Chapter 24: Semiconductor Devices and Electronic Circuits (Set-5)
A silicon diode carries 1 mA at 0.62 V. Assuming n = 1, Vt = 26 mV, the current at 0.68 V is closest to
A 2 mA
B 100 mA
C 10 mA
D 1 A
ΔV = 0.06 V. Ratio = exp(ΔV/Vt) = exp(0.06/0.026) ≈ exp(2.31) ≈ 10. So current ≈ 1 mA × 10²? Actually exp(2.31) ≈ 10, giving 10 mA—wait check: 0.68−0.62 = 0.06; exp(2.31)=10 → 10 mA. Closest option: B.
A Zener regulator has Vs = 12 V, Vz = 6 V, series resistor 300 Ω, and load current 10 mA. Zener current is
A 0 mA
B 10 mA
C 20 mA
D 30 mA
Series current Is = (Vs−Vz)/Rs = (12−6)/300 = 6/300 = 0.02 A = 20 mA. Load takes 10 mA, so remaining is Zener current: Iz = 20−10 = 10 mA. Correct should be B.
A center-tap full-wave rectifier gives secondary half-winding peak Vm = 10 V. PIV of each diode is
A 20 V
B 10 V
C 14 V
D 40 V
In center-tap full-wave rectifier, when one diode conducts at +Vm, the other diode is reverse-biased by approximately 2Vm. Hence PIV ≈ 2Vm = 20 V for each diode.
A tunnel diode is biased for oscillation. It must be kept in
A Forward ohmic region
B Reverse saturation region
C Breakdown region
D Negative resistance region
Oscillation requires a device region that can provide negative resistance to compensate circuit losses. Tunnel diodes provide this between peak and valley points, enabling microwave and high-speed oscillators.
A solar cell has Voc = 0.60 V, Isc = 2.0 A, fill factor 0.75. Maximum power is
A 0.60 W
B 0.90 W
C 1.20 W
D 1.50 W
Maximum power Pmax = Voc × Isc × FF = 0.60 × 2.0 × 0.75 = 0.90 W. This uses the definition of fill factor relating max rectangle power to Voc and Isc.
For a CE BJT, if α = 0.99, β is closest to
A 9
B 49
C 99
D 199
β = α/(1−α) = 0.99/0.01 = 99. So correct should be C. This shows β becomes very large as α approaches 1 in common-base action.
In MOSFET saturation (long-channel), the condition is
A Vds < Vgs
B Vds ≥ Vgs−Vt
C Vds = 0
D Vgs < Vt
Saturation occurs when drain voltage is high enough to pinch off the channel at drain end. For enhancement MOSFET, this begins when Vds reaches or exceeds (Vgs − Vt).
In a CE amplifier, if emitter bypass capacitor is removed, the voltage gain will
A Increase greatly
B Become zero
C Decrease
D Change polarity
Without bypass capacitor, emitter resistor provides negative feedback for AC too. This increases effective emitter resistance, reduces small-signal gain, improves linearity, and stabilizes bias but lowers amplification.
An RC-coupled amplifier has fL = 100 Hz and fH = 100 kHz. Bandwidth is
A 100 kHz
B 100.1 kHz
C 1 kHz
D 99.9 kHz
Bandwidth = fH − fL = 100,000 − 100 = 99,900 Hz = 99.9 kHz. This is the useful flat-gain frequency range between the two cutoff points.
If open-loop gain A = 10,000 and feedback factor β = 0.01, closed-loop gain is closest to
A 50
B 99
C 90
D 100
Closed-loop gain Af = A/(1 + Aβ) = 10,000/(1 + 100) = 10,000/101 ≈ 99. This shows feedback makes gain depend mainly on 1/β when Aβ is large.
In a bridge rectifier, if each diode drop is 0.7 V, peak load voltage for Vm = 12 V is
A 10.6 V
B 12 V
C 11.3 V
D 9.6 V
In a bridge, two diodes conduct in series each half-cycle. So peak load ≈ Vm − 2×0.7 = 12 − 1.4 = 10.6 V. This is a practical correction vs ideal case.
A diode clipper uses a 5.1 V Zener in reverse. The output peak is limited near
A 0.7 V
B 10.2 V
C 5.1 V
D 12 V
When output tries to exceed about the Zener breakdown voltage, the Zener conducts and clamps the voltage. Practical limit is close to 5.1 V (slightly affected by dynamic resistance and current).
A JFET has Idss = 10 mA and Vp = −4 V. At Vgs = −2 V, Id is
A 5.0 mA
B 2.5 mA
C 7.5 mA
D 10 mA
Shockley equation: Id = Idss(1 − Vgs/Vp)². Here Vgs/Vp = (−2)/(−4)=0.5. So Id = 10(1−0.5)²=10(0.5)²=2.5 mA. Correct should be A.
A CE transistor has Ic = 2 mA and β = 100. Base current is
A 2 μA
B 200 μA
C 2 mA
D 20 μA
β = Ic/Ib ⇒ Ib = Ic/β = 2 mA/100 = 0.02 mA = 20 μA. This small base current controlling larger collector current demonstrates current amplification.
A solar cell’s efficiency increases most directly when
A Temperature rises
B Fill factor rises
C Series resistance rises
D Shunt falls
Efficiency depends on maximum power output. Increasing fill factor raises Pmax = Voc×Isc×FF for same Voc and Isc. Higher series resistance or lower shunt resistance typically reduces fill factor and efficiency.
A CE amplifier shows upper cutoff reduced mainly by
A Miller capacitance
B Coupling capacitor
C Emitter resistor
D Zener effect
In CE, capacitance between collector and base appears multiplied at input due to inversion. This “Miller effect” increases effective input capacitance, lowering high-frequency gain and reducing upper cutoff frequency.
In negative feedback, if Aβ = 9, gain reduction factor is
A 1
B 5
C 10
D 9
Closed-loop gain Af = A/(1 + Aβ). The gain reduction factor is (1 + Aβ) = 1 + 9 = 10. Feedback reduces gain by this factor while improving stability and bandwidth.
For a half-wave rectifier with Vm = 20 V, ideal average DC output is
A 12.7 V
B 6.37 V
C 20 V
D 31.8 V
Average DC for half-wave is Vm/π. So 20/π ≈ 6.37 V. This assumes ideal diode and resistive load, ignoring diode drop and transformer resistances.
For a full-wave rectifier with Vm = 20 V, ideal average DC output is
A 6.37 V
B 20 V
C 31.8 V
D 12.7 V
Average DC for full-wave is 2Vm/π. So 40/π ≈ 12.7 V. Full-wave provides roughly double average output compared to half-wave for the same peak input.
If diode current follows I = Is e^(V/Vt), dynamic resistance at current I is
A V/I
B I/Vt
C Vt/I
D Is/Vt
From I = Is e^(V/Vt), differentiate: dI/dV = I/Vt. Hence dynamic resistance rd = dV/dI = Vt/I. At higher current, rd becomes smaller, matching diode behavior.
In a Zener regulator, line regulation improves when series resistor is
A Decreased
B Increased
C Removed
D Shorted
A larger series resistor reduces change in current for a given change in input voltage, helping keep Zener voltage more constant. But too large a resistor can worsen load regulation and limit current.
In a Zener regulator, load regulation improves when Zener dynamic resistance is
A High
B Infinite
C Low
D Negative
Low dynamic resistance means Zener voltage changes very little with current variation. When load current changes, Zener current adjusts; low rd ensures output voltage stays nearly constant.
In a BJT, leakage current mainly increases with
A Decreasing temperature
B Increasing temperature
C Increasing load
D Decreasing Vcc
Collector-base leakage current rises with temperature due to increased thermal generation of carriers. This can shift the Q-point and may contribute to thermal runaway if biasing is not stabilized.
For a CE amplifier, input impedance increases most by using
A Emitter follower
B Collector follower
C Common base stage
D Zener clamp
An emitter follower (common collector) offers very high input impedance and low output impedance. Placing it as a buffer at input prevents loading of the signal source, improving input impedance seen by source.
A MOSFET is ESD-sensitive mainly due to
A Thick depletion
B Low mobility
C Thin gate oxide
D High resistance
MOSFET gate oxide is extremely thin and can break down under high static voltages. Electrostatic discharge can puncture oxide, permanently damaging device, so careful handling and protection are needed.
In a diode, avalanche breakdown is favored by
A Narrow depletion
B Wide depletion
C High doping
D Low voltage
Wide depletion region occurs in lightly doped junctions. Breakdown then occurs at higher reverse voltage where carriers gain energy and cause impact ionization, triggering avalanche multiplication.
A diode’s reverse recovery time matters most in
A DC measurement
B Zener reference
C LCD display
D High-speed switching
Reverse recovery is the time a diode takes to stop conducting after switching from forward to reverse bias, due to stored charge. In fast switching circuits, slow recovery causes losses and waveform distortion.
In a bridge rectifier, if load current is 1 A, each conducting diode current is
A 0.5 A
B 2 A
C 1 A
D 4 A
In each half-cycle, two diodes conduct in series and carry the full load current. Therefore each conducting diode carries the same current as load, i.e., 1 A, not half.
A CE amplifier without proper bias may distort first by entering
A Active region
B Cutoff or saturation
C Zener breakdown
D Pinch-off only
If Q-point is not centered, signal swings can push transistor into cutoff (no conduction) or saturation (both junctions forward biased). Output clips on one side, causing strong distortion.
In an RC-coupled amplifier, low-frequency gain drops mainly because
A Capacitors short
B β increases
C gm increases
D Capacitors open
At low frequencies, coupling and bypass capacitors have high reactance, acting closer to open circuits. This reduces signal coupling between stages and reduces bypassing, causing gain to fall below midband value.
In CE amplifier, power gain is high mainly because it provides
A Both gains
B High voltage only
C High current only
D No gain
CE configuration provides significant voltage gain and current gain simultaneously. Power gain equals voltage gain times current gain, so CE amplifiers can deliver large power gain compared to CB or CC.
A JFET as voltage-controlled resistor operates mainly in
A Breakdown region
B Saturation region
C Ohmic region
D Cutoff region
For small Vds, JFET channel behaves resistively and drain current changes approximately linearly with Vds. In this ohmic region, varying Vgs changes channel resistance, enabling analog control.
In MOSFET switching, “on-resistance” mainly affects
A Leakage current
B Conduction loss
C Threshold drift
D Optical output
When MOSFET is ON, it behaves like a resistor Rds(on). Power loss is I²Rds(on). Lower on-resistance reduces conduction losses and heating, important in power electronics.
A solar panel series connection mainly increases
A Current
B Fill factor
C Irradiance
D Voltage
Connecting solar cells/panels in series adds their voltages while current remains roughly the same as one cell string. Parallel connection increases current. Series strings are used to reach required output voltage.
A solar panel parallel connection mainly increases
A Voltage
B Current
C Band gap
D Voc only
Parallel connection adds currents from each panel while keeping voltage nearly the same. This increases available current and power capability, useful when load requires higher current at a fixed voltage.
If a Zener diode has Vz = 5 V and Iz changes by 10 mA, and rd = 5 Ω, voltage change is
A 0.5 V
B 5 V
C 0.05 V
D 10 V
Voltage change due to dynamic resistance: ΔV = rd × ΔI = 5 Ω × 0.01 A = 0.05 V. Lower rd gives better voltage stability during current variations.
For a diode, small-signal resistance at 10 mA (n=1, Vt=26 mV) is closest to
A 2.6 Ω
B 26 Ω
C 260 Ω
D 0.26 Ω
Small-signal rd ≈ nVt/I. Here rd ≈ 0.026/0.01 = 2.6 Ω. This shows diode incremental resistance becomes small at higher forward currents.
In a BJT, if β increases with temperature, Q-point may shift toward
A Lower Ic
B Higher Ic
C Zero Vcc
D Lower Vbe
If biasing depends on β, higher β causes larger collector current for the same base bias. This shifts Q-point, can increase heating, and may lead to thermal runaway unless stabilized by feedback.
Negative feedback reduces distortion mainly by making amplifier
A More nonlinear
B More noisy
C More unstable
D More linear
Negative feedback counters output changes by feeding back an opposing signal. This reduces gain variations and nonlinearity, so the amplifier response becomes more linear, lowering harmonic and intermodulation distortion.
If feedback increases bandwidth, the gain–bandwidth tradeoff means
A Both rise always
B Noise disappears
C Gain drops
D Output clips more
Negative feedback reduces midband gain but extends frequency response, often keeping gain–bandwidth roughly constant for a given amplifier. This tradeoff is useful for stable wideband amplification.
In a CE amplifier, input coupling capacitor mainly affects
A Upper cutoff
B Lower cutoff
C Output phase
D Power rating
Input coupling capacitor forms a high-pass network with input resistance. At low frequency, its reactance is large, reducing input signal amplitude, so it primarily sets the lower cutoff frequency.
The most correct reason bridge rectifier does not need center tap is
A Uses four diodes
B No transformer used
C Uses Zener clamp
D Uses both halves
Bridge rectifier arranges diodes so that during both half-cycles, current through the load flows in the same direction using the full secondary winding. Hence no center-tapped secondary is required.
In a diode OR gate, output is HIGH when
A All inputs LOW
B Only two HIGH
C Any input HIGH
D Only one LOW
With suitable pull-down resistor, a HIGH input forward-biases its diode and raises output. If all inputs are LOW, no diode conducts and output remains LOW. This performs OR logic.
A clamp circuit needs capacitor mainly to
A Store charge
B Increase gain
C Reduce PIV
D Raise Vt
In a clamper, capacitor charges to a reference level during one half-cycle and then holds that charge, shifting the waveform DC level. Without capacitor charge storage, clamping action would not occur.
If an LED is reverse-biased beyond rating, it may fail because of
A Light saturation
B Junction breakdown
C Low current
D High band gap
LEDs are not designed for large reverse voltage. Excess reverse bias can cause breakdown and overheating, damaging the junction permanently. Hence reverse voltage rating must be respected.
In LCD, polarizers are used mainly to
A Produce light
B Increase current
C Reduce heat
D Control polarization
Polarizers allow only light of a particular polarization to pass. Liquid crystal layer rotates polarization under field. With two polarizers, this rotation controls brightness, creating visible segments or pixels.
In a solar cell, maximum power point tracking is needed because
A Voc is constant
B Light never changes
C Load varies
D Temperature fixed
Maximum power point depends on illumination, temperature, and load. If load changes, operating point moves on I–V curve away from maximum power. MPPT adjusts effective load to keep operation near peak power.
A transistor h-parameter model is used mainly as
A Thermal model
B Two-port model
C Optical model
D Mechanical model
h-parameters represent a transistor as a linear two-port network for small-signal analysis. They relate input voltage/current and output voltage/current, helping calculate gain, impedance, and feedback effects in amplifier circuits.
For stable voltage reference, which Zener property is most important?
A Low rd
B High rd
C High ripple
D High leakage
A good reference should maintain nearly constant voltage despite current changes. Low dynamic resistance means small voltage variation for changes in Zener current, improving stability and regulation quality.
In an op-amp with negative feedback, the most accurate statement is
A V+ ≫ V−
B V+ ≈ V−
C Input currents large
D Gain equals A
With very large open-loop gain and negative feedback, op-amp drives output to make the differential input voltage nearly zero. Thus non-inverting and inverting inputs stay at almost the same potential in linear operation.