if considering ideal diode drops or a misprint in standard problem sets where Vdrop = 1V$V_{\text{drop}} = 1\mathrm{V}$ total across the network, but strictly for Ge Vb = 0.3V$V_b = 0.3\mathrm{V}$, let's check standard solution behavior... Wait, the standard PDF solution explicitly uses 15 ~V - 1 ~V = 14 ~V$15 \mathrm{~V} - 1 \mathrm{~V} = 14 \mathrm{~V}$? No, wait. Let's look at the source PDF: i = 14 / 4 = 3.5 ~mA$i = 14 / 4 = 3.5 \mathrm{~mA}$. This implies a total diode drop of 1 ~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 = 3.5mA$i = 14/4 = 3.5\mathrm{mA}$, it means the voltage drop across the diode was taken as 1V$1\mathrm{V}$ (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

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 = 14.7V$15 - 0.3 = 14.7\mathrm{V}$, the provided solution implies an effective 1V$1\mathrm{V}$ drop is used to reach the 14V$14\mathrm{V}$ net. Follow the specific provided calculation.

Chapter Mix

Class 12 Physics: Semiconductor Electronics: Materials, Devices and Simple Circuits

Related Formula

Core Logic

Step 1: Calculate Currents

Current across the load resistor (500 Ω$500 \,\Omega$):

Voltage across the series resistor (200 Ω$200 \,\Omega$) is 20 - 10 = 10 V$20 - 10 = 10 \mathrm{V}$. Current through the series resistor:

Step 2: Extract Zener Current

Applying Kirchhoff's Current Law (KCL) at the junction: I₁ = I₂ + I₃$I_1 = I_2 + I_3$ I₂ = I₁ - I₃$I_2 = I_1 - I_3$

Pattern Recognition

Always perform the unregulated voltage check first. If Vᵢₙ (RL / (RL + RS)) > VZ$V_{in} (R_L / (R_L + R_S)) > V_Z$, the diode behaves like a constant VZ$V_Z$ battery. Apply nodal analysis.

Chapter Mix

Class 12 Physics: Semiconductor Electronics