Consider a n-type semiconductor in which nₑ$n_{e}$ and nh$n_{h}$ are number of electrons and holes, respectively. (A) Holes are minority carriers (B) The dopant is a pentavalent atom (C) nₑnh≠ nᵢ²$n_{e}n_{h}\ne n_{i}^{2}$ (where nᵢ$n_{i}$ is number of electrons or holes in semiconductor when it is in intrinsic form) (D) nₑnh≥ nᵢ²$n_{e}n_{h}\ge n_{i}^{2}$ (E) The holes are not generated due to the donors Choose the correct answer from the options given below:

Related Formula

PZ = VZ IZ$P_Z = V_Z I_Z$

Core Logic

For the Zener diode, the maximum power dissipated is:

Since PD = VZ IZ$P_D = V_Z I_Z$:

Step 1: Find Protective Series Resistance

The voltage drop across the series protective resistance R$R$ is:

Using Ohm's law, R = (VR)/(IZ)$R = \frac{V_R}{I_Z}$:

Pattern Recognition

To secure a Zener against burning out, calculate its max safe current via Pmax / Vz$P_{max} / V_z$. Feed this current into the required voltage drop (Vᵢₙ - Vz$V_{in} - V_z$) to get the exact protective resistance.

Chapter Mix

Class 12 Physics: Semiconductor Electronics

Related Formula

Core Logic

Let's trace the logic line by line. Top branch: A passes through an AND gate with both inputs tied to A, so it remains A$A$. Bottom branch: A and B pass through a NAND gate, yielding A · B$\overline{A \cdot B}$. Then it passes through an AND gate with inputs tied together, so it remains A · B$\overline{A \cdot B}$.

Step 1: Boolean Expression

Finally, the inputs A$A$ and A · B$\overline{A \cdot B}$ are fed into a final AND gate.

Applying De Morgan's Law:

Step 2: Simplify

Since A · A = 0$A \cdot \overline{A} = 0$, we have:

Checking values: If A=1, B=0$A=1, B=0$, then Y = 1 · 1 = 1$Y = 1 \cdot 1 = 1$. For all other combinations, Y = 0$Y = 0$.

Pattern Recognition

An AND gate acting on A$A$ and NAND(A, B)$\text{NAND}(A, B)$ inherently acts as a "strictly A and NOT B" checker. The boolean algebra instantly simplifies A( AB)$A(\overline{AB})$ to A B$A\overline{B}$.

Chapter Mix

Class 12 Physics: Semiconductor Electronics

Related Formula

Core Logic

Check the polarity of the battery to determine which diodes are forward-biased. Diodes D₁$D_1$ and D₂$D_2$ are forward-biased, while D₃$D_3$ is reverse-biased (acts as an open circuit). Because D₁$D_1$ and D₂$D_2$ are in parallel, the total voltage drop across the parallel combination is just 0.7 ~V$0.7 \mathrm{~V}$.

Step 1: Loop Equation

Applying KVL to the main loop containing the forward-biased diodes:

Step 2: Total Current

Step 3: Current Division

Since D₁$D_1$ and D₂$D_2$ are identical and in parallel, the total current divides equally between them.

Rounding to nearest option gives 18.8 ~mA$18.8 \mathrm{~mA}$.

Pattern Recognition

Identical diodes in parallel share the current equally. The voltage drop across the entire parallel diode bank is just the drop of one diode (0.7 ~V$0.7 \mathrm{~V}$).

Chapter Mix

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

Core Logic

The circuit contains two diodes with their n-sides connected to the inputs A and B, and their p-sides tied together and connected to +5V$+5\text{V}$ through a resistor R$R$. The output C is taken from the common p-side junction.

Step 1: Analyzing the Truth Table

If either A = 0$A = 0$ (ground) or B = 0$B = 0$ (ground), the corresponding diode becomes forward biased. Current flows through the resistor R$R$, dropping the voltage at C to near 0V$0\text{V}$ (Logic 0). If both A = 1$A = 1$ (+5V$+5\text{V}$) and B = 1$B = 1$ (+5V$+5\text{V}$), both diodes are reverse biased. No current flows through R$R$, so the voltage at C remains at +5V$+5\text{V}$ (Logic 1).

Step 2: Conclusion

The output C is 1 ONLY when both inputs A AND B are 1. This corresponds exactly to the truth table of an AND Gate.

Pattern Recognition

Diodes pointing away from the inputs with a pull-up resistor (connected to +Vcc$+V_{cc}$) form an AND gate. If diodes point towards the inputs with a pull-down resistor to ground, it's an OR gate.

Chapter Mix

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

Related Formula

Truth relations of basic logic operations:

• NAND operation: Y = A · B$Y = \overline{A \cdot B}$
• NOR operation: Y = A + B$Y = \overline{A + B}$
• NOT operation: Y = A$Y = \overline{A}$
• OR operation: Y = A + B$Y = A + B$

Core Logic

The inputs are:

• Top input = 1$1$
• Bottom input = 1$1$

Let's analyze step-by-step from left to right:

• First Gate (NAND gate at the top-left):
• Inputs are 1$1$ and 1$1$.
• Output = 1 · 1 = 0$\overline{1 \cdot 1} = 0$.
• Bottom-left path with NOT gates:
• Top input (1$1$) goes to a NOT gate, producing 0$0$.
• Bottom input (1$1$) goes to a NOT gate, producing 0$0$.
• These two 0$0$ values feed into the OR gate:
• Output = 0 + 0 = 0$0 + 0 = 0$.

Step 1: Calculate output P

Now trace the path to P$P$:

• The inputs to the top-right AND gate are:
• Output of the top-left NAND gate = 0$0$
• Output of the bottom-left OR gate = 0$0$
• Therefore, output P$P$ is:

Step 2: Calculate output Q

Now trace the path to Q$Q$:

• The gate at the bottom-right is a NOR gate with two inputs:
• Input 1: Output of the top-left NAND gate (0$0$) inverted by a NOT gate = 0 = 1$\overline{0} = 1$.
• Input 2: Output of the bottom-left OR gate (0$0$).
• Passing these inputs (1$1$ and 0$0$) through the final NOR gate:

Thus, both P = 0$P = 0$ and Q = 0$Q = 0$.

Pattern Recognition

Sees: Combinational trace with inverted nodes. Trap: Missing bubbles (NOT gates) representing inversion on internal circuit paths. Shortcut: The first NAND gate output is 0$0$ (since both inputs are 1). This 0$0$ directly goes to the upper AND gate, immediately guaranteeing output P = 0$P = 0$ (eliminates options 1 and 4). Now, you only need to evaluate Q$Q$ to choose between options 2 and 3.

Chapter Mix

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