Question 1

Flip-flop timing integrity: What is a race condition in the context of an edge-triggered flip-flop and its input timing?

Accepted Answer

Introduction / Context: A race condition in sequential logic occurs when signals that should be stable during the sampling event change too close to the active clock edge, potentially violating setup or hold requirements and causing unpredictable outputs or metastability. Given Data / Assumptions: Edge-triggered flip-flop with defined setup time ts and hold time th. Inputs intended to be stable around the sampling edge. Clock with finite transition time. Concept / Approach: When inputs transition during the aperture window around the clock edge, internal latches can capture inconsistent states, leading to raced or metastable outcomes. Ensuring ts and th margins avoids races. Step-by-Step Solution: 1) Identify the aperture: from (edge − ts) to (edge + th).2) If input changes inside this window, the device may not resolve deterministically.3) Design practice: constrain combinational delays and synchronize asynchronous signals.4) Add proper timing constraints and CDC synchronizers where needed. Verification / Alternative check: Static timing analysis should show positive slack for ts and th; oscilloscope capture reveals glitches if violated. Why Other Options Are Wrong: Changing slightly before the edge: If it is earlier than ts, it is acceptable. Changing slightly after the edge: If it is later than th, it is acceptable. Static and unchanged: This is the desired condition, not a race. Common Pitfalls: Feeding asynchronous inputs directly to synchronous logic without synchronization, and neglecting clock skew while budgeting ts/th. Final Answer: The inputs to a trigger device are changing at the same time as the active trigger edge.

Question 2

Flip-flop timing parameters: Which parameters specify the propagation delay from the clock (Cp) input to the Q output?

Accepted Answer

Introduction / Context: Datasheets list several timing parameters for sequential devices. Correctly identifying propagation delays from clock to output is essential for timing closure and performance prediction. Given Data / Assumptions: Edge-triggered flip-flops with defined delays. tPHL and tPLH denote propagation from input to output with polarity. Other parameters include setup/hold, pulse width, and maximum frequency. Concept / Approach: Propagation delay measures time for an output transition after a triggering input event. For clocked devices, the relevant labels are tPLH (LOW-to-HIGH at output) and tPHL (HIGH-to-LOW at output) after a defined clock edge. Step-by-Step Solution: 1) Identify the cause: active clock edge at Cp.2) Map to effects: Q transitions occur after a delay.3) Read tPLH/tPHL as the delays for the two transition polarities. Verification / Alternative check: Scope measurements of Q relative to Cp confirm datasheet values within tolerances. Why Other Options Are Wrong: ts, th: Setup and hold requirements, not propagation delay. tw(L), tw(H): Minimum pulse widths, not output delays. fmax: Maximum toggling frequency, a derived bound, not a delay parameter. Common Pitfalls: Confusing tpd with setup/hold and ignoring process/voltage/temperature effects on delays. Final Answer: tPHL, tPLH

Question 3

Pull-down resistor caution: What is the primary concern when biasing an input with a pull-down resistor?

Accepted Answer

Introduction / Context: Pull-down resistors bias an input to logic LOW when undriven. While useful, they can increase steady-state current draw and power loss, especially when the node is driven HIGH for significant time. Given Data / Assumptions: VCC is present and the signal may be HIGH for long durations. Pull-down connects the node to ground through a resistor. Input leakage is small but nonzero. Concept / Approach: When the node is driven HIGH, current flows from VCC through the driver into the resistor to ground, dissipating power P = V^2 / R approximately (neglecting driver drop). Choosing too small a resistor leads to unnecessary power and thermal stress. Step-by-Step Solution: 1) Consider the worst case: node HIGH continuously with pull-down to ground.2) Compute power in resistor: P ≈ (VCC^2) / R.3) Select R large enough to limit current yet small enough to provide noise immunity when undriven.4) Evaluate logic-family input currents to set the allowable R range. Verification / Alternative check: Thermal check of resistor and supply current measurements validate the choice. Why Other Options Are Wrong: Low power dissipation: The concern is the opposite—excess dissipation. Keeps a floating terminal LOW: That is a benefit, not a concern. False triggering: Possible with poor values, but the primary design worry is wasted power. Common Pitfalls: Copying pull-up values without considering asymmetry, using very small resistances, and ignoring battery life impacts. Final Answer: the high power dissipation of the resistor

Question 4

J-K versus S-R flip-flop: What is the key advantage of the J-K flip-flop when compared to the S-R latch/flip-flop?

Accepted Answer

Introduction / Context: The S-R latch has an invalid input condition when both S and R are HIGH simultaneously. The J-K flip-flop extends this behavior to allow a defined toggle action when both inputs are asserted, making it more versatile. Given Data / Assumptions: J-K is edge-triggered in most logic families. Inputs J = 1 and K = 1 cause toggling on each active clock edge. S-R with S = R = 1 is invalid or forbidden. Concept / Approach: The toggle state enables divide-by-2 counters and flexible state machines without additional gating to avoid illegal states, simplifying designs. Step-by-Step Solution: 1) Map inputs: J=1, K=0 → set; J=0, K=1 → reset; J=K=0 → hold; J=K=1 → toggle.2) Compare to S-R: S=R=1 is undefined; hardware must prevent it.3) Conclude advantage: J-K provides a clean, useful toggle mode. Verification / Alternative check: Construct a divide-by-2 using a single J-K with J=K=1; observe Q toggling each clock. Why Other Options Are Wrong: Much faster: Speed depends on implementation, not a fundamental advantage. No propagation delay problems: All real devices have propagation delays. Two outputs: Many flip-flops provide Q and not-Q; this is not unique to J-K. Common Pitfalls: Assuming J-K eliminates all illegal transitions automatically; race-around must be mitigated by edge triggering or master-slave design. Final Answer: The J-K flip-flop has a toggle state.

Question 5

Digital electronics — input jitter and false switching mitigation In digital systems, which circuit specifically overcomes erratic switching caused by noisy or jittery input edges by introducing hysteresis?

Accepted Answer

Introduction / Context: Schmitt trigger circuits are widely used in digital electronics to clean up slow, noisy, or jittery input signals. They provide two distinct threshold voltages (one for rising and another for falling inputs), a behavior known as hysteresis. This prevents multiple unwanted transitions when an input hovers near a single threshold, thereby ensuring stable logic-level decisions. Given Data / Assumptions: We are dealing with a digital input that exhibits jitter or noise around its switching point. Goal is to avoid false triggering or multiple transitions near the threshold. Standard logic families expect clean, fast transitions. Concept / Approach: The key concept is hysteresis. A Schmitt trigger implements two thresholds: V_T+ for LOW to HIGH transitions and V_T- for HIGH to LOW transitions. Once the input crosses V_T+, the output switches and will not switch back until the input falls below V_T-. This input hysteresis window filters slow edges and noise. Step-by-Step Solution: 1) Identify the problem: input jitter near a single threshold causes repeated toggling.2) Require a device with separate rising and falling thresholds to add immunity.3) Recognize that a Schmitt trigger provides hysteresis by design.4) Conclude the correct choice for robust switching is the Schmitt trigger. Verification / Alternative check: Oscilloscope measurements show that without hysteresis, a noisy slow edge produces multiple transitions. With a Schmitt trigger, the output flips once on crossing V_T+ and remains stable until the signal clearly crosses V_T-, eliminating chatter. Why Other Options Are Wrong: Astable multivibrator: generates a free-running square wave, not a noise-immune input buffer. Monostable multivibrator: produces one pulse per trigger, not continuous edge cleanup. Bistable multivibrator: a latch or flip-flop with no inherent hysteresis window on an analog input. Common Pitfalls: Confusing edge shaping (debouncing) with pulse generation. Monostables can time-limit responses but do not fix a noisy threshold. The Schmitt trigger specifically addresses noisy transitions via hysteresis. Final Answer: Schmitt trigger

Question 6

TTL NAND driving an LED indicator LOW (active-LOW sink configuration) A standard TTL NAND output sinks current through a 470 Ω series resistor and LED to VCC = 5 V. What current flows when the LED is ON (output LOW)?

Accepted Answer

Introduction / Context: When a TTL gate output pulls LOW, it can sink current from a load tied to VCC. For an LED indicator with a series resistor connected to VCC, the LED current is set by the voltage across the resistor, which depends on VCC, the LED forward drop, and the LOW-level output voltage of the gate (VOL). The question tests understanding of active-LOW LED driving and basic current calculation. Given Data / Assumptions: Logic family: standard TTL NAND, VCC ≈ 5 V. Series resistor: R = 470 Ω in series with the LED, tied to VCC. Gate output is LOW when LED is ON, so current flows from VCC → resistor → LED → gate output to ground. Use typical textbook values for LED forward drop and VOL to obtain the closest numerical option. Concept / Approach: Ohm's law sets current I = V_R / R, where V_R is the voltage across the resistor. V_R = VCC − V_LED − VOL. Typical small red LED forward drop is about 2.0 V to 2.2 V under a few milliamps. A TTL VOL under modest sink currents is typically a few hundred millivolts. The closest standard value among the options generally corresponds to around 5 mA to 6 mA. Step-by-Step Solution: 1) Assume VCC = 5.0 V.2) Assume V_LED ≈ 2.2 V (typical) and VOL ≈ 0.3 V (representative for TTL at a few mA).3) Compute resistor voltage: V_R = 5.0 − 2.2 − 0.3 = 2.5 V.4) Compute current: I = V_R / R = 2.5 / 470 ≈ 0.00532 A = 5.32 mA. Verification / Alternative check: Trying nearby typical values (for example, V_LED between 2.0 and 2.3 V and VOL between 0.2 and 0.4 V) yields currents in the 5 mA to 6.5 mA range. The given options include 5.32 mA, which matches the calculation using mainstream assumptions. Why Other Options Are Wrong: 7.02 mA: implies a much lower LED drop or VOL, not typical at this operating point. 8.51 mA: requires an unrealistically low combined drop across LED and VOL for 470 Ω. 10.63 mA: far too high for the stated component values and typical TTL levels. Common Pitfalls: Forgetting to include VOL. Assuming the LED is tied to ground rather than VCC in an active-LOW sink configuration leads to the wrong reference polarities. Final Answer: 5.32 mA

Question 7

Use-case of a Schmitt trigger in practical digital circuits Which application best illustrates the advantage of hysteresis for cleaning noisy, bouncing inputs?

Accepted Answer

Introduction / Context: Mechanical pushbuttons and switches do not make a clean single transition; they bounce, producing rapid ON-OFF edges. A Schmitt trigger adds hysteresis, ensuring only one clean digital transition despite noisy or bouncing inputs. This question checks recognition of the most common real-world use-case for Schmitt triggers. Given Data / Assumptions: Input is from a mechanical switch that exhibits contact bounce. We want a single, stable digital-level transition at the logic output. Hysteresis eliminates multiple triggers near the threshold. Concept / Approach: Schmitt triggers implement two thresholds: V_T+ and V_T-. An input must cross V_T+ to be registered as HIGH and later fall below V_T- to register LOW. This window rejects intermediate chatter due to bounce or small noise spikes. Step-by-Step Solution: 1) Identify the symptom: multiple spurious transitions from a bouncing switch.2) Choose an input conditioning method that requires a clear overdrive past a threshold window.3) A Schmitt trigger naturally provides hysteresis to clean the signal.4) Therefore, the best application is switch debouncing. Verification / Alternative check: Compare scope traces: a raw switch shows multiple edges; after a Schmitt trigger, only one clean edge remains per actuation. Why Other Options Are Wrong: Racer: not a defined digital building block and does not address bounce. Astable oscillator: generates periodic signals rather than debouncing an input. Transition pulse generator: shapes edges but does not inherently solve bounce without hysteresis. Common Pitfalls: Confusing simple RC filtering with guaranteed single-edge switching. RC alone may still allow chatter; hysteresis provides deterministic thresholds. Final Answer: switch debouncer

Question 8

Flip-flop behavior near the active clock edge What is the condition called when flip-flop inputs change at the same instant the active clock edge occurs, risking an indeterminate or unintended output?

Accepted Answer

Introduction / Context: Flip-flops sample inputs at an active clock edge. If inputs change too close to that edge, setup or hold time requirements can be violated, leading to uncertain behavior. Many curricula refer to the resulting hazard as a race or race condition, closely tied to metastability in practical devices. Given Data / Assumptions: Inputs are changing as the clock edge arrives. Device requires finite setup and hold time margins. Violation of these margins triggers unpredictable or oscillatory outcomes. Concept / Approach: If data transitions occur within the aperture around the sampling edge, the storage nodes can enter a metastable state or produce unexpected toggling. This conflict between changing data and the latching action is commonly described as a race condition in many introductory texts on sequential logic. Step-by-Step Solution: 1) Recognize that flip-flops need stable inputs around the edge (setup and hold).2) Inputs changing at the sampling instant violate those requirements.3) Such timing leads to a race between data settling and latching.4) Hence, the condition is termed racing (race condition). Verification / Alternative check: Manufacturers specify setup and hold in datasheets. Timing analysis ensures input edges avoid the sampling aperture; otherwise, failures consistent with racing or metastability appear in hardware tests. Why Other Options Are Wrong: Toggling: normal behavior of a T or JK flip-flop, not a timing hazard description. Slave loading: not the standard name for this hazard. Pulse timing: generic phrase, not the accepted term for the race issue. Common Pitfalls: Using asynchronous input signals without synchronization, or ignoring clock-to-output, setup, and hold in timing closure. Final Answer: racing

Question 9

A series RC circuit can be used to generate a power-up (automatic) reset signal.

Accepted Answer

True

Question 10

A Schmitt trigger has a positive feedback circuit and experiences a phenomenon called hysteresis.

Accepted Answer

False

Question 11

TTL requires a constant supply voltage of 8.0 V.

Accepted Answer

False

Question 12

Pull-up resistors and pull-down resistors are used to keep a floating terminal HIGH.

Accepted Answer

False

Question 13

The output of a phototransister is determined by the presence or absence of light at its base input.

Accepted Answer

True

Question 14

The term race condition describes the time that an active signal must be present at the input to a flip-flop before an active clock edge is applied.

Accepted Answer

False

Question 15

A transfer function graph illustrates the most important specifications for the Schmitt trigger devices.

Accepted Answer

True

Question 16

There is no way to eliminate the effects of a switch bounce.

Accepted Answer

False

Question 17

The 7805 will get very hot if your circuit draws more than 0.5 A.

Accepted Answer

True

Question 18

The input levels to a flip-flop must be maintained for a minimum time period both before and after the edge of the clock signal is applied.

Accepted Answer

True

Digital Design Questions