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BJT & Biasing (Electronics Engineering) Solved Questions & Notes (2026) - Apex Rankers

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BJT & Biasing

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Q. 1 Electronics Engineering
Difficulty: Easy (1 Mark)
In a Bipolar Junction Transistor (BJT), the three terminals and their relative doping concentrations are:
A
Emitter (heavily doped), Base (lightly doped and very thin), Collector (moderately doped and physically largest)
✓ Correct
B
Base (heavily doped), Emitter (lightly doped), Collector (thin)
C
Collector (heavily doped), Emitter (moderately doped), Base (wide)
D
All three regions have identical doping concentrations
💡 Step-by-Step Explanation & Concept Rationale
The Emitter is heavily doped to inject large numbers of majority carriers. The Base is lightly doped and extremely thin to minimize carrier recombination (< 1-2%). The Collector is moderately doped and physically largest to dissipate heat generated by carrier collection.
Q. 2 Electronics Engineering
Difficulty: Easy (1 Mark)
For a BJT to operate in the Active (Linear Amplification) region, the junction biasing conditions must be:
A
Emitter-Base junction forward-biased, and Collector-Base junction reverse-biased
✓ Correct
B
Both junctions forward-biased
C
Both junctions reverse-biased
D
Emitter-Base reverse-biased, Collector-Base forward-biased
💡 Step-by-Step Explanation & Concept Rationale
In Active mode: EB junction is forward-biased to inject carriers into the base, and CB junction is reverse-biased to sweep carriers into the collector. Saturation mode requires both forward-biased. Cutoff mode requires both reverse-biased.
Q. 3 Electronics Engineering
Difficulty: Easy (1 Mark)
In a BJT operating in the Cutoff region, the transistor behaves as a(n):
A
Closed switch (short circuit)
B
Open switch (off state) with collector current near zero (IC = ICEO)
✓ Correct
C
Linear amplifier with unity gain
D
Constant current source
💡 Step-by-Step Explanation & Concept Rationale
When both EB and CB junctions are reverse-biased, majority carrier injection ceases completely. The transistor acts as an open circuit with only microamps/nanoamps of reverse leakage current flowing, representing digital logic '0'.
Q. 4 Electronics Engineering
Difficulty: Easy (1 Mark)
In a BJT operating in the Saturation region, the collector-emitter saturation voltage (VCE_sat) for a Silicon transistor is approximately:
A
0.2 V
✓ Correct
B
0.7 V
C
5.0 V
D
Zero strictly
💡 Step-by-Step Explanation & Concept Rationale
In saturation, both junctions are forward-biased (VBE ≈ 0.7 V, VBC ≈ 0.5 V). Collector-emitter voltage drops to VCE_sat = VBE - VBC ≈ 0.7 - 0.5 = 0.2 V. The transistor acts as a closed switch carrying maximum collector current limited only by the external load.
Q. 5 Electronics Engineering
Difficulty: Easy (1 Mark)
The common-base DC current gain (alpha) and common-emitter DC current gain (beta) of a BJT are related by:
A
beta = alpha / (1 - alpha)
✓ Correct
B
alpha = beta / (1 - beta)
C
beta = alpha / (1 + alpha)
D
beta = 1 / alpha
💡 Step-by-Step Explanation & Concept Rationale
Since IE = IB + IC, dividing by IC gives 1/alpha = 1/beta + 1, which rearranges to beta = alpha / (1 - alpha), and inversely alpha = beta / (1 + beta). For example, if alpha = 0.99, beta = 0.99 / (1 - 0.99) = 99.
Q. 6 Electronics Engineering
Difficulty: Medium (1 Mark)
The Early Effect (base-width modulation) in a BJT refers to the phenomenon where:
A
Increasing reverse bias across the Collector-Base junction widens the depletion layer into the lightly doped base, reducing effective neutral base width
✓ Correct
B
Emitter current drops at early morning temperatures
C
Base resistance doubles with frequency
D
Transistor breaks down prematurely
💡 Step-by-Step Explanation & Concept Rationale
As reverse voltage VCB increases, the CB space-charge region penetrates deeper into the thin base. This decreases the effective neutral base width Wb, reducing recombination, slightly increasing alpha and beta, and giving the output characteristics a finite upward slope corresponding to the Early Voltage (VA).
Q. 7 Electronics Engineering
Difficulty: Hard (1 Mark)
Punch-through (or reach-through) breakdown in a BJT occurs when:
A
Collector-base reverse voltage increases to the point where the CB depletion layer expands across the entire base width and touches the EB depletion layer
✓ Correct
B
Base current exceeds collector current
C
Emitter lead is detached
D
Thermal runaway burns the package
💡 Step-by-Step Explanation & Concept Rationale
When VCB is excessively high, the CB depletion region completely consumes the neutral base width, contacting the emitter space-charge boundary. The barrier collapses, and direct collector-emitter punch-through current surges, destroying transistor action.
Q. 8 Electronics Engineering
Difficulty: Medium (1 Mark)
Thermal runaway in a BJT is triggered because:
A
An increase in junction temperature causes collector leakage current (ICBO) to increase, which increases total collector current IC and power dissipation, causing further temperature rise cumulatively
✓ Correct
B
Base current drops with temperature
C
Emitter resistance increases
D
VCE rises to infinity
💡 Step-by-Step Explanation & Concept Rationale
Collector current IC = beta*IB + (1 + beta)*ICBO. As temperature rises, ICBO increases exponentially. The higher IC increases collector power dissipation (P = IC * VCE), raising junction temperature further in an unstable positive-feedback loop that destroys the transistor.
Q. 9 Electronics Engineering
Difficulty: Hard (1 Mark)
To prevent thermal runaway in a common-emitter BJT amplifier, the circuit design must satisfy which stability condition?
A
VCE < VCC / 2
B
d(Pc)/d(Tj) < 1 / theta (rate of heat generation less than rate of heat dissipation through thermal resistance theta)
✓ Correct
C
Beta must be greater than 200
D
Collector resistance must equal zero
💡 Step-by-Step Explanation & Concept Rationale
Thermal stability requires that the rate at which heat is generated at the collector junction d(Pc)/dTj be strictly less than the rate at which heat is dissipated to the ambient environment (1 / theta_th). Maintaining VCE < VCC / 2 also naturally prevents thermal runaway.
Q. 10 Electronics Engineering
Difficulty: Easy (1 Mark)
Which BJT biasing configuration provides the HIGHEST stability factor against variations in temperature and transistor beta?
A
Fixed-bias (base bias) circuit
B
Voltage Divider Bias (Self-Bias / Potential Divider) with an emitter resistor RE
✓ Correct
C
Collector-to-base feedback bias without emitter resistor
D
Direct coupling without resistors
💡 Step-by-Step Explanation & Concept Rationale
Voltage divider bias with an unbypassed/bypassed emitter resistor provides exceptional negative feedback. If IC tries to rise due to temp or high beta, the voltage drop across RE (VE = IE*RE) rises, reducing VBE (VBE = VB - VE), which chokes base current IB and restores IC to its stable Q-point.
Q. 11 Electronics Engineering
Difficulty: Medium (1 Mark)
In a voltage divider bias circuit with R1, R2, RC, and RE, the condition for the Q-point to be virtually independent of transistor beta is:
A
Rth = (R1 || R2) << (1 + beta) * RE (typically Rth <= 0.1 * beta * RE)
✓ Correct
B
Rth >> beta * RE
C
RE = 0
D
R1 = R2 = RC
💡 Step-by-Step Explanation & Concept Rationale
Base current IB = (Vth - VBE) / [Rth + (beta + 1)*RE]. If Rth << (beta + 1)*RE, then IE ≈ (Vth - VBE) / RE. Because Vth is fixed purely by the R1-R2 divider ratio, collector operating current is completely immune to transistor beta variations.
Q. 12 Electronics Engineering
Difficulty: Medium (1 Mark)
The Stability Factor S (or S(ICBO)) of a transistor bias circuit is defined as:
A
dIC / dICBO (at constant VBE and beta)
✓ Correct
B
dIC / dbeta
C
dIC / dVBE
D
VCC / IC
💡 Step-by-Step Explanation & Concept Rationale
Stability factor S measures how sensitive collector current IC is to changes in reverse leakage current ICBO: S = dIC / dICBO. In an ideal circuit, S = 1. In a fixed-bias circuit, S = 1 + beta (very poor, ~100-200). In voltage-divider bias, S approaches 1.
Q. 13 Electronics Engineering
Difficulty: Easy (1 Mark)
The Common Emitter (CE) transistor configuration is the most widely used amplifier configuration because it provides:
A
Both high voltage gain and high current gain, yielding the highest power gain of all three configurations
✓ Correct
B
Unity voltage gain and zero phase shift
C
Extremely low input impedance (< 1 ohm)
D
Zero output impedance
💡 Step-by-Step Explanation & Concept Rationale
CE provides moderate input impedance (~1-2 kΩ), moderate output impedance (~50 kΩ), high voltage gain, and high current gain (beta), resulting in the highest power gain (Ap = Av * Ai). It produces a 180-degree phase reversal between input and output.
Q. 14 Electronics Engineering
Difficulty: Easy (1 Mark)
The phase difference between the input voltage and output voltage in a Common Emitter (CE) BJT amplifier is:
A
0 degrees (in phase)
B
90 degrees
C
180 degrees (inversion)
✓ Correct
D
270 degrees
💡 Step-by-Step Explanation & Concept Rationale
An increase in input base voltage increases base current IB and collector current IC. The increased drop across collector resistor RC (IC * RC) causes output collector voltage Vo = VCC - IC*RC to drop, producing an exact 180-degree phase inversion.
Q. 15 Electronics Engineering
Difficulty: Easy (1 Mark)
The Common Collector (CC) configuration, also known as the Emitter Follower, is characterized by:
A
High input impedance, low output impedance, and voltage gain slightly less than unity (Av ≈ 1), with 0-degree phase shift
✓ Correct
B
Low input impedance and high voltage gain
C
180-degree phase inversion and high voltage gain
D
Zero current gain
💡 Step-by-Step Explanation & Concept Rationale
The Emitter Follower has high input impedance (~hundreds of kΩ) and very low output impedance (~tens of ohms) with Av ≈ 1. It is universally used as an impedance-matching buffer between a high-impedance source and low-impedance load.
Q. 16 Electronics Engineering
Difficulty: Medium (1 Mark)
The Common Base (CB) configuration is primarily employed in:
A
High-frequency RF and VHF amplifiers because of low input impedance, absence of Miller effect capacitance multiplication, and good high-frequency response
✓ Correct
B
Audio power output stages driving 8-ohm speakers
C
Low-frequency operational amplifiers
D
Digital CMOS inverters
💡 Step-by-Step Explanation & Concept Rationale
CB has low input impedance (~20-50 Ω), high output impedance (> 1 MΩ), current gain alpha < 1, and non-inverting voltage gain. Because the grounded base shields input from output, feedback capacitance (Miller effect) is suppressed, enabling operation at very high RF frequencies.
Q. 17 Electronics Engineering
Difficulty: Easy (1 Mark)
A Darlington Pair consists of two cascaded BJTs connected such that their collectors are tied together and the emitter of the first drives the base of the second. Its overall current gain is approximately:
A
beta_total = beta1 + beta2
B
beta_total = beta1 * beta2
✓ Correct
C
beta_total = (beta1 + beta2) / 2
D
beta_total = beta1 / beta2
💡 Step-by-Step Explanation & Concept Rationale
Total current gain beta_total = beta1 + beta2 + (beta1 * beta2) ≈ beta1 * beta2. This compound configuration achieves ultra-high current gain (thousands) and very high input impedance, but has a higher total base-emitter drop VBE_total ≈ 2 * 0.7 = 1.4 V.
Q. 18 Electronics Engineering
Difficulty: Medium (1 Mark)
The Sziklai Pair (complementary feedback pair) uses one NPN and one PNP transistor. Compared to a Darlington pair, its primary advantage is:
A
Lower turn-on voltage (VBE ≈ 0.7 V instead of 1.4 V) and superior thermal stability
✓ Correct
B
Zero output impedance
C
Current gain exceeding 100,000
D
It requires no DC supply
💡 Step-by-Step Explanation & Concept Rationale
The Sziklai pair requires only a single diode drop across the input NPN base-emitter junction (0.7 V), making it linear over a wider dynamic range with less crossover distortion in Class-AB audio output stages.
Q. 19 Electronics Engineering
Difficulty: Easy (1 Mark)
In the small-signal hybrid-pi (or re) model of a BJT, the dynamic emitter resistance re is given at room temperature (300 K) by:
A
re = 26 mV / IE (where IE is DC emitter current)
✓ Correct
B
re = IE / 26 mV
C
re = beta * 26 mV
D
re = VCE / IC
💡 Step-by-Step Explanation & Concept Rationale
Dynamic resistance of the forward-biased EB junction is re = dVBE / dIE = V_T / IE. At room temperature where V_T ≈ 26 mV, re = 26 mV / IE (e.g., if IE = 2 mA, re = 13 ohms).
Q. 20 Electronics Engineering
Difficulty: Medium (1 Mark)
The transconductance (gm) of a BJT in terms of DC collector current IC and thermal voltage V_T is:
A
gm = IC / V_T
✓ Correct
B
gm = V_T / IC
C
gm = beta * IC
D
gm = 1 / re^2
💡 Step-by-Step Explanation & Concept Rationale
Transconductance gm = dIC / dVBE = IC / V_T. At 300 K with IC = 1 mA, gm = 1 mA / 26 mV ≈ 38.5 mS (or mA/V).
Q. 21 Electronics Engineering
Difficulty: Easy (1 Mark)
In a Common Emitter amplifier, connecting an emitter bypass capacitor CE in parallel with RE serves to:
A
Short-circuit AC signal currents to ground, preventing negative feedback at signal frequencies and significantly increasing AC voltage gain
✓ Correct
B
Filter DC supply ripple
C
Increase the DC input resistance
D
Prevent high-frequency oscillation
💡 Step-by-Step Explanation & Concept Rationale
RE provides essential DC bias stabilization. Without CE, RE also introduces negative AC feedback that lowers voltage gain to -RC / (re + RE). Adding CE bypasses AC signals around RE directly to ground, restoring voltage gain to -RC / re.
Q. 22 Electronics Engineering
Difficulty: Medium (1 Mark)
The upper cutoff frequency (fH) of a BJT amplifier is primarily limited by:
A
Internal junction capacitances (Cpi and Cmu) and stray wiring capacitances
✓ Correct
B
External coupling and bypass capacitors (CC1, CC2, CE)
C
DC supply voltage
D
Load resistor value only
💡 Step-by-Step Explanation & Concept Rationale
External coupling and bypass capacitors have large values (~microfarads) and determine the lower cutoff frequency fL. High-frequency rolloff fH is determined by internal depletion/diffusion junction capacitances (picofarads) magnified by the Miller effect.
Q. 23 Electronics Engineering
Difficulty: Medium (1 Mark)
The Miller Effect in an inverting amplifier of voltage gain Av transforms a feedback capacitance Cf connected between input and output into an equivalent input capacitance Cin_Miller equal to:
A
Cin_Miller = Cf * (1 - Av) = Cf * (1 + |Av|)
✓ Correct
B
Cin_Miller = Cf / (1 + Av)
C
Cin_Miller = Cf * Av^2
D
Cin_Miller = Cf
💡 Step-by-Step Explanation & Concept Rationale
By Miller's Theorem, feedback impedance between inverting nodes creates an equivalent input capacitance Cin = Cf * (1 - Av). For an inverting amplifier with Av = -100, Cin = Cf * (1 - (-100)) = 101 * Cf, severely loading the input and restricting high-frequency bandwidth.
Q. 24 Electronics Engineering
Difficulty: Hard (1 Mark)
A Cascode amplifier configuration consists of a Common Emitter (CE) stage feeding directly into a Common Base (CB) stage. Its primary benefit is:
A
Eliminating the Miller effect on the input stage, providing very high bandwidth and high input-to-output reverse isolation
✓ Correct
B
Operating with zero DC supply voltage
C
High current gain with unity voltage gain
D
Eliminating the need for transistors
💡 Step-by-Step Explanation & Concept Rationale
The low input impedance of the CB stage (rin_CB ≈ re) acts as the load for the CE stage, keeping the voltage gain of the CE stage at unity (Av_CE ≈ -1). Thus, Miller multiplication is eliminated (Cin ≈ 2*Cmu), allowing the cascode to amplify up to gigahertz frequencies.
Q. 25 Electronics Engineering
Difficulty: Medium (1 Mark)
The unity-gain cutoff frequency (fT) of a BJT is defined as the frequency at which the common-emitter short-circuit current gain (|hfe|) drops to:
A
Unity (1.0 or 0 dB)
✓ Correct
B
0.707 (3 dB down)
C
Beta / 2
D
Zero
💡 Step-by-Step Explanation & Concept Rationale
fT (transition frequency or gain-bandwidth product) is the frequency at which short-circuit current gain |beta(f)| drops to 1 (0 dB). It is related to transconductance and internal capacitances by fT = gm / [2 * pi * (Cpi + Cmu)].
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