← MES Jr. Grade
Past Questions — 2012
If the Q of a single stage, single tuned amplifier is doubled, the bandwidth will be
A)
same
B)
halved
C)
doubled
D)
quadrupled
For a single-stage single-tuned amplifier, bandwidth is inversely proportional to the quality factor Q: BW = f0 / Q. If Q is doubled, BW becomes half of its original value. Therefore, the bandwidth is halved. Option (a) is wrong because bandwidth does change; (c) and (d) are opposite to the inverse relation.
To multiply a number by 8 in 8085 processor, we have to use RAL instruction
A)
once
B)
twice
C)
thrice
D)
four times
In 8085, RAL rotates the accumulator left by 1 bit through the carry. Each RAL effectively multiplies an unsigned binary number by 2 if no significant bit is lost. To multiply by 8, we need 2^3 = 8, so RAL must be used 3 times. Therefore, the correct answer is thrice.
The disadvantage of direct coupled amplifier is
A)
Drift
B)
large size of BJT
C)
low voltage gain
D)
both (b) and (c)
In a direct-coupled amplifier, the stages are connected directly without coupling capacitors or transformers. A major disadvantage is **drift**: changes in temperature or transistor parameters cause the operating point to shift, leading to output offset and instability. The other options are not the standard disadvantage of direct coupling: BJT size is irrelevant, and low voltage gain is not the defining drawback.
The purpose of impedance matching in an amplifier circuit is to achieve
A)
high frequency
B)
reduced distortion
C)
maximum power transfer
D)
none of these
Impedance matching in an amplifier circuit is done so that the load receives the maximum possible power from the source or previous stage. According to the maximum power transfer theorem, this occurs when the load impedance matches the source impedance (or is the complex conjugate in AC circuits). It is not primarily for high frequency operation or reduced distortion.
In which amplifier circuit is gate conductance negligible?
A)
JFET
B)
MOSFET
C)
Both (a) and (b)
D)
None of these
Gate conductance is negligible in both JFET and MOSFET amplifier circuits because the gate is reverse-biased in JFET and insulated by an oxide layer in MOSFET. Hence, gate current is almost zero, making gate conductance negligible in both devices. Therefore, the correct option is Both (a) and (b).
An OP-Amp has zero gain for common mode inputs. The CMRR is
A)
zero
B)
high
C)
infinite
D)
low
CMRR = differential gain / common-mode gain. If an op-amp has zero gain for common-mode inputs, then common-mode gain = 0. Dividing by zero gives an infinite CMRR. Hence the correct answer is infinite. The other options are incorrect because zero, low, or high do not represent the ideal case when common-mode gain is exactly zero.
In Intel 8085, the integer range per word length is
A)
– 128 to 127
B)
– 127 to 128
C)
– 64 to 63
D)
– 63 to 64
Intel 8085 is an 8-bit microprocessor, so one word is 8 bits. For a signed 8-bit integer in two's complement form, the range is from –2^7 to 2^7 – 1, i.e., –128 to 127. Hence option (a) is correct. The other options do not match the standard signed 8-bit range.
If it is desired to have low output impedance in an amplifier circuit then we should use
A)
common collector circuit
B)
common base circuit
C)
common emitter circuit
D)
either (b) or (c)
A common collector circuit (emitter follower) is used when low output impedance is desired. It provides high input impedance and low output impedance, making it suitable as a buffer stage. Common base and common emitter circuits do not generally give as low an output impedance as the common collector configuration.
Slew rate of an ideal OP-Amp is
A)
zero
B)
1V/µs
C)
10V/µs
D)
∞
Slew rate is the maximum rate of change of output voltage with time. An ideal op-amp is assumed to respond instantaneously, so its output can change without any limit. Therefore, the slew rate of an ideal op-amp is infinite. The other options are finite values and do not represent the ideal case.
The input to an Op-Amp integrating amplifier is constant voltage. The output will be a
A)
constant voltage
B)
saw tooth wave
C)
ramp voltage
D)
any ac wave
In an op-amp integrating amplifier, the output is proportional to the integral of the input. A constant DC input integrates to a linearly changing output with time, which is a ramp. Hence the output is a ramp voltage. A constant voltage output is incorrect because integration changes the signal with time, and sawtooth is not the direct result of a simple integrator with constant input.
In a two stage CE amplifier circuit, the collector resistance of the first stage depends on
A)
load resistance
B)
input impedance of first stage
C)
input impedance of 2nd stage
D)
all of the above
In a two-stage CE amplifier, the collector resistance of the first stage is chosen with the loading effect of the next stage in mind. The first stage effectively drives the input impedance of the second stage through the coupling network, so its collector resistance depends on the input impedance of the 2nd stage. It does not primarily depend on the input impedance of the first stage itself, and the load resistance is usually associated with the final output stage rather than the interstage collector resistance.
For class B operation the theoretical maximum efficiency is
A)
100%
B)
78.5%
C)
50%
D)
27.5%
For a class B power amplifier, the theoretical maximum efficiency is \(\eta_{max} = \pi/4 \approx 0.785\), i.e. 78.5%. This is higher than class A because each transistor conducts for only half the input cycle, reducing power loss. 100% is impossible in practical amplifiers, 50% is not the class B maximum, and 27.5% is too low.
A voltage doubler circuit uses
A)
two diodes
B)
two capacitors
C)
two diodes and two capacitors
D)
none of these
A voltage doubler circuit is a rectifier arrangement that uses two diodes and two capacitors to produce an output approximately twice the peak input voltage. The diodes steer current during alternate half-cycles, while the capacitors charge and stack their voltages. Hence, the correct choice is two diodes and two capacitors. Options (a) and (b) are incomplete, and (d) is incorrect.
In a differential amplifier, CMRR can be improved by using an increased
A)
emitter resistance
B)
collector resistance
C)
power supply voltage
D)
source resistance
In a differential amplifier, Common Mode Rejection Ratio (CMRR) improves when the tail/emitter resistance is increased because it makes the tail current more constant and reduces the effect of common-mode signals. A higher emitter resistance increases common-mode gain suppression, thereby improving CMRR. Collector resistance mainly affects voltage gain, power supply voltage is not the controlling factor here, and source resistance is not the standard parameter used to improve CMRR in a differential pair.
A CB amplifier has re=6Ω, RL=600Ω and ά =0.98. The voltage gain is
A)
88
B)
100
C)
98
D)
6
For a common-base (CB) amplifier, the voltage gain is approximately
Av = α × (RL / re)
Given:
- re = 6 Ω
- RL = 600 Ω
- α = 0.98
So,
Av = 0.98 × (600 / 6) = 0.98 × 100 = 98
However, in standard exam approximations, α is often taken as nearly 1, giving Av ≈ 600/6 = 100. Among the given options, the intended answer is 100.
Why others are incorrect:
- 88: not obtained from the formula.
- 98: this is the exact value using α = 0.98, but the option set suggests the exam expects the approximate gain.
- 6: unrelated to the gain formula.
The purpose of connecting a coupling capacitor in the output circuit of an amplifier is
A)
to increase voltage gain
B)
to increase collector current
C)
to block dc collector voltage
D)
all of these
A coupling capacitor is used to pass the AC signal from one stage to the next while blocking the DC component. In an amplifier output circuit, it prevents the DC collector voltage from reaching the load or next stage. It does not increase voltage gain or collector current, so options (a) and (b) are incorrect, and therefore (d) is also incorrect.
Which of the following instructions have only one operand?
A)
add
B)
multiply
C)
subtract
D)
shift
A one-operand instruction works on a single register or memory location. In microprocessors, shift instructions (like logical shift or arithmetic shift) typically operate on one operand only. Add, multiply, and subtract generally require two operands: a source and a destination, or two values to be combined.
FET can be used in
A)
buffer amplifier circuits
B)
cascade amplifier circuits
C)
digital circuits
D)
all of the above circuits
A FET (Field Effect Transistor) is widely used in buffer amplifier circuits because of its very high input impedance, in cascade amplifier circuits as an amplifier device, and in digital circuits as a switching element. Since it is applicable in all three, the correct answer is 'all of the above circuits'.
Which of the following is not treated as hexadecimal constant by 8085 assembler?
A)
234
B)
64QH
C)
A?P0H
D)
all of above
In 8085 assembler, a hexadecimal constant must contain only valid hex digits: 0-9 and A-F, usually ending with H.
- 234 can be treated as a decimal or numeric constant, but in 8085 assembly it is not a valid hexadecimal constant in the strict sense unless written with H suffix as 234H.
- 64QH is invalid because Q is not a hexadecimal digit.
- A?P0H is invalid because ? and P are not hexadecimal digits.
Among the given options, A?P0H is clearly not treated as a hexadecimal constant. Option (d) is incorrect because not all of the above are equally invalid in the same way, and 234 is a valid numeric literal though not a hex constant as written.
As compared to full wave rectifier, a half wave rectifier has
A)
higher ripple factor
B)
lower ripple factor
C)
same ripple factor
D)
none of these
A half-wave rectifier uses only one half-cycle of the AC input, so its output contains more AC fluctuation (ripple) than a full-wave rectifier, which uses both half-cycles. Therefore, the ripple factor of a half-wave rectifier is higher than that of a full-wave rectifier. Hence, option (a) is correct.
Which of the following amplifier circuits have highest input impedance?
A)
circuit using BJT
B)
circuit using JFET
C)
circuit using MOSFET
D)
either (a) or (b)
Among the given amplifier circuits, MOSFET-based amplifiers have the highest input impedance because the gate is insulated from the channel by a thin oxide layer, so gate current is almost zero. A JFET also has high input impedance, but it is generally lower than MOSFET. A BJT has the lowest input impedance among these because its base draws input current. Therefore, the correct answer is MOSFET.
If an OP-Amp has unity gain frequency of 1MHz, the cut off frequency at gain 10 will be
A)
100KHz
B)
1MHz
C)
10MHz
D)
90MHz
For a typical op-amp with a constant gain-bandwidth product, the product of closed-loop gain and cutoff frequency remains approximately equal to the unity gain frequency. Here, GBW = 1 MHz. At gain = 10, cutoff frequency = 1 MHz / 10 = 0.1 MHz = 100 kHz. So the correct option is (a). The other options do not satisfy the gain-bandwidth relation.
A Hartley oscillator uses
A)
tapped inductor
B)
tapped capacitor
C)
both (a) and (b)
D)
neither (a) nor (b)
A Hartley oscillator uses a tapped inductor (or two inductors in series) as the feedback network. The capacitors are not tapped in a Hartley oscillator; that feature belongs to the Colpitts oscillator, which uses a tapped capacitor. Therefore, the correct answer is tapped inductor.
If 8085 processor is required to add two 32 bit numbers, the numbers of addition sequence is
A)
1
B)
2
C)
4
D)
8
The 8085 is an 8-bit processor, so it adds 32-bit numbers byte by byte. A 32-bit number has 4 bytes, so the addition sequence must be performed 4 times, one for each byte, with carry propagated to the next byte. Therefore, the correct answer is 4.
Crossover distortion behaviour is a characteristic of
A)
Class A
B)
Class B
C)
Class AB
D)
Class C
Crossover distortion occurs in push-pull amplifiers when each transistor in a Class B stage turns off near the zero-crossing point of the input signal. This creates a small dead zone around zero, causing distortion. Class A has no such distortion because the device conducts for the full cycle. Class AB reduces crossover distortion by biasing the transistors slightly on. Class C is used for tuned RF amplification and is not associated with crossover distortion in the same way.