Question 1

BJT bias requirement: For proper active-region operation of a bipolar transistor, the base–emitter (BE) junction must be forward biased. Which other junction must be reverse biased?

Accepted Answer

Introduction: Correct DC biasing of a BJT establishes the operating (Q) point in the active region. This enables linear amplification. Knowing which junctions must be forward or reverse biased is a foundational concept in analog electronics. Given Data / Assumptions: BJT operated in active region for amplification. Base–emitter (BE) junction forward biased. Second PN junction bias must be identified. Concept / Approach: In active mode: the BE junction is forward biased to inject carriers from emitter into base; the collector–base (CB) junction is reverse biased to sweep carriers from the base into the collector. This separation of biases provides transistor action with current gain. Step-by-Step Solution: Active-mode conditions: VBE > 0 (typically ~0.7 V for Si), VCB > 0 (reverse bias).Reverse-biased CB junction creates a strong electric field that collects carriers.Thus, proper operation requires CB reverse bias along with BE forward bias.Therefore the correct junction is the collector–base junction. Verification / Alternative check: Examining output characteristics (IC vs VCE) shows that with BE forward and CB reverse bias, the device exhibits near-constant IC for a given IB, the signature of active mode. Why Other Options Are Wrong: collector–emitter junction: VCE is a terminal-to-terminal potential; active-mode condition is specifically CB reverse bias. base–emitter junction: This is already forward biased; reverse biasing it would move the device to cutoff or reverse-active region. emitter–substrate junction: Not a standard discrete BJT terminal pair in circuit bias discussions. base–collector junction (wording): Equivalent to collector–base; among duplicates, we choose the explicit correct phrasing. Common Pitfalls: Confusing saturation (both junctions forward) with active region. Overlooking that reverse-active mode swaps roles and is rarely used in designs. Final Answer: collector–base junction

Question 2

Bias stability in BJT amplifiers: What does a voltage-divider bias network generally provide for a transistor stage, considering variations in transistor current gain (β)?

Accepted Answer

Introduction: Transistor β (current gain) can vary widely between devices and with temperature. A good biasing scheme should hold the operating point (Q point) nearly constant despite such variations. Voltage-divider bias (also called self-bias or emitter-stabilized bias) is the most common approach in discrete BJT amplifiers for this reason. Given Data / Assumptions: BJT in common-emitter configuration. Bias via two resistors forming a divider to set base voltage, plus an emitter resistor for feedback. Objective: stable quiescent collector/emitter currents and voltages. Concept / Approach: The base voltage is set largely by the divider and is relatively insensitive to β. The emitter resistor provides negative feedback: if collector current increases, emitter voltage rises, reducing base–emitter voltage and counteracting the change. Together, these mechanisms stabilize the Q point. Step-by-Step Solution: Set base bias using R1–R2 divider aiming for Vb ≈ Ve + VBE.Include emitter resistor Re to develop Ve = IE * Re, providing feedback.Design divider current ≫ base current to reduce β sensitivity.Result: Q point varies modestly with β, enhancing stability. Verification / Alternative check: Small-signal bias analyses show collector current depends far less on β with divider bias than with simple fixed-bias networks. SPICE sweeps over β confirm limited Q-point drift. Why Other Options Are Wrong: an unstable Q point / no defined Q point: Opposite of divider bias behavior. a Q point that easily varies with β: Describes fixed-bias (base resistor only), not divider bias. a Q point that is stable and easily varies: Self-contradictory statement. Common Pitfalls: Choosing divider current too small; then base current perturbs the divider and stability suffers. Omitting emitter resistor or bypassing it fully at DC, which reduces stabilization. Final Answer: a stable Q point

Question 3

Transistor current gain calculation: A BJT has a DC current gain β = 250. If the base current is IB = 20 µA, what is the collector current IC (assume active region)?

Accepted Answer

Introduction: In the active region of a BJT, the collector current is approximately proportional to base current via the DC current gain β (also called hFE). This relationship is used to estimate required base drive and to size bias networks for desired collector currents. Given Data / Assumptions: Current gain β = 250. Base current IB = 20 µA. Transistor operating in active region (not saturated). Concept / Approach: The basic relation is IC = β * IB. Multiply the given IB by β, keeping careful track of microampere-to-ampere unit conversions, then express the result in milliamperes for clarity. Step-by-Step Solution: Write IC = β * IB.Substitute values: IC = 250 * 20 µA.Compute: 250 * 20 µA = 5,000 µA.Convert units: 5,000 µA = 5 mA. Verification / Alternative check: Reverse check: IB = IC / β = 5 mA / 250 = 0.02 mA = 20 µA, matching the given value. If the transistor were in saturation, IC would be lower than β * IB, but the problem specifies normal operation. Why Other Options Are Wrong: 500 µA: Corresponds to β = 25 or IB = 2 µA, not the given numbers. 50 mA: Implies β = 2,500 for IB = 20 µA, unrealistically high. 5 A: Six orders of magnitude too large; unit error. 2.5 mA: Would require β = 125, not 250. Common Pitfalls: Miscalculating microampere to milliampere conversions. Forgetting that β varies with operating conditions; use as a nominal design estimate unless specified. Final Answer: 5 mA

Question 4

In a silicon bipolar junction transistor (BJT) operated in a common-emitter (C–E) configuration, what is the typical forward-bias base–emitter junction voltage V_BE at room temperature when the device is properly turned on?

Accepted Answer

Introduction: The base–emitter junction of a silicon BJT behaves like a diode. In a common-emitter (C–E) amplifier or switch, the transistor conducts when this junction is forward biased to a characteristic voltage. Knowing the typical V_BE is essential for bias design, operating point prediction, and troubleshooting. Given Data / Assumptions: Device: Silicon BJT (not germanium). Room-temperature operation (around 25 °C). Forward-biased base–emitter junction during normal conduction. Moderate collector current typically in the milliampere range. Concept / Approach: For silicon junctions, the forward conduction knee is near 0.6–0.7 V at room temperature. Practical V_BE rises slightly with current (approximately 60 mV per decade change) and falls about 2 mV/°C with increasing temperature. Designers therefore use 0.7 V as a convenient nominal value in hand calculations for silicon BJTs in conduction. Step-by-Step Solution: Recognize V_BE(silicon) ≈ 0.6–0.7 V under typical bias.For common-emitter operation in forward active or saturation, the base–emitter junction is forward biased near 0.7 V.Select the closest typical nominal value: 0.7 V. Verification / Alternative check: Datasheets often show V_BE(on) around 0.65–0.75 V for currents of a few milliamperes. Lab measurements with a DMM across base–emitter while the device conducts corroborate this range. At very low currents, V_BE may be nearer 0.55–0.6 V; at higher currents, nearer 0.8 V—0.7 V remains the standard design heuristic. Why Other Options Are Wrong: voltage-divider bias: A biasing method, not a V_BE value. 0.4 V: Too low for silicon; more typical of germanium devices. emitter voltage: A node quantity, not the junction drop. Common Pitfalls: Confusing silicon with germanium values, ignoring temperature dependence, or assuming V_BE is constant regardless of current. Always treat 0.7 V as a nominal starting point and refine with datasheet curves when precision matters. Final Answer: 0.7 V

Question 5

On a transistor output characteristic family, the two endpoints of the DC load line correspond to which operating conditions of the device?

Accepted Answer

Introduction: The DC load line graphically represents all possible operating points (I_C, V_CE) for a given supply and load on a transistor's output characteristic curves. Understanding its endpoints is crucial for biasing, allowable signal swing, and avoiding distortion in amplifiers or achieving robust switching in digital applications. Given Data / Assumptions: Fixed supply voltage V_CC and collector load resistor R_C. Transistor plotted with I_C vs. V_CE characteristics. Static (DC) analysis; no AC signal applied when drawing the line. Concept / Approach: The load line is defined by V_CE + I_C * R_C = V_CC. When I_C = 0, V_CE = V_CC (cutoff). When V_CE ≈ 0, I_C ≈ V_CC / R_C (saturation). These two intercepts are the physical extremes; every practical quiescent point (Q-point) lies somewhere between them when the device conducts linearly. Step-by-Step Solution: At I_C = 0 → V_CE = V_CC: transistor off → cutoff.At V_CE = 0 → I_C = V_CC / R_C: maximum current → saturation.The line between these points shows allowed combinations under the DC constraint. Verification / Alternative check: Plotting the line on the characteristic family shows it intersecting the axes at the cutoff and saturation corners. Simulations or bench measurements (varying base drive) move the Q-point along this line, confirming the endpoint interpretations. Why Other Options Are Wrong: the operating point: The Q-point is one chosen point on the line, not an endpoint. the power curve: Power contours are separate overlays, not the load line ends. the amplification factor: A small-signal parameter, not indicated by the intercepts. Common Pitfalls: Confusing AC load lines with DC, or mislabeling the intercepts. Remember cutoff at I_C = 0 and saturation at V_CE ≈ 0. Final Answer: saturation and cutoff

Question 6

Common-base (CB) configuration: Given emitter current IE = 4.2 mA and collector current IC = 4.0 mA, what is the current gain α (alpha = IC / IE)?

Accepted Answer

Introduction: In a common-base BJT configuration, the relevant current gain is α (alpha), defined as the ratio of collector current to emitter current. This parameter is typically close to but less than 1. Calculating α from given currents reinforces current relationships in BJTs and complements the more familiar β in common-emitter circuits. Given Data / Assumptions: IE = 4.2 mA (emitter current). IC = 4.0 mA (collector current). Steady-state operation, neglecting leakage for simplicity. Concept / Approach: By definition, α = IC / IE. The conservation relation IE = IB + IC holds; thus α = 1 − (IB / IE). For a healthy transistor, IB is small compared to IE, so α approaches 1 but remains slightly less. We simply substitute the given values to compute α numerically. Step-by-Step Solution: α = IC / IE.Substitute numbers: α = 4.0 mA / 4.2 mA.Compute: 4.0 / 4.2 ≈ 0.95238.Round to two decimal places: α ≈ 0.95. Verification / Alternative check: From IE = IB + IC → IB = IE − IC = 0.2 mA. Then α = 1 − IB/IE = 1 − 0.2/4.2 ≈ 1 − 0.04762 = 0.95238, confirming the result. Why Other Options Are Wrong: 16.80: Not dimensionally meaningful for this ratio; grossly incorrect. 1.05: α cannot exceed 1 in a normal BJT. 0.20: Would imply unrealistic base current comparable to emitter current. Common Pitfalls: Confusing α with β (β = IC / IB) or mixing up which currents define the ratio. Always use α = IC / IE for common-base analysis. Final Answer: 0.95

Question 7

Why is a common-collector (emitter follower) stage often used as the final stage before a low-resistance load, and what is its principal function?

Accepted Answer

Introduction: The common-collector (CC) configuration, also known as an emitter follower, is widely used as an output buffer in amplifier chains. Its hallmark is high input impedance, low output impedance, and near-unity voltage gain, making it ideal for driving low-resistance loads without loading previous gain stages. Given Data / Assumptions: Load has relatively low resistance compared to the preceding amplifier's optimal load. Signal integrity (amplitude and bandwidth) must be preserved. Standard BJT operation with appropriate biasing. Concept / Approach: The CC stage provides impedance transformation: it accepts a signal with minimal loading (high input impedance) and delivers current into the load (low output impedance). Voltage gain is approximately 1, but current gain is high, enabling power transfer to the load. This protects earlier voltage-amplifying stages and improves overall efficiency and linearity when driving speakers, cables, or other low-resistance loads. Step-by-Step Solution: Note that A_v ≈ 1; A_i ≫ 1 for a CC stage.High input impedance prevents loading of prior voltage gain stages.Low output impedance drives the load effectively, improving power delivery and bandwidth under load.Therefore, the primary function is buffering and impedance matching to the load. Verification / Alternative check: Small-signal models show output resistance roughly Re || (re' + (β+1) * emitter network), which is low; input resistance is approximately (β+1) times the emitter network, which is high—quantitatively confirming buffering behavior. Why Other Options Are Wrong: provide voltage gain: CC has near-unity voltage gain by design. provide phase inversion: CC is non-inverting; CE inverts. provide a high-frequency path: Frequency response may be good, but this is not the defining role. Common Pitfalls: Expecting large voltage gain from a CC stage or overlooking the importance of proper bias and emitter bypass choices for linear operation. Final Answer: buffer the voltage amplifiers from the low-resistance load and provide impedance matching for maximum power transfer

Question 8

Basic gain calculation: If a 2 mV input signal results in a 2 V output from an amplifier, what is the voltage gain A_v (ratio of output voltage to input voltage)?

Accepted Answer

Introduction: Voltage gain is fundamental in amplifier analysis. It is defined as the ratio of output voltage to input voltage and is often expressed as a pure ratio or in decibels. Quick mental math on small signals and unit conversions is a practical skill when reviewing specifications or debugging circuits. Given Data / Assumptions: V_in = 2 mV. V_out = 2 V. Linear operation (no clipping or compression). Concept / Approach: Compute A_v = V_out / V_in, taking care to convert millivolts to volts to avoid errors. Because 1 mV = 10^-3 V, 2 mV = 0.002 V. The ratio then becomes straightforward division. Step-by-Step Solution: Convert input: 2 mV = 0.002 V.Compute ratio: A_v = 2 V / 0.002 V.Perform division: 2 / 0.002 = 1000.Therefore the voltage gain is 1000 (or 60 dB if expressed as 20 * log10(1000)). Verification / Alternative check: Cross-check by proportional reasoning: increasing input by a factor of 1000 (2 mV * 1000 = 2 V) matches the output, confirming the gain. Why Other Options Are Wrong: 0.001 and 0.004: These invert the ratio or mix units incorrectly. 100: Underestimates the true ratio by a factor of 10. Common Pitfalls: Forgetting unit conversion (mV to V), or accidentally calculating V_in / V_out. Always normalize units first to avoid magnitude errors. Final Answer: 1000

Question 9

For a bipolar junction transistor, which control quantity primarily determines the collector current in normal operation?

Accepted Answer

Introduction: Transistor operation involves interrelated currents and voltages. In a BJT, the collector current is largely controlled by the base–emitter junction and the resulting base current, with device gain β relating I_C and I_B. Recognizing the primary control input is crucial for amplifier design and switching analysis. Given Data / Assumptions: Forward-active region of operation (not breakdown). Constant temperature and neglecting second-order effects for first-pass analysis. Reasonable transistor β (e.g., 50–200) so that I_C ≈ β * I_B. Concept / Approach: In forward-active region, collector current is approximately proportional to base current: I_C ≈ β * I_B, where β is the common-emitter current gain. Although collector–emitter voltage influences I_C via the Early effect, the primary control knob is base current. Collector resistance influences voltage drops but does not directly set I_C absent changes in bias/current drive. Step-by-Step Solution: Drive base–emitter junction into forward bias → base current flows.Relate I_C to I_B: I_C ≈ β * I_B under normal bias.Conclude: Adjusting I_B primarily modulates I_C; other parameters are secondary or indirect. Verification / Alternative check: Load-line experiments show that varying base current shifts the Q-point vertically (I_C axis) along the load line, confirming I_B as the main determinant of collector current for a given device. Why Other Options Are Wrong: collector voltage: Affects I_C modestly (Early effect) but is not the primary control variable. collector resistance: Sets V_CE with I_C but does not directly control I_C without changing I_B. all of the above: Overstates secondary effects; the dominant control is base current. Common Pitfalls: Confusing MOSFET gate-voltage control with BJT base-current control, or assuming I_C is independent of V_CE (it is not perfectly independent due to Early effect). Final Answer: base current

Question 10

In a common-emitter (C–E) amplifier, why is an emitter resistor commonly included in the bias network?

Accepted Answer

Introduction: Bias stability is critical in analog amplifier design. Incorporating an emitter resistor in a common-emitter stage introduces negative feedback that counters variations in temperature, β, and supply, improving operating-point stability and making designs more robust. Given Data / Assumptions: Single-transistor CE amplifier with resistive bias network. Emitter resistor Re present; optional bypass capacitor may be used. Small-signal linear operation around a chosen Q-point. Concept / Approach: The emitter resistor creates emitter degeneration. As collector/emitter current rises, the voltage across Re increases, effectively reducing V_BE and counteracting the increase in current—this negative feedback stabilizes I_C and V_CE. A bypass capacitor may be added to maintain AC gain while preserving DC stability, but the resistor's fundamental role is stabilization, not raw gain. Step-by-Step Solution: Include Re to develop an emitter voltage Ve = I_E * Re.If temperature or β change increases I_E, Ve rises → V_BE = V_B − V_E decreases → current is pushed back down.Hence the DC operating point becomes less sensitive to parameter variations. Verification / Alternative check: Small-signal models show transconductance reduction and increased linearity with emitter degeneration; bias simulations show reduced drift over β spread and temperature. Why Other Options Are Wrong: ac signal bypass: This refers to a capacitor placed across Re to restore AC gain, not the resistor's purpose. collector bias: Collector biasing uses collector resistors and supply; Re is not for collector bias. higher gain: Re without bypass reduces voltage gain; it improves linearity and stability instead. Common Pitfalls: Assuming Re increases gain (it usually lowers it unless bypassed) or forgetting that the bypass capacitor affects AC only, leaving DC stabilization intact. Final Answer: stabilization

Question 11

Definition check: In a BJT, which current ratio defines the common-emitter current gain β (beta)?

Accepted Answer

Introduction: Two current gains are frequently used to characterize BJTs: α (common-base) and β (common-emitter). Correctly identifying which ratio defines β is essential for circuit calculations, bias design, and interpreting datasheets. Given Data / Assumptions: BJT operating in forward-active region. Steady-state currents IB, IC, and IE well-defined. Neglect leakage currents for simplicity. Concept / Approach: By definition, β (beta) is the ratio of collector current to base current in the common-emitter configuration: β = IC / IB. It relates directly to how base drive controls collector current. The related common-base gain is α = IC / IE. These quantities are linked approximately by β = α / (1 − α) and α = β / (β + 1). Step-by-Step Solution: Recall definitions: β = IC/IB; α = IC/IE.Select the option that matches β's definition: IC/IB.Note relationship among currents: IE = IB + IC (useful for translating between α and β). Verification / Alternative check: From IE = IB + IC, derive α and β relationships to confirm consistency: β = α/(1 − α), demonstrating that the chosen ratio is coherent with transistor current laws. Why Other Options Are Wrong: IC/IE: That is α, not β. IB/IE or IE/IB: Neither corresponds to standard current gains. Common Pitfalls: Mixing up α and β or forgetting that β can vary widely with current, temperature, and device type; design should not rely on a single β value without margin. Final Answer: IC/IB

Question 12

Digital switching with BJTs: When bipolar transistors are used as logic switches in digital circuits, which operating regions do they alternate between?

Accepted Answer

Introduction: In digital logic, transistors act as two-state devices rather than linear amplifiers. BJTs configured as switches toggle between non-conducting and fully conducting states to represent logic levels. Recognizing the correct operating regions helps understand switching speed, power dissipation, and noise margins. Given Data / Assumptions: BJT used as a saturated switch in common-emitter configuration. Logic levels encoded by off and on states. Supply and load chosen to ensure reliable drive to the intended regions. Concept / Approach: Cutoff corresponds to no base drive and essentially zero collector current; the transistor is off. Saturation occurs when base drive is sufficient to forward-bias both base–emitter and base–collector junctions, minimizing V_CE(sat) and providing a low-resistance path. Switching between these two regions yields robust logic 0/1 behavior. The forward-active (linear) region is used for analog amplification, not for crisp digital switching, and breakdown must be avoided. Step-by-Step Solution: Apply zero or insufficient base current → cutoff (logic off).Apply strong base current → drive into saturation (logic on) with low V_CE(sat).Alternate base drive to switch states as required by logic waveforms. Verification / Alternative check: Static transfer characteristics show distinct regions; timing diagrams confirm that digital families (e.g., TTL) exploit saturation and cutoff for defined logic thresholds and noise margins. Why Other Options Are Wrong: active region/linear region: Intended for analog amplification; produces ambiguous logic levels. breakdown region: Damaging; not a normal operating region. Common Pitfalls: Insufficient base drive causing the device to linger in the active region (slow edges, more power dissipation) or accidentally exceeding breakdown voltages. Final Answer: saturation and cutoff regions

Question 13

Current relationships in a BJT: Which expression correctly states the total emitter current in terms of the other two transistor currents?

Accepted Answer

Introduction: Understanding the current relationships in a BJT is foundational for both DC biasing and small-signal analysis. Kirchhoff’s current law at the transistor's base–emitter–collector junctions gives a simple identity linking the three terminal currents: emitter, base, and collector. Given Data / Assumptions: Conventional current directions: currents entering/leaving as defined in standard BJT models. Steady-state operation, neglecting leakage for clarity. Device not in breakdown. Concept / Approach: By device topology and charge flow, the emitter current equals the sum of base and collector currents. This holds across operating regions (cutoff excepted where all are approximately zero). The identity is frequently used to translate between α and β formulations and to compute bias currents. Step-by-Step Solution: Apply KCL at the emitter node: I_E = I_B + I_C.Rearrange as needed for design: I_C = β * I_B and I_E = (β + 1) * I_B.Use the relationship to verify bias calculations and power estimates. Verification / Alternative check: From α = I_C / I_E and β = I_C / I_B, it follows that I_E = I_B + I_C consistently. Experimental measurement of the three currents will show this sum to within leakage and measurement tolerance. Why Other Options Are Wrong: IE – IC: This equals IB, not IE. IC + IE: Double counts IE; not a meaningful identity. IB – IC: Not a standard relation and often negative in forward-active operation. Common Pitfalls: Sign convention mistakes when using simulated current probes, or forgetting that leakage currents can slightly perturb the equality at very low currents. Final Answer: IB + IC

Question 14

In transistor amplifier theory, compare the input–output phase relationship of two classic single-transistor stages: the common-collector (emitter follower) and the common-base amplifier. Considering a small-signal sinusoidal input, what overall pha…

Accepted Answer

Introduction: The phase relationship between an amplifier's input and output is a foundational concept in analog electronics. It determines whether a stage inverts the signal or preserves its polarity. This question focuses on two important single-transistor configurations—the common-collector (emitter follower) and the common-base—and asks you to identify their net input–output phase shift under small-signal sinusoidal operation. Given Data / Assumptions: Bipolar junction transistor (BJT) operated in its linear region (active mode). Small-signal sinusoidal excitation so that incremental models apply. Common-collector (input at base, output at emitter) and common-base (input at emitter, output at collector) topologies. Biasing is correct so the device does not clip, saturate, or cut off within the signal swing. Concept / Approach: For small signals, we analyze the instantaneous relationship of input and output voltages. A stage is noninverting if the output rises when the input rises (0-degree phase shift) and inverting if the output falls when the input rises (180-degree shift). The common-emitter is the classic inverting amplifier. By contrast, the common-collector and common-base stages are noninverting in their usual voltage transfer sense and therefore exhibit approximately 0-degree phase shift. Step-by-Step Solution: Common-collector: input at base, output at emitter ≈ base minus a nearly constant V_BE. When the base voltage increases slightly, the emitter follows that increase (emitter follower). Hence the output and input move in phase, implying ~0 degrees.Common-base: input current enters the emitter, and the output is taken at the collector. For a small increase in emitter input voltage (raising emitter current), collector current increases and the collector voltage also moves in the same direction once load effects are considered in voltage gain; the configuration provides noninverting voltage transfer (small-signal output voltage is in phase with the effective input signal).Therefore, both stages are effectively noninverting for voltage, yielding an overall 0-degree phase shift between input and output. Verification / Alternative check: Small-signal hybrid-pi analysis confirms that the common-collector has voltage gain slightly less than 1 with positive sign, while the common-base has voltage gain typically greater than 1 with positive sign. Positive sign indicates noninversion, i.e., ~0-degree phase shift. Why Other Options Are Wrong: 180 degrees: describes the common-emitter, not common-collector or common-base.90 degrees and 270 degrees: phase quadrature does not apply to these real-valued voltage transfer stages. Common Pitfalls: Assuming all single-transistor stages invert. Only the common-emitter reliably inverts; the common-collector and common-base are noninverting for voltage transfer. Final Answer: 0 degrees

Question 15

Transistor junction check (silicon NPN): With a digital ohmmeter's positive probe placed on the base terminal, what type of reading should you expect when measuring from base to the other two terminals (base–emitter and base–collector)? Assume a sta…

Accepted Answer

Introduction: A quick health check for BJTs uses an ohmmeter or diode tester to evaluate the two internal PN junctions. Understanding expected readings helps technicians identify open or shorted junctions without powering the circuit. This question targets the expected readings for a silicon NPN transistor when the positive meter lead is on the base. Given Data / Assumptions: Device under test: NPN BJT (silicon). Meter positive probe on base; negative probe touching emitter or collector. Meter in ohms/diode test mode with small forward test current. Transistor not connected to a powered circuit during measurement. Concept / Approach: An NPN contains two PN junctions: base–emitter (P–N) and base–collector (P–N). With the positive probe on the P-type base and negative probe on N-type emitter or collector, each junction is forward-biased. A forward-biased diode shows a low resistance (or about 0.6–0.7 V on diode scale). Reversing the leads should show high resistance, indicating a good junction. Step-by-Step Solution: Place positive probe on base and negative on emitter: forward-biased P–N → low resistance/diode drop.Move negative probe to collector: again forward-biased P–N → low resistance/diode drop.Reverse polarity (negative on base, positive on emitter/collector): reverse-biased → high resistance/open on typical ohms range.Consistent low forward readings and high reverse readings indicate healthy junctions. Verification / Alternative check: On a diode tester, a good silicon NPN shows ~0.6–0.7 V from base to emitter and base to collector with the positive lead on base, confirming low forward impedance in that direction. Why Other Options Are Wrong: Open / infinite / high resistance: these correspond to reverse bias or a damaged/open junction; not expected when forward-biasing with positive on base. Common Pitfalls: Touching probes improperly or testing in-circuit (sneak paths can mislead). Always isolate the transistor or at least lift leads for reliable junction checks. Final Answer: low resistance

Question 16

On a DC load line for a BJT amplifier, three possible quiescent (Q) points are marked. Interpreting the axes as I_C (vertical) versus V_CE (horizontal), what does the Q point located near the high V_CE and low I_C end (upper-right region) physically…

Accepted Answer

Introduction: The DC load line graphically shows all possible operating points for a given collector load and supply. Choosing the Q point determines headroom for signal swings and sets the amplifier's linear range. This question asks you to interpret a Q point located toward the high V_CE/low I_C end of the load line. Given Data / Assumptions: Axes: vertical axis is collector current I_C; horizontal axis is collector–emitter voltage V_CE. Load line drawn from V_CE = V_CC at I_C = 0 (cutoff end) to I_C = V_CC / R_C at V_CE ≈ 0 (saturation end). Three Q points indicate different bias choices, one near the high V_CE/low I_C end. Concept / Approach: At the high V_CE, low I_C end of the load line, the transistor conducts little to no collector current. This is the cutoff region, where the base–emitter junction is not forward-biased sufficiently and the device is effectively off. Moving along the line toward higher I_C and lower V_CE approaches saturation. Midline choices target symmetrical swing. Step-by-Step Solution: Identify the endpoints: (V_CE = V_CC, I_C ≈ 0) is cutoff; (V_CE ≈ 0, I_C = V_CC / R_C) is saturation.Locate the specified Q point: high V_CE with small I_C.Match region: this corresponds to cutoff or near-cutoff operation.Conclude that the upper-right Q point represents the cutoff point on the load line. Verification / Alternative check: Kirchhoff's voltage law around the collector loop at cutoff shows essentially the full V_CC across the collector–emitter terminals because I_C ≈ 0, confirming the interpretation. Why Other Options Are Wrong: Minimum/intermediate/maximum current gain: current gain is primarily a transistor parameter (beta) and bias dependent, but the endpoint in question specifically indicates the cutoff region. Common Pitfalls: Confusing the two ends of the load line. Remember: high V_CE with low I_C is cutoff; low V_CE with high I_C is saturation. Final Answer: cutoff point

Question 17

In bipolar transistor terminology, the small-signal current ratio alpha is defined as the ratio of collector current to emitter current. What is this parameter called when expressed as IC / IE (a value slightly less than 1 under active operation)?

Accepted Answer

Introduction: BJT operation is often summarized by two core current ratios: alpha and beta. Engineers use them to relate terminal currents and to translate between common-base and common-emitter viewpoints. This question asks you to identify the parameter corresponding to IC / IE. Given Data / Assumptions: BJT operated in active region. Emitter current splits into collector current (majority) and base current (minority). We focus on the ratio IC / IE. Concept / Approach: By definition, alpha (α) = IC / IE in the common-base perspective and is slightly less than 1 because IE = IC + IB and IB is small but nonzero. The common-emitter current gain is beta (β) = IC / IB, which can be large (tens to hundreds). The two are related by β = α / (1 − α) and α = β / (β + 1). Step-by-Step Solution: Identify ratio: IC / IE.Recall definition: α = IC / IE.Note magnitude: since IE = IC + IB, α = IC / (IC + IB) < 1.Relate to β if needed: α = β / (β + 1). Verification / Alternative check: With β = 100, α = 100 / 101 ≈ 0.990, illustrating that α is close to but less than unity, consistent with physical intuition. Why Other Options Are Wrong: Beta: equals IC / IB, not IC / IE.Theta and omega: not standard current-gain symbols in BJTs. Common Pitfalls: Interchanging α and β or assuming α equals 1 exactly; small base current prevents α from being exactly unity. Final Answer: alpha

Question 18

Carrier flow in an NPN transistor: When biased in the active region, most of the electrons injected from the emitter into the thin, lightly doped base will ultimately flow to which terminal?

Accepted Answer

Introduction: Understanding carrier flow in BJTs clarifies why the device can amplify current. In an NPN transistor, the emitter injects electrons into the base. How these electrons travel determines the resulting collector current and overall device behavior. Given Data / Assumptions: NPN BJT biased in active region. Emitter–base junction forward-biased; collector–base junction reverse-biased. Base is thin and lightly doped. Concept / Approach: In active operation, the forward-biased emitter–base junction injects majority carriers (electrons) from the N-type emitter into the P-type base. Because the base is thin and lightly doped, only a small fraction recombines with holes (creating base current). Most electrons diffuse across the base and are swept into the collector region by the reverse-biased collector–base junction's electric field, forming the collector current IC. Step-by-Step Solution: Electrons injected from emitter enter base.Base is engineered to be thin/lightly doped so recombination is minimized.Remaining electrons reach the collector–base depletion region.Reverse-bias field sweeps them into the collector, producing IC ≈ α * IE with α close to 1. Verification / Alternative check: The relation IE = IC + IB shows that IB is small. Therefore, most of IE must appear as IC, confirming that most injected electrons end up collected at the collector terminal. Why Other Options Are Wrong: Out of the base lead / into the base supply: would imply large base current, contradicting transistor design (IB is small).Into the emitter: reverse flow would contradict the forward-bias injection and device physics. Common Pitfalls: Overestimating base recombination. High transistor gain relies on making the base thin and lightly doped to minimize recombination and maximize collection. Final Answer: into the collector

Question 19

Datasheet notation: The small-signal common-emitter current gain denoted by hfe (lowercase h, f, e) corresponds to which familiar transistor parameter?

Accepted Answer

Introduction: Transistor datasheets use the h-parameter model for small-signal behavior, where symbols like hfe, hie, and hoe appear frequently. Knowing how these map to everyday terms such as beta (β) avoids confusion when comparing devices or calculating bias and gain. Given Data / Assumptions: BJT characterized in common-emitter configuration. Small-signal linearization around a bias point. Standard h-parameter notation is used. Concept / Approach: In the hybrid h-parameter set for a common-emitter amplifier, hfe is the small-signal forward current gain, essentially ΔIC / ΔIB at a given operating point. This is the same quantity commonly called beta (β). While β can also denote DC current gain (sometimes labeled hFE with uppercase FE), hfe (lowercase) refers to incremental (small-signal) gain; context in datasheets usually makes the distinction clear. Step-by-Step Solution: Identify symbol: hfe.Match to definition: small-signal forward current gain in common-emitter.Map to familiar term: beta (β), i.e., change in collector current per change in base current.Note practical ranges: β may vary widely with device, current, and temperature. Verification / Alternative check: Consult any BJT datasheet: parameters listed as hfe or hFE correspond to current gain figures; test conditions specify collector current and VCE, confirming the mapping to β. Why Other Options Are Wrong: Alpha: equals IC / IE in common-base, not CE current gain.Hi-fi: unrelated slang for audio quality.Transconductance gm: equals ΔIC / ΔVBE and has units of siemens, not a unitless current ratio. Common Pitfalls: Confusing hfe (small-signal) with hFE (DC). Both are forms of β but measured differently; design calculations must match the appropriate definition. Final Answer: beta (common-emitter current gain)

Question 20

Common-base (CB) amplifier usage: Among the classic single-transistor configurations, which type of gain is primarily provided by the common-base stage, making it useful for wide-band, low-input-impedance applications?

Accepted Answer

Introduction: The common-base configuration is less common than the ubiquitous common-emitter, but it serves distinct purposes. Recognizing what kind of gain it excels at clarifies when to choose it for RF front ends or impedance transformation tasks. Given Data / Assumptions: BJT operated in linear region, common-base configuration. Signal applied at emitter; output taken at collector; base is AC-grounded. We focus on the type of gain most associated with CB. Concept / Approach: Common-base amplifiers have very low input impedance at the emitter, relatively high output impedance at the collector, and current gain approximately equal to alpha (close to 1). Thus, they offer little current gain but can provide substantial voltage gain (Av often greater than 1), leading to useful power gain even though current gain is near unity. Step-by-Step Solution: Assess current gain: Ai ≈ α ≈ 1 → minimal current amplification.Evaluate voltage gain: with a high collector load and low input impedance, Av can be large.Infer power gain: since voltage gain > 1 and current gain ≈ 1, power gain is achieved.Conclude CB is primarily a voltage-gain stage with wide bandwidth. Verification / Alternative check: Small-signal models demonstrate low input resistance r_e and large collector load resistance reflected at the emitter, yielding Av ≈ gm * RC (in magnitude), which is often several times unity. Why Other Options Are Wrong: Current: CB provides ~unity current gain; CE is chosen when current gain is needed.Resistance: not a gain type; CB has low input resistance but that is not a 'gain'.Power: CB can have power gain, but the primary distinguishing gain type is voltage. Common Pitfalls: Assuming every amplifier provides both voltage and current gain. CB is mainly used where voltage gain and wide bandwidth with low input impedance are required. Final Answer: voltage

Bipolar Junction Transistors (BJT) Questions