Digital Electronics and Logic Design MCQs 2026

60 questions with detailed answers · 21 from past papers · 6 quiz batches available

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Page 1 of 1Questions 1–10 of 60
  1. Q1Past Paper · PPSC/FPSC/NTSeasy

    Asynchronous (ripple) counter has

    1. Aall FFs clocked by common edge simultaneously
    2. Bflip-flops clocked by previous stage output
    3. Cno flip-flops only combinational logic
    4. Donly shift register without counting
    💡 Explanation:

    Ripple propagation causes cumulative delay.

  2. Q2hard

    Clock skew between flip-flops can cause

    1. Aguaranteed faster operation always
    2. Belimination of metastability without limit
    3. Cinfinite fan-out capability
    4. Dtiming violations if effective setup/hold margins are reduced
    💡 Explanation:

    Unequal clock arrival times stress timing budget.

  3. Q3hard

    Scan flip-flop design aids

    1. Aonly analog amplification
    2. Bonly power factor correction
    3. Ctesting by serially shifting test vectors through the chain
    4. Donly rectifier ripple reduction
    💡 Explanation:

    Scan path improves controllability and observability in IC test.

  4. Q4Past Paper · PPSC/FPSC/NTSmedium

    Level-sensitive latch differs from edge-triggered FF in that it

    1. Ais transparent while enable is active
    2. Bchanges only on clock edge always
    3. Ccannot store any state
    4. Drequires JK inputs only
    💡 Explanation:

    Latch passes input through when enabled; FF captures on edge.

  5. Q5easy

    Register is constructed from

    1. Aonly combinational gates without memory
    2. Bmultiple flip-flops storing parallel bits on common clock
    3. Csingle diode rectifier
    4. Dop-amp integrator only
    💡 Explanation:

    Array of FFs holds multi-bit word synchronously.

  6. Q6Past Paper · PPSC/FPSC/NTSmedium

    Flip-flop propagation delay tpd is

    1. Asetup plus hold only without output delay
    2. Btime from clock edge to valid output change at Q
    3. Cclock period definition only
    4. Drise time of power supply only
    💡 Explanation:

    tpd limits maximum operating frequency with margin.

  7. Q7hard

    Metastability occurs when

    1. Ainputs are stable with large setup margin
    2. Bpower supply is perfectly regulated
    3. Conly in combinational logic without storage
    4. Dsetup or hold time is violated and output dwells between valid logic levels
    💡 Explanation:

    Unresolved state may persist unpredictably before settling.

  8. Q8hard

    Master-slave flip-flop reduces

    1. Aneed for setup time entirely
    2. Bclock frequency to DC only
    3. Crace conditions by separating capture and output update
    4. Dpower supply decoupling
    💡 Explanation:

    Two-stage isolation prevents transparent window glitches.

  9. Q9Past Paper · PPSC/FPSC/NTSmedium

    Asynchronous preset and clear on flip-flops

    1. Aoverride clocked operation immediately when asserted
    2. Bonly work on clock edge like D input
    3. Cdisable all outputs to high-Z always
    4. Dreplace the need for power supply
    💡 Explanation:

    Direct set/reset forces known state independent of clock.

  10. Q10medium

    Hold time of flip-flop is

    1. Atime before clock edge only
    2. Bpropagation delay from clock to Q only
    3. Cminimum time data must remain stable after active clock edge
    4. Dfrequency of oscillator only
    💡 Explanation:

    Hold violation can corrupt latched value.

  11. Q11medium

    Setup time of flip-flop is

    1. Aminimum time data must be stable before active clock edge
    2. Btime after clock edge for output change
    3. Chold time after edge only without setup
    4. Dclock period only
    💡 Explanation:

    Violating setup time causes metastable or wrong capture.

  12. Q12Past Paper · PPSC/FPSC/NTSmedium

    T flip-flop toggles output when

    1. AT is 0 at clock edge
    2. Breset is deasserted only
    3. CT input is 1 at active clock edge
    4. Dpreset alone without clock
    💡 Explanation:

    T=1 complements Q; T=0 holds state.

  13. Q13medium

    JK flip-flop toggles when

    1. AJ=0 and K=0 always
    2. BJ and K are both 1 at active clock edge
    3. Conly asynchronous reset is asserted
    4. Dclock is held LOW indefinitely
    💡 Explanation:

    JK combines S-R with toggle when both inputs high.

  14. Q14easy

    Positive-edge-triggered flip-flop changes state on

    1. Afalling edge only always
    2. Brising edge of clock
    3. Cany DC level without transition
    4. Dboth edges simultaneously
    💡 Explanation:

    Edge triggering samples input at clock transition.

  15. Q15Past Paper · PPSC/FPSC/NTSeasy

    D flip-flop eliminates

    1. Aneed for clock entirely
    2. Ball propagation delay
    3. Cindeterminate condition of SR latch by using single data input
    4. Dreset pin unconditionally
    💡 Explanation:

    D input maps to next state; S/R derived internally.

  16. Q16medium

    SR latch built with NOR gates has invalid state when

    1. Aboth inputs are 0
    2. Bonly S is 0
    3. Conly R is 0 with S pulsed
    4. DS and R are both 1 simultaneously (active-high inputs)
    💡 Explanation:

    S=R=1 can force both outputs to 0 in NOR latch (race).

  17. Q17hard

    K-map method is limited compared to Quine-McCluskey for

    1. Abeing less accurate always
    2. Binability to handle don′t-cares ever
    3. Crequiring software always
    4. Dmanual convenience on few variables rather than algorithmic scalability
    💡 Explanation:

    QM tabular method scales better for many variables.

  18. Q18Past Paper · PPSC/FPSC/NTSmedium

    Overlapping groups in K-map are allowed because

    1. Aeach minterm can appear in only one group always
    2. Boverlap invalidates simplification
    3. Ca minterm may participate in multiple groups to form larger coverings
    4. Donly diagonal overlap is permitted
    💡 Explanation:

    Shared minterms help achieve minimal total term count.

  19. Q19hard

    Five-variable K-map is often implemented as

    1. Asingle row of five cells only
    2. Bcircular map without Gray code
    3. Ctruth table without grouping
    4. Dtwo four-variable maps with shifting variable
    💡 Explanation:

    Fifth variable selects between two 4-var submaps.

  20. Q20easy

    SOP form obtained from K-map is a sum of

    1. Aonly single literals without products
    2. BPOS maxterms only
    3. CNAND operations without OR
    4. Dproduct terms (AND terms ORed together)
    💡 Explanation:

    Each group becomes a product term; terms are ORed.

  21. Q21Past Paper · PPSC/FPSC/NTShard

    Redundant group in K-map is one that

    1. Ais the smallest possible group always
    2. Bcovers only don′t-cares without 1s
    3. Cdoes not cover any minterm not already covered by essential groups
    4. Dcannot be formed on the map
    💡 Explanation:

    Non-essential groups may be omitted from minimal cover.

  22. Q22easy

    Number of cells in K-map for n variables is

    1. An × 2 only
    2. B2^n
    3. Cn^2 without exponent
    4. D2n only
    💡 Explanation:

    Each input combination maps to one cell.

  23. Q23hard

    For POS minimization using K-map, groupings are made on

    1. Azeros (or maxterms) to obtain sum terms
    2. Bonly 1s without zeros ever
    3. Cdon′t-cares only without zeros
    4. Doutput column of truth table only
    💡 Explanation:

    Grouping 0s yields simplified product-of-sums form.

  24. Q24easy

    Digital clock application uses counters for

    1. Aonly DAC conversion without timing
    2. Bonly rectifier filtering
    3. Cfrequency division and time display multiplexing
    4. Donly motor slip measurement
    💡 Explanation:

    Counters divide crystal frequency to 1 Hz time base segments.

  25. Q25hard

    Cascade counters achieve

    1. Aonly single-digit BCD without linkage
    2. Belimination of clock signal
    3. Chigher modulus by connecting carry/borrow between stages
    4. Dpurely analog integration without FFs
    💡 Explanation:

    Multiple chips or stages extend counting range.

  26. Q26Past Paper · PPSC/FPSC/NTSmedium

    Synchronous counter typically has

    1. Ahigher cumulative delay than ripple always
    2. Bno clock input requirement
    3. Conly one flip-flop regardless of modulus
    4. Dlower propagation delay uncertainty than ripple counter of same width
    💡 Explanation:

    Glitch-free outputs and predictable timing favor synchronous design.

  27. Q27medium

    Terminal count output of counter indicates

    1. Aalways midpoint count only
    2. Bmaximum count reached in the sequence
    3. Crandom logic level without definition
    4. Danalog ramp completion only
    💡 Explanation:

    TC signal often triggers reset or external action.

  28. Q28easy

    Frequency division by 2 per stage in binary counter means

    1. Aeach successive FF toggles at half the previous stage frequency
    2. Ball stages share identical frequency
    3. Cfrequency doubles each stage
    4. Dno relation to clock frequency
    💡 Explanation:

    Ripple/binary chain divides clock by 2 per bit.

  29. Q29Past Paper · PPSC/FPSC/NTSmedium

    Presettable counter can be loaded with

    1. Aonly zero without load capability
    2. Banalog voltage on op-amp input only
    3. Conly serial shift without parallel load
    4. Dparallel data to start from a specific initial value
    💡 Explanation:

    Load enable sets starting count synchronously or asynchronously.

  30. Q30medium

    Glitch problem in asynchronous ripple counter occurs because

    1. Astages change at slightly different times producing brief false counts
    2. Ball bits change simultaneously always
    3. Cclock is perfectly synchronous by definition
    4. Doutputs are analog without discrete levels
    💡 Explanation:

    Ripple delay creates momentary invalid output codes.

  31. Q31medium

    Counter modulus reduction uses

    1. Aonly increasing clock frequency without feedback
    2. Bfeedback logic to skip states and achieve desired modulo
    3. Cremoving all flip-flops
    4. Danalog RC timing without digital logic
    💡 Explanation:

    Gating or load value forces reset at terminal count.

  32. Q32Past Paper · PPSC/FPSC/NTShard

    Johnson counter (twisted ring) has sequence length of

    1. An states only
    2. B2^n states always like binary
    3. C2n for n flip-flops
    4. Done state only
    💡 Explanation:

    Inverted feedback doubles states vs ring counter.

  33. Q33medium

    Ring counter is implemented with

    1. Aone hot circulating bit among flip-flops
    2. Bbinary weighted sum only without circulation
    3. CJohnson sequence inverted feedback only
    4. DK-map groups only
    💡 Explanation:

    Single 1 shifts through register forming sequence.

  34. Q34medium

    Up-down counter can

    1. Aonly count up without down mode
    2. Bincrement or decrement count on clock depending on control
    3. Conly operate asynchronously without clock
    4. Dstore analog samples without digital states
    💡 Explanation:

    Direction pin selects increment vs decrement.

  35. Q35Past Paper · PPSC/FPSC/NTSeasy

    Decade counter (BCD) counts

    1. A0 through 15 in binary always
    2. Bonly even numbers
    3. Canalog voltage levels without digits
    4. D0 through 9 then returns to 0
    💡 Explanation:

    BCD decade counter skips 10–15 in four-bit segment.

  36. Q36medium

    Modulo-N counter repeats count sequence every

    1. A2^N states always without exception
    2. BN squared states
    3. CN states
    4. Done clock period only
    💡 Explanation:

    Counter cycles through N distinct states before wrap.

  37. Q37easy

    Synchronous counter has all flip-flops clocked by

    1. Aprevious stage Q output only
    2. Bcommon clock signal
    3. Casynchronous preset only without clock
    4. Drandom noise transitions
    💡 Explanation:

    Common clock updates all stages in parallel logic plan.

  38. Q38Past Paper · PPSC/FPSC/NTSmedium

    Wrap-around adjacency on K-map means

    1. Aonly interior cells can group
    2. Bedge cells are considered adjacent across map boundaries
    3. Cdiagonal cells always adjacent
    4. Dcells two steps apart are adjacent
    💡 Explanation:

    Toroidal adjacency on top/bottom and left/right edges.

  39. Q39hard

    Essential prime implicant is

    1. Aa prime implicant that covers at least one minterm not covered by any other prime implicant
    2. Bany group of two cells only
    3. Calways the largest group without uniqueness
    4. Doutput of OR gate without minimization
    💡 Explanation:

    Essential PIs must appear in any minimal expression.

  40. Q40Past Paper · PPSC/FPSC/NTSmedium

    Don′t-care conditions in K-map are marked as

    1. Aalways 0 without choice
    2. Balways 1 without choice
    3. CX and may be assigned 0 or 1 to maximize groups
    4. Dignored without affecting simplification ever
    💡 Explanation:

    Exploiting don′t-cares can yield simpler cover.

  41. Q41medium

    A group of 8 adjacent 1s on a 4-variable K-map eliminates

    1. Athree variables
    2. Bone variable only
    3. Call four variables always without remainder
    4. Dno variables
    💡 Explanation:

    Eight-cell group covers three changing variables as don′t-cares.

  42. Q42medium

    A group of 4 adjacent 1s in a K-map eliminates

    1. Aone variable only
    2. Bthree variables always
    3. Ctwo variables from the product term
    4. Dno simplification
    💡 Explanation:

    Four-cell group corresponds to two eliminated variables.

  43. Q43Past Paper · PPSC/FPSC/NTSmedium

    A group of 2 adjacent 1s in K-map eliminates

    1. Atwo variables always
    2. Bno variables ever
    3. Cone variable from the product term
    4. Dall variables in the expression
    💡 Explanation:

    Each doubling of group size removes one literal.

  44. Q44easy

    Adjacent cells in a K-map differ by

    1. Atwo variables always
    2. Bone variable only (Gray code adjacency)
    3. Call variables simultaneously
    4. Dno systematic rule
    💡 Explanation:

    Single-bit change ensures valid grouping for simplification.

  45. Q45easy

    Karnaugh map is used for

    1. Ameasuring propagation delay in oscilloscope
    2. Bcalculating Fourier coefficients only
    3. Cdetermining ADC resolution only
    4. Dsimplifying Boolean expressions graphically
    💡 Explanation:

    K-map groups adjacent 1s to find minimal SOP/POS.

  46. Q46Past Paper · PPSC/FPSC/NTSeasy

    Buffer gate is used to

    1. Ainvert all inputs always
    2. Bstore one bit of memory
    3. Cisolate loads and restore signal levels without inversion
    4. Ddivide clock frequency by two without flip-flop
    💡 Explanation:

    Buffer provides drive capability and isolation.

  47. Q47medium

    TTL logic HIGH level is typically

    1. A0 V always for HIGH
    2. B12 V without regulation
    3. Cfloating undefined as HIGH without pull-up
    4. Daround 2.4 V minimum at output for 5 V family
    💡 Explanation:

    TTL thresholds defined relative to 5 V supply standards.

  48. Q48hard

    CMOS logic dissipates significant power primarily during

    1. Asteady state with inputs fixed at valid levels
    2. Bonly when output is high-Z
    3. Conly at DC with no capacitive load
    4. Dswitching transitions when both PMOS and NMOS briefly conduct
    💡 Explanation:

    Dynamic power arises from charging/discharging load capacitance and shoot-through.

  49. Q49Past Paper · PPSC/FPSC/NTSmedium

    Noise margin indicates

    1. Amaximum clock frequency only
    2. Bchip area in mm² only
    3. Conly DC supply current
    4. Dallowable noise on input before logic level is misinterpreted
    💡 Explanation:

    High noise margin improves reliability in noisy environments.

  50. Q50medium

    Propagation delay of a logic gate is

    1. Atime between input transition and corresponding output transition
    2. BDC noise margin only
    3. Cpower dissipation in watts only
    4. Dnumber of pins on package
    💡 Explanation:

    tpd limits maximum clock frequency in synchronous systems.

  51. Q51medium

    Tri-state buffer third state is

    1. Aalways logic 1
    2. Bhigh-impedance output disconnected from logic levels
    3. Calways logic 0
    4. Dshort circuit to ground
    💡 Explanation:

    Enable control allows bus sharing with high-Z when disabled.

  52. Q52Past Paper · PPSC/FPSC/NTSmedium

    Fan-out of a logic gate specifies

    1. Amaximum number of similar gate inputs it can drive reliably
    2. Bpropagation delay only
    3. Csupply voltage tolerance only
    4. Dnumber of product terms in SOP
    💡 Explanation:

    Excessive fan-out degrades logic levels and speed.

  53. Q53easy

    Positive logic defines HIGH as

    1. Alogic 1 and LOW as logic 0
    2. Blogic 0 as HIGH always
    3. Canalog voltage without digital levels
    4. Dfloating input as definite 1
    💡 Explanation:

    Positive logic associates higher voltage with binary 1.

  54. Q54medium

    De Morgan theorem states (A · B)′ equals

    1. AA′ + B′
    2. BA′ · B′
    3. CA + B
    4. DA ⊕ B
    💡 Explanation:

    Complement of AND equals OR of complements.

  55. Q55Past Paper · PPSC/FPSC/NTSeasy

    XNOR gate is also called

    1. Ainverter only
    2. Bequivalence gate
    3. Cbuffer only
    4. Ddemultiplexer
    💡 Explanation:

    XNOR is 1 when inputs are equal.

  56. Q56easy

    XOR gate output is 1 when

    1. Ainputs are different
    2. Binputs are equal
    3. Call inputs are 0 only
    4. Dboth inputs are 1 only
    💡 Explanation:

    Exclusive OR is 1 for odd number of 1s on two inputs.

  57. Q57medium

    NOR gate as a universal gate can implement

    1. Aonly AND functions without NOT
    2. Bonly buffers without inversion
    3. Conly XOR without other gates
    4. Dany Boolean function given enough gates
    💡 Explanation:

    NAND and NOR are both functionally complete.

  58. Q58Past Paper · PPSC/FPSC/NTSeasy

    OR gate output is HIGH when

    1. Aall inputs are LOW only
    2. Bany input is HIGH
    3. Cinputs are equal always
    4. Donly when inputs differ
    💡 Explanation:

    OR sums inputs logically; 1 if any input is 1.

  59. Q59easy

    NAND gate is functionally equivalent to

    1. Ainverted OR only without AND
    2. Bdirect AND without inversion
    3. Cexclusive OR of inputs
    4. Dinverted AND of its inputs
    💡 Explanation:

    NAND = NOT(AND); universal gate.

  60. Q60Past Paper · PPSC/FPSC/NTSeasy

    The Boolean expression for output of AND gate is

    1. AY = A · B
    2. BY = A + B
    3. CY = A ⊕ B
    4. DY = A′
    💡 Explanation:

    AND output is 1 only when all inputs are 1.