৪৯তম বিসিএস ⎯ তথ্য ও যোগাযোগ প্রযুক্তি [২৮১]

In the reverse bias region, the diode ideally blocks current flow, and the current remains very small (except for a small reverse leakage current). If the reverse voltage exceeds the breakdown voltage, the diode enters breakdown and allows significant current to flow (e.g., Zener breakdown or Avalanche breakdown).

Example:
For a silicon diode in reverse bias, the current is negligible until the reverse voltage exceeds the breakdown voltage, at which point the diode starts conducting in the reverse direction.

Source: Electric Circuits by James W. Nilsson and Susan A. Riedel

A diode in a rectifier circuit is used to allow current to flow in only one direction. This converts AC (alternating current) into DC (direct current), as it blocks the reverse current and only allows the forward current to pass.

Example:
In a half-wave rectifier, the diode allows the positive half of the AC input to pass through while blocking the negative half, resulting in a pulsating DC output.

Source: Electric Circuits by James W. Nilsson and Susan A. Riedel

A Zener diode is designed to allow reverse current to flow when the reverse voltage exceeds a certain Zener breakdown voltage. It is commonly used for voltage regulation in circuits.

Example:
In a voltage regulator circuit, a Zener diode with a breakdown voltage of 5V will clamp the voltage to 5V in reverse bias, allowing current to flow in the reverse direction after this voltage.

Source: Electric Circuits by James W. Nilsson and Susan A. Riedel

In a half-wave rectifier, the diode only conducts during the positive half of the input AC cycle, allowing current to pass through. During the negative half of the cycle, the diode is reverse-biased and does not conduct.

Example:
For a 10V peak-to-peak AC signal, the diode will only conduct when the input voltage is positive, resulting in a pulsating DC signal at the output.

Source: Electric Circuits by James W. Nilsson and Susan A. Riedel

A diode breaks down and allows a significant increase in reverse current when it is subjected to a reverse voltage greater than a certain value called the reverse breakdown voltage. This occurs in the reverse bias condition of the diode.

Forward bias: In this condition, the diode conducts current easily and allows current to flow through in the direction of forward conduction (with a typical forward voltage of around 0.7V for silicon diodes).
Reverse bias: In reverse bias, if the voltage exceeds the breakdown voltage, the diode will enter reverse breakdown, and the reverse current will increase drastically. This phenomenon is often used in Zener diodes (for voltage regulation) or avalanche diodes.

At this point, the diode behaves like a short circuit and allows current to flow in the reverse direction, which can potentially damage the diode if the current is not limited.

Source: Electric Circuits by James W. Nilsson and Susan A. Riedel

In the DC model of a diode, the diode is often modeled as an open circuit when it is in reverse bias (i.e., not conducting). In forward bias, the diode is modeled as a short circuit when the voltage exceeds the threshold voltage (typically 0.7V for a silicon diode), allowing current to flow freely.

Example:
For a silicon diode, when the forward voltage is less than 0.7V, it behaves like an open circuit (no current flows). When the voltage exceeds 0.7V, it behaves like a short circuit, allowing current to pass.

Source: Electric Circuits by James W. Nilsson and Susan A. Riedel

In a diode clipping circuit, the diode is used to limit the amplitude of the signal by conducting when the voltage exceeds a certain threshold (either positive or negative), thus "clipping" off parts of the waveform. This is often used to protect circuits from voltage spikes or to shape the waveform.

Example:
A positive clipping circuit with a diode will limit the signal’s positive voltage to the threshold voltage of the diode, clipping off any part of the signal that exceeds this voltage.

Source: Electric Circuits by James W. Nilsson and Susan A. Riedel

Given that, 
The change in voltage across the diode, dV =0.1V
The change in current through the diode, dI = 1mA = 0.001A

Now,
The dynamic resistance ( r_d ) of a diode is calculated using the formula, 
r_d = dV/dI
    = 0.1/0.001
   = 100
∴ r_d =100Ω

Source: Electric Circuits by James W. Nilsson and Susan A. Riedel

In the forward bias region, the dynamic resistance of a diode is typically very small once the voltage exceeds the threshold (approximately 0.7V for silicon diodes). The diode behaves like a short circuit in this region after the threshold voltage is reached.

Example:
For a silicon diode, the dynamic resistance decreases significantly as the voltage exceeds 0.7V. This means once the diode is forward-biased, it offers minimal resistance to current flow.

Source: Fundamentals of Electric Circuits by Alexander and Sadiku

In reverse bias, as the reverse voltage increases, the depletion region widens, which reduces the effective area for charge storage and therefore decreases the capacitance. This is because the depletion region behaves like a capacitor, and the capacitance C is inversely proportional to the width of the depletion region.

Example:
If a diode is reverse-biased, increasing the reverse voltage will cause the depletion region to widen, and the capacitance will decrease accordingly.

Source: Fundamentals of Electric Circuits by Alexander and Sadiku

Under forward bias, the diode’s diffusion (or storage) capacitance dominates.
This capacitance comes from stored minority carriers in the depletion region and neutral regions.

Mathematically:
C_d =dQ/dV

where,
Q is the stored minority carrier charge.
Therefore, the small-signal capacitance in forward bias depends directly on how charge storage changes with voltage.

Source: Fundamentals of Electric Circuits by Alexander and Sadiku

In small-signal analysis, the piecewise linear model of the diode is often used, where the diode is modeled as a small-signal resistance in the forward bias region and an open circuit in the reverse bias region. This simplifies the analysis of circuits with small AC signals superimposed on a DC bias.

Example:
A diode in a small-signal AC analysis might be modeled as a small resistor (dynamic resistance) for voltages above the threshold voltage.

Source: Fundamentals of Electric Circuits by Alexander and Sadiku

The load line is a graphical representation that shows the relationship between the diode's voltage and current in a forward biased diode circuit. It is drawn by plotting the diode’s current-voltage characteristic along with the load line determined by the load resistance R_L. The intersection of the load line with the diode's characteristic curve gives the operating point of the diode.

Example:
For a circuit with a 12V DC source and a load resistor 1kΩ, the load line will be represented by I=V/R_L. As the diode's I-V characteristic curve is plotted, the intersection of the two will give the diode’s operating point.

Source: Fundamentals of Electric Circuits by Alexander and Sadiku

The operating point of the diode is determined by the intersection of the load line (which is determined by the external circuit, i.e., the load resistance) and the diode’s I-V characteristic curve (which describes the diode’s voltage-current relationship). This intersection gives the voltage across the diode and the current flowing through it.

Example:
For a diode with a 0.7V threshold, and a load resistor 1kΩ with a 12V source, the load line will intersect the diode's characteristic curve at the operating point, where the voltage across the diode is 0.7V and the current is 12mA.

Source: Electric Circuits by James W. Nilsson and Susan A. Riedel

When the load resistance R_L increases, the current decreases according to Ohm’s Law, I = V/R_L. As the load resistance increases, for a given source voltage, the current flowing through the load decreases. This results in a change in the operating point, which shifts to a lower current along the load line.

Example:
If the load resistance is increased, for example, from 1kΩ to 2kΩ in a circuit with a 12V source, the current will decrease, and the operating point will shift accordingly.

Source: Fundamentals of Electric Circuits by Alexander and Sadiku

To calculate the power dissipated in the diode, we can use the formula for electrical power,
P=V×I

Given that,
The current through the diode, I=10mA=0.01A
The voltage across the diode, V=0.7V
The power dissipated in the diode, P = ?

Substitute these values into the formula,

P = 0.7V×0.01A
   = 0.007 W

So, the power dissipated in the diode is 0.007 W.
The answer is 0.007 W.

Source: Electric Circuits by James W. Nilsson and Susan A. Riedel

he power dissipated in a diode (or any electrical component) is calculated by the formula,
P=I×V

Where,
P is the power dissipated (in watts, W),
I is the current through the diode (in amperes, A),
V is the voltage across the diode (in volts, V).

This formula is a direct application of the definition of electrical power, which is the rate at which energy is used or converted in a circuit.

Example:
For example, if a diode has a voltage of 0.7 V across it and a current of 100 mA (0.1 A) flows through it, the power dissipated in the diode can be calculated as
P= 0.7V×0.1A = 0.07W = 70mW

So, the diode dissipates 70 milliwatts (mW) of power in the form of heat.

Source: Electric Circuits by James W. Nilsson and Susan A. Riedel

The load line represents the constraints imposed by the external circuit on the diode. It is a straight line on the I–V graph of the diode and shows all the possible combinations of diode voltage (V_D) and current (I) that satisfy Kirchhoff’s Voltage Law (KVL) in the circuit.
For a simple series circuit with a diode and resistor R connected to a supply voltage V_S then according to Kirchhoff’s Voltage Law, 
V_S = IR + V_D
This is the load line equation. It is a straight line when plotted on the diode’s I–V characteristic graph. The point where the load line intersects the diode’s characteristic curve is the operating point (Q-point) of the diode.

Source: Electric Circuits by James W. Nilsson and Susan A. Riedel​

In a Zener diode voltage regulator, the Zener diode is connected in reverse bias and in parallel with the load. The Zener diode maintains the output voltage by allowing current to flow through it when the input voltage exceeds the Zener breakdown voltage.

Example:
In a 12V input circuit, a 5.1V Zener diode will regulate the output voltage across the load to 5.1V.

Source: Electric Circuits by James W. Nilsson and Susan A. Riedel​

The Zener breakdown voltage is the reverse voltage at which the Zener diode starts to conduct in reverse direction and maintains a constant voltage across it. This is the voltage at which the diode enters the Zener breakdown region, and the voltage across the diode remains stable, irrespective of further increases in the reverse voltage.

Example:
For a Zener diode with a breakdown voltage of 5.1V, once the reverse voltage reaches 5.1V, the diode will start conducting and the voltage across the load will be clamped to 5.1V, even if the input voltage increases further.

Source: Electric Circuits by James W. Nilsson and Susan A. Riedel​

In a Zener diode voltage regulator, the Zener diode maintains a constant output voltage across the load, even if the load current increases. This is because the Zener diode adjusts the amount of current it conducts to maintain the voltage. The output voltage will remain at the Zener breakdown voltage, as long as the Zener diode is not driven into its breakdown region and the source can provide enough current.

Example:
If a 5.1V Zener diode is used in a voltage regulator, the output voltage will remain stable at 5.1V even if the load draws more current, as long as the source voltage is sufficient.

Source: Electric Circuits by James W. Nilsson and Susan A. Riedel

The average DC output voltage of a half-wave rectifier is given by,
V_avg= V_peak /π

Now, Substituting the given value of V_peak=12V:

∴ V_avg= 12/π
 ​          ≈9.45V

This is the average voltage that appears at the output after rectification.
The right answer is 9.45V.

Source: Electric Circuits by James W. Nilsson and Susan A. Riedel

A full-wave rectifier uses both halves of the AC input cycle, which leads to a higher average DC output voltage compared to a half-wave rectifier, which only uses one half of the cycle. This results in higher efficiency in converting AC to DC.

Example:
A full-wave rectifier with a peak voltage of 12V will have an average DC voltage of approximately 7.64V, whereas a half-wave rectifier with the same peak voltage will have an average DC voltage of only 3.82V.

Source: Electric Circuits by James W. Nilsson and Susan A. Riedel

The ripple factor is a measure of the variation (ripple) in the output DC voltage of a rectifier. The full-wave rectifier produces less ripple in the output compared to a half-wave rectifier because it uses both halves of the AC waveform. The ripple factor for a full-wave rectifier is lower, meaning the DC output is more constant.

Example:
A full-wave rectifier produces a smoother DC signal with less ripple compared to a half-wave rectifier, which produces more noticeable fluctuations in the output.

Source: Electric Circuits by James W. Nilsson and Susan A. Riedel

A voltage multiplier is a circuit that increases the voltage across the load by using diodes and capacitors in a specific arrangement. It multiplies the input voltage by a factor greater than 1, typically for use in applications that require higher voltage.

Example:
A doubler circuit can double the input AC voltage, while a quadrupler can quadruple it. For example, a 12V AC input can be stepped up to 24V DC using a voltage doubler.

Source: Electric Circuits by James W. Nilsson and Susan A. Riedel

Voltage multipliers are commonly used in high-voltage applications like CRT monitors and X-ray machines where high voltage is required from a relatively low voltage source. These circuits are capable of stepping up the voltage without needing large, bulky transformers.

Example:
In an X-ray machine, a voltage multiplier might be used to provide the high voltage (tens of kilovolts) required for the X-ray tube, using an input voltage of around 120V.

Source: Electric Circuits by James W. Nilsson and Susan A. Riedel

A voltage doubler is a type of voltage multiplier that steps up the input voltage by a factor of two. It does this by using two diodes and two capacitors. During the first half of the AC cycle, one capacitor charges up to the peak input voltage, and during the second half-cycle, the second capacitor adds to this voltage, effectively doubling it.

Example:
A 12V AC input into a voltage doubler circuit would result in an output of approximately 24V DC.

Source: Electric Circuits by James W. Nilsson and Susan A. Riedel

A voltage multiplier is a circuit that increases the input voltage by a certain factor, typically used in applications where high voltage is needed from a low-voltage AC or DC source. However, the efficiency of a voltage multiplier is always less than 100% because some energy is lost in the form of heat, and due to the inherent limitations of the components (such as diodes and capacitors) used in the multiplier.
The efficiency of a voltage multiplier depends on the number of stages in the multiplier and the load resistance, but in practice, it is always less than 100%.

This loss is primarily due to:
1. Voltage drop across diodes: Each diode in the multiplier causes a voltage drop (typically 0.7V for silicon diodes), which reduces the output voltage.
2. Leakage losses in capacitors: Capacitors used in voltage multipliers have leakage currents, which cause energy dissipation over time.
3. Load-dependent losses: As the load increases, the efficiency typically decreases, especially in higher-stage multipliers.

Example:
In a Cockcroft-Walton multiplier, which is a common type of voltage multiplier, the efficiency might be about 50% to 70% for small loads but decreases with increased load. The exact efficiency depends on the design of the multiplier and the components used.

Source: Electric Circuits by James W. Nilsson and Susan A. Riedel

Voltage multipliers are primarily used in applications requiring high voltage but relatively low current, such as X-ray machines, CRT displays, and other high-voltage equipment. They are ideal for stepping up the voltage in situations where transformers would be too large or expensive.

Example:
In a CRT (Cathode Ray Tube) display, a voltage multiplier is used to generate the high voltage needed to drive the electron gun, which is essential for the operation of the display.

Source: Electric Circuits by James W. Nilsson and Susan A. Riedel

A voltage multiplier increases the output voltage by a factor of the number of stages, but it typically decreases the current in proportion to the voltage increase. This happens because, according to the law of conservation of energy, the power (product of voltage and current) remains roughly constant in the circuit, and the current must decrease as the voltage increases.

Example:
In a voltage doubler circuit, if the input voltage is 10V and the output voltage is doubled to 20V, the output current will be halved, assuming ideal conditions.

Source: Electric Circuits by James W. Nilsson and Susan A. Riedel

A voltage quadrupler circuit is made by combining two stages of a voltage doubler. Each stage doubles the input voltage, resulting in a total output voltage that is four times the input voltage.

Example:
If the input voltage to the quadrupler is 10V, the output will be 40V (4 times the input).

Source: Electric Circuits by James W. Nilsson and Susan A. Riedel

The efficiency of a voltage multiplier is typically low because of the ripple in the output voltage. Ripple is caused by incomplete rectification, resulting in fluctuations in the DC output. These fluctuations mean that the voltage multiplier cannot maintain a smooth output voltage, thus reducing its efficiency.

Example:
In a voltage doubler, while the output voltage is increased, it is not perfectly DC but has periodic variations (ripple) that reduce the overall efficiency.

Source: Electric Circuits by James W. Nilsson and Susan A. Riedel

In a voltage multiplier, as the output voltage increases, the output current generally decreases. This is because the power must remain constant (ignoring losses), and since power = voltage × current, an increase in voltage results in a decrease in current.

Example:
If the input voltage is increased to double the value, the output voltage can be doubled, but the output current will be halved, assuming ideal conditions.

Source: Electric Circuits by James W. Nilsson and Susan A. Riedel

In a positive clamper circuit, the diode is arranged to add a DC bias to the input signal. The output signal is shifted by a fixed amount determined by the DC voltage provided by the circuit, typically determined by the voltage across the capacitor and diode configuration.

Example:
If the input signal ranges from 0V to 5V, and the clamper adds a 5V DC bias, the output will range from 5V to 10V, effectively shifting the entire signal upwards.

Source: Electric Circuits by James W. Nilsson and Susan A. Riedel

A Bipolar Junction Transistor (BJT) is made of three layers of semiconductor material: emitter, base, and collector. It is a three-layer, two-junction device, consisting of either n-p-n or p-n-p layers. The middle layer is the base, and the outer layers are the collector and emitter. In an n-p-n transistor, the emitter and collector are made of n-type material, and the base is made of p-type material.

Example:
In an n-p-n BJT, the emitter is n-type, the base is p-type, and the collector is n-type.

Source: Electric Circuits by James W. Nilsson and Susan A. Riedel

In a BJT, the collector current (I_C) is related to the base current (I_B) by the current gain β.
The relationship is given by

I_C = β×I_B
 ​
Given that, 
The base current, I_B = =1mA=0.001A
The current gain of the transistor, β = 100
The collector current, I_C = ?

Substitute these values into the equation,
I_C =100×0.001A=0.1A=10mA

So, the collector current is 10 mA.
The right answer is 10mA.

Source: Electric Circuits by James W. Nilsson and Susan A. Riedel

In the saturation region, both the emitter-base and collector-base junctions are forward biased, meaning both junctions allow current to flow freely. This results in the BJT being fully "on" with the maximum possible current flowing from the emitter to the collector.

Example:
In a switching application, when the BJT is used as a switch, it operates in saturation mode to allow the maximum current to flow from the emitter to the collector.

Source: Electric Circuits by James W. Nilsson and Susan A. Riedel

In a common-emitter amplifier configuration, the BJT operates in the active region. The transistor is biased such that the emitter-base junction is forward biased and the collector-base junction is reverse biased, which allows the transistor to amplify the input signal.

Example:
A small AC input signal is applied to the base, and the BJT amplifies it to produce a larger AC output signal at the collector.

Source: Electric Circuits by James W. Nilsson and Susan A. Riedel

In a common-emitter amplifier, the output signal is taken from the collector. The input signal is applied to the base, and the amplified output is collected from the collector. This configuration results in inverted amplification, meaning the output signal is inverted with respect to the input.

Example:
If a sinusoidal input signal is applied to the base, the output at the collector will be a larger sinusoidal wave that is inverted.

Source: Electric Circuits by James W. Nilsson and Susan A. Riedel

In an n-channel JFET, the current flows from the source to the drain when a voltage is applied across the two. The flow of current is controlled by the gate-to-source voltage (V_GS). The current is typically carried by electrons in the n-type channel, with the p-type gate controlling the width of the conducting channel.

Example:
When a positive voltage is applied to the drain relative to the source, and the gate is kept at a negative voltage, the current flows from the source to the drain, but its magnitude depends on the gate voltage.

Source: Electric Circuits by James W. Nilsson and Susan A. Riedel

In a Junction Field-Effect Transistor (JFET), the gate is typically reverse biased with respect to the source. This reverse bias creates a depletion region within the channel, which controls the flow of current between the source and drain.

Example:
In an n-channel JFET, applying a negative voltage to the gate relative to the source narrows the channel and limits current flow, while a less negative or positive gate voltage allows more current to flow.

Source: Fundamentals of Electric Circuits by Alexander and Sadiku

The primary difference between n-channel and p-channel MOSFETs lies in the type of semiconductor material used for the channel. In an n-channel MOSFET, the channel is made of n-type semiconductor, while in a p-channel MOSFET, the channel is made of p-type semiconductor. The type of channel determines the direction of current flow and the polarity of the gate voltage required to switch the MOSFET.

Example:
In an n-channel MOSFET, electrons are the majority carriers, and a positive voltage on the gate enhances the flow of electrons from the source to the drain.
In a p-channel MOSFET, holes are the majority carriers, and a negative gate voltage allows current flow from the source to the drain.

Source: Fundamentals of Electric Circuits by Alexander and Sadiku

In a MOSFET, the gate-to-source voltage (V_GS) controls the formation of the conductive channel between the source and drain. The gate voltage creates an electric field that modulates the conductivity of the channel, allowing current to flow from the source to the drain. For n-channel MOSFETs, a positive V_GS enhances the channel conductivity, while for p-channel MOSFETs, a negative V_GS allows current flow.

Example:
When a positive V_GS is applied to an n-channel MOSFET, the electrons accumulate in the channel, allowing current to flow from source to drain.

Source: Fundamentals of Electric Circuits by Alexander and Sadiku

When the V_DS exceeds the pinch-off voltage, the MOSFET enters the saturation region (also called the constant current region). In this region, the current between the source and drain becomes nearly constant, regardless of further increases in V_DS. The current is controlled by the gate-to-source voltage.

Example:
If V_DS increases beyond the pinch-off voltage, the channel near the drain becomes pinched off, and the current remains nearly constant as V_DS increases further.

Source: Fundamentals of Electric Circuits by Alexander and Sadiku

The parameter h₁₁ in the h-parameter model represents the input resistance of the amplifier. It is defined as the ratio of the input voltage to the input current, assuming that the output is short-circuited (no output voltage).

Example:
If h₁₁=1kΩ, this means that the input resistance seen by the source is 1 kΩ when the output is short-circuited.

Source: Fundamentals of Electric Circuits by Alexander and Sadiku

The h₁₂ parameter in the h-parameter model represents the reverse voltage gain or feedback in the amplifier. It shows the impact of the output voltage on the input voltage, which is essential for understanding how changes in output affect the input in feedback systems.

Example:
If h₁₂=0.1, it means that a 1V change in the output voltage would result in a 0.1V change in the input voltage.

Source: Fundamentals of Electric Circuits by Alexander and Sadiku

Step 1: Relationship Between Input and Output:Consider an amplifier with an input voltage V_in and output voltage V_out. The voltage gain A_v is defined as the ratio of the output voltage to the input voltage:
A_v =V_out /V_in

In an amplifier circuit, the input voltage is related to the input current by the input impedance (h11):
V_in = h₁₁ × I_in

Where,
h₁₁ is the input impedance of the amplifier.

Step 2: Voltage Gain in Terms of Current:
Similarly, the output voltage is related to the output current I_out by the output impedance (though not directly part of this equation, it influences overall gain). However, using the current gain parameter h₂₁, we can express the output voltage as:
V_out = h₂₁ × V_in

Where,
h₂₁ is the voltage gain of the amplifier.

Step 3: Combining Both Relationships:
Now, combine the two relationships:
• The input voltage is related to the input current through h₁₁,
• The output voltage is related to the input voltage through h₂₁.

From this, we can express the total voltage gain of the circuit as:

Voltage Gain = V_out /V_in = h₂₁ /h₁₁

Source: Fundamentals of Electric Circuits by Alexander and Sadiku