← MES Jr. Grade
Past Questions — 2022
To address the memory 14 bits are used. Then what is the address of the last memory location?
A)
16382
B)
16383
C)
16384
D)
16385
Step-by-step calculation:
1. The number of address lines (n) is given as 14 bits.
2. The total number of addressable memory locations is calculated using the formula 2^n.
Therefore, Total Locations = 2^14 = 16,384.
3. In computer memory, addressing starts from 0. This means the first memory location is at address 0.
4. The address of the last memory location is therefore (Total Locations - 1).
Last Address = 16,384 - 1 = 16,383.
Analysis of options:
- Option (a) 16382: Incorrect, this is the second to last location.
- Option (b) 16383: Correct, as it represents the maximum index for 2^14 locations starting from 0.
- Option (c) 16384: Incorrect, this is the total count of locations, not the address of the last one.
- Option (d) 16385: Incorrect, this exceeds the addressable range of 14 bits.
In d.c choppers, per unit ripple is maximum when duty cycle a is :
A)
0.2
B)
0.5
C)
0.7
D)
0
In a DC chopper (specifically a step-down or buck converter), the output current ripple (ΔI) is given by the formula: ΔI = (V_s * α * (1 - α)) / (f * L), where V_s is the source voltage, α is the duty cycle, f is the switching frequency, and L is the inductance. The per unit ripple is defined as the ratio of the peak-to-peak ripple current (ΔI) to the average output current (I_avg). Since I_avg = α * V_s / R (or proportional to α), the per unit ripple is proportional to (1 - α). However, in the context of the ripple magnitude itself (ΔI), the expression α(1 - α) is a quadratic function. To find the maximum value of α(1 - α), we take the derivative with respect to α and set it to zero: d/dα (α - α²) = 1 - 2α = 0, which gives α = 0.5. At α = 0.5, the current ripple reaches its maximum value. Therefore, the ripple is maximum when the duty cycle is 0.5.
The output of a JK flip-flop toggles when :
A)
J=1 and K=0
B)
J=0 and K=1
C)
J=1 and K=1
D)
J=0 and K=0
A JK flip-flop is a universal flip-flop that improves upon the SR flip-flop by eliminating the invalid state. Its operation is defined by the following truth table:
1. J=0, K=0: No Change (Hold state). The output remains the same as the previous state.
2. J=0, K=1: Reset state. The output (Q) becomes 0 regardless of the previous state.
3. J=1, K=0: Set state. The output (Q) becomes 1 regardless of the previous state.
4. J=1, K=1: Toggle state. The output (Q) changes to its complement (if it was 0, it becomes 1; if it was 1, it becomes 0).
Therefore, the condition for the output to toggle is when both inputs J and K are high (1).
In a commutation circuit employed to turn-off an SCR, satisfactory turn-off is obtained when :
A)
Circuit turn-off time < device turn-off time
B)
Circuit turn-off time > device turn-off time
C)
Circuit time constant > device turn-off time
D)
Circuit time constant < device turn-off time
To successfully turn off a Silicon Controlled Rectifier (SCR), two conditions must be met: first, the current must be reduced below the holding current, and second, a reverse voltage must be applied across the device for a sufficient duration to allow the internal charge carriers to recombine.
1. **Device Turn-off Time (tc):** This is a characteristic property of the SCR. It is the minimum time required for the device to regain its forward-blocking capability after the current has dropped to zero.
2. **Circuit Turn-off Time (tc'):** This is the time provided by the external commutation circuit during which the SCR is reverse-biased.
For a satisfactory turn-off (to prevent the device from accidentally re-triggering or failing to block forward current), the time provided by the circuit (tc') must be greater than the time required by the device (tc). If tc' < tc, the device will not have fully recovered its blocking state before the forward voltage is reapplied, leading to an immediate conduction (failure to turn off). Therefore, the condition for successful commutation is: Circuit turn-off time > device turn-off time.
Which one of the following instructions is a 3-byte instruction?
A)
MVI A
B)
LDAX B
C)
JMP 2050
D)
MOV A, M
The question refers to the 8085 Microprocessor instruction set. Instructions in the 8085 are categorized by their size into 1-byte, 2-byte, or 3-byte instructions.
1. **MOV A, M (Option d):** This is a register-to-register/memory move instruction. It requires only the opcode to specify the source and destination. It is a **1-byte** instruction.
2. **LDAX B (Option b):** This is 'Load Accumulator Indirect'. It uses the BC register pair to find the address. Since the address is already in the registers, only the opcode is needed. It is a **1-byte** instruction.
3. **MVI A (Option a):** This is 'Move Immediate to Accumulator'. It requires one byte for the opcode and one byte for the 8-bit data to be loaded into the accumulator. It is a **2-byte** instruction.
4. **JMP 2050 (Option c):** This is an unconditional jump to a specific 16-bit memory address (2050H). It requires one byte for the opcode (JMP) and two bytes to store the 16-bit address (low-order byte and high-order byte). Therefore, it is a **3-byte** instruction.
In a RC coupled amplifier, the gain decreases in the frequency response due to the :
A)
coupling capacitor at low frequency and bypass capacitor at high frequency
B)
coupling capacitor at high frequency and bypass capacitor at low frequency
C)
coupling junction capacitance at low frequency and coupling capacitor at high frequency
D)
device junction capacitor at high frequency and coupling capacitor at low frequency
In an RC coupled amplifier, the frequency response is characterized by a band-pass behavior where the gain is constant in the mid-band but drops at low and high frequencies.
1. **Low-Frequency Response:** At low frequencies, the reactance of the coupling capacitors ($X_C = 1 / 2\pi fC$) and bypass capacitors becomes very high. This causes a significant voltage drop across these capacitors, reducing the signal reaching the next stage or the emitter, thereby decreasing the overall gain. Thus, the coupling capacitor is responsible for the gain drop at low frequencies.
2. **High-Frequency Response:** At high frequencies, the internal parasitic capacitances of the transistor (junction capacitances, such as $C_{be}$ and $C_{bc}$) and stray wiring capacitances become significant. Their reactance ($X_C = 1 / 2\pi fC$) decreases as frequency increases, effectively shunting the signal to ground. This 'loading effect' reduces the voltage gain of the amplifier. Thus, the device junction capacitors are responsible for the gain drop at high frequencies.
Comparing this to the options:
- Option (a) is incorrect because bypass capacitors affect low frequency, not high frequency.
- Option (b) is incorrect because coupling capacitors affect low frequency, not high frequency.
- Option (c) is incorrect because junction capacitance affects high frequency, not low frequency.
- Option (d) correctly identifies that device junction capacitors cause the drop at high frequency and coupling capacitors cause the drop at low frequency.
In high speed TDM, the channels are separated in the receiver employing :
A)
OR gate
B)
AND gate
C)
NAND gate
D)
NOR gate
In Time Division Multiplexing (TDM), multiple signals are transmitted over a single channel by dividing the time into slots. At the receiver end, a demultiplexer is used to separate these channels. In high-speed digital TDM, the demultiplexing process involves 'gating' the incoming composite signal. An AND gate acts as a synchronization switch: one input is the composite TDM signal, and the other input is a synchronization pulse (clock signal) specific to that channel. When the synchronization pulse is 'HIGH' (1), the AND gate allows the signal for that specific channel to pass through; when it is 'LOW' (0), the output is zero. This effectively 'picks out' the time slots belonging to a specific channel, separating it from the others. OR, NAND, and NOR gates do not perform this specific 'windowing' or 'gating' function required for signal separation in this context.
If a diode is connected in anti-parallel with a thyristor, then :
A)
both turn-off power loss and turn-off time decreases
B)
turn-off power loss and turn-off time increase
C)
turn-off power loss increase, but turn-off time decreases
D)
none of the above
In a thyristor (SCR), the turn-off process involves the removal of charge carriers from the junction. When a diode is connected in anti-parallel (reverse parallel) with the thyristor, it provides an alternative path for the current to flow during the commutation process. Specifically, during the turn-off phase, the anti-parallel diode allows the reverse recovery current to flow more efficiently and helps in faster removal of the stored charges from the thyristor's gate and junctions. This reduces the 'turn-off time' (tc). Because the device spends less time in the transition region where both voltage and current are present, the energy dissipated during the turn-off process is reduced, thereby decreasing the 'turn-off power loss'. Therefore, both the turn-off time and the associated power loss decrease.
The channel capacity of a noiseless channel is equal to :
A)
rate at which information is transmitted
B)
signaling speed
C)
bandwidth
D)
bandwidth-SNR product
In communication theory, the channel capacity (C) is defined as the maximum rate at which information can be transmitted over a communication channel with an arbitrarily small error probability. For a noiseless channel, according to the Nyquist formula, the capacity is determined by the bandwidth and the number of discrete signaling levels. Specifically, for a noiseless channel, the capacity is the maximum information rate (bits per second) that the channel can support.
Analysis of options:
- Option (a) is correct because channel capacity is fundamentally the maximum rate of information transmission.
- Option (b) 'signaling speed' refers to the baud rate (symbols per second), which is not the same as the information rate (bits per second) unless each symbol carries exactly one bit.
- Option (c) 'bandwidth' is a physical property of the channel (measured in Hz), not the capacity (measured in bps).
- Option (d) 'bandwidth-SNR product' refers to the Shannon-Hartley theorem (C = B log2(1 + SNR)), which specifically defines the capacity of a NOISY channel. In a noiseless channel, SNR is effectively infinite, making this formula inapplicable in its standard form.
Push-pull amplifier circuit is used as :
A)
power amplifier
B)
audio amplifier
C)
RF amplifier
D)
none of these
A push-pull amplifier is a specific configuration of a power amplifier where two complementary transistors (typically one NPN and one PNP) are used to amplify the positive and negative halves of the input signal separately.
1. Why it is a Power Amplifier: The primary goal of a push-pull configuration is to deliver a large amount of power to a low-impedance load (like a loudspeaker) while maintaining high efficiency. By splitting the signal, it reduces the power dissipation in the transistors compared to a single-ended Class A amplifier.
2. Why 'audio amplifier' (b) is not the best answer: While push-pull amplifiers are very common in audio systems, 'audio amplifier' describes the application/frequency range, whereas 'power amplifier' describes the functional category of the circuit. A push-pull circuit is a method of implementing a power amplifier, regardless of whether the signal is audio or another low-frequency signal.
3. Why 'RF amplifier' (c) is incorrect: RF (Radio Frequency) amplifiers typically use different topologies (like cascaded stages or specific tuned circuits) to handle high-frequency stability and gain, although push-pull can be used in RF power stages, it is the defining characteristic of power amplification in general electronics textbooks.
Which of the following is a non vectored hardware interrupt of 8085?
A)
TRAP
B)
RST 7.5
C)
INTR
D)
All of these
In the 8085 microprocessor, interrupts are categorized into two types: Vectored and Non-Vectored.
1. Vectored Interrupts: These are interrupts where the CPU automatically knows the address of the Interrupt Service Routine (ISR) to jump to. The 8085 has five vectored interrupts: TRAP, RST 7.5, RST 6.5, RST 5.5, and RST 0.
- TRAP is the highest priority non-maskable vectored interrupt.
- RST 7.5 is a maskable vectored interrupt.
2. Non-Vectored Interrupts: These are interrupts where the CPU does not have a predefined address for the ISR. The external device providing the interrupt must provide the address of the ISR on the data bus during the interrupt acknowledge cycle.
- INTR (Interrupt Request) is the only non-vectored hardware interrupt in the 8085. When INTR is activated, the microprocessor responds by issuing an INTA (Interrupt Acknowledge) signal, and the external hardware must then place an instruction (usually a RST n or CALL instruction) on the data bus to tell the CPU where to go.
Therefore, TRAP and RST 7.5 are vectored, making INTR the correct choice.
In a single-phase full converter, the output voltage during overlap is equal to :
A)
zero
B)
source voltage
C)
source voltage minus the inductance drop
D)
none of these
In a single-phase full converter (fully controlled bridge), 'overlap' occurs due to the source inductance (L_s). During the overlap period, two thyristors in the same leg (or opposite legs depending on the commutation phase) are conducting simultaneously. Specifically, during the commutation interval, the outgoing thyristor and the incoming thyristor both conduct. Because they are connected in parallel across the AC source during this brief transition, the load is effectively connected directly across the AC source. Therefore, the output voltage during the overlap period is equal to the instantaneous source voltage (v_s).
- Option (a) is incorrect because the voltage does not drop to zero; it follows the source waveform.
- Option (c) is incorrect because the 'inductance drop' typically refers to the voltage drop across the source inductor that causes the delay in commutation, but the net voltage delivered to the load during the overlap period is the source voltage itself.
- Option (d) is incorrect as option (b) is the standard theoretical result for power electronics converters.
In a three phase half-wave rectifier, each diode conducts for a duration of
A)
180o
B)
30o
C)
60o
D)
120o
In a three-phase half-wave rectifier, there are three diodes, each connected to one phase of the three-phase supply. The diodes conduct based on which phase has the highest instantaneous voltage.
1. The three-phase supply voltages are displaced by 120° from each other.
2. A diode will conduct only when its anode voltage is higher than the cathode voltage and higher than the voltages of the other two phases.
3. Each phase voltage remains the most positive for a duration of 120° (specifically, from 30° before its peak to 30° after its peak relative to the zero crossing, or more accurately, from the point where it intersects the previous phase to where it intersects the next phase).
4. Therefore, each diode conducts for exactly 1/3 of the total cycle (360° / 3 = 120°).
Comparison with other options:
- 180° is the conduction angle for diodes in a single-phase full-wave rectifier or a three-phase full-wave bridge rectifier.
- 60° and 30° do not correspond to the standard conduction intervals of a half-wave three-phase configuration.
A microprocessor with a 12-bit address bus will be able to access :
A)
16 kilobytes of memory
B)
4 kilobytes of memory
C)
8 kilobytes of memory
D)
none of these
Step-by-step calculation:
1. The number of addressable memory locations is determined by the formula 2^n, where 'n' is the number of bits in the address bus.
2. Given n = 12 bits, the total number of locations = 2^12.
3. 2^12 = 4096 locations.
4. In standard microprocessor architecture, each memory location typically stores 1 byte (8 bits) of data.
5. Therefore, the total memory capacity = 4096 bytes.
6. Since 1 kilobyte (KB) = 1024 bytes, we calculate: 4096 / 1024 = 4 KB.
Comparing with options:
- Option (a) 16 KB would require a 14-bit address bus (2^14 = 16384).
- Option (b) 4 KB matches our calculation (2^12 = 4096).
- Option (c) 8 KB would require a 13-bit address bus (2^13 = 8192).
The disadvantage of FM over AM is that
A)
noise is very high for high frequency signals
B)
larger bandwidth is required
C)
high modulating power is required
D)
high output power is required
To determine the disadvantage of Frequency Modulation (FM) compared to Amplitude Modulation (AM), we must compare their fundamental characteristics:
1. **Bandwidth**: In AM, the bandwidth is simply twice the highest modulating frequency (BW = 2fm). In FM, the bandwidth is determined by Carson's Rule (BW ≈ 2(Δf + fm)), where Δf is the frequency deviation. Because Δf is typically much larger than the modulating frequency to achieve a good signal-to-noise ratio, FM requires a significantly larger bandwidth than AM. This is the primary disadvantage of FM.
2. **Noise (Option a)**: This is incorrect. One of the main advantages of FM over AM is its superior noise immunity. Since noise primarily affects the amplitude of a signal, an FM receiver can use a 'limiter' to clip amplitude variations without losing the frequency-encoded information.
3. **Modulating Power (Option c)**: This is incorrect. The power required for modulation is not the defining disadvantage of FM; rather, it is the complexity of the circuitry.
4. **Output Power (Option d)**: This is incorrect. In FM, the transmitted power remains constant regardless of the modulation index, whereas in AM, the power varies with the modulation depth. FM does not inherently require 'higher' output power to function; it simply uses its power more efficiently to combat noise.
The thermal noise is due to random :
A)
motion of atoms and molecules
B)
motion of free electrons
C)
vibration of atoms about their mean position, inside the conducting medium
D)
none of these
Thermal noise, also known as Johnson-Nyquist noise, is the electronic noise generated by the thermal agitation of the charge carriers (usually free electrons) inside an electrical conductor at finite temperature.
1. **Mechanism**: In a conductor, electrons are in constant, random motion due to thermal energy. Even in the absence of an applied voltage, these random fluctuations in the position and velocity of the free electrons create a small, fluctuating voltage across the terminals of the conductor.
2. **Mathematical Relation**: The root-mean-square (RMS) noise voltage is given by the formula V = sqrt(4kTRB), where k is Boltzmann's constant, T is absolute temperature, R is resistance, and B is bandwidth. This shows that the noise is directly proportional to the temperature (T), confirming it is a thermal effect.
3. **Why other options are incorrect**:
- Option (a) and (c) refer to the motion or vibration of atoms and molecules. While thermal energy does cause atomic vibration (phonons), the electrical noise specifically measured as 'thermal noise' in communication systems is the result of the movement of the *charge carriers* (electrons) which creates the electric current/voltage fluctuations, not the vibration of the neutral atomic nuclei themselves.
How many machine cycle does 8085 microprocessor has?
A)
11
B)
7
C)
5
D)
16
In the 8085 microprocessor, a machine cycle is the time required to complete one operation of accessing memory or an I/O device. There are 5 basic types of machine cycles used to execute instructions: 1. Opcode Fetch (used to fetch the instruction from memory), 2. Memory Read (used to read data from memory), 3. Memory Write (used to write data to memory), 4. I/O Read (used to read data from an input port), and 5. I/O Write (used to write data to an output port). Every instruction is composed of one or more of these machine cycles. Options a, b, and d are incorrect as they do not correspond to the standard architectural definition of machine cycles in the 8085.
How many flags do 8085 microprocessor has?
A)
5
B)
9
C)
11
D)
None of these
The 8085 microprocessor has a 8-bit Flag Register (also known as the Program Status Word). Although the register is 8 bits wide, only 5 bits are used as status flags to indicate the results of arithmetic and logical operations. These five flags are:
1. Carry Flag (CY): Set if there is a carry out from the most significant bit (MSB).
2. Parity Flag (P): Set if the result contains an even number of 1s.
3. Auxiliary Carry Flag (AC): Set if there is a carry from bit D3 to D4 (used for BCD arithmetic).
4. Zero Flag (Z): Set if the result of an operation is zero.
5. Sign Flag (S): Set if the MSB of the result is 1 (indicating a negative number in 2's complement).
The other 3 bits in the flag register are undefined/unused. Therefore, the correct number of flags is 5.
In double sideband suppressed carrier modulation, the modulated wave undergoes phase reversal, whenever :
A)
modulating signal’s amplitude decreases
B)
modulating signal’s amplitude increases
C)
modulating signal crosses zero
D)
carrier signal crosses zero
In Double Sideband Suppressed Carrier (DSB-SC) modulation, the modulated wave is represented by the equation s(t) = m(t) * c(t), where m(t) is the modulating (message) signal and c(t) is the carrier signal.
1. The carrier signal c(t) is typically a high-frequency sine or cosine wave: c(t) = Ac cos(2πfc t).
2. The resulting modulated wave is s(t) = m(t) * Ac cos(2πfc t).
3. The phase of the carrier is determined by the sign of the modulating signal m(t).
4. When m(t) is positive, the carrier maintains its original phase.
5. When m(t) becomes negative, the term m(t) * Ac cos(2πfc t) becomes equivalent to |m(t)| * (-Ac cos(2πfc t)), which is a 180-degree phase shift (phase reversal) of the carrier.
6. Therefore, every time the modulating signal m(t) crosses the zero axis (changing from positive to negative or vice versa), the phase of the modulated wave reverses.
Comparing the options:
- Option (a) and (b): Changes in amplitude affect the envelope of the signal, not the phase.
- Option (d): The carrier signal crosses zero naturally many times per cycle; this does not cause a phase reversal of the modulation envelope.
- Option (c): This is the correct condition for phase reversal in DSB-SC.
Which of the following amplifier operations has the highest efficiency?
A)
Class A amplifier
B)
Class B amplifier
C)
Class C amplifier
D)
None of these
Efficiency in power amplifiers is defined as the ratio of the AC power delivered to the load to the DC power drawn from the supply.
1. **Class A Amplifier**: The transistor conducts for the full 360° of the input cycle. Because it draws current even when there is no input signal, it has the lowest efficiency. The maximum theoretical efficiency for a series-fed Class A amplifier is 25%, and 50% for a transformer-coupled one.
2. **Class B Amplifier**: The transistor conducts for only 180° (half) of the input cycle. This significantly reduces power wastage. The maximum theoretical efficiency is approximately 78.5%.
3. **Class C Amplifier**: The transistor conducts for less than 180° (a small fraction) of the input cycle. Because the conduction angle is very small, the DC power consumption is minimized while the output is maintained using a tuned LC resonant circuit. This allows Class C amplifiers to achieve the highest theoretical efficiency, often exceeding 80% and reaching up to 90% or more.
Comparing the three, Class C is the most efficient, although it introduces significant distortion and is primarily used in RF applications.
In a 3-phase controlled bridge rectifier, with an increase of overlap angle the output DC voltage :
A)
decreases
B)
increases
C)
does not change
D)
depends upon load inductance
In a 3-phase controlled bridge rectifier, the 'overlap angle' (denoted as μ) occurs due to the source inductance (commutation inductance). In an ideal scenario (zero source inductance), the current transfers instantaneously from one thyristor to the next. However, in practical circuits, the source inductance prevents the current from changing instantaneously. During the overlap period, two thyristors in the same group are conducting simultaneously. This creates a short-circuit path across the DC output for a brief duration, effectively averaging the output voltage between the outgoing and incoming phases. The mathematical expression for the average output voltage (Vdc) of a 3-phase fully controlled bridge rectifier considering overlap is: Vdc = (3√3 Vm / π) * [cos(α) - (μ/2π)sin(α)] (approximately, or more precisely involving the integral of the voltage during overlap). As the overlap angle μ increases, the voltage drop across the source inductance increases, which reduces the average DC output voltage. Therefore, an increase in the overlap angle leads to a decrease in the output DC voltage.
Low frequency response of an RC coupled amplifier can be improved by :
A)
increasing the coupling capacitors only
B)
increasing the by-pass capacitors only
C)
increasing the by-pass capacitor as well as coupling capacitor
D)
decreasing the by-pass capacitor
In an RC coupled amplifier, the low-frequency response is limited by the reactance of the coupling capacitors (Cc) and the bypass capacitors (Ce).
1. **Coupling Capacitors (Cc):** These capacitors are used to block DC and allow AC signals to pass from one stage to the next. The capacitive reactance is given by Xc = 1 / (2πfC). At low frequencies (f), the reactance (Xc) increases, causing a significant voltage drop across the capacitor and reducing the gain of the amplifier. To lower this reactance and allow low-frequency signals to pass more effectively, the value of the coupling capacitor must be increased.
2. **Bypass Capacitors (Ce):** These capacitors are used to bypass the emitter (or source) resistor to prevent a reduction in gain due to negative feedback (degeneration). Similarly, at low frequencies, the reactance of the bypass capacitor increases, which means the emitter resistor is no longer fully bypassed. This increases the negative feedback and decreases the voltage gain. Increasing the bypass capacitor reduces its reactance at low frequencies, maintaining the gain.
Therefore, to improve the low-frequency response (i.e., to extend the lower cutoff frequency to a lower value), both the coupling and bypass capacitors should be increased.
Each diode of a 3-phase half-wave diode rectifier conducts for :
A)
60o
B)
120o
C)
180o
D)
90o
In a 3-phase half-wave rectifier, there are three diodes, each connected to one phase of the 3-phase supply. The circuit is designed such that the diode connected to the phase with the highest instantaneous voltage conducts, while the others are reverse-biased.
1. The phase voltages are displaced by 120° from each other.
2. A specific diode starts conducting when its phase voltage becomes the most positive among the three phases. This occurs at 30° relative to the phase's own zero-crossing (or -30° relative to the peak).
3. The diode continues to conduct until the next phase voltage exceeds its own voltage.
4. Since the phases are separated by 120°, each diode conducts for exactly one-third of the total cycle (360° / 3 = 120°).
Therefore, the conduction angle for each diode is 120°.
Comparison with other options:
- 60°: Incorrect; this is not the conduction period for a standard half-wave rectifier.
- 180°: Incorrect; this is the conduction angle for diodes in a single-phase full-wave bridge rectifier.
- 90°: Incorrect; this does not correspond to the symmetry of a 3-phase system.
Turn-on and turn-off times of transistor depend on :
A)
static characteristic
B)
junction capacitances
C)
current gain
D)
voltage injection
The turn-on and turn-off times of a transistor (switching speed) are primarily determined by the time it takes to charge and discharge the internal capacitances of the device. In a Bipolar Junction Transistor (BJT), these are the depletion region capacitances of the emitter-base and collector-base junctions.
1. **Turn-on time**: This involves charging the junction capacitances to the required threshold voltage to move the transistor from the cutoff region to the saturation region.
2. **Turn-off time**: This is dominated by the 'storage time,' which is the time required to remove the excess minority carriers stored in the base region and discharge the junction capacitances to move the transistor from saturation back to cutoff.
Why other options are incorrect:
- **Static characteristics (a)**: These describe the DC behavior (like V-I curves) and do not account for the time-dependent (transient) behavior of the device.
- **Current gain (c)**: While current gain ($\beta$) affects the amount of current flowing, it does not directly dictate the speed of the charging/discharging process of the junctions.
- **Voltage injection (d)**: This is not a standard term used to define the intrinsic switching speed of a transistor; the speed is a property of the device's physical construction and capacitance.
The drawback of FM relative to AM is that :
A)
noise is very high for high modulation frequencies
B)
larger bandwidth is required
C)
higher modulating power is required
D)
higher output power is required
To determine the drawback of Frequency Modulation (FM) relative to Amplitude Modulation (AM), we compare their fundamental characteristics:
1. **Bandwidth**: In AM, the bandwidth is simply twice the highest modulating frequency (BW = 2fm). In FM, the bandwidth is determined by Carson's Rule (BW ≈ 2(Δf + fm)), where Δf is the frequency deviation. Because Δf is typically much larger than the modulating frequency to achieve a good signal-to-noise ratio, FM requires a significantly larger portion of the frequency spectrum than AM. This is the primary technical drawback of FM.
2. **Noise Performance**: FM is actually superior to AM in terms of noise. Most noise (like atmospheric or electrical noise) affects the amplitude of a signal. Since FM carries information in the frequency variations and uses 'limiters' to clip amplitude spikes, it is much more resistant to noise than AM. Therefore, option (a) is incorrect.
3. **Power Requirements**: In FM, the amplitude of the carrier remains constant regardless of the modulation index. This means the transmitted power is constant and does not increase with modulation depth, unlike AM where power increases with the modulation index. Therefore, options (c) and (d) are incorrect as FM does not inherently require 'higher' power to operate; in fact, it is more power-efficient in terms of signal quality.
Conclusion: The trade-off for the superior noise immunity and audio quality of FM is the requirement for a much larger bandwidth.