A half adder adds two single-bit inputs ( and ) and produces a Sum and a Carry output.
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2The Sum output of a half adder is generated using which logic gate?
Adders
Easy
A.XOR gate
B.OR gate
C.NAND gate
D.AND gate
Correct Answer: XOR gate
Explanation:
In a half adder, Sum (XOR) and Carry (AND).
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3A full adder adds how many input bits together?
Adders
Easy
A.Three bits
B.Two bits
C.Four bits
D.Five bits
Correct Answer: Three bits
Explanation:
A full adder adds three bits: , , and a carry-in (), producing Sum and Carry-out.
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4The difference output of a half subtractor is produced by which gate?
Subtractors
Easy
A.AND gate
B.NOR gate
C.OR gate
D.XOR gate
Correct Answer: XOR gate
Explanation:
In a half subtractor, Difference and Borrow .
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5In a half subtractor, the borrow output is generated when:
Subtractors
Easy
A. and
B. and
C. and
D. and
Correct Answer: and
Explanation:
A borrow is needed when subtracting a larger bit from a smaller one, i.e. Borrow , which is 1 when .
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6A digital comparator is used to:
Comparators
Easy
A.Compare two binary numbers
B.Store binary data
C.Convert binary to decimal
D.Add two binary numbers
Correct Answer: Compare two binary numbers
Explanation:
A comparator compares two binary numbers and indicates whether they are equal, or one is greater or less than the other.
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7How many output conditions does a magnitude comparator typically produce?
Comparators
Easy
A.Two (, )
B.One (equality only)
C.Three (, , )
D.Four
Correct Answer: Three (, , )
Explanation:
A magnitude comparator produces three outputs indicating , , and .
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8A multiplexer (MUX) performs which function?
Multiplexers and Demultiplexers
Easy
A.Stores multiple bits
B.Adds multiple inputs together
C.Selects one of many inputs to a single output
D.Splits one input into many outputs
Correct Answer: Selects one of many inputs to a single output
Explanation:
A multiplexer is a data selector that routes one of several input lines to a single output based on select lines.
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9How many select lines are required for an multiplexer?
Multiplexers and Demultiplexers
Easy
A.3
B.4
C.2
D.8
Correct Answer: 3
Explanation:
For inputs, select lines are needed. Since , an MUX needs 3 select lines.
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10A demultiplexer (DEMUX) performs which function?
Multiplexers and Demultiplexers
Easy
A.Generates a parity bit
B.Compares two numbers
C.Routes one input to one of many outputs
D.Combines many inputs into one output
Correct Answer: Routes one input to one of many outputs
Explanation:
A demultiplexer takes a single input and directs it to one of several output lines based on select lines. It is the reverse of a multiplexer.
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11A -to- line decoder has how many output lines?
Decoders
Easy
A.16
B.6
C.8
D.3
Correct Answer: 8
Explanation:
A decoder with inputs has outputs. For , there are outputs.
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12The main function of a decoder is to:
Decoders
Easy
A.Combine multiple outputs into one
B.Convert coded input into a unique output
C.Compare two inputs
D.Add two binary numbers
Correct Answer: Convert coded input into a unique output
Explanation:
A decoder converts an -bit binary code into one active output line among possible outputs.
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13An encoder performs the opposite operation of which circuit?
Encoders
Easy
A.Adder
B.Flip-flop
C.Decoder
D.Comparator
Correct Answer: Decoder
Explanation:
An encoder converts input lines into an -bit code, which is the reverse of a decoder's operation.
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14How many input lines does an -to- encoder have?
Encoders
Easy
A.4
B.8
C.16
D.3
Correct Answer: 8
Explanation:
An -to- encoder accepts 8 input lines and produces a 3-bit binary output code.
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15A parity bit is added to data primarily for:
Parity circuits
Easy
A.Data compression
B.Faster processing
C.Data encryption
D.Error detection
Correct Answer: Error detection
Explanation:
A parity bit is an extra bit added to detect single-bit errors during data transmission.
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16Which gate is commonly used to generate a parity bit?
Parity circuits
Easy
A.XOR gate
B.AND gate
C.OR gate
D.NOT gate
Correct Answer: XOR gate
Explanation:
XOR gates are used in parity generators because their output reflects whether the number of 1s is odd or even.
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17Which of the following is a bipolar logic family?
Introduction to different logic families
Easy
A.TTL
B.PMOS
C.CMOS
D.NMOS
Correct Answer: TTL
Explanation:
TTL (Transistor-Transistor Logic) is a bipolar family using BJTs, while CMOS, NMOS, and PMOS are MOS-based families.
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18CMOS logic uses which type of transistors?
Structure and operations of TTL, MOS and CMOS logic families
Easy
A.Only bipolar junction transistors
B.Only NMOS
C.Both NMOS and PMOS
D.Only PMOS
Correct Answer: Both NMOS and PMOS
Explanation:
CMOS (Complementary MOS) uses both NMOS and PMOS transistors together, giving very low static power consumption.
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19Which logic family is known for the lowest static power consumption?
Structure and operations of TTL, MOS and CMOS logic families
Easy
A.ECL
B.DTL
C.CMOS
D.TTL
Correct Answer: CMOS
Explanation:
CMOS draws negligible current in a steady state, giving it the lowest static power consumption among common logic families.
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20The basic building block of TTL logic is the:
Structure and operations of TTL, MOS and CMOS logic families
Easy
A.Bipolar junction transistor (BJT)
B.Vacuum tube
C.Diode only
D.MOSFET
Correct Answer: Bipolar junction transistor (BJT)
Explanation:
TTL circuits are built using bipolar junction transistors (BJTs), often with multi-emitter input transistors.
Incorrect! Try again.
21A full adder adds three input bits , , and . What is the sum and carry-out when , , and ?
Adders
Medium
A.Sum , Carry
B.Sum , Carry
C.Sum , Carry
D.Sum , Carry
Correct Answer: Sum , Carry
Explanation:
Adding in binary is . Therefore Sum and Carry-out .
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22To construct a 4-bit ripple carry adder, how many full adders are required?
Adders
Medium
A.
B.
C.
D.
Correct Answer:
Explanation:
Each bit position requires one full adder, so a 4-bit adder needs full adders connected in cascade.
Incorrect! Try again.
23The main disadvantage of a ripple carry adder compared to a carry look-ahead adder is:
Adders
Medium
A.More power efficiency
B.Higher propagation delay
C.Fewer logic gates
D.Simpler carry generation
Correct Answer: Higher propagation delay
Explanation:
In a ripple carry adder, each carry must propagate through all preceding stages, causing cumulative delay that grows with the number of bits.
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24In a full subtractor computing , what are the Difference and Borrow when , , and ?
Subtractors
Medium
A.Difference , Borrow
B.Difference , Borrow
C.Difference , Borrow
D.Difference , Borrow
Correct Answer: Difference , Borrow
Explanation:
Computing requires a borrow, giving Difference and Borrow-out .
Incorrect! Try again.
25A full adder can be converted into a full subtractor (using 2's complement) by:
Subtractors
Medium
A.Adding a NOT gate at the sum output only
B.Inverting the subtrahend bits and setting
C.Inverting the minuend bits and setting
D.Inverting both inputs and setting
Correct Answer: Inverting the subtrahend bits and setting
Explanation:
Subtraction in 2's complement equals , so the subtrahend is inverted and the initial carry is set to .
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26A 1-bit magnitude comparator outputs , , and . What is the logic expression for the output?
Comparators
Medium
A.
B.
C.
D.
Correct Answer:
Explanation:
Equality occurs when both bits match, which is the XNOR function .
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27For two 4-bit numbers, how many comparison outputs (, , ) does a standard magnitude comparator produce?
Comparators
Medium
A.
B.
C.
D.
Correct Answer:
Explanation:
A magnitude comparator produces exactly three mutually exclusive outputs: greater than, equal to, and less than.
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28When comparing two 4-bit numbers and , which output of the comparator is high?
Comparators
Medium
A.None of the outputs
B.
C.
D.
Correct Answer:
Explanation:
and . Since , the output is high.
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29How many select lines are required for a 16-to-1 multiplexer?
Multiplexers and Demultiplexers
Medium
A.
B.
C.
D.
Correct Answer:
Explanation:
A -to-1 multiplexer needs select lines. For 16 inputs, , so select lines are required.
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30A demultiplexer with 3 select lines can route a single input to how many output lines?
Multiplexers and Demultiplexers
Medium
A.
B.
C.
D.
Correct Answer:
Explanation:
A demultiplexer with select lines has outputs. With select lines, outputs are available.
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31To implement a Boolean function of 3 variables using a single multiplexer with the variables on the select lines, what size multiplexer is needed?
Multiplexers and Demultiplexers
Medium
A.-to-1
B.-to-1
C.-to-1
D.-to-1
Correct Answer: -to-1
Explanation:
With 3 variables driving the select lines, data inputs are needed, requiring an -to-1 multiplexer.
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32A 3-to-8 line decoder activates how many output lines at a time for a given valid input?
Decoders
Medium
A.
B.
C.
D.
Correct Answer:
Explanation:
A decoder activates exactly one output line corresponding to the applied binary input combination.
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33How many 2-to-4 decoders (with enable) are needed to build a 4-to-16 decoder?
Decoders
Medium
A.
B.
C.
D.
Correct Answer:
Explanation:
Four 2-to-4 decoders generate the 16 outputs, and one additional 2-to-4 decoder acts as the enable selector, giving total.
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34A BCD-to-7-segment decoder is primarily used to:
Decoders
Medium
A.Generate parity bits
B.Drive display segments for decimal digits
C.Convert binary to Gray code
D.Perform arithmetic addition
Correct Answer: Drive display segments for decimal digits
Explanation:
A BCD-to-7-segment decoder converts a 4-bit BCD input into signals that illuminate the correct segments of a seven-segment display.
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35In a priority encoder, when multiple inputs are active simultaneously, the output corresponds to:
Encoders
Medium
A.The sum of all active inputs
B.The highest-priority active input
C.The lowest-priority active input
D.An undefined value
Correct Answer: The highest-priority active input
Explanation:
A priority encoder resolves multiple simultaneous active inputs by encoding only the input with the highest assigned priority.
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36An octal-to-binary (8-to-3) encoder has how many output lines?
Encoders
Medium
A.
B.
C.
D.
Correct Answer:
Explanation:
An -input encoder produces a binary code with output lines, since .
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37For the 7-bit data word , what even parity bit should be appended?
Parity circuits
Medium
A.
B.
C.Cannot be determined
D.
Correct Answer:
Explanation:
The data has four 1's (an even count). For even parity the total number of 1's must stay even, so the parity bit is .
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38Which gate is fundamental to building parity generator and checker circuits?
Parity circuits
Medium
A.OR gate
B.AND gate
C.XOR gate
D.NAND gate
Correct Answer: XOR gate
Explanation:
XOR gates output the parity (odd count of 1's) of their inputs, making them the basic building block of parity circuits.
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39Which characteristic best describes CMOS logic compared to TTL?
Structure and operations of TTL, MOS and CMOS logic families
Medium
A.Faster switching in all conditions
B.Very low static power consumption
C.Higher static power consumption
D.Lower noise margin
Correct Answer: Very low static power consumption
Explanation:
CMOS draws negligible static current because one transistor in each complementary pair is always off, giving very low standby power.
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40A standard TTL NAND gate uses which device at its input stage?
Structure and operations of TTL, MOS and CMOS logic families
Medium
A.A pair of complementary FETs
B.A multi-emitter bipolar transistor
C.A single MOSFET
D.A Schottky diode array
Correct Answer: A multi-emitter bipolar transistor
Explanation:
Standard TTL gates use a multi-emitter bipolar junction transistor at the input to perform the logic AND function of the inputs.
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41A ripple-carry adder is built from full adders each having a propagation delay of for both sum and carry outputs. For an -bit adder, the worst-case time to produce a valid final sum is best approximated by:
Adders
Hard
A.
B.
C.
D.
Correct Answer:
Explanation:
The carry must ripple through stages ( for the last carry-in), but the final sum bit needs one more after its carry-in arrives. The carry chain contributes to reach the last stage's carry-in plus for its sum, giving in the worst case.
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42In a carry-lookahead adder, the block generate and block propagate for two bits are defined using and . The group carry-out (with carry-in ) equals:
Adders
Hard
A.
B.
C.
D.
Correct Answer:
Explanation:
A carry emerges from stage 1 if it is generated there (), or generated at stage 0 and propagated through stage 1 (), or the input carry is propagated through both stages ().
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43A 4-bit adder/subtractor uses XOR gates and a mode line ( for subtraction) feeding . When computing in 2's complement, an overflow for signed numbers is detected by:
Subtractors
Hard
A.
B. alone
C.
D.
Correct Answer:
Explanation:
Signed overflow occurs when the carry into the MSB () differs from the carry out of the MSB (). Thus overflow , independent of whether the operation is add or subtract.
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44Using 2's complement in a 4-bit subtractor to compute (unsigned inputs), the result bits and the borrow/carry-out are:
Subtractors
Hard
A.Result , carry-out (borrow occurred)
B.Result , carry-out (no borrow)
C.Result , carry-out
D.Result , carry-out
Correct Answer: Result , carry-out (borrow occurred)
Explanation:
with carry-out . In 2's complement subtraction, carry-out signals a borrow, meaning (), consistent with the true result .
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45For a 1-bit magnitude comparator with inputs and , the output is and is . When cascading to compare multi-bit numbers, the overall output for two bits ( vs ) is:
Comparators
Hard
A.
B.
C.
D.
Correct Answer:
Explanation:
when the MSBs already decide it (), or the MSBs are equal (, XNOR) and the LSB comparison gives ().
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46An IC 7485 4-bit comparator has cascading inputs. To correctly cascade multiple 7485s, the cascading inputs of the least-significant stage must be set to:
Comparators
Hard
A. respectively
B. respectively
C. respectively
D. respectively
Correct Answer: respectively
Explanation:
The least-significant stage must assume equality before it examines its own bits, so its cascade inputs are set , , . This lets the equality propagate correctly up the chain.
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47A function is implemented using an 8:1 MUX with as select lines. If , the data inputs through must be:
Multiplexers and Demultiplexers
Hard
A., rest
B., rest
C., rest
D.All inputs
Correct Answer: , rest
Explanation:
With all three variables as select lines, each minterm maps directly to one data input. The minterms present in (0,2,5,7) get logic 1; all others get logic 0.
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48To realize using an 8:1 MUX with as selects and available, the input (select , i.e., ) should be connected to:
Multiplexers and Demultiplexers
Hard
A.
B.
C.
D.
Correct Answer:
Explanation:
Select corresponds to minterms 4 () and 5 (). Here is present but is not, so the output equals ... rechecking: present, absent means . Wait—minterm 4 has giving correct, minterm 5 () absent giving correct, so . The intended standard mapping for present-at- cases yields ; with only present the value tracks . Choose based on which minterm ( or ) is present: since only is present, connect .
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49A demultiplexer is built by cascading two demultiplexers. The minimum number of additional select and enable connections required is best described as:
Multiplexers and Demultiplexers
Hard
A.Two extra select bits and two enables
B.One extra select bit driving the two enables (via true/complement)
C.One extra enable only, no new select bit
D.Four extra select bits
Correct Answer: One extra select bit driving the two enables (via true/complement)
Explanation:
A 1:16 DEMUX needs 4 select bits; two 1:8 units already use 3 shared selects. The 4th select bit (and its complement) enables one 1:8 block at a time, routing the input to the correct half.
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50An active-low output decoder with an active-high enable is used to implement . Since outputs are active-low, the correct gate to combine the relevant decoder outputs is:
Decoders
Hard
A.A NOR gate over outputs
B.An AND gate over outputs
C.A NAND gate over outputs
D.An OR gate over outputs
Correct Answer: A NAND gate over outputs
Explanation:
For active-low outputs, the selected minterm line goes low. NANDing the three relevant active-low lines produces a high output whenever any one of them is asserted (low), correctly realizing the sum of minterms.
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51A BCD-to-7-segment decoder receives an invalid input (decimal 10). In a typical non-ripple-blanking decoder without explicit handling of invalid codes, the segment outputs are:
Decoders
Hard
A.Always all segments ON
B.Always all segments OFF
C.Undefined/don't-care unless the design specifies otherwise
D.Guaranteed to display the digit 0
Correct Answer: Undefined/don't-care unless the design specifies otherwise
Explanation:
BCD codes 1010–1111 are invalid and normally treated as don't-cares during K-map minimization. The resulting outputs depend on how the designer assigned those don't-cares, so behavior is not guaranteed unless explicitly defined.
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52In an 8-to-3 priority encoder, inputs are active with and all others 0. Assuming highest-index priority, the 3-bit output and valid bit are:
Encoders
Hard
A., valid
B., valid
C., valid
D., valid
Correct Answer: , valid
Explanation:
A priority encoder outputs the binary code of the highest-index active input. Among , the highest is , giving output and asserting the valid (group-signal) output.
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53Why is a plain (non-priority) -to- encoder unsuitable when two or more inputs may be active simultaneously?
Encoders
Hard
A.It disables its output and floats high-impedance
B.It outputs all zeros regardless of inputs
C.It produces an output that is the OR of the individual codes, which is generally incorrect
D.It automatically selects the lowest-index input
Correct Answer: It produces an output that is the OR of the individual codes, which is generally incorrect
Explanation:
A simple encoder ORs together the codes of active inputs. With multiple active lines the result is a bitwise OR of their binary codes, which usually does not correspond to any single valid input, so a priority scheme is required.
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54A 4-bit word is transmitted with an even parity bit . During transmission and both flip. At the receiver, the parity check will:
Parity circuits
Hard
A.Pass (no error detected) because two bits flipped
B.Fail, correctly detecting an error
C.Pass and correct the error
D.Fail, but also correct the error
Correct Answer: Pass (no error detected) because two bits flipped
Explanation:
Single parity detects only an odd number of bit errors. Two simultaneous flips keep the total parity unchanged, so the check passes and the double error goes undetected.
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55A parity generator/checker for a 9-bit code (8 data + 1 parity) is built from a tree of 2-input XOR gates. The minimum number of gate delays (tree depth) to compute the parity of 8 data bits is:
Parity circuits
Hard
A.
B.
C.
D.
Correct Answer:
Explanation:
XORing 8 bits in a balanced binary tree needs levels of 2-input XOR gates, giving a propagation depth of 3 gate delays rather than the 7 delays of a linear chain.
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56Two logic families are compared: Family X has and ; Family Y has and . Comparing their speed-power products:
Introduction to different logic families
Hard
A.Family Y is better with
B.Both have equal speed-power product of
C.Family X is better with
D.Family X is better with
Correct Answer: Both have equal speed-power product of
Explanation:
Speed-power product . Family X: ; Family Y: . They are equal, illustrating the classic speed-vs-power trade-off.
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57For a logic family, , , , . The high-state and low-state noise margins are:
Introduction to different logic families
Hard
A.,
B.,
C.,
D.,
Correct Answer: ,
Explanation:
and .
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58In a standard TTL NAND gate, the totem-pole output stage includes a diode between the pull-up transistor's emitter and the pull-down transistor's collector. Its primary purpose is to:
Structure and operations of TTL, MOS and CMOS logic families
Hard
A.Ensure the upper transistor is off when the output is low, preventing conduction overlap
B.Reduce the input clamping threshold
C.Increase the output current sinking capability
D.Provide reverse-polarity protection at the input
Correct Answer: Ensure the upper transistor is off when the output is low, preventing conduction overlap
Explanation:
The diode raises the required voltage to turn on the upper (pull-up) transistor, guaranteeing it stays off when the output is pulled low. This avoids both output transistors conducting simultaneously, which would cause large current spikes.
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59A CMOS inverter driving a capacitive load at switches at . Its dynamic power dissipation is approximately:
Structure and operations of TTL, MOS and CMOS logic families
Hard
A.
B.
C.
D.
Correct Answer:
Explanation:
Dynamic power .
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60A designer connects an unused input of a CMOS NAND gate by leaving it floating. The most likely consequence is:
Structure and operations of TTL, MOS and CMOS logic families
Hard
A.Guaranteed logic-high behavior with no side effects
B.The gate is permanently damaged immediately
C.The output is forced to a stable logic low
D.Erratic output and increased power due to the input drifting near the switching threshold
Correct Answer: Erratic output and increased power due to the input drifting near the switching threshold
Explanation:
CMOS inputs have extremely high impedance, so a floating input can drift to the mid-supply region where both transistors conduct. This causes unpredictable switching and elevated static current, so unused inputs must be tied to or ground.
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