Q1 easy
Digital clock application uses counters for
A only DAC conversion without timing ✓ B only rectifier filtering ✓ C frequency division and time display multiplexing ✓ D only motor slip measurement ✓ Show Answer 💡 Explanation: Counters divide crystal frequency to 1 Hz time base segments.
Q2 hard
Cascade counters achieve
A only single-digit BCD without linkage ✓ B elimination of clock signal ✓ C higher modulus by connecting carry/borrow between stages ✓ D purely analog integration without FFs ✓ Show Answer 💡 Explanation: Multiple chips or stages extend counting range.
Q3 Past Paper · PPSC/FPSC/NTS medium
Synchronous counter typically has
A higher cumulative delay than ripple always ✓ B no clock input requirement ✓ C only one flip-flop regardless of modulus ✓ D lower propagation delay uncertainty than ripple counter of same width ✓ Show Answer 💡 Explanation: Glitch-free outputs and predictable timing favor synchronous design.
Q4 medium
Terminal count output of counter indicates
A always midpoint count only ✓ B maximum count reached in the sequence ✓ C random logic level without definition ✓ D analog ramp completion only ✓ Show Answer 💡 Explanation: TC signal often triggers reset or external action.
Q5 easy
Frequency division by 2 per stage in binary counter means
A each successive FF toggles at half the previous stage frequency ✓ B all stages share identical frequency ✓ C frequency doubles each stage ✓ D no relation to clock frequency ✓ Show Answer 💡 Explanation: Ripple/binary chain divides clock by 2 per bit.
Q6 Past Paper · PPSC/FPSC/NTS medium
Presettable counter can be loaded with
A only zero without load capability ✓ B analog voltage on op-amp input only ✓ C only serial shift without parallel load ✓ D parallel data to start from a specific initial value ✓ Show Answer 💡 Explanation: Load enable sets starting count synchronously or asynchronously.
Q7 medium
Glitch problem in asynchronous ripple counter occurs because
A stages change at slightly different times producing brief false counts ✓ B all bits change simultaneously always ✓ C clock is perfectly synchronous by definition ✓ D outputs are analog without discrete levels ✓ Show Answer 💡 Explanation: Ripple delay creates momentary invalid output codes.
Q8 medium
Counter modulus reduction uses
A only increasing clock frequency without feedback ✓ B feedback logic to skip states and achieve desired modulo ✓ C removing all flip-flops ✓ D analog RC timing without digital logic ✓ Show Answer 💡 Explanation: Gating or load value forces reset at terminal count.
Q9 Past Paper · PPSC/FPSC/NTS hard
Johnson counter (twisted ring) has sequence length of
A n states only ✓ B 2^n states always like binary ✓ C 2n for n flip-flops ✓ D one state only ✓ Show Answer 💡 Explanation: Inverted feedback doubles states vs ring counter.
Q10 medium
Ring counter is implemented with
A one hot circulating bit among flip-flops ✓ B binary weighted sum only without circulation ✓ C Johnson sequence inverted feedback only ✓ D K-map groups only ✓ Show Answer 💡 Explanation: Single 1 shifts through register forming sequence.
Q11 medium
Up-down counter can
A only count up without down mode ✓ B increment or decrement count on clock depending on control ✓ C only operate asynchronously without clock ✓ D store analog samples without digital states ✓ Show Answer 💡 Explanation: Direction pin selects increment vs decrement.
Q12 Past Paper · PPSC/FPSC/NTS easy
Decade counter (BCD) counts
A 0 through 15 in binary always ✓ B only even numbers ✓ C analog voltage levels without digits ✓ D 0 through 9 then returns to 0 ✓ Show Answer 💡 Explanation: BCD decade counter skips 10–15 in four-bit segment.
Q13 medium
Modulo-N counter repeats count sequence every
A 2^N states always without exception ✓ B N squared states ✓ C N states ✓ D one clock period only ✓ Show Answer 💡 Explanation: Counter cycles through N distinct states before wrap.
Q14 easy
Synchronous counter has all flip-flops clocked by
A previous stage Q output only ✓ B common clock signal ✓ C asynchronous preset only without clock ✓ D random noise transitions ✓ Show Answer 💡 Explanation: Common clock updates all stages in parallel logic plan.
Q15 Past Paper · PPSC/FPSC/NTS medium
Wrap-around adjacency on K-map means
A only interior cells can group ✓ B edge cells are considered adjacent across map boundaries ✓ C diagonal cells always adjacent ✓ D cells two steps apart are adjacent ✓ Show Answer 💡 Explanation: Toroidal adjacency on top/bottom and left/right edges.
Q16 hard
Essential prime implicant is
A a prime implicant that covers at least one minterm not covered by any other prime implicant ✓ B any group of two cells only ✓ C always the largest group without uniqueness ✓ D output of OR gate without minimization ✓ Show Answer 💡 Explanation: Essential PIs must appear in any minimal expression.
Q17 Past Paper · PPSC/FPSC/NTS medium
Don′t-care conditions in K-map are marked as
A always 0 without choice ✓ B always 1 without choice ✓ C X and may be assigned 0 or 1 to maximize groups ✓ D ignored without affecting simplification ever ✓ Show Answer 💡 Explanation: Exploiting don′t-cares can yield simpler cover.
Q18 medium
A group of 8 adjacent 1s on a 4-variable K-map eliminates
A three variables ✓ B one variable only ✓ C all four variables always without remainder ✓ D no variables ✓ Show Answer 💡 Explanation: Eight-cell group covers three changing variables as don′t-cares.
Q19 medium
A group of 4 adjacent 1s in a K-map eliminates
A one variable only ✓ B three variables always ✓ C two variables from the product term ✓ D no simplification ✓ Show Answer 💡 Explanation: Four-cell group corresponds to two eliminated variables.
Q20 Past Paper · PPSC/FPSC/NTS medium
A group of 2 adjacent 1s in K-map eliminates
A two variables always ✓ B no variables ever ✓ C one variable from the product term ✓ D all variables in the expression ✓ Show Answer 💡 Explanation: Each doubling of group size removes one literal.
Q21 easy
Adjacent cells in a K-map differ by
A two variables always ✓ B one variable only (Gray code adjacency) ✓ C all variables simultaneously ✓ D no systematic rule ✓ Show Answer 💡 Explanation: Single-bit change ensures valid grouping for simplification.
Q22 easy
Karnaugh map is used for
A measuring propagation delay in oscilloscope ✓ B calculating Fourier coefficients only ✓ C determining ADC resolution only ✓ D simplifying Boolean expressions graphically ✓ Show Answer 💡 Explanation: K-map groups adjacent 1s to find minimal SOP/POS.
Q23 Past Paper · PPSC/FPSC/NTS easy
Buffer gate is used to
A invert all inputs always ✓ B store one bit of memory ✓ C isolate loads and restore signal levels without inversion ✓ D divide clock frequency by two without flip-flop ✓ Show Answer 💡 Explanation: Buffer provides drive capability and isolation.
Q24 medium
TTL logic HIGH level is typically
A 0 V always for HIGH ✓ B 12 V without regulation ✓ C floating undefined as HIGH without pull-up ✓ D around 2.4 V minimum at output for 5 V family ✓ Show Answer 💡 Explanation: TTL thresholds defined relative to 5 V supply standards.
Q25 hard
CMOS logic dissipates significant power primarily during
A steady state with inputs fixed at valid levels ✓ B only when output is high-Z ✓ C only at DC with no capacitive load ✓ D switching transitions when both PMOS and NMOS briefly conduct ✓ Show Answer 💡 Explanation: Dynamic power arises from charging/discharging load capacitance and shoot-through.
Q26 Past Paper · PPSC/FPSC/NTS medium
Noise margin indicates
A maximum clock frequency only ✓ B chip area in mm² only ✓ C only DC supply current ✓ D allowable noise on input before logic level is misinterpreted ✓ Show Answer 💡 Explanation: High noise margin improves reliability in noisy environments.
Q27 medium
Propagation delay of a logic gate is
A time between input transition and corresponding output transition ✓ B DC noise margin only ✓ C power dissipation in watts only ✓ D number of pins on package ✓ Show Answer 💡 Explanation: tpd limits maximum clock frequency in synchronous systems.
Q28 medium
Tri-state buffer third state is
A always logic 1 ✓ B high-impedance output disconnected from logic levels ✓ C always logic 0 ✓ D short circuit to ground ✓ Show Answer 💡 Explanation: Enable control allows bus sharing with high-Z when disabled.
Q29 Past Paper · PPSC/FPSC/NTS medium
Fan-out of a logic gate specifies
A maximum number of similar gate inputs it can drive reliably ✓ B propagation delay only ✓ C supply voltage tolerance only ✓ D number of product terms in SOP ✓ Show Answer 💡 Explanation: Excessive fan-out degrades logic levels and speed.
Q30 easy
Positive logic defines HIGH as
A logic 1 and LOW as logic 0 ✓ B logic 0 as HIGH always ✓ C analog voltage without digital levels ✓ D floating input as definite 1 ✓ Show Answer 💡 Explanation: Positive logic associates higher voltage with binary 1.
Q31 medium
De Morgan theorem states (A · B)′ equals
A A′ + B′ ✓ B A′ · B′ ✓ C A + B ✓ D A ⊕ B ✓ Show Answer 💡 Explanation: Complement of AND equals OR of complements.
Q32 Past Paper · PPSC/FPSC/NTS easy
XNOR gate is also called
A inverter only ✓ B equivalence gate ✓ C buffer only ✓ D demultiplexer ✓ Show Answer 💡 Explanation: XNOR is 1 when inputs are equal.
Q33 easy
XOR gate output is 1 when
A inputs are different ✓ B inputs are equal ✓ C all inputs are 0 only ✓ D both inputs are 1 only ✓ Show Answer 💡 Explanation: Exclusive OR is 1 for odd number of 1s on two inputs.
Q34 medium
NOR gate as a universal gate can implement
A only AND functions without NOT ✓ B only buffers without inversion ✓ C only XOR without other gates ✓ D any Boolean function given enough gates ✓ Show Answer 💡 Explanation: NAND and NOR are both functionally complete.
Q35 Past Paper · PPSC/FPSC/NTS easy
OR gate output is HIGH when
A all inputs are LOW only ✓ B any input is HIGH ✓ C inputs are equal always ✓ D only when inputs differ ✓ Show Answer 💡 Explanation: OR sums inputs logically; 1 if any input is 1.
Q36 easy
NAND gate is functionally equivalent to
A inverted OR only without AND ✓ B direct AND without inversion ✓ C exclusive OR of inputs ✓ D inverted AND of its inputs ✓ Show Answer 💡 Explanation: NAND = NOT(AND); universal gate.
Q37 Past Paper · PPSC/FPSC/NTS easy
The Boolean expression for output of AND gate is
A Y = A · B ✓ B Y = A + B ✓ C Y = A ⊕ B ✓ D Y = A′ ✓ Show Answer 💡 Explanation: AND output is 1 only when all inputs are 1.
Q38 Past Paper · PPSC/FPSC/NTS easy
Asynchronous (ripple) counter has
A all FFs clocked by common edge simultaneously ✓ B flip-flops clocked by previous stage output ✓ C no flip-flops only combinational logic ✓ D only shift register without counting ✓ Show Answer 💡 Explanation: Ripple propagation causes cumulative delay.
Q39 hard
Clock skew between flip-flops can cause
A guaranteed faster operation always ✓ B elimination of metastability without limit ✓ C infinite fan-out capability ✓ D timing violations if effective setup/hold margins are reduced ✓ Show Answer 💡 Explanation: Unequal clock arrival times stress timing budget.
Q40 hard
Scan flip-flop design aids
A only analog amplification ✓ B only power factor correction ✓ C testing by serially shifting test vectors through the chain ✓ D only rectifier ripple reduction ✓ Show Answer 💡 Explanation: Scan path improves controllability and observability in IC test.
Q41 Past Paper · PPSC/FPSC/NTS medium
Level-sensitive latch differs from edge-triggered FF in that it
A is transparent while enable is active ✓ B changes only on clock edge always ✓ C cannot store any state ✓ D requires JK inputs only ✓ Show Answer 💡 Explanation: Latch passes input through when enabled; FF captures on edge.
Q42 easy
Register is constructed from
A only combinational gates without memory ✓ B multiple flip-flops storing parallel bits on common clock ✓ C single diode rectifier ✓ D op-amp integrator only ✓ Show Answer 💡 Explanation: Array of FFs holds multi-bit word synchronously.
Q43 Past Paper · PPSC/FPSC/NTS medium
Flip-flop propagation delay tpd is
A setup plus hold only without output delay ✓ B time from clock edge to valid output change at Q ✓ C clock period definition only ✓ D rise time of power supply only ✓ Show Answer 💡 Explanation: tpd limits maximum operating frequency with margin.
Q44 hard
Metastability occurs when
A inputs are stable with large setup margin ✓ B power supply is perfectly regulated ✓ C only in combinational logic without storage ✓ D setup or hold time is violated and output dwells between valid logic levels ✓ Show Answer 💡 Explanation: Unresolved state may persist unpredictably before settling.
Q45 hard
Master-slave flip-flop reduces
A need for setup time entirely ✓ B clock frequency to DC only ✓ C race conditions by separating capture and output update ✓ D power supply decoupling ✓ Show Answer 💡 Explanation: Two-stage isolation prevents transparent window glitches.
Q46 Past Paper · PPSC/FPSC/NTS medium
Asynchronous preset and clear on flip-flops
A override clocked operation immediately when asserted ✓ B only work on clock edge like D input ✓ C disable all outputs to high-Z always ✓ D replace the need for power supply ✓ Show Answer 💡 Explanation: Direct set/reset forces known state independent of clock.
Q47 medium
Hold time of flip-flop is
A time before clock edge only ✓ B propagation delay from clock to Q only ✓ C minimum time data must remain stable after active clock edge ✓ D frequency of oscillator only ✓ Show Answer 💡 Explanation: Hold violation can corrupt latched value.
Q48 medium
Setup time of flip-flop is
A minimum time data must be stable before active clock edge ✓ B time after clock edge for output change ✓ C hold time after edge only without setup ✓ D clock period only ✓ Show Answer 💡 Explanation: Violating setup time causes metastable or wrong capture.
Q49 Past Paper · PPSC/FPSC/NTS medium
T flip-flop toggles output when
A T is 0 at clock edge ✓ B reset is deasserted only ✓ C T input is 1 at active clock edge ✓ D preset alone without clock ✓ Show Answer 💡 Explanation: T=1 complements Q; T=0 holds state.
Q50 medium
JK flip-flop toggles when
A J=0 and K=0 always ✓ B J and K are both 1 at active clock edge ✓ C only asynchronous reset is asserted ✓ D clock is held LOW indefinitely ✓ Show Answer 💡 Explanation: JK combines S-R with toggle when both inputs high.
Q51 easy
Positive-edge-triggered flip-flop changes state on
A falling edge only always ✓ B rising edge of clock ✓ C any DC level without transition ✓ D both edges simultaneously ✓ Show Answer 💡 Explanation: Edge triggering samples input at clock transition.
Q52 Past Paper · PPSC/FPSC/NTS easy
D flip-flop eliminates
A need for clock entirely ✓ B all propagation delay ✓ C indeterminate condition of SR latch by using single data input ✓ D reset pin unconditionally ✓ Show Answer 💡 Explanation: D input maps to next state; S/R derived internally.
Q53 medium
SR latch built with NOR gates has invalid state when
A both inputs are 0 ✓ B only S is 0 ✓ C only R is 0 with S pulsed ✓ D S and R are both 1 simultaneously (active-high inputs) ✓ Show Answer 💡 Explanation: S=R=1 can force both outputs to 0 in NOR latch (race).
Q54 hard
K-map method is limited compared to Quine-McCluskey for
A being less accurate always ✓ B inability to handle don′t-cares ever ✓ C requiring software always ✓ D manual convenience on few variables rather than algorithmic scalability ✓ Show Answer 💡 Explanation: QM tabular method scales better for many variables.
Q55 Past Paper · PPSC/FPSC/NTS medium
Overlapping groups in K-map are allowed because
A each minterm can appear in only one group always ✓ B overlap invalidates simplification ✓ C a minterm may participate in multiple groups to form larger coverings ✓ D only diagonal overlap is permitted ✓ Show Answer 💡 Explanation: Shared minterms help achieve minimal total term count.
Q56 hard
Five-variable K-map is often implemented as
A single row of five cells only ✓ B circular map without Gray code ✓ C truth table without grouping ✓ D two four-variable maps with shifting variable ✓ Show Answer 💡 Explanation: Fifth variable selects between two 4-var submaps.
Q57 easy
SOP form obtained from K-map is a sum of
A only single literals without products ✓ B POS maxterms only ✓ C NAND operations without OR ✓ D product terms (AND terms ORed together) ✓ Show Answer 💡 Explanation: Each group becomes a product term; terms are ORed.
Q58 Past Paper · PPSC/FPSC/NTS hard
Redundant group in K-map is one that
A is the smallest possible group always ✓ B covers only don′t-cares without 1s ✓ C does not cover any minterm not already covered by essential groups ✓ D cannot be formed on the map ✓ Show Answer 💡 Explanation: Non-essential groups may be omitted from minimal cover.
Q59 easy
Number of cells in K-map for n variables is
A n × 2 only ✓ B 2^n ✓ C n^2 without exponent ✓ D 2n only ✓ Show Answer 💡 Explanation: Each input combination maps to one cell.
Q60 hard
For POS minimization using K-map, groupings are made on
A zeros (or maxterms) to obtain sum terms ✓ B only 1s without zeros ever ✓ C don′t-cares only without zeros ✓ D output column of truth table only ✓ Show Answer 💡 Explanation: Grouping 0s yields simplified product-of-sums form.