Q: Negative feedback stability: For an amplifier with loop gain Aβ, which value of |Aβ| among the given choices yields the most stable closed-loop operation (largest stability margins)?

Introduction / Context: Stability of feedback amplifiers depends on the loop gain magnitude |Aβ| and loop phase. As the Nyquist/Bode criteria indicate, getting too close to the critical point (|Aβ| = 1 with −180° phase) risks oscillation. This question checks conceptual understanding of how loop-gain magnitude influences robustness. Given Data / Assumptions: Negative feedback is used. We compare only the magnitudes |Aβ| at the frequency where phase lag approaches −180°. Other factors (phase margin, pole locations) are comparable. Concept / Approach: Greater stability margin occurs when the locus stays farther from the critical point. Therefore, for similar phase conditions, a smaller |Aβ| provides larger gain margin and usually better phase margin. Step-by-Step Solution: Identify the smallest |Aβ| among options: 0.70.At |Aβ| = 0.70, the distance from the critical value 1 is largest, improving gain margin.Options with |Aβ| > 1 (1.20, 1.50) are potentially unstable if the phase lag nears −180°. Verification / Alternative check: Bode plot reasoning: with the unity-gain crossover well below the −180° phase point, phase margin is increased; reducing |Aβ| helps achieve that condition. Why Other Options Are Wrong: 0.95: closer to unity → smaller margins. 1.20 and 1.50: loop magnitude > 1 at critical phase → risk of oscillation. Exactly 1.00: marginal stability. Common Pitfalls: Assuming larger |Aβ| always improves performance; it improves accuracy but can harm stability if phase margin is limited. Final Answer: 0.70.

Question 1

An RC-coupled amplifier has open-loop midband gain A = 200 and a lower cutoff frequency of 50 Hz. Negative feedback with feedback factor β = 0.1 is applied. Estimate the new lower cutoff frequency with feedback.

Accepted Answer

Introduction / Context: Negative feedback reshapes an amplifier’s frequency response, typically trading gain for improved bandwidth and more controlled cutoff characteristics. In RC-coupled voltage amplifiers, applying negative feedback reduces the midband gain but extends both the low- and high-frequency bandwidth, often by approximately the desensitivity factor (1 + Aβ) for dominant single-pole regions. Given Data / Assumptions: Open-loop midband gain A = 200. Lower cutoff (−3 dB) frequency f_L = 50 Hz before feedback. Feedback factor β = 0.1, negative (regenerative effects are excluded). Single-pole dominant behavior around f_L so textbook bandwidth scaling applies. Concept / Approach: For a simple first-order low-frequency pole, negative feedback reduces the effective time constant by approximately the factor (1 + Aβ). Thus the lower cutoff shifts downward by the same factor, improving low-frequency response. Closed-loop gain also reduces to A_cl ≈ A / (1 + Aβ). Step-by-Step Solution: Compute desensitivity: 1 + Aβ = 1 + 200 * 0.1 = 21.New lower cutoff: f_L,new = f_L / (1 + Aβ) = 50 / 21 ≈ 2.38 Hz.Closed-loop midband gain (for reference): A_cl ≈ 200 / 21 ≈ 9.52. Verification / Alternative check: If you sketch Bode plots, the low-frequency pole moves left by a factor of ~21, consistent with the calculated ~2.4 Hz. Why Other Options Are Wrong: About 50 Hz: ignores bandwidth extension from feedback.About 5 Hz: uses a factor of 10 instead of 21.About 70.5 Hz: would be a narrowing of bandwidth (contrary to negative feedback benefits).About 21 Hz: misapplies (1 + Aβ) as a multiplier rather than a divider. Common Pitfalls: Forgetting that negative feedback broadens bandwidth while reducing midband gain; mixing up low- and high-frequency corner shifts. Final Answer: about 2.38 Hz

Question 2

In a common-emitter BJT configuration with DC current gain β_dc = 100, determine the collector current when the base current is 10 μA. Assume forward-active operation.

Accepted Answer

Introduction / Context: Small-signal and bias calculations in bipolar junction transistors (BJTs) often start from the simple relationship between collector current and base current: I_C = β I_B in the forward-active region. This question asks for a direct substitution, reinforcing basic DC analysis in common-emitter configurations. Given Data / Assumptions: β_dc (also denoted h_FE) = 100. Base current I_B = 10 μA. Transistor operates in forward-active region (not saturated or cut off). Concept / Approach: In forward-active region, collector current is proportional to base current by β_dc. The emitter current is approximately I_E ≈ (1 + β) I_B, but here only I_C is requested, so we use I_C = β_dc * I_B directly. Step-by-Step Solution: I_C = β_dc * I_B = 100 * 10 μA.Compute: 100 * 10 μA = 1000 μA = 1 mA.Thus, the collector current is 1 mA under the stated conditions. Verification / Alternative check: If V_CE is sufficiently large to avoid saturation (e.g., V_CE > 0.2–0.3 V typical), the forward-active assumption holds and the relation is valid. Why Other Options Are Wrong: 10 μA: This equals I_B, which would imply β = 1 (not the given 100).100 μA or 0.5 mA: Do not satisfy the β = 100 relationship.10 mA: Would require I_B = 100 μA or β = 1000 instead. Common Pitfalls: Confusing β with α; α ≈ β / (β + 1) is used in common-base analysis, not needed here. Final Answer: 1 mA

Question 3

Assertion–Reason (BJT Amplifiers): The common-emitter (CE) amplifier is the most widely used BJT amplifier configuration. Reason: A CE amplifier introduces zero phase shift between input and output.

Accepted Answer

Introduction / Context: Common-emitter (CE) stages are the workhorses of BJT amplifier design. They provide substantial voltage gain, moderate input impedance, and reasonably low output impedance, making them very versatile. This assertion–reason item checks your understanding of both the popularity of CE and its phase characteristics. Given Data / Assumptions: Linear small-signal CE amplifier with resistive biasing and load. Standard midband analysis (coupling/bypass capacitors treated as AC shorts). No special phase compensation network added. Concept / Approach: CE amplifiers offer high voltage gain due to collector load and transistor transconductance, alongside convenient biasing. However, a hallmark of CE is signal inversion: the small-signal output at the collector is 180° out of phase with the input at the base. Therefore, any claim of zero phase shift is false for a typical CE stage. Step-by-Step Solution: Assess Assertion (A): CE is most widely used — correct due to favorable gain and impedance properties.Assess Reason (R): CE has zero phase difference — false; CE introduces approximately 180° phase inversion in the passband.Match: A true, R false ⇒ the correct choice is 'A is correct R is wrong'. Verification / Alternative check: Small-signal hybrid-π model shows v_o ≈ −g_m * v_π * R_C/parallel load, where the negative sign indicates inversion (≈180°). Why Other Options Are Wrong: If R were true, CE would exhibit no inversion, contradicting standard textbook results. Other combinations misstate either the popularity or the phase property. Common Pitfalls: Confusing CE (inverting) with emitter follower (common-collector), which has ~0° phase shift but unity gain. Final Answer: A is correct R is wrong

Question 4

Evaluate the statement: 'The maximum theoretical efficiency of a class B power amplifier is 50%.' Indicate whether this statement is true or false and justify based on ideal analysis.

Accepted Answer

Introduction / Context: Class B push-pull power amplifiers are ubiquitous in audio and RF power stages because they significantly improve efficiency compared to class A. Understanding their ideal efficiency limit is a core topic in analog electronics and power amplifier design. Given Data / Assumptions: Ideal class B push-pull stage (complementary devices or transformer-coupled), no crossover distortion and sinusoidal output. No quiescent current (each device conducts over 180° of the cycle). Resistive load, ideal supplies, and devices with zero saturation voltage. Concept / Approach: Efficiency η is defined as P_out / P_in. For an ideal class B amplifier delivering a sinusoid to a resistive load, the average input power from the supply is related to the average of the rectified output current waveforms, while the output power equals V_rms^2 / R_L. The well-known ideal limit is η_max = π/4 ≈ 0.785, i.e., 78.5%, which is significantly higher than 50%. Step-by-Step Solution: For a sinusoidal output v_o(t) = V_m sin(ωt) across R_L, the rms value is V_m/√2.Output power: P_out = (V_m^2/2)/R_L.Average supply current per half cycle leads to total input power P_in = (2/π) * (V_m^2/R_L).Efficiency: η = P_out/P_in = [(V_m^2/2R_L)] / [(2/π)(V_m^2/R_L)] = π/4 ≈ 0.785. Verification / Alternative check: Textbook derivations for class B consistently yield η_max = 78.5% under ideal conditions; practical values are lower due to device drops and crossover effects. Why Other Options Are Wrong: 'True': contradicts the standard 78.5% ideal limit.Other qualifiers (transformer coupling, low power, regulated supply) do not change the fundamental ideal limit; they affect practical realizations, not the theoretical maximum. Common Pitfalls: Confusing class A (25–50% ideal) with class B; mixing up conduction angle effects on efficiency. Final Answer: False

Question 5

Common-emitter (CE) amplifier with a series R–L load: considering that a CE stage inverts phase by about 180°, and the load impedance Z_L consists of a resistor R in series with an inductor L (which introduces additional positive phase lead between …

Accepted Answer

Introduction / Context: In analog electronics, understanding phase relationships through active devices and reactive loads is essential for correct signal interpretation and stability analysis. A common-emitter (CE) amplifier inherently produces a phase inversion of approximately 180° between its small-signal input (base) and output (collector). When the load is not purely resistive but contains inductance, the output voltage across that load gains an additional phase lead with respect to the collector current. Given Data / Assumptions: Topology: single-stage CE amplifier. Load impedance: Z_L = R + jωL (series R–L). Small-signal operation; transistor internal phase shift around midband is ~180°. The inductor causes the load voltage to lead the load current by an angle φ where 0° < φ < 90°. Concept / Approach: In a CE stage, v_out (collector) is inverted with respect to v_in (base), giving 180°. The R–L series load has voltage that leads its current by the load angle φ = tan^-1(ωL/R). Since the collector current (and hence load current) largely follows the input with the 180° inversion already accounted for, the final load voltage leads the inverted current by φ, yielding a total phase shift of 180° + φ where 0° < φ < 90°. Step-by-Step Solution: Start with CE inversion: phase ≈ 180°.Compute load angle φ = tan^-1(ωL/R), which lies between 0° and 90°.Total phase difference = 180° + φ.Therefore the result must lie between 180° and 270° (exclusive). Verification / Alternative check: If L = 0 (purely resistive), φ = 0° and the phase is 180°. If R → 0 (nearly pure L), φ → 90° and the limit approaches 270°, confirming the range. Why Other Options Are Wrong: 0°: contradicts CE inversion. 180°: valid only for purely resistive load. More than 90° but less than 180°: underestimates the CE inversion. Exactly 270°: would require ideal pure inductance (impractical midband). Common Pitfalls: Forgetting that the phase of voltage across an inductive load leads the current. Confusing transistor internal small-signal delay with the fixed 180° inversion of CE configuration. Final Answer: More than 180° but less than 270°.

Question 6

Negative feedback stability: For an amplifier with loop gain Aβ, which value of |Aβ| among the given choices yields the most stable closed-loop operation (largest stability margins)?

Accepted Answer

Introduction / Context: Stability of feedback amplifiers depends on the loop gain magnitude |Aβ| and loop phase. As the Nyquist/Bode criteria indicate, getting too close to the critical point (|Aβ| = 1 with −180° phase) risks oscillation. This question checks conceptual understanding of how loop-gain magnitude influences robustness. Given Data / Assumptions: Negative feedback is used. We compare only the magnitudes |Aβ| at the frequency where phase lag approaches −180°. Other factors (phase margin, pole locations) are comparable. Concept / Approach: Greater stability margin occurs when the locus stays farther from the critical point. Therefore, for similar phase conditions, a smaller |Aβ| provides larger gain margin and usually better phase margin. Step-by-Step Solution: Identify the smallest |Aβ| among options: 0.70.At |Aβ| = 0.70, the distance from the critical value 1 is largest, improving gain margin.Options with |Aβ| > 1 (1.20, 1.50) are potentially unstable if the phase lag nears −180°. Verification / Alternative check: Bode plot reasoning: with the unity-gain crossover well below the −180° phase point, phase margin is increased; reducing |Aβ| helps achieve that condition. Why Other Options Are Wrong: 0.95: closer to unity → smaller margins. 1.20 and 1.50: loop magnitude > 1 at critical phase → risk of oscillation. Exactly 1.00: marginal stability. Common Pitfalls: Assuming larger |Aβ| always improves performance; it improves accuracy but can harm stability if phase margin is limited. Final Answer: 0.70.

Question 7

Bridge rectifier with capacitor filter at 50 Hz: For good smoothing, the ripple period is 1/(2f). What is the minimum recommended time constant R_L C so that the output waveform remains acceptably smooth?

Accepted Answer

Introduction / Context: In a full-wave (bridge) rectifier, the charging pulses arrive at twice the mains frequency. The reservoir capacitor must be large enough relative to the load resistance so that the discharge between peaks is small, giving low ripple. Designers often target a time constant at least an order of magnitude larger than the ripple period. Given Data / Assumptions: Supply frequency f = 50 Hz → ripple frequency f_r = 2f = 100 Hz. Ripple period T_r = 1/f_r = 10 ms. Good waveform shape typically requires R_L C ≥ 10 * T_r. Concept / Approach: Capacitor discharge between charging peaks is approximately ΔV ≈ I_load * T_r / C. To make ΔV small for a given I_load, choose C large (or equivalently R_L C large). A common rule-of-thumb is R_L C ≥ 10 T_r so that the exponential decay over T_r is modest. Step-by-Step Solution: Compute T_r = 1 / (2f) = 1 / (100) = 10 ms.Apply guideline: R_L C ≥ 10 * T_r = 100 ms.Select the nearest option: 100 ms. Verification / Alternative check: With R_L C = 100 ms, the fractional voltage drop over 10 ms is approximately 1 − exp(−T_r/(R_L C)) ≈ 1 − exp(−0.1) ≈ 9.5%, which is acceptable for many supplies; larger time constants reduce ripple further. Why Other Options Are Wrong: 10 ms equals only one ripple period; ripple would be high. 20 ms and 50 ms are improvements but typically still not smooth for many loads. Common Pitfalls: Using the mains period (20 ms) instead of the ripple period (10 ms) for full-wave rectification. Final Answer: 100 ms.

Question 8

Op-amp function blocks: An exponential (anti-log) amplifier configuration uses which component placement relative to the operational amplifier?

Accepted Answer

Introduction / Context: Log and antilog (exponential) amplifier building blocks are classic op-amp applications that exploit the exponential I–V characteristic of a diode or transistor junction. Distinguishing between the two is frequently tested in analog electronics. Given Data / Assumptions: Ideal op-amp behavior around its linear region. Use of a diode (or BJT junction) obeying i = I_s * exp(v/(nV_T)). Concept / Approach: Log amplifier: place the diode/junction in the feedback path and drive the input through a resistor; output is proportional to ln of the input magnitude. Antilog (exponential) amplifier: place the junction at the input (in series from source to the summing node) and keep a resistor in the feedback path; the output then becomes proportional to the exponential of the input voltage. Step-by-Step Solution: Recognize that the prompt asks about an exponential amplifier (antilog).For antilog behavior, the nonlinear element must translate an input voltage into an exponential current into the summing node.Use a linear feedback resistor to convert that current into a proportional output voltage. Verification / Alternative check: Standard analog design texts show: log amplifier → diode in feedback; antilog amplifier → diode at input. This matches option (b). Why Other Options Are Wrong: (a) describes a log amplifier, not exponential. (c), (d), and (e) do not produce exponential transfer characteristics. Common Pitfalls: Confusing log and antilog configurations. Ignoring temperature compensation requirements for precision designs. Final Answer: A diode at the input and a resistor in feedback.

Question 9

Clampers (DC restorers): Which statements correctly describe what a clamper circuit does to a waveform?

Accepted Answer

Introduction / Context: Clampers are widely used for baseline restoration, level shifting, and AC-coupled signal processing. They translate an entire waveform up or down without altering its peak-to-peak value (ideally), enabling diode detectors and CRT/ADC front-ends to operate within range. Given Data / Assumptions: Ideal diode and large capacitor so that steady-state clamping is achieved. Load resistance much larger than the source resistance to minimize droop. Concept / Approach: A clamper adds a DC component by charging its capacitor on one half-cycle and then referencing subsequent cycles to a new DC level through the diode action. The instantaneous shape is preserved except for small ripple due to the capacitor discharge; the peak-to-peak amplitude remains the same. Step-by-Step Solution: During the charging interval, the diode conducts and sets a reference level.On the opposite half-cycle, the diode is off and the capacitor holds the charge, shifting the entire waveform by a DC amount.Thus, only the DC level changes; the AC shape is essentially unchanged. Verification / Alternative check: Measure the output: the difference between maximum and minimum remains the input peak-to-peak value, confirming that no amplification occurs. Why Other Options Are Wrong: Amplification is not a function of a clamper; it is a passive level shifter. Common Pitfalls: Confusing clampers with clippers (which alter shape by limiting peaks). Neglecting capacitor droop under finite load, which slightly distorts the ideal behavior. Final Answer: Both (a) and (b) are correct.

Question 10

Input impedance of a CE amplifier: By definition, the small-signal input impedance is the ratio of which two AC quantities at the input port?

Accepted Answer

Introduction / Context: Input impedance is a fundamental parameter that dictates how strongly a stage loads its source. In a common-emitter (CE) BJT amplifier, we measure it at the base port. Given Data / Assumptions: Small-signal linear operation. Port of interest: base node relative to signal ground. Concept / Approach: By definition, input impedance Z_in equals the applied small-signal voltage divided by the small-signal current entering the same port. For a CE amplifier, those quantities are the base voltage v_b and base current i_b. Step-by-Step Solution: Apply a small test AC source at the base.Measure the resulting AC base current.Compute Z_in = v_b / i_b. Verification / Alternative check: Differential hybrid-π model directly shows Z_in ≈ r_π || biasing network at midband, which derives from v_b/i_b. Why Other Options Are Wrong: Options (b)–(d) use currents or voltages at other terminals and thus do not represent the input port definition. Common Pitfalls: Confusing input transresistance (v/i at different ports) with input impedance (same port). Final Answer: AC base voltage to AC base current.

Question 11

Voltage gain of a common-base (CB) amplifier: Given r_e = 6 Ω, R_L = 600 Ω and current gain α = 0.98, compute the midband voltage gain.

Accepted Answer

Introduction / Context: The CB amplifier is known for a high voltage gain and low input impedance. A quick estimation formula relates the voltage gain to the transconductance (1/r_e), the load resistance, and the current gain α. Given Data / Assumptions: Small-signal emitter resistance r_e = 6 Ω. Collector load resistance R_L = 600 Ω. Common-base current gain α = 0.98. Concept / Approach: For a midband CB stage, the approximate magnitude of voltage gain is A_v ≈ α * (R_L / r_e). This comes from v_out ≈ i_c * R_L with i_c ≈ α * i_e and i_e ≈ v_in / r_e. Step-by-Step Solution: Compute the ratio R_L / r_e = 600 / 6 = 100.Multiply by α: A_v ≈ 0.98 * 100 = 98.Thus, the midband voltage gain is approximately 98 (in magnitude). Verification / Alternative check: Using a hybrid-π model: g_m ≈ 1/r_e; v_out ≈ −α * g_m * R_L * v_in. The magnitude equals α * (R_L/r_e), matching the quick formula. Why Other Options Are Wrong: 100 ignores α < 1. 600 × 0.98 omits division by r_e. 6 and 0.98 are unrelated to the required composite expression. Common Pitfalls: Confusing CB with CE gain expressions; CE uses −g_m * R_C when r_o is neglected. Final Answer: 98.

Analog Electronics Questions