RBSE Solutions Class 12 Physics Chapter 16 Electronics

RBSE Class 12 Physics Chapter 16 Multiple Choice Type Questions

 

Question 1. At absolute zero temperature intrinsic germanium and intrinsic silicon; are :
(a) More conductor
(b) Good semiconductors
(c) Ideal insulator
(d) Conductors
Answer: (c) Ideal insulator
In simple words: At very cold temperatures (absolute zero), pure germanium and silicon do not conduct electricity. They act like materials that block all electric flow. This is because their electrons are tightly bound and cannot move freely.

🎯 Exam Tip: Remember that intrinsic semiconductors behave as insulators at absolute zero due to the absence of free charge carriers. This is a fundamental property to recall.

 

Question 2. In insulators the forbidden energy gap between valence band and conduction band is of .......... value :
(a) (Options are cut off in the source text.)
Answer: Very large value
In simple words: Insulators have a very big gap between the energy levels where electrons sit (valence band) and where they can move freely (conduction band). This large gap means electrons cannot easily jump to the conduction band to carry current. The energy gap determines how much energy is needed for an electron to become free.

🎯 Exam Tip: For insulators, the forbidden energy gap is typically greater than 3 eV, which is why they do not conduct electricity easily.

 

Question 3. At room temperature in intrinsic silicon the number of charge carriers per unit volume is \( 1.6 \times 10^{16}/m^3 \). If mobility of electrons is \( 0.150 m^2V^{-1}s^{-1} \) and mobility of holes is \( 0.05 m^2V^{-1}s^{-1} \) then conductivity of silicon in \( (\Omega^{-1}m^{-1}) \) is :
(a) \( 1.28 \times 10^{-4} \)
(b) \( 3.84 \times 10^{-4} \)
(c) \( 5.12 \times 10^{-4} \)
(d) \( 2.14 \times 10^{-4} \)
Answer: (c) \( 5.12 \times 10^{-4} \)
In simple words: To find how well silicon conducts electricity, we use a formula that adds up how easily electrons and holes move. This involves multiplying the number of charge carriers, their charge, and their mobility. The calculation shows the silicon conducts at a level of \( 5.12 \times 10^{-4} \).

🎯 Exam Tip: The conductivity \( \sigma \) of an intrinsic semiconductor is given by \( \sigma = n_i e (\mu_e + \mu_h) \), where \( n_i \) is the intrinsic carrier concentration, \( e \) is the electronic charge, \( \mu_e \) is electron mobility, and \( \mu_h \) is hole mobility. Be careful with unit conversions.

 

Question 4. If an NPN transistor is used as an amplifier then;
(a) Electrons move from base to collector
(b) Holes move from emitter to base
(c) Holes move from base to emitter
(d) Electrons move from emitter to base
Answer: (d) Electrons move from emitter to base
In simple words: When an NPN transistor makes a signal stronger (amplifies), the main flow of electric current happens when electrons move from the emitter to the base. This initial movement creates a base current that helps control a much larger collector current. The emitter is designed to release many electrons.

🎯 Exam Tip: In an NPN transistor, electrons are the majority carriers. For amplification, the emitter-base junction is forward biased, causing electrons to flow from the emitter to the base.

 

Question 5. The Boolean equation for given circuit will be :
(a) \( Y = A + B \)
(b) \( Y = A + B \)
(c) \( Y = A + B \)
(d) \( Y = A \cdot B \)
Answer: (c) \( Y = A + B \)
In simple words: The picture shows an OR gate. This gate gives a "true" (1) output if either of its inputs, A or B, is "true" (1), or if both are "true". The plus sign in Boolean algebra means the OR operation.

🎯 Exam Tip: Identify the standard logic gate symbol first. An OR gate symbol has a curved input side and a curved output side, often like a shield, and its Boolean expression is \( Y = A + B \).

 

Question 6. If the three inputs are A, B and C then the output Y will be;
(The options are cut off in the source text.)
Answer: (d) \( Y = A \cdot B \cdot C \) (Assuming the intended question was for an AND gate with three inputs)
In simple words: If the question is about an AND gate with three inputs, the output is only "true" (1) when all three inputs (A, B, and C) are "true" (1) at the same time. This is like a series circuit where all switches must be closed for the light to turn on.

🎯 Exam Tip: For logic gate questions, clearly understand the type of gate and the number of inputs. The dot symbol \( (\cdot) \) in Boolean algebra represents the AND operation.

 

Question 7. The current amplification factor for common base circuit of a transistor is 0.95. When emitter current is 2mA then base current is :
(a) 0.1 mA
(b) 0.2 mA
(c) 0.19 mA
(d) 1.9 mA
Answer: (a) 0.1 mA
In simple words: In a transistor, current amplification factor \( \alpha \) tells us how much collector current we get for a given emitter current. We know the emitter current is the sum of collector and base currents. By using the given values, we can calculate the base current to be 0.1 mA. This small base current controls a larger collector current.

🎯 Exam Tip: Remember the relationship between currents in a transistor: \( I_E = I_C + I_B \). Also, the common-base current gain \( \alpha = \frac{I_C}{I_E} \). Use these formulas to find the unknown current.

 

Question 8. The forbidden energy gap in germanium is 0.7 eV. The wavelength whose absorption is done by germanium is :
(a) 35000 Å
(b) 17700 Å
(c) 25000 Å
(d) 51600 Å
Answer: (b) 17700 Å
In simple words: Germanium can absorb light if the light's energy is enough to let electrons jump across its energy gap. We can find the longest wavelength of light that has this minimum energy. This wavelength is about 17700 Å. The energy of light is related to its wavelength, so a smaller energy gap allows for absorption of longer wavelengths.

🎯 Exam Tip: The energy of a photon \( E = \frac{hc}{\lambda} \). To find the maximum wavelength that can be absorbed, use the forbidden energy gap \( E_g \) for E. Remember to convert electron volts to Joules and Angstroms to meters, or use constants with appropriate units.

 

Question 9. The logic gate obtained by two NAND gate is :
(a) AND gate
(b) OR gate
(c) XOR gate
(d) NOR gate
Answer: (a) AND gate
In simple words: If you connect two NAND gates in a specific way, they can act like an AND gate. This means the final output will only be "true" (1) if both inputs are "true" (1), just like a regular AND gate. This shows how one type of logic gate can be built from another.

🎯 Exam Tip: A NAND gate is a universal gate, meaning any other logic gate (AND, OR, NOT, NOR, XOR) can be constructed using only NAND gates. Two NAND gates with inverted inputs act as an OR gate, and three NAND gates act as an AND gate.

 

RBSE Class 12 Physics Chapter 16 Short Answer Type Questions

 

Question 1. Answer: It flows from P region to N region.
Answer: The current in a forward-biased p-n junction flows from the P-region to the N-region. This direction is considered due to the movement of holes from P to N and electrons from N to P, which together constitute a conventional current direction. This flow is established when an external voltage pushes carriers across the depletion layer.
In simple words: When a p-n junction is set up correctly (forward biased), electricity flows from the P-side to the N-side. This is the common way we describe how current moves in these devices.

🎯 Exam Tip: Always remember the conventional current direction in a forward-biased p-n junction (P to N) and that it's due to both hole and electron movement.

 

Question 2. Write down the relation between \( \alpha \) and \( \beta \) (current amplification factor) of the transistor.
Answer: The relationships between the common-base current gain \( \alpha \) and the common-emitter current gain \( \beta \) for a transistor are given by:
\( \beta = \frac{\alpha}{1-\alpha} \)
OR
\( \alpha = \frac{\beta}{\beta+1} \)
These formulas allow you to convert between the two different current gain factors, which describe how much a transistor amplifies current in different circuit configurations. They are key to understanding transistor operation.
In simple words: These are two ways to show how a transistor makes current bigger. Alpha (\( \alpha \)) is for one type of circuit, and Beta (\( \beta \)) is for another. The formulas help us change from one value to the other.

🎯 Exam Tip: Be able to derive or recall both forms of the relationship between \( \alpha \) and \( \beta \). Also, remember that \( \beta \) is typically much larger than 1, while \( \alpha \) is always less than 1 but very close to 1.

 

Question 3. Can in any forward biased PN junction present potential difference can be measured by a voltmeter connected at the ends of the junction?
Answer: No, the potential difference across a forward-biased PN junction cannot be directly measured by connecting a voltmeter at the ends of the junction. This is because a voltmeter measures the total voltage drop, which would include the voltage drop across any internal resistance of the junction and not just the barrier potential itself. The voltmeter itself would also draw some current. The potential barrier is an internal property.
In simple words: No, you cannot simply use a voltmeter to measure the exact internal voltage across a PN junction when it's forward biased. The voltmeter measures the total external voltage, not just the tiny internal barrier voltage.

🎯 Exam Tip: Understand that a voltmeter measures potential difference between two points, but for a p-n junction, the barrier potential is an internal electric field that helps establish equilibrium, not a voltage easily measurable externally in isolation.

 

Question 4. Make truth table for OR gate.
Answer: An OR gate produces a "true" (1) output if at least one of its inputs is "true" (1). If both inputs are "false" (0), then the output is "false" (0). This behavior is summarized in the truth table below. The OR gate works like two switches connected in parallel, where closing either switch (or both) completes the circuit.

In simple words: The OR gate truth table shows that if A is 1 or B is 1 (or both), then the output Y is 1. If both A and B are 0, then Y is 0.

🎯 Exam Tip: Clearly list all possible input combinations (usually 00, 01, 10, 11 for two inputs) and then determine the output based on the OR gate's function: output is 1 if any input is 1.

 

Question 5. Write the name of that logic gate in which the output is 1 when all the inputs are 1.
Answer: The logic gate in which the output is 1 (true) only when all its inputs are 1 (true) is called an **AND gate**. For any other combination of inputs, the output of an AND gate is 0 (false). This gate acts as a logical "and" condition. It's like having multiple switches in a row; all must be closed for electricity to flow.
In simple words: This gate is called an AND gate. It only turns "on" (output 1) if all its "on" (input 1) signals are on at the same time.

🎯 Exam Tip: Remember that the AND gate requires all conditions to be met simultaneously for a true output, making it suitable for situations needing multiple conditions to be true.

 

Question 7. What will be the value of \( \alpha \) for that transistor for which \( \beta = 19 \)?
Answer: We can find the value of \( \alpha \) using the relationship between \( \alpha \) and \( \beta \):
\( \alpha = \frac{\beta}{\beta+1} \)
Given \( \beta = 19 \), we substitute this value into the formula:
\( \alpha = \frac{19}{19+1} \)
\( \alpha = \frac{19}{20} \)
\( \alpha = 0.95 \)
Thus, for a transistor with a common-emitter current gain \( \beta \) of 19, its common-base current gain \( \alpha \) is 0.95. This value is always less than 1, as \( \alpha \) represents the fraction of emitter current that reaches the collector.
In simple words: We use a formula that links \( \alpha \) and \( \beta \). Since \( \beta \) is 19, we put that into the formula, and we find that \( \alpha \) is 0.95.

🎯 Exam Tip: Always recall the formula \( \alpha = \frac{\beta}{\beta+1} \). Ensure you substitute the \( \beta \) value correctly and perform the division to get the \( \alpha \) value, which should always be between 0 and 1.

 

Question 8. The diode shown in the figure is in which biasing?
Answer: The diode shown in the figure is **reversed-biased**. This is because the N-region (connected to \( +10V \)) is at a higher potential than the P-region (connected to \( +5V \)). In reverse bias, the depletion region widens, and only a very small leakage current flows through the diode. This setup actively prevents the flow of majority carriers.
In simple words: The N-side has a higher voltage (+10V) than the P-side (+5V). When this happens, the diode is connected backwards, which is called reverse bias.

🎯 Exam Tip: A p-n junction diode is forward-biased when the P-side is at a higher potential than the N-side, and reverse-biased when the N-side is at a higher potential than the P-side.

 

RBSE Class 12 Physics Chapter 16 Short Answer Type Questions

 

Question 1. What is rectification? Draw the figure of bridge wave rectifier.
Answer: Rectification is the process of converting an alternating current (AC), which periodically reverses direction, into a direct current (DC), which flows in only one direction. A rectifier is an electronic device, typically employing diodes, that achieves this conversion. Full-wave bridge rectifiers are highly efficient as they utilize both halves of the AC input cycle. The diagram for a full-wave bridge rectifier is shown below:

In simple words: Rectification is changing AC power into DC power. A bridge rectifier uses four special components called diodes to make sure the current only flows in one direction, giving a smooth DC output from an AC input.

🎯 Exam Tip: When drawing a bridge rectifier, ensure the diodes are correctly oriented to form the bridge configuration and that the output is taken across the load resistor. A full-wave bridge rectifier is more efficient than a half-wave rectifier because it utilizes both positive and negative halves of the AC input.

 

Question 3. Make the full I-V characteristic curve for ideal PN junction diode. Define dynamic resistance in forward biased state.
Answer: The I-V characteristic curve of an ideal PN junction diode shows its current-voltage relationship. In the forward-biased state, the diode allows current to flow easily after a certain voltage (knee voltage) is reached. In the reverse-biased state, it blocks current until a breakdown voltage is met. Dynamic resistance is defined as the change in voltage divided by the change in current ( \( r_d = \frac{\Delta V_F}{\Delta I_F} \) ) in the forward-biased state. It describes how much the voltage changes for a small change in current when the diode is actively conducting. Dynamic resistance is typically low in forward bias.

In simple words: The graph shows how current changes with voltage for a diode. Dynamic resistance in forward bias means how much the voltage needs to change for a small change in current, usually when the diode is letting current flow easily.

🎯 Exam Tip: When drawing I-V curves, remember to label axes correctly (I and V) and mark key points like knee voltage and breakdown voltage. Dynamic resistance is specific to the conducting region of the diode's curve.

 

Question 4. What do you understand by logic gate? Draw the symbol for XOR gate and also the truth table.
Answer: A logic gate is a basic building block of a digital circuit that takes one or more binary inputs (0 or 1) and produces a single binary output based on a specific logical rule. Logic gates are the foundation of all digital electronics and computing. They help control the flow of information. The symbol for an XOR (Exclusive OR) gate and its truth table are:

The Boolean expression for XOR gate is \( Y = A \oplus B \). The output of an XOR gate is 1 if and only if its inputs are different (one is 0 and the other is 1). If both inputs are the same (both 0 or both 1), the output is 0.

In simple words: Logic gates are like tiny decision-makers in electronics. An XOR gate gives a "yes" (1) only when its two inputs are different from each other. If they are the same, it gives a "no" (0).

🎯 Exam Tip: For XOR gates, remember the key characteristic: output is high (1) only when an odd number of inputs are high. Practice drawing the symbol with the distinguishing curved line at the input.

 

Question 5. Draw the transistor based circuit diagram for NOT gate and also give its truth table.
Answer: A NOT gate, also known as an inverter, is a fundamental logic gate with a single input and a single output. It produces an output that is the inverse of its input. If the input is 0, the output is 1, and if the input is 1, the output is 0. A transistor can be used to construct a NOT gate circuit. When the input signal to the transistor's base is high (1), it turns on, causing the output voltage to drop to low (0). Conversely, when the input is low (0), the transistor turns off, allowing the output voltage to rise to high (1).
The Boolean expression for a NOT gate is \( Y = \overline{A} \), where \( \overline{A} \) means "NOT A".
The symbol for a NOT gate and a basic electrical switch circuit that behaves like a NOT gate are shown below:

Fig. 16.55. Electrical switch of NOT gate

Fig. 16.56. Symbol of NOT gate
A transistor-based NOT gate circuit typically uses an NPN transistor where the input is applied to the base, and the output is taken from the collector with a load resistor, as shown in the diagram. When the input 'A' is 0V (low), the transistor is in a cut-off state, no collector current flows, and the output 'Y' (potential difference at the collector) is high (e.g., Vcc). When 'A' is +5V (high), the transistor saturates, causing the collector current to be maximum, and the output 'Y' becomes low (close to 0V).

Fig. 16.57. Transistor NOT Gate

In simple words: A NOT gate flips the input: if you put in a 0, you get a 1; if you put in a 1, you get a 0. You can build it with a transistor circuit where the output is always the opposite of the input.

🎯 Exam Tip: When drawing transistor circuits for logic gates, ensure correct biasing (forward for base-emitter, reverse for collector-base) and proper component placement for resistors to define input/output conditions.

 

Question 7. Draw the circuit used for voltage regulation in a zener diode and also explain its process in short.
Answer: A Zener diode is specially designed to operate in the reverse breakdown region, where it maintains a nearly constant voltage across its terminals, even if the current through it changes significantly. This property makes it ideal for voltage regulation. In a Zener diode voltage regulator circuit, the Zener diode is connected in parallel with the load and in series with a current-limiting resistor, as shown in the figure. If the input voltage or load current changes, the Zener diode adjusts its current to keep the output voltage constant at its Zener voltage. This helps stabilize the output voltage.

Zener Diode Symbol:

Voltage Regulation Circuit with Zener Diode:

In simple words: A Zener diode can keep the output voltage steady. It's connected so that if the input voltage changes, the Zener diode makes sure the output voltage stays the same, thanks to its special ability to maintain a constant voltage after a certain point.

🎯 Exam Tip: For Zener diode regulators, remember that the diode is always reverse-biased and connected in parallel with the load. The series resistor protects the Zener diode from excessive current.

 

Classification of Conductor, Insulator and Semiconductor

Electrical conductivity is a physical quantity whose range is very wide. On one hand we know about metals whose electrical conductivity is very high. On the other hand there is quartz, mica like insulators whose electrical conductivity is very low. Such substances are also known whose conductivity is very less than metals at normal temperatures but very high than insulators. These are called semiconductors.

The conductivity of semiconductors is not only between conductors and insulators but also the change in the conductivity with temperature is very interesting. Near absolute zero their behaviour is similar to insulators but as the temperature increases the conductivity also increases which is opposite to the projected behaviour of metals. The following questions are not answered by the principle related to electrical conductivity which you are will versed with :
1. Why are the conductivities of solid substances different?
2. Why does any substance show the behaviour of a semiconductor?
3. Why is the change in conductivity with temperature different for metals and semi-conductors?

On the basis of the principle of energy bands in solids and answering the questions, these are divided into conductors, insulators and semiconductors. Every solid substance has its own band construction which defines its electrical conductance behaviour.

Intrinsic Semiconductors

Those semiconductors in which there is no impurity are called intrinsic semiconductors. In ideal state in this type of intrinsic semiconductors there should be atoms of only that semiconductor. But in reality it is not possible to get such type of crystal. Therefore, if in a semiconductor material the number of impure atoms and the number of semiconductor atoms is in the ratio 1 : \( 10^8 \) or less than that then it is considered as an intrinsic semiconductor. To study the properties of semiconductors here we take the example of silicon and germanium.

Silicon and germanium both are the members of fourth group in the periodic table and their valency is 4. Their electronic configuration is as follows:
Ge (32) = \( 1s^2 2s^2 2p^6 3s^2 3p^6 3d^{10} 4s^2 4p^2 \)
Si (14)= \( 1s^2 2s^2 2p^6 3s^2 3p^2 \)

Fig. 16.4. Regular tetrahedron construction for Silicon (or Germanium) crystal

In the crystal construction of both these elements the atom is in an ordered array in such a way that each atom is inside a regular tetrahedron and on the four corners there are its other nearby atoms.

Figure (16.4) shows one such tetrahedron unit. Every valence electron of the atom makes a covalent bond with the valence electron of the nearby atom. In this way every atom is connected with the nearby atoms through covalent bonds. In every covalent bond there are two electrons. For convenience in study; figure (16.5) shows the two dimensional form of crystal construction of germanium; which is also applicable for silicon.

Fig. 16.5. Crystal construction of germanium at 0 K temperature

At absolute zero (0 K) also the valence electrons are bonded in covalent bonds hence there is no free electron available for electrical conductivity. Due to this at absolute zero temperature intrinsic semiconductors behave as insulators. When the temperature of the crystal is increased then some valence electron gain thermal energy and break their covalent bonds and become free. These free electrons move in the crystal freely and participate in electrical conductance when the crystal is applied external electronic energy.

When one electron moves out of the covalent bond a vacancy is created. This vacancy is called a hole. The absence of electron is equal to the same magnitude positive charge. Hence, the hole can be assumed as a positive charge same in magnitude as that of an electron. As explained earlier, holes also participate in electrical conductance in semiconductors. In any semiconductor when a covalent bond breaks then a electrons hole pair is generated. At room temperature (300K) or a temperature close to it there are many electron hole pairs available at definite temperature. The production of electron hole pair is shown in figure (16.6).

Fig. 16.6. Electron-hole pair

 

Question 2. What is P-N junction? Explain the process during its formation on the junction surface. If this junction is forward biased then explain the effect on depletion layer.
Answer: A P-N junction is formed when a P-type semiconductor is joined closely with an N-type semiconductor. It cannot be made by simply pressing them together because the crystal structure needs to be continuous. When a P-N junction is formed, holes from the P-side move to the N-side, and electrons from the N-side move to the P-side due to differences in concentration. This movement creates a diffusion current. When an electron moves from N to P, it leaves behind a positive immobile ion on the N-side. Similarly, when a hole moves from P to N, it leaves behind a negative immobile ion on the P-side. These immobile ions create a region near the junction called the depletion region, which is free of mobile charge carriers. This region acts as a barrier to further charge movement.When the P-N junction is forward biased, the P-side is connected to the positive terminal of a battery and the N-side to the negative terminal. This external voltage opposes the internal electric field of the depletion region, causing the barrier potential to decrease. As a result, the depletion layer becomes thinner. This allows majority charge carriers (holes from P-side and electrons from N-side) to easily cross the junction, leading to a significant current flow.
In simple words: A P-N junction joins two types of semiconductors. When it is forward biased, the P-side gets positive voltage and the N-side gets negative. This makes the "empty" area (depletion layer) between them thinner, allowing current to flow easily. This thin layer is critical for diode function, controlling current flow.

🎯 Exam Tip: Clearly define the P-N junction and its formation. For forward biasing, explain how the external voltage reduces the depletion layer width and barrier potential, leading to current flow.

 

Question 3. Draw the figure of full wave rectifier used to change AC to DC and explain its working.
Answer: A full wave rectifier circuit changes both positive and negative half cycles of an alternating current (AC) signal into a direct current (DC) signal. This circuit uses four junction diodes arranged in a bridge pattern. During the positive half-cycle of the AC input, one pair of diodes conducts, allowing current to flow through the load. During the negative half-cycle, the other pair of diodes conducts, directing current through the load in the same direction. This ensures that the output voltage and current are always positive, producing a pulsating DC signal. A center-tap transformer can also be used, but the bridge rectifier is more common as it doesn't require a center tap.
In simple words: A full wave rectifier turns all of an AC signal into a positive, but wavy, DC signal. It uses diodes to make sure the current flows in only one direction through the device, no matter if the input is positive or negative.

🎯 Exam Tip: When explaining rectifiers, clearly distinguish between half-wave and full-wave operation and their respective output waveforms. Mention the key components like diodes and transformers.

 

Question 4. Draw arrangement of characteristic curve for forward and reverse biasing of any P-N junction diode. Also make the figure of the curves.
Answer: The characteristic curve of a P-N junction diode shows how the current (\(I\)) through the diode changes with the voltage (\(V\)) applied across it. For forward biasing, the P-side is connected to the positive terminal and the N-side to the negative terminal of a voltage source. Initially, with a small forward voltage (around 0.3 V for Germanium or 0.7 V for Silicon), the current is very low as the applied voltage is less than the barrier potential. Once the voltage exceeds this "knee voltage," the current increases sharply, almost exponentially. For reverse biasing, the P-side is connected to the negative terminal and the N-side to the positive terminal. A very small leakage current flows initially due to minority charge carriers. As the reverse voltage increases, this leakage current remains almost constant until it reaches a "breakdown voltage." At this point, the current suddenly increases very sharply in the reverse direction, which can damage the diode if not controlled.
In simple words: The characteristic curve shows how much current flows through a diode at different voltages. In one direction (forward bias), it acts like a switch that turns on after a certain voltage. In the other direction (reverse bias), almost no current flows until the voltage gets too high, causing it to break down.

🎯 Exam Tip: Clearly label the axes, origin, knee voltage, and breakdown voltage on the characteristic curve. Emphasize the non-linear nature of the diode.

 

Question 5. What is junction transistor? Make the necessary diagram and explain the working of a p-n-p transistor.
Answer: A junction transistor is a semiconductor device that can amplify or switch electronic signals and electrical power. It is made of a single crystal of extrinsic semiconductor material (like silicon or germanium) with three distinct regions of different conductivity. These regions are called the emitter (E), base (B), and collector (C). The base region is very thin and has less doping compared to the emitter and collector. There are two main types: NPN and PNP. In a PNP transistor, two P-type semiconductor regions (emitter and collector) sandwich a thin N-type semiconductor region (base).
In a PNP transistor, for active operation, the emitter-base (E-B) junction is forward biased, and the base-collector (B-C) junction is reverse biased. Since the emitter is P-type, it has many holes. When the E-B junction is forward biased, these holes are "injected" from the emitter into the base (N-type). Because the base is very thin and lightly doped, only a few holes (about 2-5%) recombine with electrons in the base. The remaining large number of holes (95-98%) diffuse through the base and reach the collector-base junction. Since the collector-base junction is reverse biased, it acts like a strong electric field that attracts these holes, pulling them into the collector (P-type). This flow of holes constitutes the collector current, which is almost equal to the emitter current. The small recombination in the base forms the base current. This controlled flow of charge allows the transistor to amplify signals.
In simple words: A transistor is a tiny electronic device that can make small signals bigger. In a PNP transistor, holes move from the emitter to the collector through a very thin base. By controlling a small current in the base, a much larger current between the emitter and collector can be managed.

🎯 Exam Tip: When explaining transistor working, clearly state the biasing conditions for each junction (forward/reverse) and trace the path of majority charge carriers for both NPN and PNP types.

 

Question 6. Draw the circuit arrangement and also explain the characteristics curves for any common emitter configuration transistor. Also make the figures of curves and write the formula for voltage gain and current gain.
Answer: In a common emitter (CE) configuration, the emitter is shared by both the input and output circuits. The input voltage is applied between the base and emitter (\(V_{BE}\)), and the output is taken across the collector and emitter (\(V_{CE}\)). The input current is base current (\(I_B\)), and the output current is collector current (\(I_C\)).

Input Characteristics: These curves show the relationship between input current (\(I_B\)) and input voltage (\(V_{BE}\)) for different constant output voltages (\(V_{CE}\)). They are similar to a forward-biased diode characteristic, where current increases rapidly after a certain knee voltage.
Output Characteristics: These curves show the relationship between output current (\(I_C\)) and output voltage (\(V_{CE}\)) for different constant input currents (\(I_B\)). The curves typically show three regions:

• Cut-off Region: When \(I_B\) is zero, \(I_C\) is very small.
• Active Region: For a given \(I_B\), \(I_C\) increases with \(V_{CE}\) initially and then becomes almost constant. This is where amplification occurs.
• Saturation Region: For very small \(V_{CE}\), \(I_C\) is not strongly dependent on \(I_B\).

Formulas for Gain:Current Amplification Factor (\(\beta\)): This is the ratio of the change in collector current (\(\Delta I_C\)) to the change in base current (\(\Delta I_B\)) at a constant collector-emitter voltage (\(V_{CE}\)). \[ \beta = \frac{\Delta I_C}{\Delta I_B} \] Voltage Gain (\(A_V\)): This is the ratio of the change in output voltage (\(\Delta V_{out}\)) to the change in input voltage (\(\Delta V_{in}\)).
In simple words: These curves show how a transistor behaves like a switch or an amplifier. Input curves relate the tiny base current to the small voltage that turns the transistor on. Output curves show how the main collector current changes with the output voltage for different base currents. This helps us understand how well it amplifies or switches signals.

🎯 Exam Tip: Clearly define and draw both input and output characteristic curves for the common emitter configuration. Label the different operating regions (cut-off, active, saturation) and state the formulas for current gain (\(\beta\)) and voltage gain (\(A_V\)).

 

Question 7. What do you understand by amplification? Make the diagram of a p-n-p transistor common amplifier and explain the process of amplification and calculate the formula for voltage gain.
Answer: Amplification is the process of increasing the strength or amplitude of an electrical signal. An amplifier is an electronic device that takes a small input signal and produces a larger output signal of the same shape and frequency. The extra energy needed for this increase comes from a separate DC power supply connected to the amplifier circuit. Transistors are commonly used as amplifiers.
In a common emitter PNP transistor amplifier, the input signal voltage (\(V_i\)) is applied to the base-emitter junction (which is forward-biased), and the output voltage (\(V_o\)) is taken across the collector-emitter junction (which is reverse-biased). A small change in the input signal voltage causes a larger change in the base current (\(I_B\)). Due to the transistor's current amplification property (\(\beta\)), this small change in \(I_B\) leads to a significant change in the collector current (\(I_C\)). This larger change in \(I_C\) flowing through a load resistor (\(R_L\)) in the output circuit produces a much larger change in output voltage. This effectively amplifies the input signal. The common emitter configuration is widely used because it provides good voltage, current, and power gains.
The voltage gain (\(A_v\)) of an amplifier is defined as the ratio of the output signal voltage (\(V_o\)) to the input signal voltage (\(V_i\)). \[ A_v = \frac{V_o}{V_i} \] In a common emitter configuration, the output voltage is \(I_C R_L\) and input voltage is \(I_B R_{ie}\) (where \(R_{ie}\) is the input resistance). So, \(A_v = \frac{I_C R_L}{I_B R_{ie}} = \beta \frac{R_L}{R_{ie}}\).
In simple words: Amplification means making a small signal bigger, like when a microphone picks up your voice and a speaker makes it loud. A transistor amplifier uses a small change in input to create a large change in output. The voltage gain tells us how many times the voltage has been increased.

🎯 Exam Tip: Define amplification clearly. For the PNP amplifier, explain the input and output circuits, the biasing conditions, and how a small base current change leads to a large collector current change. Include the basic block diagram and the formula for voltage gain.

 

Question 7. What do you understand by amplification? Make the diagram of a p-n-p transistor common amplifier and explain the process of amplification and calculate the formula for voltage gain.
Answer: An amplifier is an electronic device that makes a small input signal bigger at its output. The output signal will have a larger amplitude (size) than the input signal, but its shape and frequency will remain the same. The extra energy needed for this increase comes from a DC power supply in the amplifier circuit. A common amplifier circuit uses a transistor, typically arranged in a common-emitter configuration, where the emitter is shared by both the input and output circuits. For a PNP transistor common-emitter setup, a battery (VBB) provides forward bias to the base-emitter junction, and another battery (VCC) provides reverse bias to the collector-emitter junction. The input signal voltage (Vi) is applied to be amplified, and the amplified output voltage (Vo) is obtained across a load resistor (RL) in the output circuit. The process involves the input signal changing the base current, which then controls a much larger collector current, leading to an amplified output.

In simple words, an amplifier takes a small electrical signal and makes it much stronger without changing its basic rhythm or tune. It uses extra power from a battery to boost the signal's size.
The amplification factor or gain is the ratio of output signals to input signals.

\( A_{ie} = \frac{\text{Output signal current}}{\text{Input signal current}} = \frac{i_c}{i_b} = \beta \)
Since \( \beta \gg 1 \), the output signal current is much larger than the input current, which means the current is amplified.

The voltage amplification or voltage gain \( A_v \) is given by:
\( A_v = \frac{\text{Output signal voltage}}{\text{Input signal voltage}} = \frac{V_o}{V_i} \)
Here, \( V_o = i_c R_L \) (voltage generated on load resistor \( R_L \))
And \( V_i = i_b R_{ie} \) (where \( R_{ie} \) is the input resistance in common-emitter configuration).

So, \( A_{ve} = \frac{i_c R_L}{i_b R_{ie}} = \beta \times \frac{R_L}{R_{ie}} \)
Here, \( \frac{R_L}{R_{ie}} \) is called the resistance gain for common-emitter configuration.
Therefore, \( A_{ve} = A_{ie} \times \text{resistance gain} \).
Since \( \beta \gg 1 \) and if \( R_L > R_{ie} \), then \( A_{ve} \gg \beta \gg 1 \), which means a large voltage amplification is possible.In simple words: Amplification makes a signal bigger. It's measured by how much the output current or voltage is compared to the input. If the output is much larger, it means good amplification.

🎯 Exam Tip: When defining amplification, emphasize that it increases signal amplitude without altering its frequency or waveform. Clearly state the role of the DC power supply.

 

Question 8. Write down the names of some diodes used for special purposes and make their circuit figures. Write about their working and uses in short.
Answer: Apart from simple rectification, P-N junction diodes are used for several special purposes. These special-purpose diodes differ from standard diodes in their properties, materials, impurities, and construction. Two such diodes are the Zener Diode, Photo Diode, and Light Emitting Diode (LED).

1. Zener Diode:
A Zener diode is designed to operate reliably in the reverse breakdown region without being damaged. This unique property makes it useful for voltage regulation. When current through the Zener diode increases significantly in reverse bias beyond a certain point (the Zener voltage), the voltage across it remains almost constant. This constant voltage makes Zener diodes ideal for stabilizing output voltage in power supplies, ensuring a steady DC voltage even if the load current changes. They are available in various breakdown voltages, from less than 1 volt to hundreds of volts.

2. Photo Diode:
A photodiode is a semiconductor device that converts light into electrical current. When light of sufficient frequency (and thus energy) falls on the photodiode, it gets absorbed. This energy causes electrons in the valence band to transition to the conduction band, creating electron-hole pairs. This process increases the semiconductor's conductivity, known as photoconductivity. The more intense the light, the more electron-hole pairs are generated, leading to a larger current. Photodiodes are used in various applications like light detection (e.g., in smoke detectors), light-operated switches, reproducing sound from films, and reading computer cards.

The graph for the photodiode shows that in reverse bias, as the intensity of light increases, the reverse current also increases. When there is no light, the current is very low, but it rises with increasing light intensity.

3. Light Emitting Diode (LED):
An LED is a P-N junction diode that emits light when it is forward-biased. In an LED, when electrons from the conduction band combine with holes in the valence band, they release energy. In certain semiconductor materials like gallium arsenide phosphide (GaAsP), this energy is released as visible light instead of heat. The color of the light depends on the semiconductor material used. LEDs are widely used as indicator lights, in seven-segment display units, in optical fiber communication, and as high-intensity LED bulbs for general lighting.
In simple words: Special diodes do special jobs. Zener diodes keep voltage steady, photodiodes turn light into electricity, and LEDs turn electricity into light. Each has a different use, like stabilizing power, sensing light, or making displays.

🎯 Exam Tip: For each special purpose diode, remember its unique characteristic (e.g., Zener: breakdown region, Photodiode: light detection, LED: light emission) and illustrate with its standard symbol.

 

Question 9. Draw circuit diagram for two input diode OR gate and AND gate. Also explain their working and give truth table.
Answer: Logic gates are fundamental building blocks of digital circuits that perform basic logical operations. They have one or more inputs and produce a single output based on logical relationships. Here, we'll look at the OR gate and AND gate, their symbols, truth tables, and diode-based circuit diagrams.

1. OR Gate:
An OR gate produces an output of 1 (high) if any of its inputs (A or B, or both) are 1. If all inputs are 0, the output is 0. This can be understood as "A OR B". The Boolean expression for a two-input OR gate is \( Y = A + B \).

OR Gate Symbol:

OR Gate using Switches:
This circuit uses two switches (A and B) connected in parallel with a battery and a bulb (Y). If either switch A or B (or both) are closed (1), current flows, and the bulb lights up (Y=1). If both switches are open (0), no current flows, and the bulb remains off (Y=0).

OR Gate using Diodes:
In a diode-based OR gate, two diodes (D1 and D2) are connected such that their anodes are the inputs (A and B), and their cathodes are joined together. This common point is connected to a resistor (R) and then to ground. The output (Y) is taken from this common cathode point. If either A or B (or both) are high (e.g., +5V), the corresponding diode becomes forward biased, allowing current to flow, and the output Y becomes high (1). If both A and B are low (0V), both diodes are reverse-biased, and the output remains low (0).

2. AND Gate:
An AND gate produces an output of 1 (high) only when all of its inputs (A AND B) are 1. If any input is 0, the output is 0. This can be understood as "A AND B". The Boolean expression for a two-input AND gate is \( Y = A \cdot B \).

AND Gate Symbol:

AND Gate using Switches:
This circuit uses two switches (A and B) connected in series with a battery and a bulb (Y). For the bulb to light up (Y=1), both switch A AND switch B must be closed (1). If either switch A or B (or both) are open (0), the circuit is broken, and the bulb remains off (Y=0).

AND Gate constructed by Diode:
In a diode-based AND gate, the anodes of two diodes (D1 and D2) are connected to the inputs (A and B). Their cathodes are joined together, and this common point is connected to ground through a resistor (R). A voltage source (e.g., +5V) is also connected to this common cathode point. The output (Y) is taken from the common cathode point. If both inputs A and B are high (+5V), both diodes are reverse-biased (or barely forward-biased if the output is high), and the output Y will be high. If any input is low (0V), that diode becomes forward-biased, pulling the common cathode point (and thus the output Y) to a low voltage (0V).
In simple words: Logic gates are like simple decision-makers in circuits. An OR gate turns "on" if any of its inputs are "on". An AND gate turns "on" only if ALL of its inputs are "on". We can build these using switches or diodes, and their behavior is shown in a truth table.

🎯 Exam Tip: When drawing circuit diagrams for logic gates, ensure correct component symbols and connections. Clearly label inputs and outputs. For truth tables, cover all possible input combinations.

 

RBSE Class 12 Physics Chapter 16 Numerical Questions

 

Question 1. Calculate the electric current generated in an intrinsic germanium plate at room temperature whose area is \( 2 \times 10^{-4} \text{ m}^2 \) and width is \( 1.2 \times 10^{-3} \text{ m} \) and a potential difference of 5V is applied. Intrinsic charge carrier density is \( 1.6 \times 10^6 \text{ /m}^3 \) for germanium at room temperature. The mobility of electrons and holes is \( 0.4 \text{ m}^2 \text{V}^{-1} \text{ s}^{-1} \) and \( 0.2 \text{ m}^2 \text{ V}^{-1} \text{ s}^{-1} \).
Answer: We need to calculate the electric current in the germanium plate. First, let's list the given values:
Area of plate, \( A = 2 \times 10^{-4} \text{ m}^2 \)
Width (length), \( l = 1.2 \times 10^{-3} \text{ m} \)
Potential difference, \( V = 5 \text{ V} \)
Intrinsic charge carrier density, \( n_i = 1.6 \times 10^6 \text{ /m}^3 \)
Mobility of electrons, \( \mu_e = 0.4 \text{ m}^2 \text{V}^{-1} \text{ s}^{-1} \)
Mobility of holes, \( \mu_h = 0.2 \text{ m}^2 \text{V}^{-1} \text{ s}^{-1} \)
The charge of an electron, \( e = 1.6 \times 10^{-19} \text{ C} \).

The formula for conductivity (\( \sigma \)) of an intrinsic semiconductor is given by:
\( \sigma = n_i e (\mu_e + \mu_h) \)
First, calculate the conductivity:
\( \sigma = (1.6 \times 10^6 \text{ m}^{-3}) \times (1.6 \times 10^{-19} \text{ C}) \times (0.4 \text{ m}^2 \text{V}^{-1} \text{ s}^{-1} + 0.2 \text{ m}^2 \text{V}^{-1} \text{ s}^{-1}) \)
\( \sigma = (1.6 \times 10^6) \times (1.6 \times 10^{-19}) \times (0.6) \)
\( \sigma = 1.536 \times 10^{-13} \text{ Sm}^{-1} \). This calculation shows how well the material conducts electricity.

Next, the current (I) can be calculated using the formula \( I = \frac{V}{R} \), where R is the resistance. Resistance (R) is related to resistivity (\( \rho \)), length (l), and area (A) by \( R = \frac{\rho l}{A} \). Since resistivity is the inverse of conductivity (\( \rho = \frac{1}{\sigma} \)), we can write the current as:
\( I = \frac{V \sigma A}{l} \)
Now, substitute the values:
\( I = \frac{5 \text{ V} \times (1.536 \times 10^{-13} \text{ Sm}^{-1}) \times (2 \times 10^{-4} \text{ m}^2)}{1.2 \times 10^{-3} \text{ m}} \)
\( I = \frac{15.36 \times 10^{-17}}{1.2 \times 10^{-3}} \)
\( I = 12.8 \times 10^{-14} \text{ A} \)
\( I = 1.28 \times 10^{-13} \text{ A} \)
In simple words: We find how easily electricity flows through the germanium by calculating its conductivity. Then, we use the voltage, conductivity, plate size, and length to figure out the total electric current flowing through it.

🎯 Exam Tip: Remember to use the correct charge of an electron \( (e = 1.6 \times 10^{-19} \text{ C}) \) when calculating conductivity, and ensure all units are consistent (SI units) throughout the problem.

 

Question 3. The current amplification is 0.99 for any transistor in common-base configuration. Calculate the change in collector current when there is 5.0 milliampere change in emitter current. What will be the change in base current?
Answer: We are given the following values for a transistor in common-base configuration:
Current amplification factor (\( \alpha \)) = 0.99
Change in emitter current (\( \Delta I_E \)) = 5.0 mA

First, we need to find the change in collector current (\( \Delta I_C \)). The definition of current amplification factor \( \alpha \) for a common-base configuration is:
\( \alpha = \frac{\Delta I_C}{\Delta I_E} \)
We can rearrange this formula to solve for \( \Delta I_C \):
\( \Delta I_C = \alpha \times \Delta I_E \)
Substitute the given values:
\( \Delta I_C = 0.99 \times 5.0 \text{ mA} \)
\( \Delta I_C = 4.95 \text{ mA} \). This means the collector current changes by 4.95 mA.

Next, we need to find the change in base current (\( \Delta I_B \)). In a transistor, the emitter current is the sum of the collector current and the base current:
\( \Delta I_E = \Delta I_B + \Delta I_C \)
We can rearrange this to solve for \( \Delta I_B \):
\( \Delta I_B = \Delta I_E - \Delta I_C \)
Substitute the calculated and given values:
\( \Delta I_B = 5.0 \text{ mA} - 4.95 \text{ mA} \)
\( \Delta I_B = 0.05 \text{ mA} \). Since 1 mA = \( 10^{-3} \) A and 1 \( \mu \)A = \( 10^{-6} \) A, this is \( 0.05 \times 10^{-3} \text{ A} = 50 \times 10^{-6} \text{ A} = 50 \, \mu\text{A} \). This small current change in the base controls the larger change in the collector current.
In simple words: We used a transistor rule to find out how much the collector current changes when the emitter current changes. Then, using another rule that connects all three currents, we found the tiny change in the base current.

🎯 Exam Tip: Remember the relationship \( I_E = I_B + I_C \) and the definitions of \( \alpha \) (for common-base) and \( \beta \) (for common-emitter) to correctly solve transistor current problems.

 

Question 4. For a PN junction the average value of a potential well is 0.1 V and the electric field \( 10^5 \text{ V/m} \) is present at junction region. What will be the width of depletion layer for this junction?
Answer: We need to find the width of the depletion layer for the PN junction. Let's write down the given information:
Average value of potential well (potential barrier), \( V_B = 0.1 \text{ V} \)
Electric field at the junction region, \( E = 10^5 \text{ V/m} \).

The relationship between the potential difference (voltage), electric field, and distance (width) in a uniform electric field is given by:
\( V_B = E \times d \)
Where \( d \) is the width of the depletion layer. We can rearrange this formula to solve for \( d \):
\( d = \frac{V_B}{E} \)
Now, substitute the given values into the formula:
\( d = \frac{0.1 \text{ V}}{10^5 \text{ V/m}} \)
\( d = 0.1 \times 10^{-5} \text{ m} \)
\( d = 1 \times 10^{-1} \times 10^{-5} \text{ m} \)
\( d = 1 \times 10^{-6} \text{ m} \)
So, the width of the depletion layer is \( 1 \times 10^{-6} \text{ m} \), which is also equal to 1 micrometer (\( 1 \, \mu\text{m} \)). This layer is a key part of how the PN junction works.
In simple words: We used the given voltage across the junction and the electric field strength to calculate how wide the depletion layer is. This layer is very thin, typically measured in micrometers.

🎯 Exam Tip: Remember the simple relationship between potential difference (voltage), electric field, and distance for calculations involving depletion layer width. Ensure consistent units.

 

Question 6. A transistor is connected in common emitter configuration. A power supply of 8V is there in the collector circuit and the potential drop of 0.5V is on the resistance of 800 \( \Omega \) connected in series with the collector. If current amplification factor is \( \alpha = 0.96 \). Then calculate base current.
Answer: We need to calculate the base current for the transistor in common emitter configuration. Let's list the given values:
Collector circuit power supply, \( V_{CC} = 8 \text{ V} \)
Potential drop across collector resistor, \( V_{out} = 0.5 \text{ V} \)
Collector resistor, \( R_L = 800 \, \Omega \)
Current amplification factor, \( \alpha = 0.96 \)

First, we can find the collector current (\( I_C \)) using Ohm's Law for the potential drop across \( R_L \):
\( V_{out} = I_C \times R_L \)
So, \( I_C = \frac{V_{out}}{R_L} \)
Substitute the values:
\( I_C = \frac{0.5 \text{ V}}{800 \, \Omega} \)
\( I_C = 0.000625 \text{ A} \)
Converting to milliamperes: \( I_C = 0.625 \text{ mA} \). This is the current flowing through the collector.

Next, we need to find the base current (\( I_B \)). We know the relationship between \( \alpha \) and \( \beta \) (current gain in common emitter configuration):
\( \beta = \frac{\alpha}{1 - \alpha} \)
Substitute the value of \( \alpha \):
\( \beta = \frac{0.96}{1 - 0.96} = \frac{0.96}{0.04} \)
\( \beta = 24 \). This value indicates how much the base current amplifies the collector current.

Now, we can use the definition of \( \beta \) to find the base current:
\( \beta = \frac{I_C}{I_B} \)
Rearranging for \( I_B \):
\( I_B = \frac{I_C}{\beta} \)
Substitute the calculated \( I_C \) and \( \beta \):
\( I_B = \frac{0.625 \text{ mA}}{24} \)
\( I_B \approx 0.02604 \text{ mA} \)
To express in microamperes: \( I_B \approx 26.04 \, \mu\text{A} \). This small base current controls a significantly larger collector current.
In simple words: We first found the collector current using the voltage drop across the resistor. Then, we calculated the current amplification factor for this type of transistor. Finally, we used this factor and the collector current to find the very small base current.

🎯 Exam Tip: Remember the relationship between \( \alpha \) (common-base current gain) and \( \beta \) (common-emitter current gain) to switch between configurations if needed. Use Ohm's Law carefully for voltage drops.

 

Question 7. A resistor \( R_L = 2 \, \text{k}\Omega \) ends has a potential difference of 15 V. The lowest working current of zener diode is 10 mA. What is the current in \( R_L \)? What is the total current in the circuit?
Answer: We need to calculate the current through the load resistor and the total circuit current. Let's list the given information:
Load resistance, \( R_L = 2 \, \text{k}\Omega = 2 \times 10^3 \, \Omega \)
Potential difference across load resistance, \( V = 15 \text{ V} \)
Lowest working current of Zener diode (Zener current), \( I_Z = 10 \text{ mA} \)

First, calculate the current in the load resistor (\( I_L \)) using Ohm's Law:
\( I_L = \frac{V}{R_L} \)
Substitute the values:
\( I_L = \frac{15 \text{ V}}{2 \times 10^3 \, \Omega} \)
\( I_L = 7.5 \times 10^{-3} \text{ A} \)
Converting to milliamperes: \( I_L = 7.5 \text{ mA} \). This is the current flowing through the load connected to the Zener diode.

Now, to find the total current in the circuit (\( I \)), we need to consider that the Zener diode and the load resistance are connected in parallel in a voltage regulator circuit. This means the total current supplied to this parallel combination is the sum of the Zener current and the load current.
\( I = I_L + I_Z \)
Substitute the calculated \( I_L \) and given \( I_Z \):
\( I = 7.5 \text{ mA} + 10 \text{ mA} \)
\( I = 17.5 \text{ mA} \). This total current indicates the power delivered to the regulated output.
The problem states that the Zener diode and load resistance are in parallel, so the potential difference across the Zener diode is also 15 V (equal to the potential across \( R_L \)).
In simple words: We first found the current through the load resistor using the voltage and its resistance. Then, knowing the Zener diode's current, we added it to the load current to get the total current flowing into the setup.

🎯 Exam Tip: For Zener diode problems, remember that in a parallel regulating circuit, the voltage across the Zener diode is the regulated output voltage, and the total current supplied splits between the Zener diode and the load.

The provided pages (43 and 44) of the document contain only navigation links, SEO titles, and copyright information, which fall under the "IGNORE AND SKIP — PAGE HEADER / SEO TITLES" and "IGNORE AND SKIP — FOOTER / NAVIGATION" rules. There are no actual questions, answers, or educational content in this specific page range to convert into HTML. Therefore, the output is empty.