Q: Naming the gain parameters for BJT versus JFET devices In low-frequency small-signal models, the current gain of a bipolar junction transistor (BJT) is typically denoted by beta (β), whereas the primary gain parameter used for a junction field-effe…

Introduction / Context: Although BJTs and JFETs are both transistors, they are modeled with different primary gain parameters. BJTs are current-controlled devices, while JFETs (and MOSFETs) are voltage-controlled devices modeled by transconductance. Recognizing these parameters helps in predicting amplifier behavior and in translating between BJT and FET design techniques. Given Data / Assumptions: BJT small-signal model uses base current and collector current with β = ic / ib. JFET small-signal model uses gate-source voltage to modulate drain current with gm = di_d / dv_gs. Low-frequency, midband, small-signal conditions are assumed. Concept / Approach: BJT: β (also h_fe) is the ratio of collector current to base current. It captures how effectively base current controls collector current. JFET: transconductance gm defines how much the drain current changes for a given small change in gate-source voltage, making it the dominant parameter for voltage-to-current conversion in gain calculations. Step-by-Step Solution: Identify BJT parameter: β (ic/ib), often large (e.g., 50–200).Identify JFET parameter: gm (di_d/dv_gs), units of siemens.Relate to gain: CS JFET voltage gain Av ≈ −gm * (RD || RL), highlighting gm’s role.Therefore, BJT uses β; JFET uses transconductance gm. Verification / Alternative check: In hybrid-pi models, BJT’s transconductance g_m,BJT = IC / VT also exists, but β remains the familiar current gain parameter. For JFETs, the equivalent “β” is not standard; gm is the central descriptor. Why Other Options Are Wrong: beta, alpha: α is the common-base current gain (ic/ie), not the JFET parameter.transconductance, beta / transconductance, transconductance / beta, channel gain: do not match standard device terminology. Common Pitfalls: Equating β across devices; FETs are controlled primarily by gate voltage and characterized by gm. Final Answer: beta, transconductance (gm)

Question 1

In a bipolar junction transistor (BJT), does the collector-to-emitter voltage satisfy VCE = VBE + VCB by Kirchhoff’s voltage law around the three-terminal loop? Choose the most accurate statement.

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Introduction / Context: Kirchhoff’s voltage law (KVL) applies to transistor junctions just as it does to passive components. Knowing the algebraic relationships between VBE, VCB, and VCE is useful for quick checks and for translating measurements between different terminal pairs. Given Data / Assumptions: BJT in any operating region where terminal voltages are well-defined. Voltages referenced consistently (polarities defined from first terminal to second). No need to know current gain beta to write KVL identities. Concept / Approach: The three terminal voltages relate by loop addition around the C–B–E loop. Tracing from collector to emitter through the base node gives VCE = VCB + VBE. This identity is purely algebraic and holds regardless of region of operation (cutoff, active, saturation), provided sign conventions are consistent. Step-by-Step Solution: Write KVL from collector → base → emitter: VCB + VBE − VCE = 0.Rearrange to obtain: VCE = VBE + VCB.Note that other pairs follow similarly, e.g., VBE = VCE − VCB. Verification / Alternative check: Check with typical values in forward-active operation: VBE ≈ 0.7 V, VCB may be a few volts reverse (positive if defined C→B). Summing gives VCE of a few volts, consistent with practical biasing. Why Other Options Are Wrong: Option b changes a sign without justification; it contradicts KVL as written with standard polarities. Option c mixes reversed polarity definitions (VBC, VEB) leading to confusion. Option d incorrectly implies dependence on beta, which affects currents, not these voltage identities. Common Pitfalls: Inconsistent polarity definitions (e.g., using VBC instead of VCB without flipping the sign) or mixing measured and defined polarities. Final Answer: Correct — VCE = VBE + VCB.

Question 2

Identify the MOSFET type: If the drain, source, and channel regions are p-type material and the device can operate in depletion as well as enhancement (depending on gate bias), which MOSFET is described?

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Introduction / Context: Understanding MOSFET channel type (n or p) and mode (depletion or enhancement) is crucial for selecting parts and predicting bias behavior. The description mentions p-type channel material and the ability to conduct at zero gate bias (depletion capability), which points to one specific device family. Given Data / Assumptions: Drain, source, and channel are p-type regions. Device can operate in depletion (channel present at VGS = 0) and also respond to gate bias with either reduction or enhancement of channel carriers. Standard MOS structure with insulated gate. Concept / Approach: Depletion-mode MOSFETs have a physical channel at zero gate bias. If that channel is p-type, it is a p-channel D-MOSFET. Applying a more negative gate relative to source depletes the channel (reducing current), while applying a more positive gate can enhance conduction. In contrast, n-channel devices have n-type channels; describing all three regions as p-type contradicts an n-channel designation. Step-by-Step Solution: Match channel doping: p-type channel ⇒ p-channel device.Mode: physical channel at VGS = 0 ⇒ depletion-mode MOSFET.Therefore: p-channel D-MOSFET. Verification / Alternative check: Device symbols for depletion MOSFETs show a solid channel line; p-channel symbols use arrow and polarity conventions indicating hole conduction. Datasheets describe ID at VGS = 0 for depletion devices, confirming conduction without gate bias. Why Other Options Are Wrong: n-channel D-MOSFET (option a) requires an n-type channel, not p-type. n-channel E-MOSFET (option c) lacks a channel at VGS = 0 and is the wrong polarity. p-channel E-MOSFET only (option d) is enhancement-only; the stem explicitly includes depletion capability. Common Pitfalls: Confusing depletion/enhancement modes with channel polarity or assuming enhancement-only behavior applies to all MOSFETs. Final Answer: Correct — p-channel D-MOSFET.

Question 3

Power gain of a transistor amplifier: Which expression correctly relates the overall power gain (Ap) to the voltage gain (Av) and current gain (Ai)?

Accepted Answer

Introduction / Context: Amplifier gain can be defined for voltage, current, or power. Linking these gains helps convert specifications and predict performance when cascading stages. Knowing the correct relationship between power gain, voltage gain, and current gain avoids design errors. Given Data / Assumptions: Small-signal linear operation so gains are meaningful ratios. Voltage gain Av = Vout / Vin. Current gain Ai = Iout / Iin. Power gain Ap = Pout / Pin. Concept / Approach: Power is proportional to the product of voltage and current. For linear operation on the same reference impedances: Pout / Pin = (Vout * Iout) / (Vin * Iin) = (Vout / Vin) * (Iout / Iin) = Av * Ai. Thus, power gain is the product of voltage and current gains. Care must be taken with load and source impedances; if they change, gains should be defined consistently. Step-by-Step Solution: Write Ap = Pout / Pin.Express power as voltage * current for input and output.Cancel terms to obtain Ap = Av * Ai. Verification / Alternative check: In decibels, Gp(dB) = 10 * log10(Ap) = 20 * log10(Av) + 20 * log10(Ai) when impedances match, consistent with adding dB gains of voltage and current. Why Other Options Are Wrong: Ai^2 (option a) only holds in special cases where Av = Ai and impedances match; it is not the general definition. Av * Rout (option c) mixes incompatible quantities and units. Ai + Av (option d) is dimensionally meaningless for power gain. Common Pitfalls: Forgetting impedance effects or adding linear gains instead of multiplying, and mixing decibel and linear domains. Final Answer: Ap = Ai * Av.

Question 4

Conduction angle identification: In a class A power amplifier, for how much of the input signal cycle does the collector (or drain) current flow?

Accepted Answer

Introduction / Context: Amplifier “class” describes the bias point and conduction angle, which in turn affects linearity and efficiency. Class A operation is the benchmark for linearity in many small-signal applications, though it sacrifices efficiency. Given Data / Assumptions: Sine-wave input signal. Proper bias so the active device never cuts off during the cycle. Focus is on classic class A (not special topologies like push-pull class A/B hybrids). Concept / Approach: In class A, the transistor is biased in the middle of its active region so that current flows continuously. The conduction angle is therefore the entire 360° of the input cycle. This maximizes linearity and minimizes crossover distortion but results in low efficiency because current still flows at zero signal. Step-by-Step Solution: Choose a quiescent point (Q-point) in the active region so the device never reaches cutoff or saturation over the signal swing.As input swings positive and negative, the collector/drain current varies around a nonzero quiescent value.Since current never drops to zero within the cycle, conduction angle = 360°. Verification / Alternative check: Compare with class B (conduction 180°) and class AB (between 180° and 360°). These reference points confirm the definition for class A. Why Other Options Are Wrong: Less than or equal to 180° describes class B or class C, not class A. “Less than 360° but more than 180°” is class AB, not class A. Common Pitfalls: Confusing classes due to push-pull implementations; even with two devices in push-pull, each class A device conducts for the full cycle. Final Answer: Full 360° of the cycle.

Question 5

Oscillator fundamentals: Which oscillator type exploits the piezoelectric effect of a quartz (or ceramic) resonator to produce a highly stable frequency?

Accepted Answer

Introduction / Context: Oscillators generate periodic waveforms without an external clock. Different topologies achieve frequency selectivity using inductors, capacitors, or electromechanical resonators. Recognizing which design uses piezoelectric materials is important for applications requiring precise and stable frequencies. Given Data / Assumptions: Seeking the topology that uses a piezoelectric resonator (e.g., quartz). Standard low-power, linear oscillators. Sinusoidal operation, not relaxation oscillators. Concept / Approach: Piezoelectric crystals mechanically resonate at sharply defined frequencies. When placed in the feedback path or as the frequency-selective element, their motional impedance models a very high Q resonator. The “crystal oscillator” leverages this property to lock the oscillation frequency to the crystal’s resonance, yielding excellent short- and long-term stability compared with LC or RC oscillators. Step-by-Step Solution: Identify the element: a quartz crystal that converts electrical energy to mechanical vibrations and back (piezoelectric effect).Use the crystal as a narrowband resonator with extremely high Q.Place it in the amplifier feedback loop to satisfy Barkhausen conditions at the crystal’s resonant frequency. Verification / Alternative check: Frequency drift, temperature coefficients, and phase noise metrics are far superior in crystal oscillators compared to typical LC (Hartley, Colpitts) or RC phase-shift oscillators, confirming the correct choice. Why Other Options Are Wrong: Hartley and Colpitts oscillators use LC tanks, not piezoelectric resonators. RC phase-shift oscillators use cascaded RC sections; no piezoelectric element is involved. Common Pitfalls: Assuming all high-Q oscillators are “crystal”; only those using piezoelectric resonators are. Final Answer: Crystal oscillator.

Question 6

Voltage-divider biased common-emitter (CE) stage: The small-signal AC input resistance Rin seen at the base is made up of which parallel combination?

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Introduction / Context: In a voltage-divider biased CE amplifier, two resistors provide a stable DC bias to the base. For AC analysis, the source “sees” not only the transistor’s base input resistance but also the bias network. Correctly identifying Rin is critical for signal source matching and gain prediction. Given Data / Assumptions: Voltage-divider bias using R1 (from VCC to base) and R2 (from base to ground). Small-signal AC analysis around the bias point. Emitter bypass capacitor, if present, does not remove the divider from the input path. r_pi (also written h_ie) models the transistor’s small-signal base resistance. Concept / Approach: From the perspective of the AC source tied to the base, R1 connects to a low-impedance supply node (AC ground), R2 connects to ground, and r_pi connects into the transistor. All three terminate to AC ground, so they appear in parallel. Therefore, the net input resistance is Rin ≈ R1 || R2 || r_pi. Step-by-Step Solution: Redraw the circuit for AC: treat VCC as AC ground.Identify base-connected elements to ground: R1, R2, and r_pi all terminate at AC ground.Combine in parallel: Rin = R1 || R2 || r_pi. Verification / Alternative check: Compute with example values (e.g., R1 = 100 kΩ, R2 = 50 kΩ, r_pi = 5 kΩ): Rin ≈ (100k || 50k || 5k) ≈ 4.2 kΩ, showing that r_pi often dominates when it is much smaller than the divider resistances. Why Other Options Are Wrong: Series combination (option b) misrepresents AC grounding of VCC. “Only r_pi” (option c) ignores the real loading of the divider. “R1 || R2 only” (option d) neglects the transistor’s base input path. Common Pitfalls: Forgetting to ground the supply in AC models and overlooking how bias resistors load the source, reducing Rin and the stage’s effective gain. Final Answer: Rin = R1 || R2 || r_pi.

Question 7

BJT saturation indicator: Approximately what is VCE when a silicon transistor is driven into saturation?

Accepted Answer

Introduction / Context: Recognizing saturation is vital in switching applications. When a BJT saturates, both junctions are forward biased and the collector-emitter voltage collapses to a small value, which sets the lower bound of the output low level and affects power dissipation. Given Data / Assumptions: Silicon BJT used as a switch. Heavy base drive sufficient to enter saturation. Typical room-temperature operation. Concept / Approach: In saturation, VBE is forward biased (around 0.7–0.9 V) and VBC is also forward biased. The collector-emitter path behaves like a low-resistance segment, and VCE(sat) is typically small, often quoted near 0.2 V for small-signal devices (exact value depends on current and device type). Step-by-Step Solution: Drive the base strongly so the collector current is limited by the load, not beta.Both junctions forward bias; the transistor leaves the active region.Observe VCE(sat) collapsing to a small residual voltage, about 0.2 V for many devices. Verification / Alternative check: Consult datasheets: VCE(sat) is specified versus IC and IB, often 0.05–0.3 V for small-signal BJTs. Power devices may exhibit higher values, but 0.2 V remains a common nominal figure. Why Other Options Are Wrong: VCC (option a) is the supply, not the drop in saturation. VB (option b) refers to base potential, not the collector-emitter drop. 0.7 V (option d) is characteristic of a base-emitter junction, not VCE(sat). Common Pitfalls: Confusing VCE(sat) with VBE or assuming a fixed 0.2 V across all currents and devices; the value varies but remains “small.” Final Answer: About 0.2 V.

Question 8

Enhancement-only channel formation: Which device type has no physically formed channel at zero gate bias and becomes conductive only when properly biased at the gate?

Accepted Answer

Introduction / Context: Choosing the correct FET requires knowing whether the channel exists at zero bias. Enhancement-mode MOSFETs are normally off, while depletion-mode devices and JFETs typically conduct at zero gate bias (subject to pinch-off with reverse gate bias). This question targets that fundamental distinction. Given Data / Assumptions: Comparing JFETs, depletion-mode MOSFETs (D-MOSFETs), and enhancement-mode MOSFETs (E-MOSFETs). Zero gate-source bias as reference condition. Silicon devices at typical operating conditions. Concept / Approach: In an E-MOSFET, no channel exists at VGS = 0. Applying a gate voltage of suitable polarity induces an inversion layer, creating a conductive channel. For an n-channel E-MOSFET, VGS must exceed a threshold VTH > 0 to conduct. By contrast, JFETs have a doped channel that conducts at VGS = 0 and is pinched off by reverse bias; D-MOSFETs also have a channel at VGS = 0 that can be depleted or enhanced by gate voltage. Step-by-Step Solution: Check channel presence at VGS = 0: E-MOSFETs — none; JFETs and D-MOSFETs — present.Assess polarity: For n-channel E-MOSFET, apply positive VGS ≥ VTH to induce channel.Conclude the correct device: n-channel E-MOSFET. Verification / Alternative check: Datasheets list threshold voltage VTH for E-MOSFETs and show negligible ID at VGS = 0. JFET and D-MOSFET characteristic curves show significant ID at VGS = 0 that falls with reverse bias. Why Other Options Are Wrong: p-channel JFET (option a) has a physical channel and conducts at VGS = 0. n-channel and p-channel D-MOSFETs (options b, c) are depletion-mode, so a channel exists at VGS = 0. Common Pitfalls: Assuming all MOSFETs behave the same at zero bias; enhancement devices are normally off. Final Answer: n-channel E-MOSFET (enhancement).

Question 9

Common-source (CS) MOSFET amplifier gain control  In small-signal analysis of a common-source amplifier (with drain load RD and an optional source resistor RS), which parameter change most directly controls the midband voltage gain magnitude |Av| un…

Accepted Answer

Introduction / Context: The common-source (CS) MOSFET amplifier is a foundational gain stage in analog electronics. Its midband small-signal voltage gain primarily depends on device transconductance and the effective load seen at the drain. This question probes your understanding of which parameter most directly scales the small-signal voltage gain when the circuit is otherwise biased and loaded consistently. Given Data / Assumptions: A common-source MOSFET stage with drain resistor RD and possibly an unbypassed source resistor RS. We are discussing small-signal midband gain (coupling and bypass capacitors treated as AC shorts unless explicitly unbypassed). Bias point and load are assumed fixed unless the chosen parameter inherently alters them. Linear operation (no clipping or strong nonlinearity). Concept / Approach: For a CS stage without source degeneration, the approximate gain is Av ≈ −gm * (RD || RL). With an unbypassed source resistor, Av ≈ −(gm * (RD || RL)) / (1 + gm * RS). In both forms, gm appears as a direct multiplicative factor that scales the magnitude of Av. Changing gm directly alters how much output voltage develops per unit input small-signal gate voltage. Step-by-Step Solution: Identify the small-signal model: a dependent current source of value gm * vgs feeding the drain load.Express gain: Av = vo / vi ≈ −gm * (effective load) with additional division by (1 + gm * RS) if RS is unbypassed.Note gm’s role: increasing gm increases the dependent current, producing a larger voltage swing across the load.Conclude: gm is the primary lever for small-signal gain scaling under fixed load conditions. Verification / Alternative check: If RD or RL is fixed, doubling gm roughly doubles |Av| (ignoring degeneration). Conversely, halving gm halves |Av|. Supply voltage VDD influences bias headroom but does not directly multiply the small-signal gain unless it shifts the operating point and thereby changes gm indirectly. Why Other Options Are Wrong: the input voltage: changes output amplitude, not the intrinsic small-signal gain.VDD: affects biasing/headroom; gain scaling is not directly proportional to VDD.RS: modifies gain through feedback (denominator 1 + gm * RS); the direct scaling factor is gm.the coupling capacitor value: midband gain is capacitor-independent (they act as shorts in midband). Common Pitfalls: Confusing amplitude with gain, or assuming VDD directly sets gain. Remember that in midband small-signal analysis, gm and the effective load dominate voltage gain. Final Answer: gm (transconductance of the MOSFET)

Question 10

Effect of removing the emitter bypass capacitor in a common-emitter (CE) BJT amplifier  Consider a CE amplifier that originally uses a bypass capacitor across the emitter resistor to maximize midband gain. If this bypass capacitor is removed (leavin…

Accepted Answer

Introduction / Context: The emitter bypass capacitor in a common-emitter (CE) amplifier shunts the emitter resistor for AC signals, reducing local feedback and boosting midband voltage gain. Removing this capacitor reintroduces emitter degeneration, significantly affecting both gain and input resistance. Understanding this trade-off is key to designing stable, linear, and predictable amplifiers. Given Data / Assumptions: Single-transistor CE amplifier with emitter resistor RE and bypass capacitor CE. Midband operation (coupling capacitors act as AC shorts; transistor in forward-active region). Small-signal analysis; bias conditions maintained. Concept / Approach: With CE installed, the small-signal emitter node is close to AC ground, minimizing degeneration and maximizing gain: Av ≈ −gm * (RC || RL) * rπ / (rπ + RSIG) in one common form. Removing CE raises the effective emitter impedance (approximately (1 + β) * RE seen at the base), increasing negative feedback. This reduces transconductance-to-load conversion and raises the input resistance since the base sees a larger resistance reflected from the emitter. Step-by-Step Solution: Originally (bypassed): small-signal emitter ≈ AC ground → high |Av|, lower Rin.Remove bypass: emitter resistor participates in AC → degeneration increases.Degeneration effect: gain decreases due to feedback; linearity and bandwidth may improve.Input resistance: Rin_base ≈ rπ + (β + 1) * RE (for unbypassed case) → increases. Verification / Alternative check: Compare measurements: you will observe a smaller midband gain and a larger input resistance after removing CE. Frequency response often flattens (more stable gain) and distortion decreases due to feedback. Why Other Options Are Wrong: increase, decrease / increase, increase: contradict the feedback effect of emitter degeneration.decrease, decrease: the input resistance does not drop; degeneration raises it.no change, no change: bypass status strongly impacts AC behavior. Common Pitfalls: Assuming the bypass only affects low frequency. In fact, at midband, bypassing is what removes the emitter resistor’s AC impact; removing it reintroduces feedback across the band. Final Answer: decrease, increase

Question 11

Identifying the JFET amplifier configuration with voltage gain less than unity  Among the standard JFET small-signal stages (common-source, common-gate, and common-drain), which configuration typically provides a voltage gain magnitude less than one…

Accepted Answer

Introduction / Context: JFET amplifiers come in three classic topologies: common-source (CS), common-gate (CG), and common-drain (CD, also called source follower). Each has distinct input/output impedances and characteristic voltage gain. Recognizing these properties is critical when choosing a stage for buffering, voltage amplification, or impedance transformation. Given Data / Assumptions: Small-signal operation of a JFET around a proper DC bias point. Idealized, midband conditions (capacitors as shorts, no loading beyond stated impedances). Focus on the intrinsic voltage gain magnitude and impedance characteristics. Concept / Approach: The common-drain (source follower) provides near-unity but slightly less than one voltage gain because the output (source) follows the input (gate) minus the small vgs required to support the drain current. It exhibits very high input impedance (at the gate) and low output impedance (at the source), making it an excellent buffer. In contrast, common-source provides significant gain (often much greater than one), and common-gate typically has a gain around or above unity with low input impedance. Step-by-Step Solution: Identify the stage needing Av < 1 and buffering properties.Recall: common-drain (source follower) → Av ≈ 1 but slightly less (due to source feedback).Confirm impedances: very high input Z, low output Z → ideal buffer.Conclude: the common-drain topology fits the specification. Verification / Alternative check: Small-signal equations show Av ≈ gm * RS / (1 + gm * RS), which approaches but remains less than 1 for finite gm * RS, confirming the “slightly under unity” behavior. Why Other Options Are Wrong: common-source: delivers gain greater than one (voltage amplification stage).common-channel: not a standard term for JFET configurations.common-gate: typically near or above unity gain with low input Z; not a buffer.None of the above: incorrect because common-drain matches the requirement. Common Pitfalls: Confusing “source follower” (CD) with “emitter follower” (BJT): they are analogous but not identical devices; both are buffers with Av slightly less than one. Final Answer: common-drain (source follower)

Question 12

Naming the gain parameters for BJT versus JFET devices  In low-frequency small-signal models, the current gain of a bipolar junction transistor (BJT) is typically denoted by beta (β), whereas the primary gain parameter used for a junction field-effe…

Accepted Answer

Introduction / Context: Although BJTs and JFETs are both transistors, they are modeled with different primary gain parameters. BJTs are current-controlled devices, while JFETs (and MOSFETs) are voltage-controlled devices modeled by transconductance. Recognizing these parameters helps in predicting amplifier behavior and in translating between BJT and FET design techniques. Given Data / Assumptions: BJT small-signal model uses base current and collector current with β = ic / ib. JFET small-signal model uses gate-source voltage to modulate drain current with gm = di_d / dv_gs. Low-frequency, midband, small-signal conditions are assumed. Concept / Approach: BJT: β (also h_fe) is the ratio of collector current to base current. It captures how effectively base current controls collector current. JFET: transconductance gm defines how much the drain current changes for a given small change in gate-source voltage, making it the dominant parameter for voltage-to-current conversion in gain calculations. Step-by-Step Solution: Identify BJT parameter: β (ic/ib), often large (e.g., 50–200).Identify JFET parameter: gm (di_d/dv_gs), units of siemens.Relate to gain: CS JFET voltage gain Av ≈ −gm * (RD || RL), highlighting gm’s role.Therefore, BJT uses β; JFET uses transconductance gm. Verification / Alternative check: In hybrid-pi models, BJT’s transconductance g_m,BJT = IC / VT also exists, but β remains the familiar current gain parameter. For JFETs, the equivalent “β” is not standard; gm is the central descriptor. Why Other Options Are Wrong: beta, alpha: α is the common-base current gain (ic/ie), not the JFET parameter.transconductance, beta / transconductance, transconductance / beta, channel gain: do not match standard device terminology. Common Pitfalls: Equating β across devices; FETs are controlled primarily by gate voltage and characterized by gm. Final Answer: beta, transconductance (gm)

Question 13

JFET operating condition for linear amplification  For a properly biased JFET small-signal amplifier intended to operate linearly (without gate current flow), what should be the polarity of the gate-to-source junction under normal operation?

Accepted Answer

Introduction / Context: Unlike BJTs, a JFET is a voltage-controlled device with a reverse-biased gate junction (for JFETs with p-n junction gates). This creates a depletion region that modulates the channel cross-section and allows linear small-signal control of drain current by gate-source voltage without significant gate current. Given Data / Assumptions: JFET amplifier intended for linear small-signal operation. Standard n-channel or p-channel JFET with a p-n junction between gate and channel. Negligible gate current is desired (high input impedance). Concept / Approach: A reverse-biased gate-source junction widens the depletion region, controlling channel conduction without injecting carriers across the junction. Small variations in vgs around the bias point vary the channel conductance, producing a proportional change in drain current id = gm * vgs (small-signal), with gm determined by the bias point. Step-by-Step Solution: Bias the gate negative relative to source (for n-channel) or positive (for p-channel) to ensure reverse bias.Set the quiescent point so that vgs is below pinch-off but not too close to cutoff, to allow signal swing.Confirm minimal gate current (practically zero in reverse bias), ensuring high input impedance.Operate within the ohmic-to-saturation region appropriate for the chosen topology. Verification / Alternative check: Measure gate current with a sensitive meter; in proper reverse bias it will be microamps to nanoamps. Any forward bias will increase gate current dramatically and distort the signal. Why Other Options Are Wrong: forward-biased / neutral: would allow significant gate current, destroying the high input impedance and linearity.either forward-biased or reverse-biased / forward-biased only at high frequency: incorrect; proper linear operation requires reverse bias regardless of frequency. Common Pitfalls: Confusing MOSFET insulated-gate behavior with JFET p-n gate junction behavior. A MOSFET gate is insulated; a JFET gate is a diode junction that must be reverse-biased. Final Answer: reverse-biased

Question 14

Barkhausen criterion for oscillators: start-up versus steady state  For a linear feedback oscillator to start and then sustain a stable sinusoidal oscillation, what condition must the loop gain satisfy at the oscillation frequency (magnitude and pha…

Accepted Answer

Introduction / Context: Oscillators rely on positive feedback to convert DC power into a sustained AC signal without an external periodic input. The Barkhausen criterion summarizes the steady-state condition required for sustained oscillation and guides practical gain setting for reliable startup and amplitude stabilization. Given Data / Assumptions: Linear time-invariant amplifier with a frequency-selective feedback network B(jω). Focus at the intended oscillation frequency ω0. Small-signal linearization applies around the operating point; nonlinearities eventually limit amplitude. Concept / Approach: The Barkhausen criterion states that at ω0 the loop gain must satisfy two conditions: magnitude |Av * B| = 1 and net phase shift 0° (or 360°). Practically, to ensure startup from noise, designers choose |Av * B| slightly greater than 1. As amplitude grows, nonlinearities reduce effective gain until the loop gain settles to unity, sustaining a constant amplitude sinusoid. Step-by-Step Solution: Ensure the frequency-selective network provides a 0° net phase at ω0 (e.g., RC phase shift totals 180° with inverting amplifier phase −180°, etc.).Choose amplifier gain Av so that |Av * B| > 1 at cold start (e.g., 1.05–1.5 depending on design).Allow device or AGC/nonlinear limiting to reduce Av as the oscillation builds until |Av * B| → 1.At steady state, oscillation is sustained with |Av * B| = 1 and net phase 0°. Verification / Alternative check: Simulations show noise seeds a small signal that grows when |Av * B| > 1. As amplitude increases, gain compression occurs and the loop gain returns to unity. Removing feedback or deviating phase from 0° kills or detunes oscillation. Why Other Options Are Wrong: |Av * B| < 1: loop attenuates; no sustained oscillation.|Av * B| ≫ 1 always: produces distortion and clipping; steady state still forces effective loop gain to 1 via nonlinearity.180° phase or 0.5 magnitude: represent negative feedback or insufficient loop gain. Common Pitfalls: Confusing start-up (slightly > 1) with steady-state (exactly 1); ignoring the phase requirement; assuming “more gain is always better,” which increases distortion. Final Answer: Set gain so |Av * B| = 1 with 0° net phase (≈ 360°); slight >1 at startup, then settles to 1

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