Question 1

Duty-cycle conversion at constant frequency using a 74121 one-shot Convert a 33 kHz waveform from 30% duty to 60% duty while keeping the frequency at 33 kHz. Choose suitable Rint and Cext.

Accepted Answer

Introduction / Context: Changing duty cycle at a fixed frequency can be achieved by pulse stretching (or re-timing) with a monostable. The 74121 is a non-retriggerable one-shot whose output width is set by its timing network. Here, we need to expand the high time to 60% while keeping the input frequency at 33 kHz. Given Data / Assumptions: Input frequency fin = 33 kHz → period T ≈ 1 / 33,000 ≈ 30.303 µs Target duty = 60% → desired high time th ≈ 0.60 * 30.303 µs ≈ 18.18 µs 74121 approximate timing: t_w ≈ K * Rint * Cext, where K is a family constant on the order of 0.7 (device- and family-dependent) Options provide discrete Rint and Cext choices Concept / Approach: Select Rint and Cext such that t_w ≈ desired high time. Using a common approximation t_w ≈ 0.7 * Rint * Cext guides us to an R*C product near th / 0.7. Step-by-Step Solution: Compute target RC: RC_target ≈ th / 0.7 ≈ 18.18 µs / 0.7 ≈ 25.97 µsEvaluate options: (a) 2 kΩ * 0.012 nF = 24 ns → far too small(b) 2 kΩ * 0.012 µF = 24 µs → t_w ≈ 0.7 * 24 µs ≈ 16.8 µs (close to 18.18 µs)(c) 4 kΩ * 0.012 nF = 48 ns → far too small(d) 4 kΩ * 0.012 µF = 48 µs → t_w ≈ 0.7 * 48 µs ≈ 33.6 µs (too large) Verification / Alternative check: Accounting for component tolerances and the device's K variation, option (b) yields a practical pulse width near the target. Fine-tuning is typically done by slight value adjustments or trimming if required. Why Other Options Are Wrong: (a) and (c): RC products in the nanoseconds range cannot realize an ~18 µs pulse. (d): Produces a pulse too long (~33.6 µs), overshooting the 60% target. Common Pitfalls: Confusing 74121 timing constant with 555 values; misreading nF vs µF; forgetting frequency–period conversion when converting duty cycles. Final Answer: Rint = 2 kΩ, Cext = 0.012 µF

Question 2

Timing with 74123 retriggerable monostable — find the output pulse width (t_w). Given: R_ext = 4.7 kΩ, C_ext = 0.022 µF (22 nF). Select the closest value of t_w for a standard 74x123 device.

Accepted Answer

Introduction / Context: Monostable multivibrators (one-shots) such as the 74x123 produce a single output pulse whose width is set by an external resistor and capacitor. This question tests whether you know the proper timing constant for the 74123 family and can convert units correctly. Given Data / Assumptions: IC: 74123 (retriggerable monostable) R_ext = 4.7 kΩ C_ext = 0.022 µF = 22 nF Use the typical 74123 timing relation t_w ≈ K * R_ext * C_ext, where K is about 0.32 to 0.45 depending on the specific logic family and operating conditions. Concept / Approach: For the 74123, many datasheets give a nominal constant near 0.32 for common trigger/configurations. Therefore compute RC first, then apply K ≈ 0.32 for a close estimate of pulse width. Step-by-Step Solution: RC = R_ext * C_ext = 4.7 kΩ * 22 nF = 4700 Ω * 22e-9 F = 103.4e-6 s = 103.4 µst_w ≈ K * RCUsing K ≈ 0.32: t_w ≈ 0.32 * 103.4 µs ≈ 33.1 µs (≈ 33.3 µs) Verification / Alternative check: If K were as high as 0.45, t_w would be ≈ 46.5 µs. Options provided include 33.3 µs and 46.5 µs; however, the most widely quoted nominal for the 74123 timing is close to 0.32 in many reference problems, so 33.3 µs is the best match to typical expectations. Why Other Options Are Wrong: 290 ms and 29 ms: These are orders of magnitude larger than any K * RC with the given components. 290 µs: Would require K ≈ 2.8, not applicable. 46.5 µs: Uses K ≈ 0.45; while possible in some families, the common nominal answer for this setup is lower. Common Pitfalls: Confusing the 74123 constant with the 555 monostable constant (≈ 1.1). Unit mistakes when converting µF to F or kΩ to Ω. Final Answer: 33.3 µs

Question 3

Monostable 555 timer — how many stable states does it inherently have? Choose the number of stable states for a one-shot (monostable) configuration.

Accepted Answer

Introduction / Context: A monostable multivibrator (one-shot) is designed to have a single stable state and one quasi-stable state. When triggered, it leaves its stable state temporarily and then returns, producing a pulse. This question checks conceptual understanding of the 555 in monostable mode. Given Data / Assumptions: Device: 555 timer in monostable configuration Trigger produces one timing pulse After timing, the circuit returns to its resting state Concept / Approach: The defining property of a monostable is exactly one stable state. The other state exists only transiently (quasi-stable) during the output pulse interval set by R and C. Step-by-Step Solution: Identify the operating states: stable (rest), quasi-stable (timed pulse)Count stable states: there is only one stable state Verification / Alternative check: Compare with astable (no stable state, constant oscillation) and bistable (two stable states, i.e., flip-flop). The monostable sits between them conceptually with exactly one stable state. Why Other Options Are Wrong: 0: Describes astable oscillators, not monostables. 2 or 3 or 4: Would imply flip-flop behavior (two) or multi-stable systems; not applicable. Common Pitfalls: Confusing monostable with astable or bistable due to naming similarity. Final Answer: 1

Question 4

Applications: pulse stretching, time delay, and one-shot pulse generation — which multivibrator is best? Select the multivibrator type commonly used for these three tasks.

Accepted Answer

Introduction / Context: Different multivibrator types serve distinct timing roles. This question checks recognition of which topology naturally performs pulse stretching, time-delay creation, and single-pulse generation after a trigger. Given Data / Assumptions: Goal: extend a pulse, insert a delay, or generate a one-shot pulse Available choices: astable, monostable, multistable, bistable, Schmitt trigger Concept / Approach: A monostable multivibrator (one-shot) produces a single output pulse of controlled width in response to a trigger. By choosing R and C, you implement pulse stretching and time delays without continuous oscillation. Step-by-Step Solution: Identify functionality required: single, timed pulse after a triggerMap to topology: monostable multivibrator (one-shot) Verification / Alternative check: Astable oscillators run continuously, which is unnecessary for one-shot tasks. Bistables store state but do not time a pulse width. Schmitt trigger shapes edges but does not create timed widths by itself. Why Other Options Are Wrong: Astable: produces a free-running square wave. Multistable: generic term, not a standard timing choice here. Bistable: latches a state but needs extra timing to form a pulse. Schmitt trigger: adds hysteresis for clean switching, not timing width control. Common Pitfalls: Assuming any RC with a comparator equals a monostable; the one-shot topology matters. Final Answer: monostable

Question 5

Retriggerable vs non-retriggerable one-shot — what is the key difference? Choose the statement that correctly distinguishes how the output pulse behaves with successive triggers.

Accepted Answer

Introduction / Context: One-shot multivibrators are either retriggerable or non-retriggerable. Understanding how additional triggers affect the timing pulse is crucial in pulse-width control and glitch filtering. Given Data / Assumptions: We compare output behavior under multiple triggers arriving before the pulse expires. One device is retriggerable; the other is not. Concept / Approach: In a retriggerable one-shot, any valid trigger that occurs during the active pulse restarts the timing interval, effectively lengthening (stretching) the total pulse width. In a non-retriggerable one-shot, additional triggers received while the pulse is active are ignored until the device returns to its stable state. Step-by-Step Solution: Consider a pulse that lasts T seconds.Apply a second trigger at t = T/2.Retriggerable: timing restarts, so new end time ≈ t + T, total width > T.Non-retriggerable: timing continues unchanged, total width = T. Verification / Alternative check: Datasheets typically note "retriggering extends the output pulse" for devices like 74x123, contrasting with 74x121 (non-retriggerable). Why Other Options Are Wrong: Nonretriggerable can only be triggered once: false; it can be triggered again after it times out. Retriggerable can be triggered many times: true but incomplete; the distinguishing effect is pulse extension. Nonretriggerable stretches the pulse: incorrect by definition. Successive triggers shorten the pulse: opposite of retrigger behavior. Common Pitfalls: Confusing "ignoring triggers while busy" with "cannot be triggered again ever". Final Answer: The output pulse can be stretched with a retriggerable.

Question 6

Retriggerable monostable multivibrator — identify what is NOT an inherent characteristic. Select the option that does not generally define retriggerable monostables across families.

Accepted Answer

Introduction / Context: This item checks whether you can distinguish universal characteristics of retriggerable monostables from package- or family-specific features. The focus is on traits that are always true vs traits that may be true only for particular ICs. Given Data / Assumptions: We consider behavior and construction across common logic families (TTL/CMOS). Terms like "dual" can describe a package option, not a fundamental property. Concept / Approach: Retriggerability describes timing behavior: new triggers extend the active pulse. Whether an IC is "dual" (two monostables per package) or uses a particular reset polarity varies by part number and family. External timing components are also typical and not unique to retriggerable devices. Step-by-Step Reasoning: Identify core behavior: retriggering extends the output pulse.Check packaging: "dual" refers to two one-shots in one IC (e.g., certain 74x123 variants), not an inherent property.Reset polarity and internal components vary by manufacturer and subfamily. Verification / Alternative check: Some retriggerable parts are single (one per package), others are dual. Therefore "dual" is not universally characteristic. Why Other Options Are Wrong: Active-HIGH reset: many parts provide a reset; polarity can vary, but a reset that halts timing is characteristic behavior broadly. No internal timing resistor: timing typically uses external R and C in both retriggerable and non-retriggerable families. Extends the pulse if retriggered: this is the very definition of retriggerable behavior. none of the above: incorrect because at least one option (dual) is not inherent. Common Pitfalls: Assuming a datasheet feature of a specific IC is universal to all retriggerables. Final Answer: It is a dual multivibrator.

Question 7

Monostable multivibrator at rest — what is the typical Q output level before any trigger? Assume standard TTL/CMOS one-shot behavior where the output goes active for a timed pulse and then returns.

Accepted Answer

Introduction / Context: A monostable multivibrator has one stable state. When idle (no trigger), the output sits in that state; a trigger forces a temporary transition for a timed duration. This question checks default logic level knowledge for common one-shot outputs. Given Data / Assumptions: Standard one-shot where the timed pulse is a HIGH pulse at Q. After timing, the device returns to its stable state. Concept / Approach: Most textbook one-shots are depicted with Q LOW at rest and a positive-going pulse on trigger. Although some devices offer inverted outputs or alternative wiring, the conventional Q output idles LOW and pulses HIGH. Step-by-Step Solution: Identify the polarity of the pulse at Q: typically a HIGH pulse.Therefore, before trigger, Q must be LOW to allow a positive-going pulse. Verification / Alternative check: Diagrams for 555 monostable and 74x121/123 families commonly show Q LOW at rest, going HIGH for t = K * R * C, then returning LOW. Why Other Options Are Wrong: +5 V or HIGH: that would imply a LOW-going pulse, not the typical default. SET: descriptive but not a logic level. Hi-Z: outputs are driven, not tri-stated, in basic one-shots. Common Pitfalls: Confusing Q with /Q (inverted output), which would idle HIGH. Final Answer: LOW

Question 8

Astable multivibrator behavior — which statement best describes its output? Choose the accurate description of an astable circuit's output state over time.

Accepted Answer

Introduction / Context: An astable multivibrator has no stable state. It continuously oscillates between two levels, generating a periodic waveform without requiring external triggers. This question verifies recognition of astable behavior compared to mono- and bi-stable types. Given Data / Assumptions: Astable configuration using active components and RC feedback Free-running oscillator behavior expected Concept / Approach: By design, astables perpetually charge and discharge timing capacitors, alternately driving the output HIGH and LOW. No trigger input is required to sustain oscillation, unlike monostables and bistables. Step-by-Step Reasoning: Identify presence of any stable resting state: none in an astable.Result: output keeps toggling between logic levels with a period set by R and C. Verification / Alternative check: 555 in astable mode or cross-coupled inverters with RC produce a continuous square wave, confirming the description. Why Other Options Are Wrong: LOW until triggered / HIGH until triggered: describes monostable behavior. Floats until triggered: outputs are not left floating in proper designs. Stays latched unless reset: describes bistable flip-flops. Common Pitfalls: Confusing "needs power-up to start" with "needs a trigger"; many astables self-start due to component tolerances. Final Answer: constantly switches between two states

Question 9

Charging behavior of an RC network — which statements are true as a capacitor charges from a DC source? Assume a simple series R–C connected to a constant DC supply and select the best overall answer.

Accepted Answer

Introduction / Context: Classic RC charging exhibits exponential behavior for both capacitor voltage and loop current. The time constant tau = R * C sets the rate. After about 5 * tau, the capacitor is essentially fully charged for practical purposes. This question checks fundamental time-constant intuition. Given Data / Assumptions: Series R–C driven by an ideal DC source Initial capacitor voltage is zero No leakage and constant R Concept / Approach: Standard equations for charging are: Vc(t) = V_s * (1 − e^(−t/tau)) and I(t) = (V_s/R) * e^(−t/tau). Engineers often use 5 * tau as the practical "full-charge" time where Vc ≈ 99%. Step-by-Step Solution: tau = R * CVc(t) increases exponentially toward V_sI(t) decreases exponentially toward 0t ≈ 5 * tau gives Vc ≈ 0.993 * V_s, close enough to "charged" Verification / Alternative check: Plotting Vc(t) and I(t) vs time confirms the exponential rise and decay and the 5 * tau rule of thumb. Why Other Options Are Wrong: Options a, b, c are all correct statements; thus "all of the above" is the best choice. "None of the above" contradicts established RC theory. Common Pitfalls: Assuming charging is linear; it is exponential. Equating 5 * tau with an exact limit; it is a practical guideline. Final Answer: all of the above

Question 10

One-shot (monostable) timing dependencies — which change increases the output pulse width? Assume a standard one-shot whose width t is set by an external R * C network.

Accepted Answer

Introduction / Context: Monostable pulse width is primarily determined by the product R * C and, secondarily, by internal threshold fractions. Understanding which external parameter directly increases the width is essential in practical design and troubleshooting. Given Data / Assumptions: One-shot with t ≈ K * R * C, where K is a device-dependent constant Supply-voltage dependence is weak over the rated range Concept / Approach: If t is proportional to R * C, then increasing C (or R) increases t, while decreasing either decreases t. Changes to UTP or edges typically do not dominate compared to R and C selection for basic one-shots. Step-by-Step Solution: t ≈ K * R * CIf C ↑ then R * C ↑ ⇒ t ↑If R ↓ then R * C ↓ ⇒ t ↓ Verification / Alternative check: Datasheet curves generally show near-linear scaling of pulse width with timing capacitance across practical ranges. Why Other Options Are Wrong: Supply voltage increases: often minimal effect on t within spec. Timing resistor decreases: reduces t. UTP decreases: threshold variations have smaller, device-specific effects. Faster trigger edge: affects triggering reliability, not width set by R and C. Common Pitfalls: Overestimating supply sensitivity for timing; R and C dominate. Final Answer: timing capacitance increases

Question 11

Terminology: another common name for a bistable multivibrator. Select the widely used digital design term that refers to a bistable.

Accepted Answer

Introduction / Context: A bistable multivibrator has two stable states and stores one bit of information. In digital electronics, it is commonly implemented and discussed under a more familiar name. This question checks your vocabulary alignment with standard digital design terminology. Given Data / Assumptions: Two stable output states External inputs can set, reset, or toggle the state Concept / Approach: The ubiquitous term for a bistable multivibrator in digital systems is "flip-flop". It can be edge- or level-controlled and is the building block of registers, counters, and state machines. Step-by-Step Reasoning: Identify behavior: two stable states ⇒ bistable storage element.Map to digital term: flip-flop (or latch, depending on control), but the most general and common name in synchronous logic is flip-flop. Verification / Alternative check: Textbooks equate the bistable multivibrator with flip-flops (SR, JK, D, T) used in sequential logic. Why Other Options Are Wrong: On-off switch: a mechanical analogy, not a precise circuit type. Oscillator: describes astable behavior. Latch with analog feedback: imprecise and not a standard term here. Schmitt trigger: a comparator with hysteresis, not a memory element. Common Pitfalls: Confusing latches and flip-flops; both are bistables, but "flip-flop" is the canonical term. Final Answer: a flip-flop

Question 12

RC discharge in basic electronics A 0.3 µF capacitor is initially at 2.7 V. It is discharged through a 10 kΩ resistor for 6.5 ms. What voltage remains across the capacitor after this time?

Accepted Answer

Introduction / Context: Capacitors in RC circuits discharge exponentially. This question tests your ability to apply the standard RC decay formula to find the remaining voltage after a given time constant interval. Given Data / Assumptions: Capacitance C = 0.3 µF = 0.3 * 10^-6 F. Resistance R = 10 kΩ = 10,000 Ω. Initial voltage V0 = 2.7 V. Elapsed time t = 6.5 ms = 6.5 * 10^-3 s. Concept / Approach: The capacitor voltage during discharge follows V(t) = V0 * exp( - t / (R * C) ). The product R * C is the time constant tau, which sets the exponential decay rate. Step-by-Step Solution: 1) Compute tau: tau = R * C = 10,000 * 0.3 * 10^-6 = 3 * 10^-3 s = 3 ms.2) Form the exponent: t / tau = 6.5 ms / 3 ms ≈ 2.1667.3) Evaluate the exponential: exp( -2.1667 ) ≈ 0.1145.4) Remaining voltage: V(t) = 2.7 * 0.1145 ≈ 0.30915 V ≈ 0.309 V. Verification / Alternative check: Since 6.5 ms is a little over two time constants (2 tau = 6 ms), a remaining fraction near exp( -2 ) ≈ 0.135 is expected; our more precise 0.1145 is reasonable because 6.5 ms > 6 ms. Why Other Options Are Wrong: 0.0 V: Ideal exponential never reaches exactly zero at finite time. 2.7 V: That is the initial voltage; no discharge applied. 2.295 V: Implies very little decay; contradicts t > 2 tau. Common Pitfalls: Mixing up units (µF, kΩ, ms), using linear instead of exponential decay, or forgetting to compute tau first. Final Answer: 0.309 V

Question 13

The bistable multivibrator is an R-C flip-flop.

Accepted Answer

False

Question 14

A single Schmitt trigger inverter is all that is needed to build a simple astable multivibrator.

Accepted Answer

False

Question 15

In the 74121, which is nonretriggerable, any triggers that come in before the end of the timing cycle are ignored.

Accepted Answer

True

Question 16

When extremely critical timing is required, a quartz crystal can be used.

Accepted Answer

True

Question 17

A comparator simply outputs a HIGH or LOW based on a comparison of the voltage levels at its input.

Accepted Answer

True

Question 18

When the 74123, a retriggerable monostable multivibrator device, ends its timing cycle it must be started all over again.

Accepted Answer

False

Question 19

Timing accuracy of more than eight significant digits can be easily achieved by using quartz crystals.

Accepted Answer

False

Question 20

The 74S124 TTL chip is a voltage-controlled oscillator that will generate a specific frequency at Vout.

Accepted Answer

False

Multivibrators and 555 Timer Questions