Bipolar Junction Transistors Questions and Answers

Q: What penalty does heavy overdrive impose on a switching transistor?

Overdrive fills the base with excess stored charge, and that charge must be removed before the device can begin to turn off, which appears as storage delay. The saturation voltage actually falls slightly with more drive, and neither leakage nor breakdown voltage is affected. The remedy is a Schottky clamp or a speed-up capacitor, both of which prevent the device from saturating deeply - a good example of designing around a limit rather than accepting it.

Q: What does a speed-up capacitor across the base resistor achieve?

The capacitor passes a burst of current at each edge, charging and then removing base charge quickly, while the resistor sets the steady drive. It does not reduce the steady current, offers no reverse protection, and cannot raise the available collector current. Giving a circuit a different response to transitions than to steady conditions is a recurring trick, and a capacitor across a resistor is its simplest form.

Q: Which breakdown rating is usually the lowest of a transistor's three?

The emitter-base junction breaks down at only five to seven volts, because the emitter is heavily doped, and exceeding it degrades beta permanently even without destroying the device. Collector-base breakdown is the highest and collector-emitter falls between. Reverse voltage on the base-emitter junction is a common and easily overlooked hazard, and it is why a protective diode often appears across it.

Q: What happens to a transistor's beta after emitter-base breakdown?

Avalanche in the emitter junction damages the surface, so beta falls permanently even if the current was limited and the device still functions. That silent degradation is what makes the rating so treacherous: nothing fails outright, but a circuit's gain drifts and the cause is invisible. Any stress that degrades rather than destroys is worth designing against explicitly, because it will not announce itself.

Q: Why does a transistor's beta vary with collector current?

Beta peaks at a moderate current: at low current recombination in the base takes a larger share, and at high current emitter crowding and high-level injection reduce the efficiency. The Early voltage and the thermal voltage are not current dependent in this way. That bell-shaped curve is why data sheets specify beta at a stated current, and why designs are placed near the peak where the parameter is least sensitive.

Q: Why does beta rise with temperature?

Longer carrier lifetime means less recombination in the base, so a larger fraction of the injected current reaches the collector and beta rises - typically half a per cent per degree. The falling base-emitter voltage is a separate temperature effect that also raises collector current, and leakage adds to the collector current rather than the base. Two independent mechanisms both pushing the current up is precisely why thermal runaway is a real risk without stabilising feedback.

Q: A transistor dissipates \( 2 \ \mathrm{W} \) with a junction-to-case resistance of \( 3 \ \mathrm{^\circ C/W} \) and a case at \( 60 \ \mathrm{^\circ C} \). Its junction reaches

Adding the rise across the junction-to-case path:\[ T_{j} = 60 + (2 \times 3) = 66 \ \mathrm{^\circ C} \]giving 66 degrees. The value 63 uses one watt and 126 doubles the case temperature. Thermal problems are all the same series circuit with temperature in place of voltage and power in place of current, which makes them easy once the analogy is accepted.

Q: A device rated \( 10 \ \mathrm{W} \) at \( 25 \ \mathrm{^\circ C} \) derates linearly to zero at \( 150 \ \mathrm{^\circ C} \). At \( 75 \ \mathrm{^\circ C} \) it may dissipate

Interpolating along the derating line:\[ \begin{aligned} P &= 10 \times \frac{150-75}{150-25} \\ &= 10 \times 0.6 = 6 \ \mathrm{W} \end{aligned} \]so 6 watts. Five watts would be the midpoint of the temperature range rather than of the power range. A rating quoted at one temperature is nearly always a straight line to zero at the limit, and reading it that way is safer than assuming the headline figure applies.

Q: Assertion (A): Assertion: A transistor's beta is a poor parameter to base a design on. Reason (R): Reason: Beta varies widely between devices and with temperature and current. Select the correct answer from the options below.

Both are true and the second is the reason: beta can vary three to one across a production batch and drifts with temperature and operating current, so any performance figure proportional to it is unpredictable. The remedy is a circuit whose behaviour depends on resistor ratios instead, which is exactly what emitter degeneration and negative feedback achieve. Designing so that the least reliable parameter cancels out is the general lesson. Checking the units of each quantity catches a wrong formula before the arithmetic begins.

41.

What penalty does heavy overdrive impose on a switching transistor?

A.A higher saturation voltage when conducting
B.A greater leakage current in the off state
C.A reduced collector-emitter breakdown voltage
D.A longer storage time at turn-off

Answer: D

Overdrive fills the base with excess stored charge, and that charge must be removed before the device can begin to turn off, which appears as storage delay. The saturation voltage actually falls slightly with more drive, and neither leakage nor breakdown voltage is affected. The remedy is a Schottky clamp or a speed-up capacitor, both of which prevent the device from saturating deeply - a good example of designing around a limit rather than accepting it.

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42.

What does a speed-up capacitor across the base resistor achieve?

A.A brief surge of drive at each transition
B.A reduction in the base current while steady
C.Protection of the base-emitter junction from reverse voltage
D.A rise in the collector current available at turn-on

Answer: A

The capacitor passes a burst of current at each edge, charging and then removing base charge quickly, while the resistor sets the steady drive. It does not reduce the steady current, offers no reverse protection, and cannot raise the available collector current. Giving a circuit a different response to transitions than to steady conditions is a recurring trick, and a capacitor across a resistor is its simplest form.

Discuss in Forum

43.

Which breakdown rating is usually the lowest of a transistor's three?

A.They are all specified equal
B.Emitter-base breakdown
C.Collector-emitter breakdown
D.Collector-base breakdown

Answer: B

The emitter-base junction breaks down at only five to seven volts, because the emitter is heavily doped, and exceeding it degrades beta permanently even without destroying the device. Collector-base breakdown is the highest and collector-emitter falls between. Reverse voltage on the base-emitter junction is a common and easily overlooked hazard, and it is why a protective diode often appears across it.

Discuss in Forum

44.

What happens to a transistor's beta after emitter-base breakdown?

A.It is unaffected provided the current was limited
B.It becomes strongly dependent on collector voltage
C.It falls and does not recover
D.It rises because carriers are multiplied

Answer: C

Avalanche in the emitter junction damages the surface, so beta falls permanently even if the current was limited and the device still functions. That silent degradation is what makes the rating so treacherous: nothing fails outright, but a circuit's gain drifts and the cause is invisible. Any stress that degrades rather than destroys is worth designing against explicitly, because it will not announce itself.

Discuss in Forum

45.

Why does a transistor's beta vary with collector current?

A.The Early voltage changes with the operating current
B.The thermal voltage falls as the current rises
C.The base spreading resistance rises with current
D.Recombination dominates at low current and crowding at high

Answer: D

Beta peaks at a moderate current: at low current recombination in the base takes a larger share, and at high current emitter crowding and high-level injection reduce the efficiency. The Early voltage and the thermal voltage are not current dependent in this way. That bell-shaped curve is why data sheets specify beta at a stated current, and why designs are placed near the peak where the parameter is least sensitive.

Discuss in Forum

46.

Why does beta rise with temperature?

A.Carrier lifetime in the base increases
B.The base-emitter voltage falls with temperature
C.Leakage current adds directly to the base current
D.The collector-base capacitance falls as it heats

Answer: A

Longer carrier lifetime means less recombination in the base, so a larger fraction of the injected current reaches the collector and beta rises - typically half a per cent per degree. The falling base-emitter voltage is a separate temperature effect that also raises collector current, and leakage adds to the collector current rather than the base. Two independent mechanisms both pushing the current up is precisely why thermal runaway is a real risk without stabilising feedback.

Discuss in Forum

47.

Which two temperature effects both tend to increase collector current?

A.Falling Early voltage and rising base resistance
B.Rising beta and falling base-emitter voltage
C.Rising beta and rising saturation voltage
D.Falling leakage and rising transconductance

Answer: B

Beta rises and the base-emitter drop falls, and both changes increase the collector current at a fixed bias - which, since more current means more heat, closes a positive feedback loop. Leakage rising is a third contributor in the same direction. That all the significant temperature effects push the same way is what makes bias stabilisation the central concern of the next topic rather than an afterthought.

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48.

A transistor dissipates \( 2 \ \mathrm{W} \) with a junction-to-case resistance of \( 3 \ \mathrm{^\circ C/W} \) and a case at \( 60 \ \mathrm{^\circ C} \). Its junction reaches

A.126 degrees
B.63 degrees
C.66 degrees
D.60 degrees

Answer: C

Adding the rise across the junction-to-case path:

\[ T_{j} = 60 + (2 \times 3) = 66 \ \mathrm{^\circ C} \]

giving 66 degrees. The value 63 uses one watt and 126 doubles the case temperature. Thermal problems are all the same series circuit with temperature in place of voltage and power in place of current, which makes them easy once the analogy is accepted.

Discuss in Forum

49.

A device rated \( 10 \ \mathrm{W} \) at \( 25 \ \mathrm{^\circ C} \) derates linearly to zero at \( 150 \ \mathrm{^\circ C} \). At \( 75 \ \mathrm{^\circ C} \) it may dissipate

A.5 W
B.10 W
C.3.3 W
D.6 W

Answer: D

Interpolating along the derating line:

\[ \begin{aligned} P &= 10 \times \frac{150-75}{150-25} \\ &= 10 \times 0.6 = 6 \ \mathrm{W} \end{aligned} \]

so 6 watts. Five watts would be the midpoint of the temperature range rather than of the power range. A rating quoted at one temperature is nearly always a straight line to zero at the limit, and reading it that way is safer than assuming the headline figure applies.

Discuss in Forum

50.

Assertion (A): Assertion: A transistor's beta is a poor parameter to base a design on.

Reason (R): Reason: Beta varies widely between devices and with temperature and current.

Select the correct answer from the options below.

A.Both A and R are true, and R is the correct explanation of A
B.Both A and R are true, but R is not the correct explanation of A
C.A is true but R is false
D.A is false but R is true

Answer: A

Both are true and the second is the reason: beta can vary three to one across a production batch and drifts with temperature and operating current, so any performance figure proportional to it is unpredictable. The remedy is a circuit whose behaviour depends on resistor ratios instead, which is exactly what emitter degeneration and negative feedback achieve. Designing so that the least reliable parameter cancels out is the general lesson. Checking the units of each quantity catches a wrong formula before the arithmetic begins.

Discuss in Forum

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