← MES Jr. Grade
Past Questions — 2012
The register set and instruction set of 8088 processor is similar to that of
A)
8085
B)
8086
C)
8087
D)
none of these
The 8088 processor is architecturally very similar to the 8086. Both have the same register set and nearly the same instruction set. The main difference is that 8088 has an 8-bit external data bus, while 8086 has a 16-bit external data bus. 8085 is an older 8-bit microprocessor with a different architecture, and 8087 is a math coprocessor, not the matching CPU architecture.
In deriving ac equivalent circuit for an amplifier circuit we short circuit all
A)
resistors
B)
inductors
C)
transistors
D)
capacitors
In the AC equivalent circuit of an amplifier, all DC sources are set to zero and coupling/bypass capacitors are treated as short circuits because their reactance is very low at signal frequencies. Inductors are not generally shorted; their treatment depends on the circuit and frequency. Resistors remain as they are, and transistors are not short-circuited.
The most commonly used bias in BJT amplifier circuit is
A)
voltage divider bias
B)
emitter bias
C)
collector bias
D)
collector feedback bias
In BJT amplifier circuits, the most commonly used biasing method is the voltage divider bias because it provides good thermal stability and keeps the operating point (Q-point) relatively stable against changes in transistor parameters like beta and temperature. Emitter bias, collector bias, and collector feedback bias are used in some circuits, but they are less common than voltage divider bias for general amplifier applications.
Which of the following is also called hand shaking programmed data transfer?
A)
synchronous data transfer
B)
asynchronous data transfer
C)
interrupt driven data transfer
D)
both (a) & (c)
Handshaking programmed data transfer means the sender and receiver coordinate each transfer using control signals, without a common clock. This is called asynchronous data transfer. In synchronous transfer, both devices use a shared clock, so it is not handshaking-based. Interrupt-driven transfer is a different method where the CPU is interrupted when the device is ready, so it is also not the same.
The element in the feedback circuit of a logarithmic amplifier using Op-Amp is a
A)
resistor
B)
capacitor
C)
inductor
D)
diode
In a logarithmic amplifier, the feedback element of the op-amp is a diode (or sometimes a transistor used in diode-connected mode). The diode’s exponential current-voltage characteristic produces an output voltage proportional to the logarithm of the input. A resistor gives linear feedback, while capacitor and inductor are not used for logarithmic action in this circuit.
Which of the following is not an interrupt line in 8085?
A)
RST5.5
B)
RST7.5
C)
RST9.5
D)
TRAP
In the 8085 microprocessor, the interrupt lines are TRAP, RST 7.5, RST 6.5, RST 5.5, and INTR. Among the given options, RST9.5 is not an 8085 interrupt line. Therefore, the correct answer is RST9.5.
A Darlington pair is used for
A)
low distortion
B)
high bandwidth
C)
high power gain
D)
high current gain
A Darlington pair consists of two BJTs connected so that the current gain of the first transistor is multiplied by the current gain of the second. This gives a very large overall current gain, which is the main purpose of using it. It is not primarily used for high bandwidth or low distortion, and while it can provide high power gain as a result of high current gain, the standard answer is high current gain.
Negative feedback in amplifier
A)
reduces gain
B)
increases frequency and phase distortion
C)
reduces bandwidth
D)
increases noise
Negative feedback in an amplifier feeds a portion of the output back in opposite phase to the input. This opposes the input signal, so the overall gain decreases. At the same time, negative feedback usually improves bandwidth, reduces distortion, and lowers noise sensitivity. Therefore, the correct option is (a) reduces gain. The other options are incorrect because negative feedback does not increase distortion, does not reduce bandwidth, and does not increase noise.
In 8085, which addressing mode is also called inherent addressing?
A)
direct
B)
register
C)
implicit
D)
immediate
In 8085, the addressing mode in which the operand is implied by the instruction itself is called **implicit (inherent) addressing**. For example, instructions like CMC, RAL, and HLT do not specify an explicit operand.
Why others are incorrect:
- **direct**: the address of the operand is given explicitly in the instruction.
- **register**: the operand is in a register specified in the instruction.
- **immediate**: the data value itself is present in the instruction.
So, inherent addressing = implicit addressing.
In 8085 stack pointer is
A)
4 bit
B)
8 bit
C)
16 bit
D)
32 bit
In the 8085 microprocessor, the stack pointer (SP) is a 16-bit register because it must hold a memory address. Since 8085 has a 16-bit address bus, the stack can be located anywhere in the 64 KB memory space. Therefore, the stack pointer is 16 bit. The other options are incorrect because 4-bit and 8-bit are too small to store a memory address, and 32-bit is not used in 8085.
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.