Question 1

JFET cutoff definition: Is VGS(OFF) the magnitude of gate-to-source voltage required to reduce the JFET drain current essentially to zero (turn the device off)?

Accepted Answer

Introduction / Context: Key JFET parameters include IDSS and VGS(OFF). Together they describe the transfer characteristic and are used in bias calculations. VGS(OFF) is the gate-to-source voltage at which the channel is fully pinched off and the drain current approaches zero (cutoff). Given Data / Assumptions: Normal operation keeps the gate-channel junction reverse-biased. VGS(OFF) has opposite polarity to IDSS sign convention (e.g., negative for n-channel). Shockley equation applies in the saturation region. “Essentially zero” means only leakage remains. Concept / Approach: The depletion regions from the gate junction widen with increasing |VGS|, narrowing the channel. At VGS(OFF), the channel is pinched off across its entire length, stopping conduction apart from leakage. This is the end-point on the transfer curve opposite IDSS (at VGS = 0 V). Step-by-Step Solution: Start from Shockley: ID = IDSS * (1 − VGS/VP)^2; with VP ≈ VGS(OFF).Set ID → 0: occurs when VGS approaches VGS(OFF).Interpret polarity: n-channel has VGS(OFF) negative; p-channel positive.Conclude: definition aligns with the statement. Verification / Alternative check: Datasheet transfer curves mark cutoff at VGS(OFF) where ID is near the leakage floor, confirming the interpretation. Why Other Options Are Wrong: Incorrect: conflicts with standard definition.Only for MOSFETs: VTH is used for MOSFETs; VGS(OFF) is standard for JFETs.Synonym for IDSS: IDSS is maximum ID at VGS = 0 V, not cutoff.Drain–gate breakdown: a separate maximum rating (BVGD or similar), unrelated to cutoff. Common Pitfalls: Confusing VGS(OFF) with MOSFET threshold; missing sign conventions; using ID ≈ 0 without accounting for leakage in precision designs. Final Answer: Correct

Question 2

Terminology: Is transconductance commonly referred to as mutual conductance in the context of FET and vacuum-tube amplifiers?

Accepted Answer

Introduction / Context: Transconductance (gm) is a central small-signal parameter in FETs and also historically in vacuum tubes. It quantifies how much output current changes for a given input voltage change. The traditional term “mutual conductance” is synonymous with transconductance and appears widely in classic texts and modern datasheets. Given Data / Assumptions: gm units are siemens (A/V). For FETs, gm = dID/dVGS at the operating point. For tubes, gm = dIP/dVG. Voltage gain depends on gm and load impedance, not gm alone. Concept / Approach: Because gm is a conductance relating current change to voltage change across different ports, “mutual conductance” aptly names the cross-port nature of the parameter. In circuit design, gm helps predict gain: Av ≈ gm * RD in a source-degeneration-free common-source stage (simplified). Step-by-Step Solution: Define gm: gm = ΔIout / ΔVin with units A/V.Recognize synonyms: mutual conductance = transconductance.Relate to gain: Av depends on gm and load; gm alone is not gain.Conclude: the statement is correct. Verification / Alternative check: Examine terminology in device literature; both terms are used interchangeably, often with gm specified in mS (millisiemens). Why Other Options Are Wrong: Incorrect: contradicts standard terminology.Only BJTs: BJTs use beta or gm differently defined (gm = IC/VT), but the synonymy is documented primarily for FETs and tubes as well.Resistance: inverse concept; gm is conductance, not resistance.Voltage gain only: gain is dimensionless; gm has units of siemens. Common Pitfalls: Confusing gm with transresistance (ohms); assuming gm alone equals stage gain; ignoring bias dependence of gm. Final Answer: Correct

Question 3

Common-drain JFET stage naming: Is a JFET configured with its output taken from the source (common-drain) also known as a source follower?

Accepted Answer

Introduction / Context: Naming conventions in amplifier stages are consistent across FET types. In a common-drain configuration, the drain is AC-grounded (or held at a fixed potential), the input is applied at the gate, and the output is taken from the source. This topology is universally called a source follower because the output voltage closely follows the input, offset by a small source-gate bias-dependent amount. Given Data / Assumptions: Three JFET terminals: gate, drain, source. Output sampled at the source node. Biasing ensures linear region operation for small signals. Load is connected to the source node (often via resistor). Concept / Approach: The source follower provides high input impedance, low output impedance, and near-unity voltage gain (slightly less than 1). The stage does not invert the signal; hence “follower.” These properties make it valuable as a buffer between high-impedance sources and heavier loads, minimizing signal loss and distortion. Step-by-Step Solution: Identify configuration: drain held relatively fixed; output at source.Analyze small-signal gain: Av ≈ gm * RS / (1 + gm * RS), typically just below 1.Check phase: non-inverting (0 degrees phase shift at low frequency).Conclude: common-drain JFET is the source follower. Verification / Alternative check: Compare with common-source (inverting gain) and common-gate (low input impedance, high bandwidth). Only the common-drain topology exhibits the follower behavior with high input and low output impedance. Why Other Options Are Wrong: Incorrect: conflicts with accepted nomenclature.Drain follower: not a standard term.Common-gate: a different topology with different impedance/gain properties.Only true for MOSFETs: the naming applies to both MOSFETs and JFETs. Common Pitfalls: Expecting voltage gain > 1 from a follower; ignoring source resistor biasing’s effect on linearity; overlooking the follower’s finite output swing limits due to VGS headroom. Final Answer: Correct

Question 4

Field-effect transistors (FETs): If the drain, source, and channel are all p-type material and the device is specified to operate in both depletion and enhancement modes, is this device correctly identified as a p-channel enhancement-mode MOSFET (E-…

Accepted Answer

Introduction / Context: MOSFET families include depletion-mode (D-MOSFET) and enhancement-mode (E-MOSFET) devices. Channel type (n-channel or p-channel) and operating mode (depletion, enhancement, or both) are fundamental identifiers. This question tests recognition of the distinction between p-channel E-MOSFETs and p-channel D-MOSFETs that can operate in both modes. Given Data / Assumptions: Drain, source, and channel are p-type material (so the device is p-channel). The device is said to operate in both depletion and enhancement modes. We are classifying the device by standard MOSFET terminology. Concept / Approach: An enhancement-mode MOSFET (E-MOSFET) has no conducting channel at V_GS = 0; the channel is induced only when gate bias exceeds a threshold (for p-channel, V_GS is sufficiently negative). A depletion-mode MOSFET (D-MOSFET) has a physically present channel at V_GS = 0 and can conduct in depletion (V_GS near 0) and also be driven into enhancement (stronger magnitude of V_GS), effectively supporting both operating regions. Thus, “operates in both modes” points to a D-MOSFET, not an E-MOSFET. Step-by-Step Solution: Identify p-type conduction path → p-channel class.Check operating modes: “both depletion and enhancement” → hallmark of D-MOSFET, not E-MOSFET.E-MOSFET requires induced channel; it does not naturally conduct at V_GS = 0.Therefore, calling it “p-channel E-MOSFET” is incorrect. Verification / Alternative check: Look at transfer curves: D-MOSFET shows I_D at V_GS = 0 and modulates for negative V_GS (p-channel). E-MOSFET shows near-zero I_D until |V_GS| exceeds threshold; it cannot be “both-mode” by definition. Why Other Options Are Wrong: Correct: Contradicts the defining property of E-MOSFETs. Ambiguous as stated / Cannot be determined: Mode wording is explicit; “both modes” uniquely indicates D-MOSFET behavior. Common Pitfalls: Confusing “p-channel” (conduction carrier type) with “enhancement” (bias mode). Assuming any p-channel device that can be turned off/on is automatically E-MOSFET. Final Answer: Incorrect

Question 5

Depletion-mode MOSFET (D-MOSFET) biasing: Evaluate the statement “A D-MOSFET cannot be biased using zero-bias (V_GS = 0).” Choose the most accurate assessment.

Accepted Answer

Introduction / Context: Biasing methods differ for depletion-mode and enhancement-mode MOSFETs. Understanding whether a device conducts at zero gate-source voltage is key for low-power analog stages and simple current sources. Given Data / Assumptions: Device type: D-MOSFET (depletion-mode). Claim: zero-bias operation (V_GS = 0) is not possible. Standard MOSFET physics and datasheet conventions apply. Concept / Approach: A D-MOSFET has a physically present channel at V_GS = 0, so it naturally conducts without gate bias. Applying appropriate V_GS of the correct polarity reduces channel carriers (depletes) to reduce I_D; reversing polarity can enhance conduction. Therefore, zero-bias is not only possible but common in simple D-MOSFET applications. Step-by-Step Solution: Identify device: depletion-mode → channel exists at V_GS = 0.At V_GS = 0, I_D is nonzero (per datasheet I_DSS spec).Varying V_GS modulates I_D around this operating point.Thus, “cannot be zero-biased” is incorrect. Verification / Alternative check: Datasheets specify I_DSS (drain current at V_GS = 0), confirming valid zero-bias operation for D-MOSFETs. Why Other Options Are Wrong: Correct: Conflicts with the presence of an I_DSS spec. Only true for n-channel / only true at very low temperatures: Device polarity and temperature change magnitudes, not the fundamental ability to conduct at V_GS = 0. Common Pitfalls: Mixing up D-MOSFET and E-MOSFET behavior. Assuming V_GS = 0 always turns a MOSFET off, which is only true for enhancement-mode types. Final Answer: Incorrect

Question 6

JFET terminology check: Are the “breakdown voltage” and the “pinch-off voltage” simply two names for the same JFET voltage? Choose the best evaluation.

Accepted Answer

Introduction / Context: JFET operation involves two important voltages: pinch-off and breakdown. Knowing the difference prevents biasing a device into destructive regions or misreading transfer curves. Given Data / Assumptions: Pinch-off voltage (V_P): channel just pinched and current becomes relatively constant with V_DS increase. Breakdown voltage (V_BR): avalanche or junction breakdown occurs at excessive reverse bias. We consider standard JFET operation with reverse-biased gate-channel junction. Concept / Approach: V_P defines the transition to the constant-current region (for a given V_GS), whereas V_BR is a high-voltage limit where the gate-channel junction avalanches. They are separated by a large margin on the output characteristic. Confusing them risks device failure. Step-by-Step Solution: Locate V_P on output curves: point where I_D saturates with rising V_DS.Locate V_BR: far to the right; current spikes due to breakdown.Different physics and vastly different magnitudes → not the same.Therefore the statement is incorrect. Verification / Alternative check: Review any JFET datasheet: V_P (or V_GS(off)) appears near a few volts; breakdown ratings are much higher (tens of volts), clearly distinct. Why Other Options Are Wrong: Correct / Ambiguous / Equivalent only for p-channel: None align with standard JFET curves where these voltages are categorically different. Common Pitfalls: Equating current saturation with breakdown, or misreading V_GS(off) as a breakdown limit. Each has separate specifications and design implications. Final Answer: Incorrect

Field Effect Transistors (FET) Questions