What is the main advantage of using CMOS technology over NMOS in digital circuits?
Answer: B
CMOS technology uses complementary pairs of transistors resulting in lower static power dissipation and better noise margins compared to NMOS.
Q.82Medium
Which of the following is the correct Boolean expression for De Morgan's Law?
Answer: C
De Morgan's Law states (AB)' = A' + B' and (A+B)' = A'B'. Option C correctly represents one form of this law.
Q.83Medium
What is the propagation delay in a ripple counter compared to a synchronous counter?
Answer: B
Ripple counters have cascading propagation delays as each stage must wait for the previous stage to settle, making them slower than synchronous counters.
Q.84Medium
In a 4:2 encoder, if input line 5 is active, what is the output?
Answer: C
A 4:2 encoder has 4 inputs and 2 outputs. However, for 8 inputs, a 3:8 encoder or 8:3 decoder is needed. Input line 5 would give binary 101.
Q.85Medium
What is the primary purpose of a Schmitt trigger in digital electronics?
Answer: B
A Schmitt trigger converts analog input to digital output with hysteresis, providing noise immunity and cleaner transitions.
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Q.86Medium
In combinational logic design, what is the sum-of-products (SOP) form?
Answer: B
Sum-of-Products form involves taking the AND (product) of variables and then OR (sum) of all such products to form the logic function.
Q.87Medium
What does the term 'fanout' mean in digital logic circuits?
Answer: B
Fanout is the maximum number of gate inputs of the same type that one gate output can reliably drive without degradation.
Q.88Medium
Which type of latch is transparent when the control signal is HIGH?
Answer: B
A D latch is transparent (passes input to output) when the Enable/Control signal is HIGH, and latches the value when it goes LOW.
Q.89Medium
What is the output frequency of a 4-bit ripple counter when driven by a 16 MHz clock?
Answer: D
In a 4-bit ripple counter, the output frequency of the last stage is f_clock / 2^4 = 16 MHz / 16 = 1 MHz.
Q.90Medium
Which of the following statements about asynchronous reset is TRUE?
Answer: B
Asynchronous reset directly resets the flip-flop without waiting for a clock pulse, providing immediate action regardless of clock state.
Q.91Medium
In a 4-bit synchronous binary counter using JK flip-flops, if the clock frequency is 10 MHz, what is the maximum counting frequency at the MSB output?
Answer: B
In a 4-bit synchronous counter, the MSB output toggles every 2^4 = 16 clock cycles. Therefore, MSB frequency = 10 MHz / 16 = 625 kHz. This is the divide-by-16 behavior of a 4-bit counter.
Q.92Medium
The bandwidth of a common emitter amplifier can be increased by
Answer: D
Emitter degeneration without bypass capacitor increases input impedance and reduces gain but significantly increases bandwidth due to negative feedback.
Q.93Medium
In a differential amplifier, the common mode rejection ratio (CMRR) is defined as
Answer: A
CMRR = Ad/Ac, where Ad is differential mode gain and Ac is common mode gain. High CMRR (>80dB typical) is desired to reject common mode noise.
Q.94Medium
The input impedance of a voltage follower (common collector) amplifier is
Answer: B
CC amplifier has very high input impedance (Zin ≈ β·re) making it suitable as a buffer stage between high impedance sources and low impedance loads.
Q.95Medium
At what frequency does the gain of a RC coupled amplifier reduce to 70.7% of its mid-band gain?
Answer: B
The -3dB cutoff frequency is where the magnitude of gain reduces to 1/√2 (≈ 70.7%) of maximum gain. This defines the bandwidth of the amplifier.
Q.96Medium
In an RC coupled amplifier, the lower cutoff frequency is primarily determined by
Answer: A
Lower cutoff frequency fL ≈ 1/(2π·CC·Rin) where CC is coupling capacitor and Rin is input impedance. Larger CC gives lower fL.
Q.97Medium
For a feedback amplifier with feedback fraction β, the loop gain is defined as
Answer: A
Loop gain = Aβ, where A is open-loop gain and β is feedback fraction. For stability, loop gain should be <1 for phase margin requirements.
Q.98Medium
In a Class A amplifier, the maximum theoretical efficiency is
Answer: A
Maximum theoretical efficiency of Class A = π/4 ≈ 78.5% when maximum swing is used, but practically 25-50% is achieved due to quiescent current and losses.
Q.99Medium
In a Colpitts oscillator, the frequency of oscillation is determined by
Answer: C
Colpitts oscillator frequency f = 1/(2π√(L·Ceq)) where Ceq = (C1·C2)/(C1+C2). Voltage divider is formed by C1 and C2 for feedback.
Q.100Medium
The gain-bandwidth product (GBP) of an op-amp is approximately constant. For a 741 op-amp with GBP ≈ 1 MHz, the open-loop gain at 100 kHz is approximately
Answer: A
GBP = Aol × f. For 741: 1 MHz = Aol × 100 kHz, therefore Aol = 1 MHz / 100 kHz = 10 V/V. GBP is approximately constant over frequency.
