Q: AC grounding in amplifier design In small-signal amplifiers, which capacitor is intentionally placed to create an AC ground at a node (while leaving DC bias unchanged)?

Introduction / Context: Many amplifier stages use capacitors for different roles: coupling signals between stages, stabilizing bias, and shaping frequency response. The term “AC ground” refers to a point that is held at nearly zero AC potential, even though it may sit at a nonzero DC voltage. Creating an AC ground strategically can increase gain and reduce noise. Given Data / Assumptions: Small-signal amplifier with bias networks. Capacitors large enough in value to behave as low reactance at the signal frequency of interest. DC operating point must remain undisturbed. Concept / Approach: A bypass capacitor connects a node (often the emitter/source or a bias divider midpoint) to ground for AC. Since capacitors block DC, the biasing resistors still set the DC voltage. For AC, the low reactance path shunts signal variations to ground, reducing degeneration and increasing stage gain. Coupling capacitors, by contrast, pass AC between stages but do not by themselves create an AC ground at a fixed node. Step-by-Step Solution: Identify node where AC variations are unwanted (e.g., emitter resistor).Place a sufficiently large capacitor from that node to ground.At signal frequency, XC ≈ 1/(2πfC) is small → node held near AC ground.DC bias unaffected due to capacitor’s open-circuit behavior at DC. Verification / Alternative check: Frequency sweep shows gain rising when the emitter/source resistor is bypassed at midband, with a roll-off at low frequencies where XC increases and bypassing becomes ineffective. Why Other Options Are Wrong: Coupling capacitor: transfers AC between stages, not an AC ground creator.DC open / AC open: descriptive phrases, not specific capacitor types in this context. Common Pitfalls: Using too small a bypass value, resulting in partial bypass and unintended low-frequency attenuation. Final Answer: bypass capacitor

Q: Depletion-mode MOSFET gate biasing capability Which statement correctly describes allowable gate bias polarities for a depletion-mode MOSFET (of appropriate type)?

Introduction / Context: Depletion-mode MOSFETs (both n-channel and p-channel) conduct at zero gate-source voltage, unlike enhancement-mode devices that require a threshold to turn on. Understanding how positive and negative gate biases affect channel conductivity is essential for analog switches, constant-current sources, and amplifier loads. Given Data / Assumptions: Depletion-mode device with an existing channel at VGS = 0. Proper polarity is referenced to the device type (n-channel or p-channel). Safe operating limits are observed. Concept / Approach: Because a depletion-mode MOSFET already conducts at zero bias, applying a reverse-polarity gate voltage relative to the source can “deplete” the channel (reducing current), while applying the opposite polarity can “enhance” it (increasing current). Thus, it can be operated with both positive and negative gate voltages (relative to the source) within ratings to modulate conductance over a wide range. Step-by-Step Solution: At VGS = 0: device conducts nominally (IDSS-like behavior).Apply gate voltage that increases depletion: channel narrows; current decreases toward cutoff.Apply gate voltage that enhances: channel widens; current increases above the zero-bias level.Therefore, both polarities (with correct reference for n or p type) are usable. Verification / Alternative check: Transfer characteristics ID vs VGS for depletion-mode devices show conduction at VGS = 0 and modulation for both signs of VGS within limits. Why Other Options Are Wrong: Only positive or only negative: contradicts depletion-mode behavior.Cannot operate in ohmic region: false; MOSFETs operate in ohmic (linear) region when VDS is small relative to VGS − VTH. Common Pitfalls: Confusing depletion- and enhancement-mode devices; most power MOSFETs are enhancement-mode, but depletion-mode parts exist for special uses. Final Answer: can operate with positive as well as negative gate voltages

Q: Class B push-pull amplifier biasing In a class B (or class AB) push-pull audio power stage, transistors are biased slightly above cutoff primarily to avoid which problem?

Introduction / Context: Class B push-pull amplifiers split the waveform between complementary devices so that each handles approximately half of the cycle. Without proper biasing, the handoff near the zero-crossing is non-linear, causing audible distortion. Biasing slightly above cutoff moves operation toward class AB to smooth the transition. Given Data / Assumptions: Complementary emitter followers or source followers in a push-pull topology. Goal is reduced distortion with reasonable efficiency. Small standing quiescent current is acceptable. Concept / Approach: Transistor conduction begins only after a threshold (base–emitter or gate–source) is exceeded. Near zero crossings, neither device may conduct in pure class B, creating a dead zone that clips the waveform. A small forward bias makes both devices slightly on around zero, eliminating the dead zone and minimizing crossover distortion while keeping efficiency high compared to class A. Step-by-Step Solution: Identify the non-linear region around zero due to device thresholds.Introduce bias (e.g., diode-connected junctions, VBE multipliers) to establish a small quiescent current.Ensure the transfer characteristic is continuous through zero.Result: significantly reduced crossover distortion with modest efficiency penalty. Verification / Alternative check: Oscilloscope measurement of an undriven output reveals a slight warm bias current. A low-level sine test shows the notch distortion vanishing once proper bias is applied. Why Other Options Are Wrong: Unusually high efficiency: biasing above cutoff slightly reduces efficiency, it does not create a new problem.Negative feedback: independent of class; feedback may help but is not the bias goal.Low input impedance: dictated by stage topology, not this bias choice. Common Pitfalls: Over-biasing can push the stage toward class A and waste power; under-biasing leaves residual crossover notch. Final Answer: crossover distortion

Q: Purpose of biasing in transistor amplifiers What is the primary role of the bias network in a transistor amplifier stage with supply voltage VCC?

Introduction / Context: Transistor amplifiers require a carefully chosen operating point, or Q-point, so that input signals can swing without clipping or distortion. The bias network sets this operating point and keeps it stable against device variations and temperature changes, enabling predictable small-signal gain and linearity. Given Data / Assumptions: A single-stage amplifier powered by a DC supply VCC. Biasing may use resistor dividers, emitter/source resistors, and bypass capacitors. Small-signal operation around the Q-point is desired. Concept / Approach: The bias network establishes the DC voltages and currents (IC, VCE, etc.) at the transistor terminals so that the device remains in the active region. With degenerating resistors and proper design, the Q-point is held relatively constant despite beta spread and temperature drift. Once the Q-point is fixed, the AC gain is determined by device transconductance and load impedances, not by the bias network directly. Step-by-Step Solution: Choose target Q-point (e.g., VCE ≈ VCC/2 for maximum symmetric swing).Design divider and emitter resistor to set base–emitter current and stabilize against beta changes.Verify thermal stability and ensure the transistor stays in active region for input range.AC couple signals so DC bias remains undisturbed. Verification / Alternative check: Measure DC operating values with no signal; verify they match the design target. Apply a small AC signal and confirm the output waveform is centered and undistorted. Why Other Options Are Wrong: Stable at VCC or vin: those are supply or input values, not operating points at the device terminals.Ensure proper gain: gain is a consequence of device parameters and load; bias enables linear operation but does not set gain alone. Common Pitfalls: Relying on base current estimates without emitter degeneration can make the Q-point very sensitive to beta variations. Final Answer: hold the circuit stable at the designed Q-point

Q: Small-signal emitter diode resistance (re) approximation At room temperature, which formula gives the approximate AC resistance of the base–emitter junction (often denoted re) in a BJT small-signal model?

Introduction / Context: In small-signal BJT analysis, the dynamic resistance of the base–emitter junction, re, is a key parameter. It captures how the junction voltage changes with current for small AC perturbations. A quick mental formula at room temperature simplifies hand calculations for gain and input resistance in common-emitter and emitter-degeneration stages. Given Data / Assumptions: Room temperature so VT ≈ 25 mV (thermal voltage). Emitter current IE is the DC operating current through the junction (IC ≈ IE for large beta). Small-signal linearization around the Q-point. Concept / Approach: The diode small-signal resistance is r = dV/dI. For a PN junction, r ≈ VT / I at the operating point. In BJTs, re is taken as VT/IE (or approximately VT/IC for large beta). Thus, at 25 °C, re ≈ 25 mV / IE, yielding ohms when IE is in amperes. Step-by-Step Solution: Use VT ≈ 25 mV at room temperature.Write re ≈ VT / IE.If IE = 1 mA, re ≈ 25 mV / 0.001 A = 25 Ω (illustrative).Therefore, the correct proportionality is division by current, not multiplication. Verification / Alternative check: Compare to transconductance gm ≈ IC/VT. Note that re ≈ 1/gm (using IC ≈ IE), confirming re decreases as current increases. Why Other Options Are Wrong: 25 mV × IE or × IC: implies re increases with current, which is opposite to physics.25 mV / IC: close numerically when beta is large, but the canonical expression uses IE. Common Pitfalls: Forgetting that currents must be in amperes when using 25 mV to get re in ohms; mixing milliamps directly yields results off by 1000. Final Answer: re ≈ 25 mV / IE

Q: Correct DC bias conditions for an NPN small-signal amplifier Which biasing condition is required to operate an NPN transistor in its linear (active) region for amplification?

Introduction / Context: For linear amplification, a bipolar transistor must be biased in the active region where collector current is controlled by base–emitter junction voltage while the collector–base junction remains reverse biased. Getting the junction polarities right is fundamental for predictable gain and low distortion. Given Data / Assumptions: NPN device in common-emitter configuration. Single positive supply with appropriate resistive biasing. Objective is small-signal linear operation. Concept / Approach: The base–emitter junction behaves like a diode that must be forward biased (~0.6–0.7 V for silicon) to inject carriers into the base. The collector–base junction must be reverse biased to sweep carriers into the collector and prevent saturation. This establishes the transistor in the forward-active region where IC ≈ beta * IB and small-signal models apply. Step-by-Step Solution: Ensure VBE is forward (~0.6–0.7 V), establishing emitter and collector currents.Maintain VCB > 0 (collector more positive than base for NPN) to keep the collector–base junction reverse biased.Verify that VCE remains well above saturation to allow signal swings.Operate around a stable Q-point for linear amplification. Verification / Alternative check: Measure DC voltages: for NPN in active region, expect VCE typically near mid-supply and VBE around 0.7 V, with collector voltage higher than base voltage. Why Other Options Are Wrong: Reversing junction biases forces cutoff or saturation, not linear region.Applying “positive to n and negative to p” is an oversimplified and often incorrect rule for three-terminal devices.Large base voltage alone may overdrive or damage the device; proper polarities and currents matter. Common Pitfalls: Allowing the collector–base junction to forward bias under large signals, which pushes the device toward saturation and distortion. Final Answer: Forward bias the base/emitter junction and reverse bias the base/collector junction.

Question 1

Common-collector (emitter follower) as the final stage: What is the primary function of this stage before driving a low-resistance load?

Accepted Answer

Introduction / Context: The common-collector (emitter follower) stage is ubiquitous as the last stage of small-signal amplifiers. It offers high input resistance and low output resistance, making it ideal to interface sensitive voltage-gain stages with a relatively heavy load such as a speaker, cable, or subsequent module input. Given Data / Assumptions: BJT operated in its linear region. Emitter follower biased for approximately unity voltage gain. Load resistance is comparatively low. Concept / Approach: Because the emitter follower has voltage gain near +1 but can supply higher current due to its low output resistance, it ”buffers” a preceding high-gain, high-impedance stage from the loading effects of R_L. This preserves the voltage gain of earlier stages and improves overall drive capability without additional voltage amplification. Step-by-Step Solution: Recognize that A_v ≈ +1 for a CC stage; not useful for extra voltage gain.Note high input and low output resistances → excellent buffer behavior.Conclude its main function is impedance matching and isolation between stages. Verification / Alternative check: Small-signal analysis shows R_out ≈ (re + emitter resistance)/(1 + β) in simplified models, which is low compared with typical load resistances, confirming buffering action. Why Other Options Are Wrong: Provide voltage gain: Emitter followers have near-unity gain. Phase inversion: CC output is in phase with the input. Provide a high-frequency path: Not the defining purpose; compensation networks address HF response. Common Pitfalls: Expecting significant voltage amplification from a CC stage; its strength is current drive and isolation. Final Answer: buffer the voltage amplifiers from the low-resistance load

Question 2

JFET control characteristic: A junction field-effect transistor (JFET) is best described as a device that is:

Accepted Answer

Introduction / Context: FETs differ fundamentally from BJTs. In a JFET, the gate-source voltage controls the channel conductivity, and because the gate junction is reverse-biased in normal operation, the input current is extremely small—leading to very high input resistance compared with BJTs. Given Data / Assumptions: JFET used in its linear region for amplification. Gate-channel junction reverse-biased (except at breakdown). Signal applied as gate-source voltage variation. Concept / Approach: The drain current I_D is primarily a function of V_GS (and V_DS), summarized by transfer characteristics I_D = f(V_GS). Since V_GS sets the depletion width and channel cross-section, the device is voltage-controlled and presents very high input impedance at the gate, making it attractive for sensor front-ends and high-impedance nodes. Step-by-Step Solution: Identify controlling quantity: V_GS.Recognize gate junction is reverse-biased → minimal I_G.Conclude: JFET is a voltage-controlled, high-input-impedance device. Verification / Alternative check: Datasheets specify gate leakage currents in the nanoampere range and input resistance in megaohms to gigaohms, consistent with voltage-controlled behavior. Why Other Options Are Wrong: Current-controlled: Describes BJTs (base current controls collector current). Low input resistance: Opposite of the JFET's hallmark property. Always forward-biased: Gate is normally reverse-biased; forward bias risks damage. Common Pitfalls: Driving the gate positive on an n-channel JFET and inadvertently forward-biasing the junction; always respect V_GS(max) ratings. Final Answer: is a voltage-controlled device

Question 3

Self-biasing a JFET: In a self-biased JFET amplifier, which component ensures the gate-source voltage V_GS becomes negative (for an n-channel device) without a separate negative supply?

Accepted Answer

Introduction / Context: Self-bias is a simple and robust way to bias JFET amplifiers. It avoids an extra negative supply by exploiting the voltage drop across a source resistor, which raises the source potential relative to the (near-zero) gate potential, thereby creating a negative V_GS for an n-channel JFET. Given Data / Assumptions: n-channel JFET operated in the active region. Gate connected to ground via a large resistor (negligible gate current). A source resistor R_S develops a voltage due to drain-source current. Concept / Approach: Because gate current is approximately zero, the gate sits near 0 V through a high-value resistor. The channel current flowing through R_S creates V_S = I_D * R_S > 0. Therefore V_GS = V_G − V_S ≈ 0 − (I_D * R_S) = −I_D * R_S, which is negative without needing a dedicated negative supply. This negative V_GS self-adjusts with I_D, providing stabilizing feedback against device parameter variations. Step-by-Step Solution: Connect gate to ground via a large resistor (e.g., 1 MΩ).Include a source resistor R_S in series with the source lead.Allow I_D to generate V_S = I_D * R_S → V_GS = −I_D * R_S (negative). Verification / Alternative check: Measure V_G, V_S in a lab circuit. With I_D ≈ 2 mA and R_S = 1.5 kΩ, V_S ≈ 3 V, so V_GS ≈ −3 V, consistent with theory and device transfer curves. Why Other Options Are Wrong: Voltage divider: Often used for BJTs or MOSFET gates to set a fixed bias, but alone does not guarantee negative V_GS without R_S feedback. Ground: Provides reference only; it does not create a bias voltage. Negative gate supply: Works, but contradicts the ”self-biased” requirement. Common Pitfalls: Omitting the source bypass capacitor analysis; bypassing R_S increases gain but reduces local feedback—trade-offs must be considered. Final Answer: source resistor

Question 4

Source follower (common-drain) small-signal gain: What is the voltage gain A_v of a source follower stage (assume linear operation and typical device parameters)?

Accepted Answer

Introduction / Context: The source follower (common-drain for FETs, analogous to the BJT emitter follower) is a go-to stage when you need high input impedance, low output impedance, and near-unity voltage gain. It is widely used as a buffer between a high-impedance source and a heavier load, preserving signal amplitude while improving drive capability. Given Data / Assumptions: Small-signal linear operation around a proper DC bias point. Transconductance g_m is finite and positive. Load connected at the source node together with any source resistor. Concept / Approach: For a small-signal model, the source follower gain is A_v ≈ g_m * (R_S || R_L) / [1 + g_m * (R_S || R_L)]. Since the numerator and denominator share the same term, A_v approaches +1 as g_m * (R_S || R_L) becomes large; practical circuits yield a value slightly less than unity (for example 0.9–0.99). Importantly, the stage is non-inverting: the output at the source follows the input at the gate in phase. Step-by-Step Solution: Write approximate gain: A_v ≈ g_m * R_eq / (1 + g_m * R_eq), where R_eq = R_S || R_L.For typical values (e.g., g_m * R_eq ≫ 1), A_v → 1 from below.Conclude that the best descriptive choice is ”approximately +1 (slightly less than unity).” Verification / Alternative check: SPICE simulations of a JFET/MOSFET source follower with g_m around a few mS and R_eq of a few kΩ yield 0.95–0.99 gain, consistent with the formula and bench measurements. Why Other Options Are Wrong: A_V = g_m * R_D: This pertains to common-source stages; source followers do not use drain load for gain setting in this manner. A_V = g_m * R_S: Not a correct expression for source follower voltage gain. Approximately −1: That is the behavior of a common-source (or common-emitter) inverter, not a follower. Common Pitfalls: Expecting voltage gain > 1 from a follower; its purpose is buffering, not voltage amplification. To obtain gain, use a common-source stage or add a preceding gain block. Final Answer: approximately +1 (slightly less than unity)

Question 5

DC load line interpretation in a transistor amplifier Three distinct operating points are marked on a transistor's DC load line (collector current versus collector–emitter voltage). Which region does the upper point on the load line represent?

Accepted Answer

Introduction / Context: The DC load line of a transistor amplifier is a straight line that shows all possible combinations of collector current and collector–emitter voltage for a fixed supply and load resistor. Understanding what the endpoints mean helps you quickly diagnose bias problems and choose a correct quiescent operating point, or Q-point, for linear amplification. Given Data / Assumptions: A standard bipolar junction transistor amplifier (common emitter) with a resistive collector load. The “upper point” on the load line refers to the high-voltage, low-current end (near the VCC axis intercept). Idealized device behavior is assumed for conceptual identification. Concept / Approach: The DC load line spans between two extremes: cutoff and saturation. At cutoff, the transistor conducts essentially no collector current, so the full supply appears across the collector–emitter terminals. At saturation, the transistor conducts heavily and the collector–emitter voltage falls near a small saturation value. The Q-point is chosen somewhere between these extremes to allow symmetrical signal swing. Step-by-Step Solution: Identify the upper endpoint: IC ≈ 0, VCE ≈ VCC.No collector current implies the transistor is not conducting in the collector path.By definition, that operating region is cutoff.Therefore, the upper point corresponds to the cutoff point. Verification / Alternative check: Plot the load line using VCE + IC * RC = VCC. When IC = 0, VCE = VCC, which is the high-voltage axis intercept, confirming it is the cutoff end. When VCE ≈ 0, IC ≈ VCC/RC, the lower point is saturation. Why Other Options Are Wrong: Minimum current gain: current gain depends on device and bias, not the geometric endpoint label.Quiescent point: that is a design choice inside the load line, not an endpoint by itself.Saturation point: located at the opposite (lower-voltage, higher-current) end. Common Pitfalls: Confusing the graphical position: upper on the VCE axis means high voltage and near zero current, which is cutoff, not saturation. Remember: cutoff → “off,” saturation → “on.” Final Answer: cutoff point

Question 6

Practical feedback oscillators: required gain behavior For a feedback oscillator to be practically useful and produce a stable waveform, the loop gain must be controlled so that the overall gain is:

Accepted Answer

Introduction / Context: Feedback oscillators (e.g., Wien bridge, Colpitts, Hartley, phase-shift) rely on positive feedback and gain to sustain sinusoidal output without an external input. The Barkhausen criteria specify the conditions at the oscillation frequency, but practical circuits must also control amplitude and gain to avoid distortion or dying out. Given Data / Assumptions: Linear sinusoidal oscillator using positive feedback. Goal is a steady, low-distortion sinusoid. Amplifier gain can be adjusted by automatic or designed mechanisms. Concept / Approach: At startup, loop gain slightly above unity allows oscillations to build from noise. For steady operation, the loop gain should settle to exactly unity in magnitude with the correct phase shift. This requires some form of gain stabilization (automatic level control, AGC element, lamp, thermistor, diodes, or nonlinear resistance) so the amplitude neither grows into clipping nor decays to zero. Step-by-Step Solution: Design the amplifier such that initial loop gain > 1 to start oscillation.Introduce an amplitude control mechanism (e.g., lamp in Wien bridge or FET/diodes as variable resistors).As amplitude grows, the control element reduces gain until the loop gain magnitude ≈ 1.Oscillation amplitude stabilizes with low distortion. Verification / Alternative check: Observe on an oscilloscope: with proper stabilization the waveform becomes a clean sinusoid and remains constant. Without stabilization the output either clips (gain too high) or decays (gain too low). Why Other Options Are Wrong: < 1: loop gain less than unity cannot sustain oscillation.Self-adjusting: too vague; what is required is explicit gain stabilization around unity.Nonlinear: although nonlinear elements may be used to stabilize, the target is a linear, low-distortion sinusoid with stabilized gain. Common Pitfalls: Misreading Barkhausen as “loop gain must always be greater than one.” That is only for startup; steady state needs unity magnitude with precise phase. Final Answer: stabilized

Question 7

AC grounding in amplifier design In small-signal amplifiers, which capacitor is intentionally placed to create an AC ground at a node (while leaving DC bias unchanged)?

Accepted Answer

Introduction / Context: Many amplifier stages use capacitors for different roles: coupling signals between stages, stabilizing bias, and shaping frequency response. The term “AC ground” refers to a point that is held at nearly zero AC potential, even though it may sit at a nonzero DC voltage. Creating an AC ground strategically can increase gain and reduce noise. Given Data / Assumptions: Small-signal amplifier with bias networks. Capacitors large enough in value to behave as low reactance at the signal frequency of interest. DC operating point must remain undisturbed. Concept / Approach: A bypass capacitor connects a node (often the emitter/source or a bias divider midpoint) to ground for AC. Since capacitors block DC, the biasing resistors still set the DC voltage. For AC, the low reactance path shunts signal variations to ground, reducing degeneration and increasing stage gain. Coupling capacitors, by contrast, pass AC between stages but do not by themselves create an AC ground at a fixed node. Step-by-Step Solution: Identify node where AC variations are unwanted (e.g., emitter resistor).Place a sufficiently large capacitor from that node to ground.At signal frequency, XC ≈ 1/(2πfC) is small → node held near AC ground.DC bias unaffected due to capacitor’s open-circuit behavior at DC. Verification / Alternative check: Frequency sweep shows gain rising when the emitter/source resistor is bypassed at midband, with a roll-off at low frequencies where XC increases and bypassing becomes ineffective. Why Other Options Are Wrong: Coupling capacitor: transfers AC between stages, not an AC ground creator.DC open / AC open: descriptive phrases, not specific capacitor types in this context. Common Pitfalls: Using too small a bypass value, resulting in partial bypass and unintended low-frequency attenuation. Final Answer: bypass capacitor

Question 8

Depletion-mode MOSFET gate biasing capability Which statement correctly describes allowable gate bias polarities for a depletion-mode MOSFET (of appropriate type)?

Accepted Answer

Introduction / Context: Depletion-mode MOSFETs (both n-channel and p-channel) conduct at zero gate-source voltage, unlike enhancement-mode devices that require a threshold to turn on. Understanding how positive and negative gate biases affect channel conductivity is essential for analog switches, constant-current sources, and amplifier loads. Given Data / Assumptions: Depletion-mode device with an existing channel at VGS = 0. Proper polarity is referenced to the device type (n-channel or p-channel). Safe operating limits are observed. Concept / Approach: Because a depletion-mode MOSFET already conducts at zero bias, applying a reverse-polarity gate voltage relative to the source can “deplete” the channel (reducing current), while applying the opposite polarity can “enhance” it (increasing current). Thus, it can be operated with both positive and negative gate voltages (relative to the source) within ratings to modulate conductance over a wide range. Step-by-Step Solution: At VGS = 0: device conducts nominally (IDSS-like behavior).Apply gate voltage that increases depletion: channel narrows; current decreases toward cutoff.Apply gate voltage that enhances: channel widens; current increases above the zero-bias level.Therefore, both polarities (with correct reference for n or p type) are usable. Verification / Alternative check: Transfer characteristics ID vs VGS for depletion-mode devices show conduction at VGS = 0 and modulation for both signs of VGS within limits. Why Other Options Are Wrong: Only positive or only negative: contradicts depletion-mode behavior.Cannot operate in ohmic region: false; MOSFETs operate in ohmic (linear) region when VDS is small relative to VGS − VTH. Common Pitfalls: Confusing depletion- and enhancement-mode devices; most power MOSFETs are enhancement-mode, but depletion-mode parts exist for special uses. Final Answer: can operate with positive as well as negative gate voltages

Question 9

Class B push-pull amplifier biasing In a class B (or class AB) push-pull audio power stage, transistors are biased slightly above cutoff primarily to avoid which problem?

Accepted Answer

Introduction / Context: Class B push-pull amplifiers split the waveform between complementary devices so that each handles approximately half of the cycle. Without proper biasing, the handoff near the zero-crossing is non-linear, causing audible distortion. Biasing slightly above cutoff moves operation toward class AB to smooth the transition. Given Data / Assumptions: Complementary emitter followers or source followers in a push-pull topology. Goal is reduced distortion with reasonable efficiency. Small standing quiescent current is acceptable. Concept / Approach: Transistor conduction begins only after a threshold (base–emitter or gate–source) is exceeded. Near zero crossings, neither device may conduct in pure class B, creating a dead zone that clips the waveform. A small forward bias makes both devices slightly on around zero, eliminating the dead zone and minimizing crossover distortion while keeping efficiency high compared to class A. Step-by-Step Solution: Identify the non-linear region around zero due to device thresholds.Introduce bias (e.g., diode-connected junctions, VBE multipliers) to establish a small quiescent current.Ensure the transfer characteristic is continuous through zero.Result: significantly reduced crossover distortion with modest efficiency penalty. Verification / Alternative check: Oscilloscope measurement of an undriven output reveals a slight warm bias current. A low-level sine test shows the notch distortion vanishing once proper bias is applied. Why Other Options Are Wrong: Unusually high efficiency: biasing above cutoff slightly reduces efficiency, it does not create a new problem.Negative feedback: independent of class; feedback may help but is not the bias goal.Low input impedance: dictated by stage topology, not this bias choice. Common Pitfalls: Over-biasing can push the stage toward class A and waste power; under-biasing leaves residual crossover notch. Final Answer: crossover distortion

Question 10

Purpose of biasing in transistor amplifiers What is the primary role of the bias network in a transistor amplifier stage with supply voltage VCC?

Accepted Answer

Introduction / Context: Transistor amplifiers require a carefully chosen operating point, or Q-point, so that input signals can swing without clipping or distortion. The bias network sets this operating point and keeps it stable against device variations and temperature changes, enabling predictable small-signal gain and linearity. Given Data / Assumptions: A single-stage amplifier powered by a DC supply VCC. Biasing may use resistor dividers, emitter/source resistors, and bypass capacitors. Small-signal operation around the Q-point is desired. Concept / Approach: The bias network establishes the DC voltages and currents (IC, VCE, etc.) at the transistor terminals so that the device remains in the active region. With degenerating resistors and proper design, the Q-point is held relatively constant despite beta spread and temperature drift. Once the Q-point is fixed, the AC gain is determined by device transconductance and load impedances, not by the bias network directly. Step-by-Step Solution: Choose target Q-point (e.g., VCE ≈ VCC/2 for maximum symmetric swing).Design divider and emitter resistor to set base–emitter current and stabilize against beta changes.Verify thermal stability and ensure the transistor stays in active region for input range.AC couple signals so DC bias remains undisturbed. Verification / Alternative check: Measure DC operating values with no signal; verify they match the design target. Apply a small AC signal and confirm the output waveform is centered and undistorted. Why Other Options Are Wrong: Stable at VCC or vin: those are supply or input values, not operating points at the device terminals.Ensure proper gain: gain is a consequence of device parameters and load; bias enables linear operation but does not set gain alone. Common Pitfalls: Relying on base current estimates without emitter degeneration can make the Q-point very sensitive to beta variations. Final Answer: hold the circuit stable at the designed Q-point

Question 11

Small-signal emitter diode resistance (re) approximation At room temperature, which formula gives the approximate AC resistance of the base–emitter junction (often denoted re) in a BJT small-signal model?

Accepted Answer

Introduction / Context: In small-signal BJT analysis, the dynamic resistance of the base–emitter junction, re, is a key parameter. It captures how the junction voltage changes with current for small AC perturbations. A quick mental formula at room temperature simplifies hand calculations for gain and input resistance in common-emitter and emitter-degeneration stages. Given Data / Assumptions: Room temperature so VT ≈ 25 mV (thermal voltage). Emitter current IE is the DC operating current through the junction (IC ≈ IE for large beta). Small-signal linearization around the Q-point. Concept / Approach: The diode small-signal resistance is r = dV/dI. For a PN junction, r ≈ VT / I at the operating point. In BJTs, re is taken as VT/IE (or approximately VT/IC for large beta). Thus, at 25 °C, re ≈ 25 mV / IE, yielding ohms when IE is in amperes. Step-by-Step Solution: Use VT ≈ 25 mV at room temperature.Write re ≈ VT / IE.If IE = 1 mA, re ≈ 25 mV / 0.001 A = 25 Ω (illustrative).Therefore, the correct proportionality is division by current, not multiplication. Verification / Alternative check: Compare to transconductance gm ≈ IC/VT. Note that re ≈ 1/gm (using IC ≈ IE), confirming re decreases as current increases. Why Other Options Are Wrong: 25 mV × IE or × IC: implies re increases with current, which is opposite to physics.25 mV / IC: close numerically when beta is large, but the canonical expression uses IE. Common Pitfalls: Forgetting that currents must be in amperes when using 25 mV to get re in ohms; mixing milliamps directly yields results off by 1000. Final Answer: re ≈ 25 mV / IE

Question 12

Correct DC bias conditions for an NPN small-signal amplifier Which biasing condition is required to operate an NPN transistor in its linear (active) region for amplification?

Accepted Answer

Introduction / Context: For linear amplification, a bipolar transistor must be biased in the active region where collector current is controlled by base–emitter junction voltage while the collector–base junction remains reverse biased. Getting the junction polarities right is fundamental for predictable gain and low distortion. Given Data / Assumptions: NPN device in common-emitter configuration. Single positive supply with appropriate resistive biasing. Objective is small-signal linear operation. Concept / Approach: The base–emitter junction behaves like a diode that must be forward biased (~0.6–0.7 V for silicon) to inject carriers into the base. The collector–base junction must be reverse biased to sweep carriers into the collector and prevent saturation. This establishes the transistor in the forward-active region where IC ≈ beta * IB and small-signal models apply. Step-by-Step Solution: Ensure VBE is forward (~0.6–0.7 V), establishing emitter and collector currents.Maintain VCB > 0 (collector more positive than base for NPN) to keep the collector–base junction reverse biased.Verify that VCE remains well above saturation to allow signal swings.Operate around a stable Q-point for linear amplification. Verification / Alternative check: Measure DC voltages: for NPN in active region, expect VCE typically near mid-supply and VBE around 0.7 V, with collector voltage higher than base voltage. Why Other Options Are Wrong: Reversing junction biases forces cutoff or saturation, not linear region.Applying “positive to n and negative to p” is an oversimplified and often incorrect rule for three-terminal devices.Large base voltage alone may overdrive or damage the device; proper polarities and currents matter. Common Pitfalls: Allowing the collector–base junction to forward bias under large signals, which pushes the device toward saturation and distortion. Final Answer: Forward bias the base/emitter junction and reverse bias the base/collector junction.

Question 13

Definition of signal voltage gain In amplifier terminology, the small-signal voltage gain Av is defined as which ratio?

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Introduction / Context: Voltage gain is one of the most fundamental amplifier metrics. It relates the output signal voltage to the input signal voltage and determines how much an amplifier increases the amplitude of an applied signal. Precise definitions prevent confusion with transresistance (vo/io) or transconductance (io/vi) parameters. Given Data / Assumptions: Linear, small-signal operation around a fixed Q-point. vo and vi represent incremental (AC) voltages, not DC levels. Load and source impedances are part of the overall gain behavior. Concept / Approach: The voltage gain Av is the ratio of the output voltage to the input voltage. In a common-emitter BJT stage, for example, Av ≈ −gm * RC (magnitude considered), but the definition itself is simply vo/vi. Distinguishing between gain definitions avoids misapplication of formulas and incorrect design conclusions. Step-by-Step Solution: Identify the quantity asked: signal voltage gain, not current or mixed units.Write the definition: Av = vo / vi.Recognize that specific stage formulas (e.g., −gmRC) are implementations that yield Av but are not the definition.Conclude the correct ratio is output voltage over input voltage. Verification / Alternative check: Dimensional analysis: vo and vi both in volts; Av is dimensionless, consistent with a pure voltage ratio. Why Other Options Are Wrong: io/vi: that is transconductance, gm.IC × RC: has units of volts and relates to a stage's internal swing, not a definition of gain.vo/io: that is transresistance, Rout or a transimpedance measure. Common Pitfalls: Confusing specific gain formulas with the general definition; always start from Av = vo/vi and then substitute the stage model to compute its value. Final Answer: Av = vo / vi

Question 14

Voltage gain to decibels — quick check: An amplifier has a voltage gain Av = 210 (ratio). Evaluate the statement: “The voltage gain expressed in decibels is 46.4 dB.”

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Introduction / Context: Engineers often express gain in decibels (dB) for convenience, especially when cascading stages. Knowing the correct conversion from a simple voltage ratio to dB is essential for quick design checks and documentation. Given Data / Assumptions: Voltage gain Av = Vout/Vin = 210 (dimensionless). dB conversion for voltage ratios uses 20 * log10(Av) when impedances are matched or when just expressing ratios. Concept / Approach: The general formulas are: Gv_dB = 20 * log10(Av) for voltage, and Gp_dB = 10 * log10(Pout/Pin) for power. Here we use the voltage formula directly because Av is given as a voltage ratio. Step-by-Step Solution: Compute log10(210) ≈ 2.322.Multiply by 20: 20 * 2.322 ≈ 46.44 dB.Round reasonably to 46.4 dB.Therefore, the statement is correct. Verification / Alternative check: Sanity check: 40 dB corresponds to a ratio of 100; each additional 6 dB approximately doubles voltage. 100 → 200 is roughly +6 dB; 40 dB + 6 dB ≈ 46 dB, matching 210 ≈ 46.4 dB. Why Other Options Are Wrong: Incorrect: Contradicts the precise calculation. Only true for power gain: dB conversion for voltage explicitly uses the 20 multiplier. Insufficient information: The ratio is enough for dB conversion. Approximately 20.0 dB: Would imply a ratio near 10, not 210. Common Pitfalls: Using 10 * log10 for voltage ratios (that is for power), or mis-rounding logarithms leading to large errors. Final Answer: Correct

Question 15

CE amplifier bias intuition: In a common-emitter BJT amplifier with a collector resistor, reducing base current (and thus collector current) will cause the measured voltage between emitter and collector (VCE) to __________.

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Introduction / Context: Biasing in a common-emitter (CE) amplifier sets the quiescent operating point (Q-point). Understanding how base current affects collector current and the resulting voltages across resistors is crucial for troubleshooting and designing linear amplifier stages. Given Data / Assumptions: Single-supply CE amplifier with a collector resistor RC and small emitter resistor (or referenced to ground). Base current IB reduction reduces collector current IC (IC ≈ beta * IB in active region). Supply voltage VCC is fixed. Concept / Approach: Collector voltage VC = VCC − IC * RC. If IC decreases due to reduced IB, the drop IC * RC decreases, so VC rises toward VCC. Since emitter is near ground (or a small bias), VCE = VC − VE increases. Thus, lowering base current moves the transistor toward cutoff and raises VCE. Step-by-Step Solution: Decrease IB ⇒ decreases IC in active region.Voltage drop across RC = IC * RC decreases.Collector node voltage VC rises closer to VCC.Therefore VCE = VC − VE increases. Verification / Alternative check: Load-line analysis on the IC–VCE plane shows movement along the line toward the cutoff point (IC ≈ 0, VCE ≈ VCC) when base drive is reduced. Why Other Options Are Wrong: Decrease / drop to near 0 V: Happens when IB increases (moving toward saturation), not when it decreases. Remain unchanged / oscillate randomly: Bias responds deterministically to IB changes; no randomness assumed. Common Pitfalls: Confusing the behavior near saturation with behavior near cutoff; forgetting the sign of the voltage drop across RC. Final Answer: increase

Question 16

Why BJT base current is small: Evaluate the statement: “In a bipolar junction transistor, base current is usually small due to the very thin, lightly doped base region that allows most carriers to traverse into the collector.”

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Introduction / Context: BJT operation relies on minority carrier injection and collection across emitter–base and base–collector junctions. Device geometry and doping profiles set the current gain and explain why the base terminal requires comparatively little current in active operation. Given Data / Assumptions: Transistor is biased in forward-active mode. Base is thin and lightly doped relative to emitter. Collector–base junction is reverse-biased, sweeping carriers into the collector. Concept / Approach: Because the base is very thin and has low recombination probability, most injected carriers from the emitter diffuse across and are collected by the collector. Only a small fraction recombines in the base, which is the portion that must be replenished by base current IB. This design yields high beta (β = IC/IB) and thus small base current for a given collector current. Step-by-Step Solution: Recognize thin, lightly doped base ⇒ low recombination.Most carriers traverse to collector ⇒ IC ≫ IB.Therefore required IB is small to sustain carrier flow.Statement is correct. Verification / Alternative check: Standard BJT physics texts show β ≈ α/(1−α) with α ≈ IC/IE close to 1 due to thin base—implying small IB relative to IC. Why Other Options Are Wrong: Incorrect / only PNP / only at high temperature / depends only on supply: Base current behavior is set mainly by device structure and bias, not solely by polarity, temperature, or supply magnitude. Common Pitfalls: Assuming base current is negligible in all modes (it increases near saturation) or that IB depends primarily on VCC rather than on junction bias conditions. Final Answer: Correct

Question 17

Definition check for BJT current gain parameters: Evaluate the statement: “The beta (β) of a BJT is the ratio of collector current to emitter current.” Select the best assessment.

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Introduction / Context: Accurate definitions of transistor parameters are vital when reading datasheets and designing bias networks. Two common current gain parameters are alpha (α) and beta (β), and mixing them up causes calculation errors and bias misinterpretation. Given Data / Assumptions: BJT operated in forward-active mode. Currents IE (emitter), IC (collector), and IB (base) satisfy IE = IC + IB. Concept / Approach: By definition: β = IC/IB (common-emitter current gain). Alpha is defined as α = IC/IE (common-base current gain). Since IB is typically much smaller than IC, β is usually large, while α is close to 1. The statement confuses β with α. Step-by-Step Solution: Recall β definition: β = IC/IB.Recall α definition: α = IC/IE.Compare with the statement “β = IC/IE” and find mismatch.Therefore, the statement is incorrect. Verification / Alternative check: Using IE = IC + IB, one can show β = α/(1 − α) and α = β/(β + 1), confirming the distinct meanings of α and β. Why Other Options Are Wrong: Correct / true only for saturation / germanium-only / temperature-only: None change the formal definitions; β is never defined as IC/IE. Common Pitfalls: Memorizing formulas without variable context; confusing device-mode effects with parameter definitions. Final Answer: Incorrect

Question 18

Identifying CE amplifier terminals: Assess the statement: “In a practical common-emitter (CE) amplifier stage, the input is applied to the base and the output is taken from the emitter terminal.”

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Introduction / Context: Amplifier naming conventions indicate which terminal is common to input and output. In a CE amplifier, understanding where signals are applied and taken is basic but essential for lab wiring, simulations, and exam questions. Given Data / Assumptions: Standard single-transistor CE stage with emitter at AC ground (via resistor and bypass capacitor) or near ground. Collector has a load resistor to the supply. Concept / Approach: In the CE topology, the emitter is the common terminal for input and output (often at AC ground). The signal input is applied between base and emitter; the signal output is taken between collector and emitter. Therefore, the output is taken at the collector node, not at the emitter. Taking the output from the emitter would make it an emitter-follower (common-collector) stage, which has unity-ish gain and different characteristics. Step-by-Step Solution: Identify the topology: CE ⇒ emitter common.Input: base-to-emitter.Output: collector-to-emitter (node at collector).Hence the statement claiming output at emitter is incorrect. Verification / Alternative check: Small-signal hybrid-pi models for CE show collector voltage gain of approximately −gm * RC (phase inversion), which is measured at the collector with respect to the emitter reference. Why Other Options Are Wrong: Correct / frequency- or polarity-dependent variants: Topology, not device polarity or frequency, determines terminals. Common Pitfalls: Confusing common-emitter with emitter-follower (common-collector), leading to swapped expectations of gain and output impedance. Final Answer: Incorrect

Question 19

Barkhausen criterion — loop phase requirement: A self-sustained sinusoidal feedback oscillator must satisfy the total phase shift around the loop being an integer multiple of 360° (i.e., effectively 0°). Evaluate this requirement.

Accepted Answer

Introduction / Context: Oscillators generate steady sinusoids without an external input by feeding a portion of the output back to the input. The Barkhausen criterion provides a practical condition for sustained oscillations in linear systems: loop gain magnitude near unity and an overall loop phase of 0° modulo 360° at the oscillation frequency. Given Data / Assumptions: Linear time-invariant small-signal analysis around the intended frequency of oscillation. Negative or positive feedback networks arranged so that the net effect at one frequency is regenerative. Concept / Approach: If the loop phase shift equals 0° (or 360°), the fed-back signal reinforces the input in phase. With loop gain magnitude equal to 1 (|Aβ| = 1), oscillation can be maintained. Practical designs start with |Aβ| slightly greater than 1 so amplitude builds, then nonlinearity stabilizes it. Various oscillator types (RC phase-shift, Wien bridge, LC, crystal) all meet the same phase requirement at their respective frequencies, even though internal distributions of phase shift differ. Step-by-Step Solution: Identify loop transfer: Aβ(jω).Set phase condition: angle{Aβ(jω0)} = 0° or 360°.Set magnitude condition: |Aβ(jω0)| ≈ 1 for sustained oscillation.Conclude that the stated 360° (0°) phase requirement is correct. Verification / Alternative check: Block diagram analysis or Nyquist viewpoint: sustained oscillation occurs when the loop locus passes through −1 on the complex plane for negative-feedback sign conventions, corresponding to 0° net phase for the regenerative path at ω0. Why Other Options Are Wrong: Incorrect / only at startup / only for RC / unrelated: The phase requirement is general for linear feedback oscillators across topologies and both at startup and steady state. Common Pitfalls: Confusing individual network phase shifts with net loop phase; overlooking that sign inversions in the amplifier stage are part of the 360° total. Final Answer: Correct

Question 20

In an n-type junction field-effect transistor (JFET), under steady-state operation with the gate-to-channel junction reverse biased, will the device conduct when the drain is held at a positive potential and the source at a negative potential (so th…

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Introduction / Context: JFETs are voltage-controlled devices where the gate-to-channel junction is normally reverse biased. Understanding which terminal polarities allow conduction is essential for biasing and small-signal design. This item focuses on the polarity requirements for current to flow in an n-type JFET channel. Given Data / Assumptions: Device is an n-type JFET. Gate-to-channel junction is kept reverse biased (VGS ≤ 0 referenced to the source). Polarities in question: drain positive, source negative, implying VDS = VD − VS > 0. Normal room-temperature operation; no breakdown or forward gate conduction. Concept / Approach: In an n-type JFET, the channel is n-type semiconductor. Electrons are the majority carriers. With the gate reverse biased, depletion regions pinch the channel to a degree set by VGS. Drain current ID flows when VDS > 0, pulling electrons from source to drain (conventional current from drain to source). The current magnitude follows ID = f(VGS, VDS), saturating when VDS exceeds a threshold often called VDS(sat) in the JFET context, with control primarily through VGS. Step-by-Step Solution: Identify correct polarity: For n-type JFET, require VDS > 0 (drain at higher potential).Maintain reverse bias of the gate-channel junction: VGS ≤ 0 to avoid forward conduction.Conclude: With drain positive and source negative (making VDS positive), the channel conducts as long as VGS is not too negative to pinch off completely. Verification / Alternative check: Datasheets show characteristic curves ID versus VDS for different VGS ≤ 0; conduction increases with more positive VDS until saturation. This confirms that a positive VDS enables current for an n-channel device. Why Other Options Are Wrong: Drain negative (option b) describes p-channel behavior, not n-channel. Positive VGS (option c) forward-biases the gate junction, which is incorrect for JFET operation and risks damage. Gate shorted to source (option d) is a possible bias (VGS = 0) but not a necessity for every case; it is not a precondition for conduction. Common Pitfalls: Confusing JFET polarity with MOSFET conventions, or assuming VGS must be positive for all field-effect devices. JFET gates are reverse biased; MOSFET gates are insulated and can take positive VGS for n-channel operation. Final Answer: Correct — with VDS > 0 and the gate reverse biased, an n-type JFET conducts.

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