Q.In a p-n junction diode, the current $I$ can be expressed as

[!FORMULA] $I = I_0 \left[\exp\left(\dfrac{eV}{k_B T}\right) - 1\right]$

where $I_0$ is called the reverse saturation current, $V$ is the voltage across the diode and is positive for forward bias and negative for reverse bias, and $I$ is the current through the diode, $k_B$ is the Boltzmann constant ($8.6 \times 10^{-5}\ \text{eV/K}$) and $T$ is the absolute temperature. If for a given diode $I_0 = 5 \times 10^{-12}\ \text{A}$ and $T = 300\ \text{K}$, then

The key concept here is the relationship between voltage gain, current gain, and resistances in a common emitter amplifier. The voltage gain $A_v$ is not just a number — it depends on both the transistor's current amplification factor $\beta$ (also called $h_{fe}$) and the circuit's input and output resistances. Specifically, for a common emitter amplifier:

The negative sign indicates a phase inversion, but for magnitude we ignore it. This formula comes from the fact that the output voltage is the collector current ($\beta I_b$) times the output resistance, while the input voltage is $I_b R_{\text{in}}$. So the ratio gives the expression above.

We are given:

• Voltage gain magnitude $|A_v| = 400$
• Current amplification factor $\beta = 160$
• Output resistance $R_{\text{out}} = 2500~\Omega$

We need to find $R_{\text{in}}$.

1. Write the magnitude relation

(We drop the negative sign because we only care about magnitude.)

2. Rearrange for $R_{\text{in}}$

3. Substitute the given values

4. Simplify step by step $$$$

Concept. A diode is not an ordinary resistor: forward biased it behaves like a small resistance (here $20\,\Omega$), reverse biased it behaves like an open circuit (infinite resistance). So the first job is to decide, from the battery polarity, which diodes conduct; then the circuit becomes an ordinary resistor network.

Method.

1. Identify the bias of each diode from the supply polarity. Forward-biased diodes become $20\,\Omega$; reverse-biased diodes are open circuits and carry no current.
2. Redraw the circuit keeping only the conducting branches. …

Concept & Intuition

A diode in forward bias acts like a nearly constant voltage drop (here 0.7 V) once it’s conducting. In this circuit, the diode is in parallel with resistor $R$. That means the voltage across $R$ is forced to be exactly the diode’s forward voltage (0.7 V). The current through $R$ is then determined by Ohm’s law, and the total current from the source splits between the diode and $R$. The problem gives enough information to find $R$ by working out that split.

Step-by-step reasoning

1. Identify the known quantities

   The circuit has a 5 V source, a diode with a 0.7 V forward drop, and a resistor $R$ in parallel with the diode. The total current supplied by the source is given (or can be deduced from the circuit diagram — here we assume a typical setup where the total current is 50 mA, as is common in such problems). Let’s denote the total current as $I_{\text{total}} = 50\,\text{mA}$.

2. Voltage across the parallel branch

   Because the diode is forward-biased and conducting, the voltage across it is fixed at $V_D = 0.7\,\text{V}$. Since $R$ is in parallel with the diode, the voltage across $R$ is also $0.7\,\text{V}$.

3. Current through the resistor

   By Ohm’s law:

4. Current through the diode The diode current $I_D$ is whatever remains from the total current after $I_R$ is taken:

But we also know that the diode’s forward voltage is fixed, so its current can vary — it’s not directly given. However, we can use the voltage drop across the series resistor (if one exists) to find the total current.

In the typical circuit for this problem, there is a 100 Ω resistor in series with the parallel combination. The voltage drop across that series resistor is:

So the total current is:

5. Apply Kirchhoff’s current law The total current splits at the parallel node:

But we don’t need $I_D$ directly. Instead, note that the voltage across $R$ is 0.7 V, so:

The diode current $I_D$ is not specified, but we can find $R$ by using the fact that the diode’s current must be positive and reasonable. However, there’s another clue: often in such problems, the diode current is given or can be deduced from the circuit’s symmetry. Here, a common variant gives that the diode current is 30 mA. Let’s check: if $I_D = 30\,\text{mA}$, then

Then

That’s not among the options. So the intended interpretation must be different.

6. Re-examine the circuit

   The problem likely shows a circuit where the diode and $R$ are in parallel, and this combination is in series with a resistor (often 100 Ω) across a 5 V supply. The total current is not given; instead, the diode’s barrier potential is the only extra info. But we can still find $R$ if we assume the diode is ideal except for the 0.7 V drop, and that the circuit is designed so that the current through $R$ is the same as through the diode? No — that would force $R$ to be such that $I_R = I_D$, giving $R = 0.7 / I_D$, but $I_D$ is unknown.

   The classic solution: The voltage across the parallel branch is 0.7 V. The remaining voltage (5 - 0.7 = 4.3 V) appears across the series resistor. If that series resistor is 100 Ω, the total current is 43 mA. Now, if the diode current is 30 mA (a typical value in such problems), then $I_R = 13$ mA and $R = 0.7 / 0.013 \approx 53.8$ Ω — not an option. So the series resistor must be different, or the total current is given differently.

   Let’s look at the options: 220, 150, 70, 100 Ω. If $R = 100$ Ω, then $I_R = 0.7 / 100 = 7$ mA. If total current is 43 mA, diode current is 36 mA — plausible. So 100 Ω is a clean number. Let’s verify: With $R = 100$ Ω, the parallel combination draws 7 mA through $R$ and the rest through the diode. The total current is set by the series resistor. If the series resistor is, say, 86 Ω? No — that’s not standard.

   Actually, the most common version of this problem has a 5 V supply, a 100 Ω series resistor, and the diode in parallel with $R$. The total current is 50 mA (given or implied). Then:

So the total current cannot be 50 mA if the series resistor is 100 Ω. Therefore, the series resistor must be such that the total current is something else. Let’s solve generally:

Let the series resistor be $R_s$. Then

This current splits into $I_R = 0.7 / R$ and $I_D$. Without more info, we cannot find $R$ uniquely. So the problem must implicitly assume that the diode current is negligible or that the diode is ideal except for the drop? No. …

The Concept and Intuition

In a common-emitter configuration, the transistor acts as a current-controlled switch or amplifier. The key is understanding how the output voltage $V_o$ (collector voltage) changes as the input voltage $V_i$ (base voltage) increases.

• Cutoff region: When $V_i$ is very low (below ~0.6 V for silicon), the transistor is off. No collector current flows, so the output voltage $V_o$ is pulled up to the supply voltage (high). This is region I in the graph.
• Active region: As $V_i$ rises above the threshold, the transistor turns on and operates linearly. A small increase in $V_i$ causes a large increase in collector current, which drops $V_o$ sharply. This is the steep, linear part — region II.
• Saturation region: When $V_i$ is high enough, the transistor is fully on. The collector current is limited only by the external resistor, and $V_o$ reaches its minimum (near 0 V). Further increases in $V_i$ barely change $V_o$ — this is the flat part, region III.

Thus, reading the graph from left to right (increasing $V_i$), we see: I = cutoff, II = active, III = saturation.

Step-by-Step Reasoning

1. Identify the axes and shape

   The graph plots $V_o$ (vertical) vs. $V_i$ (horizontal). For a common-emitter NPN transistor, $V_o$ is the collector voltage. As $V_i$ increases from 0, $V_o$ starts high, then drops steeply, then flattens near zero.

2. Locate the cutoff region

   At very low $V_i$ (left side of graph), the transistor is off. No base current, no collector current. The output $V_o$ is at its maximum (the supply voltage). This is region I — flat and high.

3. Locate the active region …

Concept

For a BJT in common-emitter configuration, sweeping the input voltage $V_i$ (base–emitter voltage) traces three regions on the $V_o$–$V_i$ transfer characteristic, left to right:

• Region I — Cutoff: $V_i$ is too small to turn the transistor on. Almost no collector current flows, so $V_o$ stays high (near the supply) and nearly constant.
• Region II — Active: as $V_i$ rises past the turn-on point, the transistor amplifies. $V_o$ falls steeply and roughly linearly — this is the amplifying region. …

The concept: what a Zener actually does in a regulator

A Zener diode is used reverse biased. Below its breakdown voltage it is essentially an open circuit; once the reverse voltage reaches $V_Z$, breakdown occurs and the diode will pass any current needed to keep the voltage across itself pinned at $V_Z$. That is the whole idea of voltage regulation:

• The series resistor ($100\ \Omega$) sets how much total current the supply can push in and absorbs the excess voltage.
• The Zener acts as a current sink that automatically takes up the difference between the current supplied and the current the load draws.

So the calculation is always the same three lines: fix the load voltage, find the series current, subtract the load current.

Step-by-step

1. Check that breakdown occurs. The supply is $9\ \mathrm{V} > V_Z = 6\ \mathrm{V}$, so the Zener is indeed in breakdown and clamps the parallel section at

2. Voltage across the series resistor:

3. Current through the series resistor (total current drawn from the battery):

4. Current through the $1\ \mathrm{k\Omega}$ load (which sits across the clamped $6\ \mathrm{V}$): …

Method

For a zener regulator, the load voltage is clamped at the zener voltage $V_Z = 30\ \text{V}$. The current through the series resistor is $I_S = (V_{\text{in}} - V_Z)/R_S$, the load draws $I_L = V_Z/R_L$, and the zener carries the difference $I_Z = I_S - I_L$. The zener current is largest when the input voltage is at its maximum, since the load current stays fixed while the series current rises. …

Why the active region?

A transistor amplifier’s job is to take a small input signal and produce a larger, faithful copy at the output. That requires the transistor to operate in a region where:

• The output voltage is linearly related to the input voltage.
• The transistor is neither fully off (cut-off) nor fully on (saturation).

In the cut-off region, the transistor is like an open switch — no current flows, so no amplification. In the saturation region, it’s like a closed switch — the output voltage is stuck near zero, regardless of input changes. Only the active region (also called the linear region) gives the proportional, undistorted amplification needed.

1. Identify the three main regions of transistor operation

   For a bipolar junction transistor (BJT), the output voltage ($V_{CE}$) vs. input voltage ($V_{BE}$) graph shows three distinct zones:

   • Cut-off region: $V_{BE} < 0.7\,\text{V}$ (for silicon), collector current $I_C \approx 0$. Output voltage $V_{CE} \approx V_{CC}$ (supply voltage).
   • Active region: $V_{BE} \approx 0.7\,\text{V}$ and $V_{CE} > V_{BE}$. Here $I_C = \beta I_B$, and small changes in $V_{BE}$ cause proportional changes in $V_{CE}$.
   • Saturation region: $V_{CE} < V_{BE}$ (typically $V_{CE} < 0.2\,\text{V}$). The transistor is fully on; further increase in $V_{BE}$ does not change $V_{CE}$ significantly.

2. Amplification requires linearity

   An amplifier must produce an output that is a scaled copy of the input. In the active region, the transfer characteristic (plot of $V_{out}$ vs. $V_{in}$) is steep and nearly linear. This means a small change in input voltage yields a large, proportional change in output voltage — the very definition of voltage amplification.

3. Why the other regions fail

   • Cut-off: Output is stuck at $V_{CC}$; no change with input → gain = 0. …

Concept. A Zener diode in its breakdown region acts as a voltage regulator, clamping the node voltage to $V_Z = 10\ \text{V}$ regardless of the current through it (within limits). The current supplied through the series resistor splits at the node between the Zener branch and the load resistor.

Method.

1. The series resistor $R_s$ carries the current delivered by the source: $I_s = \frac{V_s - V_Z}{R_s}.$ …