Q.In the depletion region of a diode

Concept & Intuition

In a common-emitter transistor, the power gain ($A_P$) is the product of the voltage gain ($A_V$) and the current gain ($A_I$ or $\beta$). Here, we are given $A_P = 1800$ and $A_V = 60$. This lets us find the current gain $\beta$. Then, the change in emitter current ($\Delta I_E$) is the sum of the change in base current ($\Delta I_B$) and the change in collector current ($\Delta I_C$). Using $\beta = \Delta I_C / \Delta I_B$, we can relate all three and solve for $\Delta I_C$.

Step-by-step solution

1. Find the current gain ($\beta$) from the given gains. Power gain is defined as:

Here $A_I = \beta$ (the common-emitter current gain). So:

2. Relate the changes in currents. For a transistor:

Therefore, for small changes:

We are given $\Delta I_E = 0.62 \, \text{mA}$.

3. Use the definition of $\beta$. The current gain $\beta$ is:

So $\Delta I_B = \frac{\Delta I_C}{\beta} = \frac{\Delta I_C}{30}$.

4. Substitute into the emitter current change equation. $$0.62 = \frac{\Delta I_C}{30} + \Delta I_C …$$

The key concept here is signal inversion in a common emitter configuration. The transistor (typically an NPN) is biased in the active region, and the output is taken from the collector. When the input voltage rises, the base current increases, which increases the collector current. This larger current flows through the collector resistor, causing a larger voltage drop across it. Since the supply voltage is fixed, the collector voltage drops. So a positive-going input produces a negative-going output — a phase shift of exactly $180^\circ$.

Let’s walk through the reasoning step by step.

1. Understand the circuit’s basic operation

   In a common emitter amplifier, the emitter is grounded (or connected to ground through a small resistor and bypass capacitor). The input signal is applied to the base, and the output is taken from the collector. The transistor acts as a current-controlled current source: a small change in base current causes a much larger change in collector current.

2. Relate input voltage to collector current

   The base-emitter voltage $V_{BE}$ controls the base current. If the input voltage $V_{in}$ increases, $V_{BE}$ increases, so base current $I_B$ increases. The collector current $I_C = \beta I_B$ also increases (where $\beta$ is the current gain).

3. See how collector current affects output voltage

   The output voltage $V_{out}$ is the voltage at the collector with respect to ground. It is given by:

where $V_{CC}$ is the supply voltage and $R_C$ is the collector resistor. As $I_C$ increases, the term $I_C R_C$ increases, so $V_{out}$ decreases.

4. Compare input and output changes
   • Input rises → $V_{in}$ increases
   • Output falls → $V_{out}$ decreases …

Concept and Intuition

In a pure (intrinsic) semiconductor, the valence band is full and the conduction band is empty at absolute zero. As temperature rises, thermal energy breaks covalent bonds, creating electron-hole pairs. These charge carriers increase conductivity. The reason given—that at absolute zero no free electrons exist—is true, but it describes a static condition, not the dynamic process of increasing conductivity with temperature. The key is that conductivity rises because more carriers are generated, not because there were none at absolute zero.

Step-by-step reasoning

1. Evaluate Assertion (A):

   In a semiconductor, conductivity $\sigma = n e \mu_e + p e \mu_h$, where $n$ and $p$ are carrier concentrations. As temperature rises, more electron-hole pairs are generated, so $n$ and $p$ increase. Even though mobility $\mu$ decreases slightly, the exponential rise in carrier concentration dominates. Hence, conductivity increases with temperature.

   → Assertion (A) is true.

2. Evaluate Reason (R):

   At $T = 0$ K, all electrons are in the valence band, and the conduction band is empty. No free electrons exist because thermal energy is insufficient to excite any across the band gap.

   → Reason (R) is true.

3. Check if (R) explains (A): …

Concept & Intuition

A p‑n junction has a built‑in potential barrier. For a hole (a positive charge carrier), the barrier is “uphill” when moving from the p‑side to the n‑side: the hole must do work against the electric field, so its kinetic energy decreases by the barrier height. Conversely, when a hole moves from the n‑side to the p‑side, it is “falling downhill” and gains kinetic energy equal to the barrier height.

The problem gives the hole’s kinetic energy before crossing as 300 meV in both cases. The barrier is 200 meV. So:

• From p to n: kinetic energy after crossing = $300 - 200 = 100$ meV. That’s $E_1$.
• From n to p: kinetic energy after crossing = $300 + 200 = 500$ meV. That’s $E_2$.

Thus $\frac{E_2}{E_1} = \frac{500}{100} = 5$.

Step‑by‑step reasoning

1. Identify the barrier’s effect on kinetic energy The potential barrier is 200 meV. For a hole (positive charge), moving from p‑side to n‑side means going against the electric field, so kinetic energy decreases by 200 meV. Moving from n‑side to p‑side means going with the field, so kinetic energy increases by 200 meV. …