Question 1

Functional roles of a transistor: Beyond acting as a two-state switching element (cutoff or saturation), which linear-mode role can a transistor perform when biased appropriately, enabling analog control and signal conditioning?

Accepted Answer

Introduction: Transistors are versatile: they can switch digital signals or operate in linear regions for analog functions. Recognizing their role as controllable resistive elements helps in understanding attenuators, automatic gain control, and sensor interfaces. Given Data / Assumptions: BJT or MOSFET biased in a region where small changes in control input modulate conduction. Goal is analog control (not merely on/off). Operation within safe-area limits to avoid distortion or damage. Concept / Approach: When biased properly, a transistor's channel or junction conduction can be approximated as a resistance that varies with a control voltage or current. In BJTs, small changes in base current alter collector–emitter conductance; in MOSFETs, gate voltage modulates channel resistance. Designers exploit this for voltage-controlled resistors, variable attenuators, and linear regulators. Step-by-Step Solution: Bias device in its linear/ohmic region.Relate control input (base current or gate voltage) to effective resistance between the controlled terminals.Use feedback or calibration to linearize response if high accuracy is required.Implement within thermal limits to avoid drift and distortion. Verification / Alternative check: Measure I–V curves in the selected region: slope around the operating point gives dynamic resistance r_d, which varies with control input, confirming variable-resistor behavior. Why Other Options Are Wrong: Fixed resistor: a transistor's resistance is not fixed; it is controllable.Tuning device: while used in RF tuning with varactors and mixers, that term is vague and not the linear-mode role asked.Rectifier: diodes are preferred for rectification; transistors are not used primarily as rectifiers. Common Pitfalls: Operating too far from the linear region, causing distortion, or ignoring temperature effects that change the apparent resistance. Final Answer: variable resistor

Question 2

Bias stability and Q-point movement: A transistor amplifier is initially biased at the midpoint of its DC load line. If the device's current gain (beta) decreases while the bias network is not perfectly stabilized against beta variation, in which di…

Accepted Answer

Introduction: Real transistors exhibit significant spread and drift in current gain (β). Bias networks that depend on β are vulnerable to Q-point shifts, which harm linearity and headroom. This question explores the qualitative effect of a β decrease on the operating point when bias stabilization is imperfect. Given Data / Assumptions: Common-emitter amplifier with a DC load line defined by V_CC and R_C. Initial Q point at midline for symmetrical swing. Biasing method allows collector current to depend on β (e.g., simple base-bias or insufficient emitter degeneration). β decreases from its initial value. Concept / Approach: Collector current approximately follows IC ≈ β * IB for a given base bias. If β falls and IB is unchanged, IC decreases. On the load line, lower IC corresponds to a movement toward cutoff (higher V_CE, lower IC), often described as moving “down” in the sense of reduced current along the vertical axis. Step-by-Step Solution: Start at midline: IC0 at the center of the allowable range.β decreases → for the same IB, IC = β * IB decreases.With smaller IC, the drop across R_C is smaller, so V_CE increases toward V_CC.Thus the Q point shifts toward cutoff (lower IC), described as moving down along the current axis. Verification / Alternative check: Plot IC versus β for fixed base current: it is proportional. Any reduction translates directly to a lower IC operating point, matching the qualitative load-line movement. Why Other Options Are Wrong: Up: would imply higher IC, which contradicts a β decrease.Nowhere: only true for fully β-independent bias (e.g., ideal emitter-bias with strong feedback), not assumed here.Off the load line: operating points for linear DC conditions must lie on the load line. Common Pitfalls: Assuming emitter resistors always eliminate β dependence; they reduce but may not eliminate it unless the design is carefully stabilized. Final Answer: down

Question 3

Voltage-divider bias calculation: With VCC = +18 V, R1 = 4.7 kΩ (from VCC to base), and R2 = 1.5 kΩ (from base to ground), what is the nominal base bias voltage provided by the divider? Assume negligible base current loading for this calculation.

Accepted Answer

Introduction: Voltage-divider bias is widely used because it provides a relatively stable base voltage independent of transistor β variations (within limits). Calculating the divider's nominal base voltage is a first step before accounting for base current loading and emitter feedback. Given Data / Assumptions: Supply voltage VCC = 18 V. R1 = 4.7 kΩ from VCC to base node. R2 = 1.5 kΩ from base node to ground. Base current is negligible (ideal divider assumption). Concept / Approach: With negligible base loading, the base node voltage is given by the simple divider formula: V_B = VCC * (R2 / (R1 + R2)). This sets the DC bias point used by the transistor's emitter–base junction and any emitter resistor that follows. Step-by-Step Solution: Compute denominator: R1 + R2 = 4.7 kΩ + 1.5 kΩ = 6.2 kΩ.Form the ratio: R2 / (R1 + R2) = 1.5 / 6.2 ≈ 0.241935.Multiply by VCC: V_B = 18 V * 0.241935 ≈ 4.3548 V.Round sensibly: ≈ 4.35 V. Verification / Alternative check: Cross-check by direct proportionality: the base node is closer to ground because R2 is smaller than R1; a value around a quarter of VCC (18 V / 4 ≈ 4.5 V) is reasonable, confirming 4.35 V. Why Other Options Are Wrong: 8.70 V: would require R1 = R2 (mid-supply), not the case here.2.90 V: corresponds to a different ratio; inconsistent with 1.5 kΩ and 4.7 kΩ.0.7 V: typical V_BE drop, not the divider voltage before junction drops. Common Pitfalls: Forgetting divider loading by base current in real circuits; accurate designs ensure divider current is 5–10 times base current to minimize shift from the ideal value. Final Answer: 4.35 V

Question 4

In a bipolar junction transistor (BJT) configured as common-emitter (C-E), the collector current Ic is related to the base current Ib by the current gain beta (β). If β = 100 and the base current is Ib = 30 µA, what is the resulting collector curren…

Accepted Answer

Introduction / Context: In the common-emitter (C-E) configuration of a bipolar junction transistor (BJT), small changes in base current control much larger changes in collector current. The proportionality factor is the small-signal or DC current gain beta (β), making this relationship a staple of analog electronics and bias design. Given Data / Assumptions: Configuration: common-emitter. Current gain β = 100 (assume approximately constant in this range). Base current Ib = 30 µA. Transistor operating in active region (not saturated or cut off). Concept / Approach: The core relationship for a BJT in active mode is Ic = β * Ib. With β specified and Ib known, Ic follows by direct multiplication. This linear approximation is widely used for first-pass bias calculations before more detailed modeling (e.g., Early effect) is applied. Step-by-Step Solution: 1) Write the relationship: Ic = β * Ib.2) Substitute β = 100 and Ib = 30 µA.3) Calculate: Ic = 100 * 30 µA = 3000 µA.4) Convert units: 3000 µA = 3 mA. Verification / Alternative check: A quick unit sense-check: for β ≈ 100, collector current is roughly two orders of magnitude larger than base current. 30 µA * 100 = 3000 µA, which is 3 mA—consistent and reasonable for small-signal devices. Why Other Options Are Wrong: 30 µA and 0.3 µA: These ignore the gain β (underestimate Ic heavily). 3 MA: Megaampere is absurdly large for a small-signal transistor—unit error by a factor of 10^6. 300 µA: This corresponds to β = 10, not 100. Common Pitfalls: Confusing β with α (common-base gain), mixing up microampere and milliampere units, or applying the formula outside the active region (near saturation where Ic no longer equals β * Ib). Always verify region of operation when finalizing a bias network. Final Answer: 3 mA

Question 5

In low-power BJT transistor packages (e.g., common small signal plastic packages), which lead is most commonly assigned to the base terminal according to typical pinouts?

Accepted Answer

Introduction / Context: Practical electronics requires recognizing common package pinouts. Many low-power, through-hole transistors (such as common small plastic packages) follow a convention where the base lead occupies the center position. Knowing this speeds breadboarding and fault-finding, although datasheets should always be consulted for the exact device. Given Data / Assumptions: Low-power, common transistor packages. Typical educational/bench assumptions—generic pinout knowledge. Recognize that manufacturer variations exist; the question seeks the most common convention. Concept / Approach: For many widely used low-power transistors, the base pin is commonly placed in the middle. While end pins may be emitter or collector (varying by device family), the center-as-base convention is frequently encountered in basic lab parts kits, making “middle” the best general answer for entry-level identification without a datasheet. Step-by-Step Reasoning: 1) Identify the package style commonly used in teaching labs.2) Recall that the center pin is often the base in those devices.3) Note that exact pinouts can differ by manufacturer, so verify for specific parts when precision matters.4) Select the most commonly correct generalization: the base is the middle lead. Verification / Alternative check: Good practice is to confirm via the datasheet and a diode-test with a multimeter: the base-emitter and base-collector junctions behave like diodes from base to the other terminals (polarity depends on NPN/PNP), helping confirm which lead is base. Why Other Options Are Wrong: Tab end/right end/left end: Ends can be emitter or collector; not a robust general rule. Stud mount: Refers to power packages and mounting style, not the identification of a base lead in low-power packages. Common Pitfalls: Assuming all devices share the same pinout. Always double-check for specific part numbers to avoid reverse connections that can damage parts. Final Answer: middle

Question 6

On a transistor DC load line for a common-emitter amplifier, the quiescent operating point (Q-point) is used to directly read which operating quantity?

Accepted Answer

Introduction / Context: The DC load line graphically relates collector current IC and collector-emitter voltage VCE for a given collector resistor and supply. The Q-point is the chosen steady operating point on that line. Understanding what can be read directly from the plot is essential for bias design and headroom analysis. Given Data / Assumptions: Common-emitter stage with defined RC and supply VCC. Load line drawn on the IC–VCE axes. Q-point set by the bias network and transistor characteristics. Concept / Approach: The load line axes are IC (vertical) and VCE (horizontal). Therefore, the Q-point directly yields IC and VCE. While VC (collector voltage to ground) equals VCE + VE, and is computable, the quantity most unambiguously read is IC (and VCE). Among the listed choices, IC is a direct axis variable, hence the best single correct answer. Step-by-Step Reasoning: 1) Identify axes: vertical axis IC, horizontal axis VCE.2) The Q-point is a coordinate (IC, VCE).3) From the plot, read IC directly at the Q-point ordinate.4) Conclude that IC is a direct readout; VC and VB are not plotted axes and require additional circuit relationships. Verification / Alternative check: Once IC is known, VC can be computed by VC = VCC − IC * RC (for a grounded emitter). This shows VC is derived, not directly read from the standard load line axes, validating IC as the direct quantity. Why Other Options Are Wrong: VCC: a fixed supply parameter; not read from the Q-point. VB and VE: not plotted; depend on biasing network. VC: can be computed from IC and RC but is not directly on the IC–VCE axes. Common Pitfalls: Confusing VCE with VC, or thinking the load line directly gives node voltages to ground. Always note which variables the axes represent. Final Answer: IC (collector current)

Question 7

For a properly biased BJT in the common-collector (emitter follower) or common-emitter family, the collector-base junction must be reverse-biased. Which other junction must be forward-biased for normal active-region operation?

Accepted Answer

Introduction / Context: Proper BJT biasing places the device in its active region so it can amplify. This requires specific polarities across the internal PN junctions. Knowing which junction is forward-biased versus reverse-biased is fundamental to understanding transistor action and small-signal models. Given Data / Assumptions: BJT intended to operate in the forward-active region. Collector-base (C-B) junction is reverse-biased. We must select which other junction is forward-biased. Concept / Approach: In forward-active operation, the base-emitter (B-E) junction is forward-biased, injecting carriers from the emitter into the base. The collector-base (C-B) junction is reverse-biased to sweep carriers from the base into the collector, enabling transistor action and gain. Step-by-Step Reasoning: 1) Forward-active condition: B-E forward bias (approximately 0.6–0.7 V for silicon).2) Reverse bias C-B junction to collect injected carriers efficiently.3) Under these conditions, collector current is approximately β times base current.4) Therefore, the required forward-biased junction is the base-emitter junction. Verification / Alternative check: Examine transistor I-V curves: forward biasing B-E produces exponential base/emitter currents; reverse-biased C-B gives the large-signal region where Ic is controlled by B-E injection, consistent with active mode. Why Other Options Are Wrong: Collector-emitter: this is not a PN junction and is not forward-biased in active mode. Collector-base/base-collector: must be reverse-biased, not forward. Cathode-anode: diode terminology; not a BJT junction designation. Common Pitfalls: Confusing saturation (both junctions forward-biased) with active mode. Saturation is useful for switching but not for linear amplification. Final Answer: base-emitter

Question 8

In a BJT output characteristic family, the “collector characteristic” is the plot used to study output behavior. Which description correctly identifies what is plotted and what bias is held constant?

Accepted Answer

Introduction / Context: Transistor characteristic curves visualize how output current varies with output voltage for fixed input drive. Recognizing the correct axes and controlled variable lets you read operating regions, saturation, and the effect of base drive directly from the plot family. Given Data / Assumptions: BJT operated with an applied base drive characterized by IB or equivalently VBB (through a resistor). Output axes are standard for collector characteristics. Small-signal effects like Early voltage may tilt the curves slightly. Concept / Approach: The collector characteristic keeps base drive constant (usually fixed IB steps) and plots collector current IC against collector-emitter voltage VCE. Families of curves are obtained for different IB values, showing active, saturation, and cutoff regions. Step-by-Step Reasoning: 1) Fix base drive (IB or VBB), vary VCE.2) Measure IC as VCE changes.3) Plot IC (vertical axis) versus VCE (horizontal axis).4) Repeat for several base-drive levels to obtain the characteristic family. Verification / Alternative check: Datasheets show IC–VCE curves at stepped IB. Saturation appears where increased IB no longer increases IC proportionally due to limited VCE, validating the axis selection. Why Other Options Are Wrong: IE vs VCE: emitter current is not the standard output axis. VC or VCC: incorrect symbols; VC is a node voltage, VCC is supply, neither is the axis label for collector-emitter voltage. IC vs VBE: that is an input characteristic, not the collector characteristic. Common Pitfalls: Confusing VCE with VC or thinking the base voltage alone defines IB; the resistor and transistor characteristics together set IB. Final Answer: Collector current IC versus collector-emitter voltage VCE with base-bias voltage VBB held constant

Question 9

Among the standard BJT configurations (common-emitter, common-collector, common-base), which type of gain is generally highest for the common-emitter (C-E) amplifier when considered as an overall amplifier stage?

Accepted Answer

Introduction / Context: The three basic BJT amplifier configurations emphasize different strengths: common-emitter (C-E), common-collector (C-C/emitter follower), and common-base (C-B). Knowing which gain metric is characteristically strongest helps with topology selection for preamps, drivers, and buffers. Given Data / Assumptions: Standard small-signal operating conditions. Proper biasing within the active region. Typical resistive loads and coupling networks. Concept / Approach: The C-E stage provides substantial voltage gain (often tens to hundreds) and non-unity current gain (β-like from base to collector path), so its overall power gain—the product of voltage and current gain—is typically the highest among the three. The C-C stage has high current gain but near-unity voltage gain (excellent buffer). The C-B stage offers voltage gain but current gain ≈ 1 (useful at high frequency). Step-by-Step Reasoning: 1) C-E: significant Av and appreciable Ai → large Ap (power gain).2) C-C: Av ≈ 1 but Ai large → moderate Ap.3) C-B: Av can be high but Ai ≈ 1 → moderate Ap.4) Therefore, C-E typically exhibits the greatest power gain. Verification / Alternative check: Small-signal hybrid-π analysis shows Av ≈ −gm * RC (potentially large), while input resistance is modest, yielding notable current gain from base input to collector output. Their product produces high stage power gain relative to C-C and C-B. Why Other Options Are Wrong: Voltage or current alone: Each can be high in certain topologies, but the question asks for which gain is highest overall for C-E. Power gain best captures C-E superiority. Resistance/bandwidth: Not “gains” in the same sense; bandwidth of C-E may be lower than C-B at RF. Common Pitfalls: Confusing buffer (C-C) behavior—excellent current gain but unity voltage—with high power amplification. C-E remains the workhorse for power gain. Final Answer: power

Question 10

A transistor amplifier shows an input current varying from 20 µA to 40 µA, while the output collector current varies from 0.5 mA to 1.5 mA over the same range. What is the small-signal (ac) current gain β_ac = ΔIc / ΔIb?

Accepted Answer

Introduction / Context: Small-signal (ac) current gain β_ac captures how changes in base current translate to changes in collector current around a bias point. This is crucial for incremental analysis, gain prediction, and linearized modeling of transistor stages. Given Data / Assumptions: Input current ranges from 20 µA to 40 µA → ΔIb = 20 µA. Output current ranges from 0.5 mA to 1.5 mA → ΔIc = 1.0 mA. Operation in active region; linear approximation valid over the range. Concept / Approach: By definition, β_ac = ΔIc / ΔIb. Convert units consistently, then divide. Keep in mind that β_ac can differ slightly from the DC β, but for many educational problems they are close in value. Step-by-Step Solution: 1) Compute ΔIb = 40 µA − 20 µA = 20 µA.2) Compute ΔIc = 1.5 mA − 0.5 mA = 1.0 mA.3) Convert ΔIc to µA: 1.0 mA = 1000 µA.4) β_ac = 1000 µA / 20 µA = 50. Verification / Alternative check: Back-of-envelope: a 2× increase in Ib (20 → 40 µA) yields a 3× increase in Ic (0.5 → 1.5 mA). The absolute slope from endpoints gives β_ac = 50, consistent with these changes. Why Other Options Are Wrong: 0.05 and 5: off by orders of magnitude due to unit mistakes. 20: corresponds to ΔIc = 0.4 mA, not 1.0 mA. 500: would require ΔIc = 10 mA for the given ΔIb. Common Pitfalls: Unit mix-ups (mA vs µA) and confusing endpoint ratio (1.5/0.5 = 3) with slope ratio (ΔIc/ΔIb). Always compute the incremental change. Final Answer: 50

Question 11

In a PNP transistor circuit using conventional supply notation, which node is typically at the most positive potential in the system?

Accepted Answer

Introduction / Context: Polarity awareness is essential when biasing PNP versus NPN transistors. In many conventional schematics for PNP stages, the emitter is connected toward the most positive rail, and current flows from emitter to collector under forward-active biasing. Given Data / Assumptions: Conventional VCC naming denotes the positive DC rail. PNP device used in a standard single-supply arrangement. Ground is the 0 V reference. Concept / Approach: With a PNP transistor, the emitter is commonly tied near the most positive potential so that the base can be biased a few tenths of a volt more negative (for silicon, around 0.6–0.7 V) to forward-bias the base-emitter junction. Therefore, the most positive potential in the circuit is typically the VCC rail. Step-by-Step Reasoning: 1) Identify rails: VCC (positive), ground (0 V).2) PNP emitter sits at high potential; base slightly below; collector lower still.3) Consequently, the highest voltage node in the typical single-supply PNP circuit is VCC.4) The collector and ground are at lower potentials than VCC. Verification / Alternative check: Examine a PNP emitter follower: VE ≈ VCC − VBE; clearly VCC is the upper rail and most positive, placing other nodes below it. Why Other Options Are Wrong: Ground: defined as 0 V; not the maximum. VC: collector is normally below the emitter in PNP operation. VBE: a small junction voltage, not a node rail. VE: often near the top rail, but the rail itself (VCC) is the maximum potential. Common Pitfalls: Transposing NPN intuition directly to PNP without flipping polarities. Always re-establish which rail is positive and which node sits closest to it. Final Answer: VCC (the positive supply rail)

Question 12

BJT amplifier naming — a “common-emitter” (C-E) amplifier is so named because the emitter node is common (shared) to both the input and the output circuits. Is this statement accurate?

Accepted Answer

Introduction / Context: Bipolar junction transistor (BJT) amplifier stages are categorized by which terminal serves as a reference for both input and output: common-emitter, common-collector, and common-base. The naming convention reflects how signals are applied and measured. This question asks if the common-emitter name indicates the emitter is shared between the input and output paths. Given Data / Assumptions: Small-signal AC view: coupling capacitors and bias networks present the emitter as a reference node. Input taken at base–emitter, output taken at collector–emitter. Applies to NPN and PNP with appropriate polarity conventions. Concept / Approach: In a C-E amplifier, the emitter terminal is the common reference for signal injection and signal extraction. The input signal perturbs base–emitter voltage, and the output is observed as a voltage change at the collector with respect to the emitter. This topology provides voltage and current gain, with a 180-degree phase inversion between input and output voltages. Step-by-Step Solution: Identify input port → between base and emitter.Identify output port → between collector and emitter.Recognize the shared reference → emitter is common to both ports.Therefore, the statement is accurate. Verification / Alternative check: Comparing topologies: common-collector (emitter follower) has collector as common node for DC but the output is taken at the emitter; common-base has the base as the common node. Naming aligns with the node common to signal paths. Why Other Options Are Wrong: Incorrect: contradicts the standard definition. “Only low frequency” or “only PNP”: the naming is independent of frequency and device polarity. Common Pitfalls: Confusing “common” with “ground”; in practice, the common node is a reference that may or may not be chassis ground. Final Answer: Correct — the emitter is common to both input and output, hence “common-emitter.”

Question 13

BJT phase relationship — in a common-collector (emitter-follower) voltage amplifier, the output voltage is in phase with the input voltage. Is this statement correct?

Accepted Answer

Introduction / Context: Different BJT amplifier topologies exhibit different phase relationships between input and output. Recognizing these signatures helps during troubleshooting and when choosing stages for buffering or gain. The common-collector stage (emitter follower) is especially valued as a buffer because of its high input impedance, low output impedance, and non-inverting behavior. Given Data / Assumptions: Topology: common-collector (input at base, output at emitter). Small-signal AC operation in the passband. Biasing set to linear active region. Concept / Approach: In a common-collector amplifier, the emitter voltage follows the base voltage minus approximately one diode drop (V_BE) for BJTs. Small-signal changes at the base produce similar-polarity changes at the emitter; therefore, the stage is non-inverting (input and output are in phase). By contrast, the common-emitter stage inverts phase by about 180 degrees for voltage signals. Step-by-Step Solution: Recall emitter follower relation → v_out ≈ v_in − V_BE (with small AC swing).Sign of gain → positive, near unity (slightly less than 1).Conclude phase → output in phase with input.Therefore, the statement is correct. Verification / Alternative check: Small-signal hybrid-π models show voltage gain ≈ R_L / (r_e + R_L) which is positive; scope measurements confirm in-phase waveforms within the midband region. Why Other Options Are Wrong: Incorrect: contradicts the well-known non-inverting property of emitter followers. “Only midband” or “only infinite load”: while extreme frequency or heavy loading alters magnitude, the sign remains non-inverting within the useful band. Common Pitfalls: Assuming all transistor amplifiers invert because common-emitter does; overlooking the buffering role of the emitter follower. Final Answer: Correct — the emitter follower's output is in phase with its input.

Question 14

Bias sensitivity — in a common-emitter amplifier, if the base current decreases (with bias still in the active region), what happens to the voltage measured between the emitter and the collector (V_EC or equivalently V_CE for an NPN referenced prope…

Accepted Answer

Introduction / Context: Understanding the bias relationships in a common-emitter (C-E) amplifier is crucial for predicting signal behavior and diagnosing faults. Base current (or, equivalently, base-emitter voltage) controls collector current. Changes in collector current alter the voltage drop across the collector load resistor, which then shifts the collector-to-emitter voltage seen at the transistor. Given Data / Assumptions: C-E amplifier in active region (not saturated or cut off initially). Collector load resistor R_C connected to a positive supply (NPN case). Decrease in base current implies decrease in collector current (I_C ≈ β * I_B). Concept / Approach: Ohm’s law and transistor action set the relationship: V_C = V_CC − I_C * R_C. If I_C decreases, the drop across R_C shrinks, raising the collector voltage. Since the emitter is near a small bias above ground (or at a fixed reference), the collector-emitter voltage V_CE increases. Equivalently, measuring V_EC (emitter relative to collector) would show a decrease because V_EC = −V_CE; but in common usage, “voltage between emitter and collector increases” refers to V_CE for NPN, which indeed increases as base current drops. Step-by-Step Solution: Decrease I_B → decrease I_C (active region).Compute collector node → V_C = V_CC − I_C * R_C increases as I_C falls.With V_E approximately fixed, V_CE = V_C − V_E increases.Therefore the measured emitter–collector separation (interpreted as V_CE) increases. Verification / Alternative check: A load-line diagram on the I_C–V_CE plane shows that moving toward lower I_C shifts the operating point toward higher V_CE along the DC load line. Why Other Options Are Wrong: “V_EC decreases” or “no change”: contradict the load-line relation. “Drops to nearly zero due to saturation”: saturation occurs with excessive base drive, not reduced base current. Common Pitfalls: Confusing the sign convention between V_CE and V_EC; always define polarity before interpreting statements. Final Answer: Correct — less base current reduces I_C and increases V_CE (thus the emitter–collector voltage, referenced as V_CE, increases).

Question 15

BJT basics and leads identification: In bipolar junction transistor theory and practical electronics, a discrete BJT device provides exactly three external terminals used for biasing and signal flow. Name these standard leads.

Accepted Answer

Introduction / Context: Bipolar junction transistors (BJTs) are foundational active devices in analog amplification and digital switching. Correctly identifying the device terminals is essential for biasing, modeling, and understanding gain behavior in common-emitter, common-collector, and common-base configurations. Given Data / Assumptions: The device under discussion is a standard discrete BJT. We are not considering special integrated packages with added pins for heatsinks or substrate. Lead names must align with conventional circuit symbols and datasheets. Concept / Approach: A BJT is formed by two back-to-back P-N junctions creating three regions. Each region is brought out to a terminal: base (control region), collector (output region that collects carriers), and emitter (region that injects carriers). These three terminals enable bias conditions that set operating modes (cutoff, active, saturation) in circuits. Step-by-Step Solution: Recognize that two junctions imply three semiconductor regions.Associate each region with its electrical role: emitter injects, base controls, collector removes carriers.Therefore, the three external leads are base, collector, and emitter. Verification / Alternative check: Examine datasheets or package pinouts (TO-92, TO-220). Regardless of pin order, the names remain base, collector, emitter—three in total. Why Other Options Are Wrong: Incorrect / only PNP: NPN and PNP both have the same three leads.Depends on package: extra tabs may be tied to collector but do not create a fourth functional terminal.Cannot be determined: the device class “BJT” defines the three leads. Common Pitfalls: Confusing MOSFET terminals (gate, drain, source) with BJT naming; assuming heat tabs are separate electrical nodes when they are tied internally. Final Answer: Correct

Question 16

Transistor operating modes in practice: Is it accurate to state that “transistors are controlled to operate either as a switch or as a variable resistor”?

Accepted Answer

Introduction / Context: Engineers commonly describe transistor use cases as “switching” (digital) and “amplifying” (analog). This question probes whether calling a transistor a “variable resistor” correctly captures analog operation, particularly for BJTs versus FETs. Given Data / Assumptions: Device of interest: BJT (bipolar junction transistor). Context spans both digital switching and linear amplification. Focus is on physical behavior, not a rough two-terminal metaphor. Concept / Approach: In digital mode, a BJT is driven between cutoff and saturation, approximating an on/off switch. In analog mode, a BJT operated in forward-active region behaves primarily as a transconductance device: small changes in base-emitter voltage (or base current) modulate collector current with current gain beta and transconductance gm. While one can bias certain transistors to approximate a two-terminal “variable resistor,” that description fits MOSFETs in the ohmic (triode) region more naturally than BJTs, which are current-controlled and exhibit exponential I-V characteristics across the base-emitter junction. Step-by-Step Solution: Identify switching mode: cutoff/saturation → on/off behavior.Identify analog mode: forward-active → controlled current source, not a simple resistor.Conclude the blanket statement is inaccurate for BJTs; “amplifier” is the proper term. Verification / Alternative check: Small-signal models show r_pi and ro as parasitic resistances around a dependent current source; the core transfer is current gain or transconductance, not variable resistance per se. Why Other Options Are Wrong: “Correct”: conflates BJT analog behavior with MOSFET ohmic behavior.Only MOSFETs: partially true but question is about transistors generally and BJTs specifically.Germanium-only / cannot be generalized: modern silicon BJTs behave the same in principle. Common Pitfalls: Assuming any linear operation equals “variable resistor”; ignoring that BJTs are three-terminal current-controlled devices. Final Answer: Incorrect

Question 17

Common-base amplifier phase property: In a small-signal common-base (CB) BJT amplifier, is the output signal inverted 180 degrees relative to the input?

Accepted Answer

Introduction / Context: BJT small-signal stages come in three canonical forms: common-emitter (CE), common-collector (CC, emitter follower), and common-base (CB). Each has distinct input/output phase relationships that matter for multistage amplifier design and feedback networks. Given Data / Assumptions: Operating region: forward-active, small-signal. Biasing ensures linear operation around a quiescent point. Source drives the emitter for CB; output is taken at the collector. Concept / Approach: The CE stage inverts (≈180°) due to how collector current variations translate into collector voltage across the load. In contrast, the CB stage exhibits a noninverting voltage transfer because the input signal is applied to the emitter; an increase in emitter current leads to an increase in collector current and a corresponding output voltage change that is in phase (considering sign conventions and load orientation). Hence, the CB stage is effectively noninverting for voltage signals, though its input impedance is low and current gain is near unity. Step-by-Step Solution: Define the CB topology: base at AC ground, input at emitter, output at collector.Track current flow: rising emitter signal increases collector current.Voltage at collector changes in phase with emitter-sense when referenced properly through the load.Therefore, CB is not a 180° inverter; CE is. Verification / Alternative check: Small-signal hybrid-pi analysis yields a positive voltage transfer for CB with appropriate polarity references and confirms near-unity current gain but significant voltage gain. Why Other Options Are Wrong: “Correct — 180°” confuses CB with CE.Low frequency / PNP-only caveats are irrelevant to the fundamental topology.Dependence on beta: phase property is topological, not beta-dependent. Common Pitfalls: Misremembering which stage inverts; overlooking the emitter-driven nature of CB. Final Answer: Incorrect — CB stage is noninverting (0° phase shift)

Question 18

BJT current relationships: Is the collector current approximately equal to the base current under normal forward-active operation?

Accepted Answer

Introduction / Context: Understanding how collector current (Ic) relates to base current (Ib) is fundamental to biasing and gain calculations. This relation determines the sizing of bias networks and the feasibility of driving loads with given base-drive capabilities. Given Data / Assumptions: BJT operating in forward-active region. Beta (current gain) is reasonably large (e.g., 50–200 typical). Temperature and Early effect may slightly modulate Ic, but the first-order relation holds. Concept / Approach: In forward-active mode, Ic ≈ β * Ib, where β (h_FE) is the DC current gain. Because β is much greater than 1, Ic is far larger than Ib, not “approximately equal.” The emitter current is Ie ≈ Ic + Ib ≈ (β + 1) * Ib. In saturation, the proportionality weakens, but Ic still is not ≈ Ib; rather, V_CE collapses and current is limited by the circuit. Step-by-Step Solution: Assume β = 100 as an example.For Ib = 0.5 mA, Ic ≈ 50 mA — clearly not equal.Hence the statement “Ic ≈ Ib” is incorrect for normal operation.Bias networks must supply adequate Ib to support the desired Ic = β * Ib. Verification / Alternative check: Datasheets list h_FE vs Ic, showing substantial ratios well above unity across operating ranges. Why Other Options Are Wrong: “Correct” contradicts the very definition of current gain.“Only in saturation”: even there, equality is not the defining relation.Material- or temperature-only claims ignore the governing device physics. Common Pitfalls: Forgetting β variation with Ic and temperature; assuming a fixed β for precision without design margins. Final Answer: Incorrect — collector current is beta times the base current (Ic ≈ β * Ib)

Question 19

Single-transistor inverter behavior (repaired stem): A common-emitter BJT stage with a collector load resistor and logic-level drive at the base acts as a digital inverter (NOT gate). Is this assessment correct?

Accepted Answer

Introduction / Context: Digital logic can be realized with discrete transistors. The simplest example is a single BJT in common-emitter configuration used as a switch to create an inverting transfer characteristic, mapping input high to output low and vice versa. Given Data / Assumptions: Topology: common-emitter with a collector resistor to VCC. Base driven by a logic-level signal via a base resistor. Transistor operates between cutoff and saturation. Concept / Approach: When the base is driven high sufficiently, the BJT saturates and pulls the collector node low (near ground). When the base is driven low (or left unbiased), the transistor is cut off, and the collector resistor pulls the output high toward VCC. This input–output inversion is the hallmark of a NOT gate. TTL families historically implement similar internal structures, although with added complexity for noise margins and fan-out. Step-by-Step Solution: Base low → cutoff → output rises to logic 1 via RC pull-up.Base high → saturation → output forced near 0 V (logic 0).Thus, the stage performs logical inversion.Proper resistor sizing ensures reliable switching and limits base current. Verification / Alternative check: Plot V_out versus V_in; the transfer shows a high-to-low transition as V_in crosses V_BE(on) with adequate drive to reach saturation. Why Other Options Are Wrong: MOSFET-only: BJTs work as inverters too.Emitter follower: that topology is noninverting and unsuitable for a NOT gate.Supply-unknown: while supply value sets logic levels, inversion behavior remains. Common Pitfalls: Under-driving the base (failing to saturate), omitting a base resistor, or overlooking pull-up resistor value and load requirements. Final Answer: Correct

Question 20

Common-collector (emitter follower) I/O terminals: In a CC stage, is the input applied to the base while the output is taken from the emitter?

Accepted Answer

Introduction / Context: The common-collector (CC) configuration, also known as the emitter follower, is widely used for buffering because it presents high input impedance and low output impedance with near-unity voltage gain. Correct terminal assignment is key to analyzing performance and biasing. Given Data / Assumptions: Topology: base is the driven terminal; collector is at AC ground (often tied to the supply via decoupling); output is taken from the emitter. Operation in forward-active region. Small-signal analysis applies around the bias point. Concept / Approach: In the emitter follower, the emitter voltage follows the base voltage minus V_BE. Because the collector is the common node for AC (hence “common-collector”), the stage does not invert and provides current gain, improving drive capability for loads without significant voltage amplification. Step-by-Step Solution: Drive the base with an input signal through a source/series resistor.Bias so the transistor remains in active region.Observe V_out at the emitter ≈ V_in − V_BE, noninverting.Leverage high input impedance and low output impedance for buffering. Verification / Alternative check: Measure gain: Av ≈ 0.9–0.99 depending on emitter load and transistor parameters; phase ≈ 0°. Why Other Options Are Wrong: Incorrect/PNP-only/high-frequency-only claims contradict the topology’s general definition.Beta dependence: while beta affects impedance, terminal assignment does not. Common Pitfalls: Forgetting voltage headroom (emitter must be below collector for NPN); overloading the emitter, which reduces gain and increases distortion. Final Answer: Correct

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