In an n-channel depletion-mode MOSFET with Vto = -3 V, if VGS = 0 V, the device is in which region?
Answer: C
For n-channel depletion MOSFET, when VGS = 0 V (which is greater than Vto = -3 V), the channel is open and device conducts. With VGS > Vto and positive VDS, the device operates in saturation. Depletion-mode devices conduct at VGS = 0.
Q.42Medium
The truth table of which gate has odd number of 1s in output column for 2-input gate?
Answer: C
XOR gate has 2 ones in output (when inputs are different). OR has 3 ones. AND has 1 one. These are odd. XNOR has 2 ones (even). XOR is most commonly cited.
Q.43Medium
A combinational circuit with n inputs and m outputs is completely specified by:
Answer: D
A combinational circuit can be completely defined using truth table, Boolean expressions, or logic diagram. All three are equivalent representations.
Q.44Medium
What is the simplified form of the Boolean expression A + AB + ABC?
Answer: A
Using Boolean algebra: A + AB + ABC = A(1 + B + BC) = A(1) = A. The dominant term A absorbs all other terms.
Q.45Medium
What is the Hamming distance between binary numbers 1011101 and 1001001?
Answer: B
Hamming distance = number of positions where bits differ. 1011101 vs 1001001: positions 3, 5, 6 differ. Hamming distance = 3.
Advertisement
Q.46Medium
Which of the following Boolean expressions represents the majority function for 3 inputs?
Answer: A
Majority function outputs 1 when majority (2 or more) inputs are 1. Expression: AB + BC + AC covers all cases where at least 2 inputs are 1.
Q.47Medium
In a 4-bit binary counter, after state 1111, the next state is:
Answer: B
In a 4-bit binary counter, states cycle from 0000 to 1111 (0 to 15). After 1111, it wraps around to 0000 (overflow occurs).
Q.48Medium
For a Karnaugh map of 4 variables, what is the minimum number of cells needed to cover all minterms?
Answer: C
A 4-variable K-map has 2^4 = 16 cells, each representing one minterm. To cover all minterms, minimum 16 cells are needed.
Q.49Medium
What is the maximum frequency of a synchronous counter if each flip-flop has a propagation delay of 10 ns?
Answer: B
In synchronous counters, all flip-flops change simultaneously, so maximum frequency is determined by one flip-flop delay. f_max = 1/t_pd = 101ns = 100 MHz.
Q.50Medium
An asynchronous counter with n flip-flops counts from 0 to:
Answer: B
An n-bit counter has 2^n possible states (from 0 to 2^n - 1). For 3-bit counter: states are 0 to 7 (000 to 111).
Q.51Medium
In Mealy machine, the output depends on:
Answer: C
Mealy machine output depends on both present state and present input. Moore machine output depends only on present state. Mealy machines are more efficient.
Q.52Medium
In a synchronous counter design, what is the main advantage over asynchronous counters?
Answer: C
Synchronous counters clock all flip-flops simultaneously, eliminating propagation delay problems. This allows higher operating frequencies without cumulative timing errors present in ripple counters
Q.53Medium
In Booth's multiplication algorithm, which sequence of multiplier bits requires the fewest operations?
Answer: B
Booth's algorithm replaces consecutive 1s with a single subtraction and addition operation. Consecutive 1s (like 0111) are treated as 1000-1, reducing computation steps compared to processing each 1 individually
Q.54Medium
What is the setup time in a flip-flop?
Answer: B
Setup time (tsu) is the minimum duration the input signal must remain stable and valid before the active clock edge for reliable capture and state change
Q.55Medium
For a modulo-6 counter using JK flip-flops, how many flip-flops are required at minimum?
Answer: B
For modulo-N counter, minimum flip-flops required = ⌈log₂(N)⌉. For modulo-6: ⌈log₂(6)⌉ = ⌈2.58⌉ = 3 flip-flops (counts 0-5)
Q.56Medium
Which multiplexer configuration is used to implement any Boolean function with minimum gates?
Answer: C
An n-variable Boolean function can be efficiently implemented using a 2ⁿ-to-1 multiplexer where variables form select lines and minterms form data inputs, achieving minimal gate implementation
Q.57Medium
A 4-bit binary subtractor using 2's complement method requires:
Answer: A
2's complement subtraction (A-B) converts to addition A+2's complement of B. This requires 4 full adders (for addition) and XOR gates (for bit complementing), avoiding dedicated subtractors
Q.58Medium
In a magnitude comparator for 4-bit numbers, if A > B, A = B, and A < B outputs are provided, what is the total number of output combinations possible?
Answer: B
A magnitude comparator produces exactly 3 mutually exclusive outputs: A>B (1,0,0) or A=B (0,1,0) or A<B (0,0,1). These are the only 3 possible comparison results for any input pair
Q.59Medium
What is the hold time in a flip-flop, and why is it critical?
Answer: B
Hold time (th) is the minimum duration input must remain stable after the clock edge. Violation can cause metastability—a state where flip-flop output is uncertain or oscillates unpredictably
Q.60Medium
In a cascaded 4-bit counter system where two 4-bit counters are connected, what maximum count value can be achieved?
Answer: B
Two cascaded 4-bit counters form an 8-bit counter system. Maximum count = 2⁸ - 1 = 255 (counts from 0 to 255, total 256 states)