Question 1

Digital switching with BJTs: When a transistor is employed as a logic element, does it primarily toggle between two states called cutoff and saturation?

Accepted Answer

Introduction / Context: In digital electronics, transistors function as controlled switches. The central idea is to represent logic 0 and logic 1 by operating a device in two well-separated regions that provide robust noise margins and low static power where possible. Given Data / Assumptions: Device: BJT used as a digital switch. Supply rails define logic 0/1 thresholds via resistor networks and device characteristics. Design aims for clean transitions and stable states. Concept / Approach: The two practical states are cutoff (no base drive, no collector current; output pulled high via load) and saturation (strong base drive forces V_CE low; output near ground). Operating in the linear active region is avoided in steady state because it increases power dissipation and creates ambiguous logic levels. Proper base current sizing ensures the device reaches saturation under worst-case beta and load conditions. Step-by-Step Solution: Define logic 0/1 levels at the output node relative to supply.Choose base resistor to guarantee saturation at desired load current.Ensure the device sees near-zero base drive for cutoff.Validate switching with timing (rise/fall) and storage-time considerations. Verification / Alternative check: Scope measurements show low V_CE(sat) in the ON state and rail pull-up in the OFF state, corresponding to logic levels. Why Other Options Are Wrong: “Incorrect”: contradicts standard digital design practice.Analog-only/JFET-only/constant-beta claims misidentify the device or ignore design margins for beta variability. Common Pitfalls: Insufficient base drive (device not saturating), forgetting base-emitter resistors for quick turn-off, and neglecting storage time that slows transitions. Final Answer: Correct

Question 2

Why “junction” in BJT? Bipolar junction transistors are so named because their operation relies on current flow through two P–N junctions. Is this statement accurate?

Accepted Answer

Introduction / Context: The term “bipolar junction transistor” encodes key physics: “bipolar” reflects participation of both electrons and holes as charge carriers, and “junction” points to the two semiconductor junctions that form the device’s core structure. Given Data / Assumptions: Device families: NPN and PNP. Internal structure: emitter–base and base–collector P–N junctions. Forward-active operation with appropriate biasing of the two junctions. Concept / Approach: In forward-active mode, the emitter–base junction is forward biased to inject carriers, while the collector–base junction is reverse biased to sweep carriers into the collector. The presence and behavior of these two P–N junctions define current control and amplification mechanisms. Hence, the name “junction transistor” directly references the physical junctions involved. Step-by-Step Solution: Identify the two junctions: E–B and C–B.Set biases: V_BE forward, V_CB reverse for amplification.Observe carrier flow: emitter injects; base modulates; collector collects.Therefore, the device is rightly called a “junction” transistor. Verification / Alternative check: Energy-band and diode-equivalent models both depict two P–N junctions. Device IV curves corroborate diode-like behavior of each junction under bias. Why Other Options Are Wrong: Material- or topology-specific claims (germanium only, Darlington only) are unnecessary; all BJTs share this junction-based operation.“Name unrelated”: contradicted by device physics and history. Common Pitfalls: Confusing BJTs with MOSFETs, which use insulated gates rather than junction conduction for control. Final Answer: Correct

Question 3

Transistor terminal voltages and KVL: Is the collector–emitter voltage equal to the sum of the base–emitter voltage and the collector–base voltage (VCE = VBE + VCB)?

Accepted Answer

Introduction / Context: Voltage relationships in three-terminal devices follow directly from Kirchhoff’s Voltage Law (KVL). Understanding how terminal voltages relate helps prevent sign errors and supports accurate bias analysis of BJTs in any operating region. Given Data / Assumptions: Standard sign convention: terminal voltages referenced from the first terminal to the second (e.g., VXY = VX − VY). Device can be NPN or PNP; the identity is algebraic and independent of polarity choices. No special operating region restrictions are required. Concept / Approach: By definition, VCE = VC − VE. Also, VBE = VB − VE and VCB = VC − VB. Summing VBE + VCB yields (VB − VE) + (VC − VB) = VC − VE, which equals VCE. Thus, VCE = VBE + VCB is an identity derived from definitions, not an approximation, and holds in cutoff, active, or saturation regions as long as consistent sign conventions are used. Step-by-Step Solution: Write the three terminal voltages using node potentials: VC, VB, VE.Compute VBE + VCB = (VB − VE) + (VC − VB).Simplify to VC − VE = VCE.Conclude the equality holds generally. Verification / Alternative check: Apply KVL around the triangle formed by the three terminals; the algebraic sum is zero, leading to the same identity. Why Other Options Are Wrong: “Incorrect” contradicts algebra.Region- or device-type-only claims are unnecessary; the identity is universal.Beta/temperature do not affect this voltage sum definition. Common Pitfalls: Mixing up sign conventions (e.g., VEB vs VBE) and then concluding the identity fails; always be consistent. Final Answer: Correct

Question 4

Does transistor current gain strictly vary inversely with collector current? Evaluate the claim: “Current gain in a transistor is inversely proportional to collector current.”

Accepted Answer

Introduction / Context: Datasheets show that a BJT’s DC current gain (beta or h_FE) is not a universal constant; it varies with collector current, temperature, and device geometry. This question checks whether the variation follows a strict inverse proportionality with collector current, as sometimes mistakenly assumed. Given Data / Assumptions: Device: general-purpose silicon BJT. Operating region: forward-active (not deliberately in saturation). We consider typical beta versus Ic behavior from datasheets. Concept / Approach: Empirically, beta typically rises with Ic from very small currents to a moderate range, reaches a plateau, and then may fall at high Ic due to high-level injection and other effects. This nonmonotonic behavior contradicts “inversely proportional.” While small-signal transconductance gm = Ic / V_T does scale with Ic, beta = Ic / Ib does not follow a simple 1/Ic law. Step-by-Step Solution: Define beta = Ic / Ib; observe that both Ic and Ib are bias-dependent.Consult typical curves: beta increases from microamp to milliamp region, then rolls off.Conclude there is no strict inverse relationship with Ic.Design with margins to accommodate beta spread and variation. Verification / Alternative check: Examine manufacturer graphs (h_FE vs Ic) for popular BJTs (e.g., 2N3904, BC547); curves clearly are not a simple inverse function. Why Other Options Are Wrong: “Correct”: contradicts real device behavior.Frequency- or topology-specific claims do not resolve the fundamental dependence.VBE knowledge is unnecessary to refute the inverse law claim. Common Pitfalls: Assuming a fixed beta for precision calculations; ignoring temperature and process variation that broaden beta ranges. Final Answer: Incorrect

Question 5

BJT fundamentals (naming the three configurations): In bipolar junction transistor amplifier theory, the three standard circuit configurations are known as common-base (CB), common-emitter (CE), and common-collector (CC, also called the emitter foll…

Accepted Answer

Introduction / Context: Bipolar junction transistors (BJTs) are used in three canonical amplifier topologies: common-base (CB), common-emitter (CE), and common-collector (CC). The CC configuration is popularly called the emitter follower because its output (emitter) “follows” the input voltage at the base with approximately unity gain but lower impedance. Recognizing this standard taxonomy is foundational for understanding gain, input/output impedance, and phase inversion across BJT stages. Given Data / Assumptions: The statement lists common-base, common-emitter, and emitter-follower as the three configurations. “Emitter follower” is synonymous with “common collector.” Small-signal, linear operation around a bias point is assumed. Concept / Approach: The “common-” label indicates which terminal is shared by both input and output circuits. In CE, the emitter is common and the stage provides significant voltage and current gain with 180° phase inversion. In CB, the base is common; the stage provides low input impedance and high voltage gain with no inversion. In CC (emitter follower), the collector is common; the stage provides high input impedance, low output impedance, and near-unity voltage gain with no inversion. Together, these three cover the standard BJT amplifier families used in cascades and buffers. Step-by-Step Solution: Identify the three names: CB, CE, CC.Map “emitter follower” to “common collector.”Confirm that these constitute the standard trio used in textbooks and practice.Therefore, the statement is accurate. Verification / Alternative check: Any introductory electronics text defines the three standard BJT configurations precisely in this way, often summarizing their gain and impedance properties in a comparison table. Why Other Options Are Wrong: Limiting to NPN ignores that topology names are independent of polarity. Frequency limits and beta values do not alter the naming of configurations. Common Pitfalls: Confusing “emitter follower” with CE; forgetting that “follower” means voltage gain near one but strong buffering (low Rout, high Rin). Final Answer: Correct

Question 6

Transistor operated as a variable resistor: Does this operating mode imply a device property called “transistance,” or is that terminology incorrect in modern electronics?

Accepted Answer

Introduction / Context: Transistors can be biased so the device behaves like a controllable resistor over a limited range (for BJTs via small-signal r_e around a bias point; for MOSFETs in the ohmic/triode region). This question probes terminology: is there a recognized property called “transistance,” or is the correct language something else such as transconductance (g_m) or small-signal resistance? Given Data / Assumptions: Small-signal linearization around a fixed quiescent point. BJT or FET used as a variable resistor in an analog control context. Standard modern electronics terminology. Concept / Approach: In small-signal models, a BJT exhibits transconductance g_m = ΔI_C / ΔV_BE and a dynamic emitter resistance r_e ≈ 25 mV / I_E (at room temperature). A MOSFET in triode region has a channel resistance that varies with gate and drain-source biases. Neither case uses the term “transistance” in contemporary practice. Another concept, “transresistance,” refers to an amplifier whose output is a voltage proportional to input current (units of ohms), but that is not the same as a device property named “transistance.” Thus, claiming a transistor “has transistance” when used as a variable resistor is incorrect terminology. Step-by-Step Solution: Identify recognized parameters: g_m (A/V), r_e or r_ds (Ω), and gain β.Note that “transistance” is not a standard parameter in device data sheets or models.Therefore, the statement conflates concepts and is incorrect. Verification / Alternative check: Review of device models (Ebers–Moll, hybrid-π, MOS Level models) shows g_m, r_π, r_o for BJTs and g_m, r_ds for MOSFETs; “transistance” does not appear as a canonical parameter. Why Other Options Are Wrong: Conditioning on device type, temperature, or beta does not legitimize nonstandard terminology. Common Pitfalls: Confusing “transresistance amplifier” (a circuit type) with an intrinsic device property; mixing up g_m with resistance concepts. Final Answer: Incorrect

Question 7

BJT junction biasing sign convention: Does V_BE normally reverse-bias the base–emitter junction in a properly biased amplifier stage?

Accepted Answer

Introduction / Context: Correct biasing of the base–emitter (B–E) junction is central to BJT operation. In forward-active operation, the B–E junction is forward-biased while the collector–base (C–B) junction is reverse-biased. Misstating the polarity of V_BE leads to confusion about conduction mechanisms, emitter injection, and small-signal parameters such as g_m. Given Data / Assumptions: Linear amplifier bias (forward-active region). NPN example: base about 0.6–0.7 V above emitter at room temperature. PNP example has reversed polarities but still forward-biases the B–E junction. Concept / Approach: For conduction, the B–E junction must be forward biased so carriers inject from the emitter into the base. In an NPN, V_BE ≈ +0.7 V (base positive with respect to emitter). In a PNP, V_BE ≈ −0.7 V (base negative with respect to emitter). Either way, the magnitude biases the B–E junction forward, not reverse. Reverse biasing B–E would shut the device off (cutoff) except for small leakage currents. Step-by-Step Solution: Recall active-region condition: B–E forward, C–B reverse.Sign for NPN: V_B − V_E ≈ +0.7 V ⇒ forward bias.Sign for PNP: V_B − V_E ≈ −0.7 V ⇒ forward bias in opposite polarity.Therefore, saying V_BE reverse-biases B–E is incorrect. Verification / Alternative check: I–V curves and diode tests confirm that base–emitter behaves like a diode that must be forward biased for normal conduction. Why Other Options Are Wrong: Device polarity does not change the forward-bias requirement; cutoff and temperature do not invert the biasing rule. Common Pitfalls: Mixing up junction polarities between NPN and PNP; assuming the same sign conventions across reference points without care. Final Answer: Incorrect

Question 8

Voltage gain dependence: Is a transistor’s small-signal voltage gain inversely proportional to the change in output current, or is that claim incorrect for standard BJT amplifier stages?

Accepted Answer

Introduction / Context: Voltage gain in a BJT amplifier arises from transconductance and load impedance. In a common-emitter stage, a small change in input voltage produces a proportional change in collector current, which then develops an output voltage across the load. The relationship between gain and output current change is not “inversely proportional” as the statement claims; rather, more transconductance or more load impedance generally increases gain magnitude. Given Data / Assumptions: Linear small-signal operation around a bias point. Common-emitter example with load resistance R_C (or R_L at the collector). Hybrid-π model parameters: g_m and r_o. Concept / Approach: For a CE stage, the small-signal voltage gain is approximately A_v ≈ −g_m * R_out, where R_out includes the collector resistor in parallel with the transistor’s output resistance r_o and any external load. Since g_m = ΔI_C / ΔV_BE, a larger change in output current per unit input voltage (higher g_m) increases |A_v|, not decreases it. Therefore, voltage gain is directly related to how effectively the device converts input voltage to output current and how that current is turned into voltage by the load. Step-by-Step Solution: Express small-signal gain: A_v ≈ −g_m * R_out.Relate g_m to current: g_m ≈ I_C / 25 mV (room temperature), so higher I_C gives higher g_m.Increased ΔI_C across a fixed R_C produces larger output swing.Thus, the claim of inverse proportionality is incorrect. Verification / Alternative check: Lab measurements show that increasing bias current (within limits) raises g_m and gain magnitude; SPICE simulations confirm A_v grows with g_m and load resistance. Why Other Options Are Wrong: Special cases like saturation or emitter followers have different gain expressions but do not support the claimed inverse relationship. Common Pitfalls: Confusing current gain β with transconductance g_m; forgetting that voltage gain depends on both device transconductance and load impedance. Final Answer: Incorrect

Question 9

Node-voltage relation in a BJT: Using consistent polarities, is the collector–emitter voltage related to the other two junction voltages by V_CE = V_CB + V_BE?

Accepted Answer

Introduction / Context: Voltages between pairs of nodes in a three-terminal device are not independent; they are linked by Kirchhoff’s Voltage Law (KVL). In a BJT, the three relevant terminal-to-terminal voltages are V_BE, V_CB, and V_CE. With a consistent sign convention, these obey a simple identity that is frequently used in analysis, bias design, and measurement interpretation. Given Data / Assumptions: Terminals: base (B), collector (C), emitter (E). Voltage definitions: V_XY means V_X − V_Y. Consistent polarity across all definitions. Concept / Approach: By definition, V_CE = V_C − V_E. Also, V_CB = V_C − V_B and V_BE = V_B − V_E. Adding V_CB and V_BE gives (V_C − V_B) + (V_B − V_E) = V_C − V_E = V_CE. This is a pure algebraic identity derived from KVL and node-voltage definitions, independent of device region (cutoff, active, or saturation). It is widely used when converting between measured junction voltages and the overall collector–emitter voltage. Step-by-Step Solution: Write node-voltage definitions: V_CE = V_C − V_E.Express V_CB and V_BE: V_CB = V_C − V_B, V_BE = V_B − V_E.Add: V_CB + V_BE = (V_C − V_B) + (V_B − V_E) = V_C − V_E = V_CE.Therefore, V_CE = V_CB + V_BE holds by identity. Verification / Alternative check: Measure any two of the three voltages on a bench; computing the third with this relation matches the direct measurement (within meter resolution and wiring drops). Why Other Options Are Wrong: Device polarity (NPN/PNP), operating region, or the approximate 0.7 V diode drop do not alter the algebraic identity. Common Pitfalls: Mixing sign conventions (e.g., using V_EB instead of V_BE) and then adding without flipping signs; ensure consistent “from–to” ordering. Final Answer: Correct

Question 10

BJT current relationships: In normal forward-active operation, is the collector current approximately equal to the base current?

Accepted Answer

Introduction / Context: Understanding the relative magnitudes of base current (I_B) and collector current (I_C) is essential when biasing BJTs. In forward-active operation, the collector current is controlled by the base–emitter junction but is typically much larger than the base current, with the ratio defined by current gain β (also written h_FE). Given Data / Assumptions: Forward-active region (not cutoff or deep saturation). β is significantly greater than 1 for a typical small-signal BJT. Steady quiescent bias with small-signal excursions. Concept / Approach: By definition, β = I_C / I_B. For common BJTs, β may range from tens to hundreds depending on device and operating point. Therefore, I_C ≈ β * I_B, making I_C much larger than I_B. Saying they are “approximately equal” is incorrect except in contrived cases where β ≈ 1, which is not representative of normal BJT behavior. Design equations and bias networks always account for a small I_B compared with I_C (and with emitter current I_E ≈ I_C + I_B). Step-by-Step Solution: Recall β definition: β = I_C / I_B.Typical β ≫ 1 ⇒ I_C ≫ I_B.Hence, I_C is not approximately equal to I_B in normal operation.Conclusion: the statement is incorrect. Verification / Alternative check: Datasheets list h_FE values; bench measurements with a simple CE stage confirm that a milliamp of I_C requires only microamps of I_B for β ~ 100. Why Other Options Are Wrong: Deep saturation changes relationships but the prompt says “in a BJT” generally; using β = 1 is atypical; feedback does not force equality of I_B and I_C. Common Pitfalls: Confusing current gain with voltage gain; forgetting that I_E ≈ I_C + I_B is the only near-equality, not I_C ≈ I_B. Final Answer: Incorrect

Question 11

Ohmmeter junction test logic: If a BJT junction reads low resistance in both directions, it is shorted and the transistor is bad; and if a BJT junction reads high resistance in both directions, the junction is shorted and the transistor is good. Is …

Accepted Answer

Introduction / Context: Checking a BJT with an ohmmeter (or diode-test function) relies on the diode behavior of the base–emitter and base–collector junctions. Proper interpretation distinguishes shorts, opens, and normal forward/reverse characteristics. The statement in the prompt mixes up the conclusions for high versus low readings and even calls a device “good” when measurements indicate a fault. Given Data / Assumptions: Transistor is removed from circuit or at least isolated to avoid parallel paths. Ohmmeter polarity awareness: swapping leads changes which junction is forward biased. Silicon BJTs expected forward drop around one diode in forward direction and high resistance in reverse. Concept / Approach: A healthy junction behaves like a diode: low resistance (or ~0.6–0.7 V on diode test) in forward direction and high resistance in reverse. A low reading in both directions indicates a shorted junction (faulty). A high reading in both directions indicates an open junction (also faulty). Normal behavior requires asymmetric readings depending on meter polarity. Therefore, declaring the device “good” when a junction reads high both ways is wrong; that indicates an open junction and a bad transistor. Step-by-Step Solution: Measure base–emitter: forward low, reverse high.Measure base–collector: forward low, reverse high.If low both ways ⇒ shorted (bad). If high both ways ⇒ open (bad).Conclusion: statement is incorrect. Verification / Alternative check: Using a multimeter’s diode mode yields ~0.6–0.7 V forward drop for silicon; any reading ~0 V (both ways) indicates a short, while OL both ways indicates open. Why Other Options Are Wrong: Device material (silicon vs germanium) alters forward voltage but not the fault logic. In-circuit testing can confuse results but does not reverse the good/bad criteria. Common Pitfalls: Testing in-circuit without isolating leads; misinterpreting auto-ranging readings; not reversing meter polarity for both directions. Final Answer: Incorrect

Question 12

Biasing with fixed base resistor: In a simple “base bias” BJT configuration (base resistor from supply), what kind of Q-point stability does this method provide against β and temperature variations?

Accepted Answer

Introduction / Context: Bias networks set the quiescent operating point (Q point) of a transistor amplifier. The simplest scheme, “base bias,” uses a single resistor from the supply to the base. Although easy to implement, this approach makes the Q point highly sensitive to transistor β (h_FE) and temperature, which shift base–emitter voltage and currents. Understanding this instability motivates better biasing methods for reliable designs. Given Data / Assumptions: Base bias topology: supply → R_B → base; emitter at or near ground; collector resistor to supply. Large variation of β among parts and with operating conditions. Temperature dependence of V_BE and leakage currents. Concept / Approach: With base bias, I_B is approximately set by (V_CC − V_BE) / R_B. The collector current I_C ≈ β * I_B; thus any β change causes a proportional change in I_C. Moreover, V_BE typically decreases about 2 mV/°C, altering I_B and I_C with temperature. There is no negative feedback path to correct these shifts, so the collector voltage and operating region can drift into cutoff or saturation, distorting signals or compromising headroom. Step-by-Step Solution: Write I_B ≈ (V_CC − V_BE)/R_B.Compute I_C ≈ β * I_B ⇒ strong dependence on β.Recognize temperature lowers V_BE and raises I_C for a fixed R_B.Conclude: Q point is very unstable. Verification / Alternative check: Replacing base bias with a voltage-divider bias and emitter resistor introduces negative feedback: emitter degeneration stabilizes I_C against β and temperature changes, a standard fix in textbooks and practice. Why Other Options Are Wrong: “Very stable” is false for base bias; “no current gain” contradicts β action; “zero current” is not a biasing outcome; “perfect temperature compensation” requires specific networks, not a lone base resistor. Common Pitfalls: Designing with a single measured β and ignoring spread; omitting emitter resistors that provide necessary stabilization. Final Answer: A very unstable Q point

Question 13

Definition of signal voltage gain: Which expression correctly defines the small-signal voltage gain of an amplifier?

Accepted Answer

Introduction / Context: Amplifier “gain” must be specified by type: voltage, current, transconductance, transresistance, or power. Confusion arises when expressions appropriate to one form (e.g., current gain β = ΔI_C/ΔI_B) are mislabeled as voltage gain. For linear small-signal analysis, the signal voltage gain relates output voltage variation to input voltage variation and is dimensionless (ratio of volts to volts). Given Data / Assumptions: Linear operation around a quiescent point. Voltage gain definition requested, not current or power gain. Incremental (small-signal) quantities implied. Concept / Approach: Voltage gain is defined as A_V = V_out / V_in, often written as ΔV_out / ΔV_in for small-signal derivatives. In a common-emitter BJT stage, A_V can be approximated by −g_m * R_out, but this is a model-based computation, not the definition. Other expressions such as I_C * R_C (voltage across a resistor given a current) calculate a specific voltage, not a ratio of two voltages; ΔI_C/ΔI_B is current gain β; P_out/P_in is power gain; g_m/r_o is not a general definition of voltage gain (though it appears in specific configurations). Step-by-Step Solution: State definition: A_V = V_out / V_in (dimensionless).Relate to models: compute A_V from device parameters where needed, but the definition remains the ratio of voltages.Differentiate from other gains: β (current), g_m (A/V), transresistance (V/A), and power gain (dimensionless but based on watts).Therefore, select A_V = V_out / V_in. Verification / Alternative check: Data sheets and textbooks consistently specify small-signal voltage gain as a ratio of voltages, often in dB using 20 * log10(|A_V|). Why Other Options Are Wrong: I_C * R_C yields a voltage, not a ratio. ΔI_C/ΔI_B is β (current gain). P_out/P_in is power gain, not voltage gain. g_m/r_o is a parameter ratio, not a universal definition of A_V. Common Pitfalls: Mixing definitions and computations; using a computed expression as if it were the general definition; ignoring whether gains are specified as magnitude, signed, or in dB. Final Answer: A_V = V_out / V_in

Question 14

Common-collector (emitter follower) input resistance: in small-signal AC analysis, and assuming the emitter resistor RE represents the load seen at the emitter, the input resistance seen at the base (Rin, looking into the amplifier) is approximately…

Accepted Answer

Introduction / Context: The common-collector amplifier (also called an emitter follower) is prized for its high input resistance and low output resistance. Designers often use it as a buffer stage between a signal source and a load. This question asks you to recall the approximate analytical expression for the input resistance seen at the base, Rin, in terms of the emitter resistor RE and the transistor's current gain parameters. Getting this right helps you predict loading on the previous stage and avoid signal attenuation. Given Data / Assumptions: Small-signal, linear AC operation around a bias point. RE is the effective AC emitter resistance (it may be RE || load seen at the emitter). beta denotes the small-signal current gain (ic/ib). alpha ≈ beta/(beta + 1). Bias network and source resistance effects are not dominating (first-order estimate). Concept / Approach: The emitter follower translates base current changes into much larger collector and emitter current changes. Looking into the base, the emitter resistor is “magnified” by approximately (beta + 1). Intuitively, a small base current perturbation produces (beta + 1) times larger emitter current, so the base sees a resistance that is (beta + 1) times the emitter-side resistance. A more complete model includes the intrinsic emitter resistance re and any loading at the emitter as RE_eff = RE || R_load || re, but the leading behavior is the (beta + 1) multiplication of whatever load is present at the emitter node. Step-by-Step Solution: 1) Write the small-signal relation: Rin(base) ≈ (beta + 1) * R_e,eff. 2) For a simple estimate, take R_e,eff ≈ RE (if RE dominates the emitter load). 3) Therefore, Rin ≈ (beta + 1) * RE. 4) Recognize that adding external emitter load decreases R_e,eff and thus Rin proportionally. Verification / Alternative check: Use hybrid-pi or T-equivalent models. The base sees r_pi in series with a dependent source arrangement; transforming to the emitter side shows the (beta + 1) scaling of resistances reflected to the base. Spice simulations of an emitter follower corroborate that measuring input current into the base and dividing the applied small AC signal yields approximately (beta + 1) * RE as long as the transistor remains in its linear region. Why Other Options Are Wrong: Rin ≈ RE / (beta + 1): inverted dependence; it would predict lower input resistance for higher beta, which contradicts emitter-follower behavior. Rin = RE * alpha: alpha is slightly less than 1 and does not capture the large multiplication effect; it severely underestimates Rin. Rin ≈ RC / beta: RC is not the determining resistor in a common-collector stage; the collector is at AC ground in many implementations. Rin independent of beta and RE: incorrect because both parameters strongly influence Rin. Common Pitfalls: Forgetting to include the load at the emitter (RE || R_load); neglecting the effect of bias network resistors which can shunt Rin; assuming Rin is infinite; or confusing common-emitter and common-collector formulas. Final Answer: Rin ≈ (beta + 1) * RE

Question 15

BJT switching in saturation: when a transistor used as a switch is driven into hard saturation, the collector-to-emitter voltage VCE(sat) is approximately equal to which value under typical silicon operation?

Accepted Answer

Introduction / Context: Transistors used as on/off switches operate in two extremes: cutoff (off) and saturation (on). In saturation, both the base-emitter and base-collector junctions are forward biased, strongly reducing the collector-to-emitter voltage. Recognizing the expected magnitude of VCE(sat) is essential for estimating power dissipation, voltage margins, and logic-level compatibility in driver circuits. Given Data / Assumptions: Silicon BJT operated in hard saturation with adequate base overdrive. Moderate currents (data-sheet typicals), not extreme high-current cases. Room temperature conditions as a first approximation. Concept / Approach: In saturation, the transistor's collector-base junction is forward-biased and the device cannot support a large collector-emitter voltage. Typical data sheets specify VCE(sat) in the range of about 0.1–0.3 V (depending on device, current, and temperature). A common rule-of-thumb value used in introductory courses is around 0.2–0.3 V, which is much lower than the 0.7 V associated with a forward-biased base-emitter junction and dramatically lower than the supply voltage VC or VCC in a switching stage. Step-by-Step Solution: 1) Identify the operating region: saturation (both junctions forward biased). 2) Consult the typical range: VCE(sat) roughly 0.1–0.3 V for many small-signal BJTs. 3) Choose the nearest standard approximate value from the options: about 0.3 V. 4) Conclude that 0.3 V is an acceptable textbook estimate for VCE(sat). Verification / Alternative check: Check device datasheets (e.g., 2N3904, BC547). At IC ≈ 10 mA and IB sufficient for saturation, VCE(sat) is commonly listed around 0.1–0.2 V. At higher currents, it can increase but stays far below 0.7 V. Therefore, among the given choices, 0.3 V best represents a typical approximate value. Why Other Options Are Wrong: Approximately equal to VC or VB: these are node potentials, not fixed drops; in saturation the device drop is small, not equal to supply or base voltage. Approximately 0.7 V: that is the base-emitter forward drop magnitude, not the collector-emitter saturated drop. Approximately 1.0 V: too high for a silicon BJT in saturation under normal conditions. Common Pitfalls: Confusing VBE(on) (~0.7 V) with VCE(sat); assuming VCE(sat) is zero; ignoring that VCE(sat) increases with current and temperature; under-driving the base so the device is not actually saturated. Final Answer: Approximately 0.3 V

Question 16

Emitter follower (common-collector) gain property: for a properly biased emitter follower, what is the approximate small-signal voltage gain from base to emitter?

Accepted Answer

Introduction / Context: The emitter follower is widely used as a buffer. Although it does not provide voltage gain above unity, it offers high input resistance and low output resistance, making it ideal for impedance matching and isolation. This question checks your understanding of the emitter follower’s hallmark: a voltage gain very near, but slightly less than, unity with no phase inversion. Given Data / Assumptions: Proper biasing so the transistor remains in its active region. Small-signal linear operation about the bias point. Load not so heavy that it forces significant emitter voltage droop. Concept / Approach: In a common-collector stage, the output is taken from the emitter, which follows the base voltage minus approximately VBE. Under small-signal AC analysis, the incremental gain is Av ≈ RE / (RE + re), typically close to 1 when RE ≫ re (where re is the intrinsic emitter resistance). The output is in phase with the input (no inversion), and because of the low output resistance, the stage can drive loads better than a common-emitter stage at the same bias current. Step-by-Step Solution: 1) Recognize topology: common-collector with output at emitter. 2) Write small-signal gain: Av ≈ RE / (RE + re). 3) In typical biasing, re is small (tens of ohms) while RE may be hundreds to thousands of ohms, so Av is slightly less than 1. 4) Conclude that the gain is approximately unity, with no phase inversion. Verification / Alternative check: Simulate or calculate with a practical example: re ≈ 25 mV / IE. For IE = 2 mA, re ≈ 12.5 Ω; with RE = 1 kΩ, Av ≈ 1000 / (1012.5) ≈ 0.988, which is very close to one. Why Other Options Are Wrong: Much less than one: contradicts the high follower gain typical of a buffered stage. Greater than one: emitter followers do not provide voltage gain beyond unity. Zero: would imply no output response; incorrect. Negative and large: describes common-emitter behavior (inversion), not emitter follower. Common Pitfalls: Confusing DC level shift (input minus VBE) with AC gain; assuming the gain is exactly 1.000 regardless of load; overlooking the effect of re and load on the exact value. Final Answer: Approximately equal to one

Question 17

BJT switching in cutoff: when a transistor switch is cut off (no base drive), what is the approximate collector-to-emitter voltage VCE across the device relative to the supply?

Accepted Answer

Introduction / Context: Understanding the two extreme states of a BJT used as a switch—cutoff and saturation—is crucial for estimating power dissipation and verifying logic levels. In cutoff, the transistor is off, so there is essentially no collector current other than leakage. The collector circuit then behaves as an open switch. Given Data / Assumptions: Transistor base drive removed (IB ≈ 0), device in cutoff. Normal silicon device at room temperature; leakage currents are negligible for first-order analysis. Collector connected through a load to a positive supply VCC. Concept / Approach: With no conduction path through the transistor, the load line forces the collector voltage to rise to the supply. Thus the collector-to-emitter voltage VCE is approximately VCC (minus tiny leakage-induced drops). This is the complement of saturation, where VCE collapses to a small value. Hence, in digital switching the off-state channel blocks current and supports nearly the full supply voltage across the device. Step-by-Step Solution: 1) Identify state: cutoff implies IC ≈ 0. 2) With IC ≈ 0 through the load, the collector node pulls up to VCC via the load. 3) Therefore VCE ≈ VCC − VE; in a ground-referenced emitter, VE ≈ 0, so VCE ≈ VCC. 4) Any small deviation is due to leakage current; we neglect it in this approximation. Verification / Alternative check: Review typical switching waveforms: turning the transistor off makes VCE rise to near the rail. Datasheets quote ICBO and ICEO leakage currents in microamps or less, confirming that the drop across the load is negligible and VCE remains nearly VCC. Why Other Options Are Wrong: Approximately equal to VB: VB is undefined in cutoff and not directly indicative of VCE. Approximately 0.2 V or 0.7 V: these are characteristic forward drops (VCE(sat), VBE(on)), not applicable in cutoff. Approximately equal to zero: would imply a short circuit, which is the saturation state, not cutoff. Common Pitfalls: Confusing cutoff with saturation; assuming VBE(on) relates to VCE in the off state; forgetting the role of leakage and the load. Final Answer: Approximately equal to VCC

Question 18

Common-emitter amplifier signal phase: what is the phase difference between the small-signal input voltage and the output voltage of a properly biased common-emitter BJT amplifier?

Accepted Answer

Introduction / Context: Phase relationships are central to amplifier design, feedback analysis, and signal interpretation. The common-emitter (CE) amplifier is known for providing substantial voltage gain and inverting the signal. Knowing the expected phase inversion helps when cascading stages and when designing negative-feedback loops that must account for phase to ensure stability. Given Data / Assumptions: BJT biased in its active region (not saturating or cutting off in normal operation). Small-signal sinusoidal analysis around the bias point. Load and coupling capacitors chosen so midband behavior applies (no extra phase shift from reactive elements at midband). Concept / Approach: The CE topology places the output at the collector. An increase in base voltage (and hence base current) increases collector current, which causes a larger voltage drop across the collector resistor RC. The collector node voltage therefore drops as the input rises. This results in an output that is inverted relative to the input, which corresponds to a 180º phase shift in sinusoidal steady state for the midband region. Additional capacitive effects at low or high frequencies can add phase lag or lead, but the ideal midband CE amplifier is considered a 180º inverter. Step-by-Step Solution: 1) Input rises → base current increases. 2) Collector current increases → RC drop increases. 3) Collector voltage decreases → output moves opposite to input. 4) Sinusoidal interpretation: a sign inversion equals 180º phase shift. Verification / Alternative check: Small-signal gain Av of a CE stage is negative (Av ≈ −gm * RC in simplified form), and a negative gain is equivalent to a 180º phase inversion for sinusoidal signals. Oscilloscope measurements on lab CE amplifiers show output peaks when input is at troughs, confirming inversion. Why Other Options Are Wrong: 0º: describes common-collector or common-drain follower behavior. 90º: indicates quadrature typically due to reactive networks, not a midband CE amplifier. 360º: implies no net phase shift (equivalent to 0º) and is not characteristic of CE midband operation. Depends on load only: while loading affects magnitude, the inversion is intrinsic to CE topology. Common Pitfalls: Observing extra phase lag at frequency extremes and assuming the midband inversion is not 180º; forgetting coupling-capacitor and bypass-capacitor phase effects outside midband. Final Answer: 180º

Question 19

Emitter follower input resistance at the base: relative to typical signal sources, how does the input resistance of the base node of a properly biased emitter follower usually compare?

Accepted Answer

Introduction / Context: One of the main reasons to use an emitter follower is to avoid loading a preceding stage. The base input resistance of an emitter follower is large, so only a small input current is required to develop the needed input voltage. This permits accurate voltage transfer from high-impedance sources to lower-impedance loads downstream. Given Data / Assumptions: Emitter follower (common-collector) topology with proper bias. Finite transistor current gain beta, typically tens to hundreds. The emitter node sees an effective resistance RE_eff that includes RE and any load at the emitter. Concept / Approach: The input resistance looking into the base is approximately Rin ≈ (beta + 1) * RE_eff. Because beta is sizable and RE_eff can be in the kilo-ohm range, Rin frequently lands in the hundreds of kilo-ohms to mega-ohm range. This is not infinite, but is “very high” relative to many signal sources. The high Rin, together with low output resistance, makes the emitter follower an excellent buffer and impedance transformer with a voltage gain near unity. Step-by-Step Solution: 1) Identify RE_eff = RE || load_at_emitter || re. 2) Apply Rin(base) ≈ (beta + 1) * RE_eff. 3) For beta = 100 and RE_eff = 10 kΩ, Rin ≈ 1.01 MΩ, which is very high. 4) Conclude that the base input resistance is typically very high, minimizing loading. Verification / Alternative check: Measured with a small AC test source at the base and observing current draw, the calculated Rin aligns with the (beta + 1) scaling. Real designs also include bias resistor dividers that may lower the net input resistance; even then, the base itself presents a very high input compared to many other stages. Why Other Options Are Wrong: Very low: contradicts the buffer role of the emitter follower. Shorted to ground: only in fault conditions; not a design feature. Open (infinite): Rin is high but not infinite; base currents and bias networks exist. Approximately equal to RC: RC is not central in a common-collector stage. Common Pitfalls: Ignoring the effect of a bias divider that can reduce overall input resistance; assuming infinite Rin and miscalculating source loading; forgetting RE_eff includes the external load. Final Answer: Very high

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