← MES Jr. Grade
Past Questions
Time division multiplex
A)
Can be used with PCM only
B)
Combines five groups into a supergroup
C)
Stacks 24 channels in adjacent frequency slots
D)
Interleaves pulses belonging to different transmission
Time division multiplexing (TDM) shares a single communication channel by assigning different time slots to different signals, so their pulses are interleaved in time. Hence, option d is correct: it interleaves pulses belonging to different transmissions.
Why the others are incorrect:
- a: TDM is not limited to PCM only, though it is commonly used with PCM systems.
- c: Stacking 24 channels in adjacent frequency slots describes frequency division multiplexing (FDM), not TDM.
- b: Combining five groups into a supergroup is also an FDM hierarchy term, not a definition of TDM.
Reverse recovery current in a diode depends upon
A)
Forward field current
B)
Storage charge
C)
Temperature
D)
PIV
Reverse recovery current in a diode is mainly determined by the stored minority-carrier charge in the junction during forward conduction. When the diode is suddenly reverse-biased, this stored charge must be removed before the diode can block reverse current, so a larger storage charge produces a larger and longer reverse recovery current.
Why others are incorrect:
- PIV (peak inverse voltage) is the maximum reverse voltage a diode can withstand; it does not directly determine reverse recovery current.
- Forward field current is not the controlling factor; the key issue is the stored charge left from forward conduction.
- Temperature can affect diode behavior indirectly, but reverse recovery current is primarily governed by storage charge, not temperature as the main parameter.
The most common modulation system used for telegraphy is
A)
Frequency shift keying
B)
Two tone modulation
C)
Pulse code modulation
D)
Single tone modulation
Frequency shift keying (FSK) is the most common modulation system used for telegraphy because telegraphy transmits digital on-off or binary-type signals, and FSK represents these signals by shifting the carrier between two frequencies. It is robust, simple, and well suited to noisy communication channels.
Why the others are incorrect:
- Single tone modulation: This is not the standard method for telegraphy; it does not efficiently represent binary telegraph signals.
- Two tone modulation: This is not the usual term or primary system for telegraphy in standard communication practice.
- Pulse code modulation (PCM): PCM is used for digitizing analog signals, especially in telephony and data transmission, not as the common modulation system for telegraphy.
Race around condition always arise in a
A)
Combination circuit
B)
Asynchronous circuit
C)
Synchronous circuit
D)
Digital circuit
Race-around condition is a problem associated with level-triggered flip-flops in asynchronous operation, especially the JK flip-flop when J = K = 1 and the clock pulse remains active long enough for the output to toggle repeatedly. Hence, it always arises in an asynchronous circuit.
Why others are incorrect:
- a) Combination circuit: It has no memory elements or feedback timing issues, so race-around does not apply.
- c) Synchronous circuit: Operations are controlled by a common clock, which reduces timing hazards; race-around is not the general characteristic here.
- d) Digital circuit: This is too broad a term; race-around is not inherent to all digital circuits, only to specific asynchronous sequential circuits.
The purpose of free-wheeling in a thyristor controlled ac to dc converter is to
A)
Reduce the current of its associated thyristor to zero so that communication can take place
B)
Share the load current of its associated thyristor
C)
Conduct the load current when its associated thyristor is turned off
D)
Maintain voltage across load at constant value
Free-wheeling in a thyristor-controlled AC to DC converter provides an alternate path for the inductive load current when the main thyristor turns off or the supply voltage reverses. This allows the load current to continue flowing through the free-wheeling diode, preventing sudden interruption of current and reducing ripple in the load current. Hence, option (c) is correct.
Why the others are incorrect:
- (a) is incorrect: reducing thyristor current to zero for commutation is not the purpose of free-wheeling; commutation is the turn-off process of the thyristor itself.
- (b) is incorrect: the free-wheeling path does not share current with the thyristor during conduction; it carries current only when the thyristor is off or the source is not supplying the load.
- (d) is incorrect: free-wheeling does not maintain a constant load voltage; it mainly maintains continuity of load current and improves waveform quality.
A 50.004 MHz carrier is to be frequency modulated by a 3 KHz audio tone resulting in a narrow band FM signal. Determine the bandwidth of the FM signal
A)
2 KHz
B)
4 KHz
C)
6 KHz
D)
8 KHz
For narrowband FM, the bandwidth is approximately twice the highest modulating frequency: BW ≈ 2fm. Here the audio tone frequency is 3 kHz, so BW ≈ 2 × 3 kHz = 6 kHz. However, the carrier frequency 50.004 MHz is irrelevant to FM bandwidth; only the modulating frequency and deviation matter. Since the question states a narrowband FM signal and gives only the tone frequency, the intended exam approximation is 2 × 3 kHz = 6 kHz, but that option is not listed. Among the given choices, 4 kHz, 2 kHz, and 8 kHz do not match the standard narrowband FM approximation. Therefore, the question/options appear inconsistent; if forced to choose from the provided options, none is exactly correct. If the exam intended Carson’s rule with very small deviation, the bandwidth would still be close to 2(Δf + fm), not derivable from the data given.
The mnemonics used in writing a program is called
A)
Assembly language
B)
Fetch cycle
C)
Microinstruction
D)
Object program
Mnemonic codes used to write programs are called assembly language. Assembly language uses short symbolic instructions (mnemonics) such as MOV, ADD, and SUB instead of binary machine code. The other options are incorrect: microinstruction refers to low-level control instructions inside the CPU, fetch cycle is a step in instruction execution, and an object program is the machine-code output produced after compilation/assembly.
The noise temperature about 4306 K. The noise figure will be
A)
3 dB
B)
6 dB
C)
12 dB
D)
24 dB
Based on established principles in Electrical / Electronics Engineering, the correct and most appropriate choice is '3 dB' (Option A).
In a thyristor, ratio of latching current to holding current is
A)
0.4
B)
1.0
C)
2.5
D)
6
In a thyristor, the latching current is always greater than the holding current. The latching current is the minimum anode current required immediately after triggering so the device remains ON when the gate signal is removed, while the holding current is the minimum current needed to keep it ON once conduction is established. Typically, latching current is about 2 to 3 times the holding current, so the ratio is approximately 2.5. Hence option (c) is correct.
Why the others are incorrect:
- (a) 0.4: This is less than 1, but latching current cannot be less than holding current.
- (b) 1.0: This would mean both currents are equal, which is not the usual case in a thyristor.
- (d) 6: This is much higher than the typical ratio; while exact values vary by device, 6 is not the standard expected answer.
A boost regulator has an input voltage of 5 V and the average output voltage of 15 V. The duty cycle is
A)
3/2
B)
2/3
C)
5/2
D)
15/2
For an ideal boost converter, the output voltage is related to the input voltage by Vout = Vin / (1 - D), where D is the duty cycle. Given Vin = 5 V and Vout = 15 V: 15 = 5 / (1 - D) => 1 - D = 1/3 => D = 2/3. So option b is correct.
The other options are incorrect because they do not satisfy the boost-converter relation: a = 3/2 and c = 5/2 are greater than 1, but duty cycle must lie between 0 and 1; d = 15/2 = 7.5 is also impossible for a duty cycle.
How many I/O ports can be accessed by direct method
A)
8
B)
256
C)
32 K
D)
64 K
In direct I/O addressing, the port address field is 8 bits wide, so the CPU can access 2^8 = 256 I/O ports. Therefore, option (b) 256 is correct. The other options are incorrect because 8 is only the number of bits in the address field, not the number of ports; 32 K and 64 K are much larger address spaces associated with memory-style addressing, not direct I/O port addressing.
Which of the following flip-flop cannot be converted to D-type (delay) flip-flop?
A)
S-R flip-flop
B)
J-K flip-flop
C)
Master slave flip-flop
D)
None of these
A D-type (delay) flip-flop can be obtained from both S-R and J-K flip-flops by appropriate input connections. For an S-R flip-flop, tying S = D and R = D̅ makes it behave like a D flip-flop. For a J-K flip-flop, tying J = D and K = D̅ also gives D-type operation. A master-slave flip-flop is a form of flip-flop implementation and can likewise be arranged to realize D-type behavior. Therefore, none of the listed flip-flops is impossible to convert to D-type, so the correct answer is 'None of these'.
A tunnel diode is
A)
High resistivity p-n junction diode
B)
A slow switching device
C)
An amplifying device
D)
A very heavily doped p-n junction diode
A tunnel diode is a very heavily doped p-n junction diode. Heavy doping makes the depletion layer extremely thin, allowing quantum-mechanical tunneling of carriers and producing its characteristic negative resistance region. Option (c) is not the best description because a tunnel diode can be used as an amplifier, but that is an application, not its definition. Option (a) is incorrect because tunnel diodes are not high-resistivity devices; they are heavily doped and low-resistance. Option (b) is incorrect because tunnel diodes are known for very fast switching, not slow switching.
In the op-amp circuit shown below (assuming ideal op-amp) with resistors 22K and 2.2K and input V = 0.5V, what is the output V?
A)
V = 0
B)
V = -5 V
C)
V = 5 V
D)
V = -2 V
For an ideal op-amp with negative feedback, the inverting and non-inverting input voltages are equal (virtual short), and input current into the op-amp is zero. The circuit is an inverting amplifier with gain = -Rf/Rin = -22 kΩ / 2.2 kΩ = -10. Therefore, for input V = 0.5 V, the output is Vout = -10 × 0.5 = -5 V. Hence option (b) is correct. Option (c) is incorrect because the output is not zero for a nonzero input. Option (a) is incorrect because the gain is not +10; the sign is negative due to the inverting configuration. Option (d) is incorrect because -2 V does not match the calculated gain-based output.
Which one of the following is used as the main switching element in a switched mode power supply operating in 20 kHz to 100 kHz range?
A)
Thyristor
B)
MOSFET
C)
Triac
D)
UJT
MOSFET is the correct choice because switched-mode power supplies (SMPS) operating in the 20 kHz to 100 kHz range require a fast-switching device with low switching losses. Power MOSFETs are widely used as the main switching element in SMPS due to their high switching speed, voltage-controlled gate, and efficient operation at high frequencies. Thyristors are not suitable because they are relatively slow and are mainly used in low-frequency, high-power applications. Triacs are used primarily for AC power control, such as light dimmers, and are not used as the main high-frequency switching device in SMPS. UJTs are mainly used for triggering and oscillator circuits, not as power switching elements.
The total capacity of a ROM with 1024 rows and 4 columns is
A)
4096
B)
1024
C)
256
D)
512
A ROM organized as 1024 rows and 4 columns has 1024 × 4 = 4096 storage cells (bits). Hence its total capacity is 4096 bits, so option (a) is correct. Option (b) 1024 counts only the rows, not the total cells; option (c) 256 and option (d) 512 are both too small and do not match the row-column product.
If the accumulator of 8085 microprocessor contains 37H and the previous operation has set the carry flag, the instruction ACI 56H will result in
A)
8DH
B)
8EH
C)
17H
D)
18H
ACI adds the immediate 8-bit data and the carry flag to the accumulator. Here, A = 37H, data = 56H, and CY = 1. So the operation is 37H + 56H + 1 = 8EH. Therefore, option b is correct.
Why others are incorrect:
- a (8DH): This would be the result if the carry were 0, since 37H + 56H = 8DH.
- c (17H) and d (18H): These are not possible results of this addition; they do not match the hexadecimal sum with or without carry.
A DC link chopper as compared to AC link
A)
Efficient
B)
Less costly
C)
Bulky
D)
Both (a) and (b)
A DC link chopper is generally preferred over an AC link because it is more efficient and less costly. It avoids the extra AC conversion stages and associated losses, so overall efficiency is higher. It is also typically more compact and economical than an AC link arrangement. Hence both (a) Efficient and (b) Less costly are correct, making option (d) सही. Option (c) Bulky is incorrect because a DC link chopper is not bulky compared to an AC link.
The snubber circuit is used in thyristor circuits for
A)
Triggering dv/dt
B)
protection dv/dt
C)
protection di/dt
D)
Phase shifting
A snubber circuit in thyristor circuits is primarily used to protect the device against a high rate of rise of voltage (dv/dt). A sudden dv/dt across a thyristor can cause unintended turn-on due to the device’s internal capacitances, so the snubber limits this voltage rise and improves safe operation.
Why the others are incorrect:
- a) Triggering dv/dt: snubbers do not intentionally trigger the thyristor; they prevent false triggering caused by dv/dt.
- c) Protection di/dt: di/dt protection is usually provided by series inductance or current-limiting elements, not the snubber’s main function.
- d) Phase shifting: phase shifting is unrelated to snubber circuits; it is used in control or AC circuit applications, not for thyristor protection.
Power diodes has characteristics of
A)
High power, high speed
B)
Low power, low speed
C)
Low power, high speed
D)
High power, low speed
Power diodes are designed to handle large currents and high voltages, so they are characterized by high power capability. Compared with small-signal diodes, they generally switch more slowly because of their larger junction area and charge storage effects, so they are considered low speed. Hence, the correct choice is high power, low speed.
Why the others are incorrect:
- a (High power, high speed): power diodes are not typically high-speed devices.
- b (Low power, low speed): power diodes are not low-power devices.
- c (Low power, high speed): this describes small-signal or switching diodes, not power diodes.
Which one of the following is the software interrupt of 8085 microprocessor?
A)
TRAP
B)
INTR
C)
RST 7
D)
RST 7.5
In the 8085 microprocessor, the software interrupt instructions are the RST (restart) instructions, such as RST 0 through RST 7. These are executed by the program itself, so they are called software interrupts. Therefore, RST 7 is the correct option.
Why the others are incorrect:
- TRAP is a hardware interrupt, and it is non-maskable.
- INTR is also a hardware interrupt input, not a software interrupt.
- RST 7.5 is a hardware interrupt request line, not the software interrupt instruction RST 7.
The correct sequence of the following devices in the increasing order of turn off times is
A)
MOSFET, IGBT, BJT, Thyristor
B)
IGBT, MOSFET, BJT, Thyristor
C)
Thrysistor, BJT, MOSFET, IGBT
D)
MOSFET, BJT, IGBT, Thyristor
The increasing order of turn-off time is: MOSFET < BJT < IGBT < Thyristor. MOSFETs are majority-carrier devices, so they switch off the fastest. BJTs are slower because they are minority-carrier devices and need charge removal. IGBTs are generally slower than BJTs in turn-off due to tail current caused by stored charge in the drift region. Thyristors have the longest turn-off time because they are latching devices and require a long recovery time to regain blocking capability. Therefore, option (d) is correct.
Why others are incorrect: (a) places IGBT before BJT, but IGBT turn-off is typically slower than BJT. (b) puts IGBT first, which is incorrect since MOSFET is the fastest. (c) is completely reversed and incorrect.
What is the memory word addressing capability in 8085
A)
32 K
B)
64 K
C)
256 K
D)
512 K
8085 has a 16-bit address bus, so it can generate 2^16 = 65,536 distinct addresses. Since memory is byte-addressable, this corresponds to 64 KB of memory addressing capability. Therefore, option b (64 K) is correct. Option a (32 K) is only half of the available address space, option c (256 K) would require 18 address lines, and option d (512 K) would require 19 address lines, both beyond the 8085’s capability.
What is the minimum number of NAND gates required to implement A + AB + ABC ?
A)
0
B)
1
C)
4
D)
7
A + AB + ABC simplifies by absorption: A + AB = A, and then A + ABC = A. So the whole expression reduces to just A. A single NAND gate can be used as an inverter only if both inputs are tied together, but to realize the function A itself with NAND-only implementation, the minimum is effectively one gate if the exam is counting a direct NAND realization of the simplified single-variable form. Hence option b (1) is taken as correct. Option a (0) is impossible because some logic element is needed. Option c (4) and option d (7) are too many, since the expression collapses to a single variable and does not require multiple gates.
Triacs cannot be used in a.c. voltage regulator for a
A)
Resistive load
B)
Back emf load
C)
Inductive load
D)
Resistive inductive load
Triacs are generally not suitable for a.c. voltage regulation with a back emf load because the load’s stored energy can force current to continue or reverse in a way that prevents proper commutation and reliable turn-off of the triac. This can cause misfiring or failure to control the output voltage correctly.
Why the other options are incorrect:
- Resistive load: Triacs work well with purely resistive loads because current and voltage are in phase, so the triac turns off naturally at current zero.
- Inductive load: Triacs can be used, though special care is needed because current lags voltage and commutation becomes more difficult.
- Resistive inductive load: Triacs are commonly used here as well, again with proper triggering/commutation considerations.
Hence, the correct answer is back emf load.