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An pnp bipolar transistor has uniform doping of NE = 6 x 1017 cm-3, N= 2 x 1016  cm-3 and N= 5 x 1014 cm-3. The transistor is operating is inverse-active mode. The maximum VCB voltage, so that the low injection condition applies, is
  • a)
    0.86 V
  • b)
    0.48 V
  • c)
    0.32 V
  • d)
    0.60 V
Correct answer is option 'B'. Can you explain this answer?
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An pnp bipolar transistor has uniform doping of NE= 6 x 1017 cm-3, NB=...
Low injection limit is reached when



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An pnp bipolar transistor has uniform doping of NE= 6 x 1017 cm-3, NB=...
< b="" />Given Data:
NE = 6 x 10^17 cm^-3 (Emitter doping concentration)
NB = 2 x 10^16 cm^-3 (Base doping concentration)
NC = 5 x 10^14 cm^-3 (Collector doping concentration)

< b="" />Operating Mode:
The transistor is operating in inverse-active mode.

< b="" />Low Injection Condition:
In the inverse-active mode, the low injection condition is applied, which means that the minority carriers injected by the emitter into the base region are negligible compared to the minority carriers injected by the collector into the base region.

< b="" />Maximum VCB Voltage:
The maximum VCB voltage is the maximum voltage that can be applied between the collector and base of the transistor, while still maintaining the low injection condition.

< b="" />Analysis:
To find the maximum VCB voltage, we need to consider the doping concentrations in the base and collector regions.

In the low injection condition, the emitter-base junction is forward biased, and the collector-base junction is reverse biased.

The low injection condition is satisfied when the collector doping concentration (NC) is much smaller than the base doping concentration (NB).

From the given data, NB = 2 x 10^16 cm^-3 and NC = 5 x 10^14 cm^-3.

< b="" />Calculation:
To maintain the low injection condition, the condition NC < nb="" should="" be="" />

Comparing the given values, we have:
NC = 5 x 10^14 cm^-3 < nb="2" x="" 10^16="" />

Therefore, the low injection condition is satisfied.

< b="" />Conclusion:
In the given question, the maximum VCB voltage, so that the low injection condition applies, is 0.48 V. Therefore, the correct answer is option B.
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An pnp bipolar transistor has uniform doping of NE= 6 x 1017 cm-3, NB= 2 x1016cm-3 and NC= 5 x 1014cm-3. The transistor is operating is inverse-active mode. The maximum VCB voltage, so that the low injection condition applies, isa)0.86 Vb)0.48 Vc)0.32 Vd)0.60 VCorrect answer is option 'B'. Can you explain this answer?
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An pnp bipolar transistor has uniform doping of NE= 6 x 1017 cm-3, NB= 2 x1016cm-3 and NC= 5 x 1014cm-3. The transistor is operating is inverse-active mode. The maximum VCB voltage, so that the low injection condition applies, isa)0.86 Vb)0.48 Vc)0.32 Vd)0.60 VCorrect answer is option 'B'. Can you explain this answer? for Electronics and Communication Engineering (ECE) 2024 is part of Electronics and Communication Engineering (ECE) preparation. The Question and answers have been prepared according to the Electronics and Communication Engineering (ECE) exam syllabus. Information about An pnp bipolar transistor has uniform doping of NE= 6 x 1017 cm-3, NB= 2 x1016cm-3 and NC= 5 x 1014cm-3. The transistor is operating is inverse-active mode. The maximum VCB voltage, so that the low injection condition applies, isa)0.86 Vb)0.48 Vc)0.32 Vd)0.60 VCorrect answer is option 'B'. Can you explain this answer? covers all topics & solutions for Electronics and Communication Engineering (ECE) 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for An pnp bipolar transistor has uniform doping of NE= 6 x 1017 cm-3, NB= 2 x1016cm-3 and NC= 5 x 1014cm-3. The transistor is operating is inverse-active mode. The maximum VCB voltage, so that the low injection condition applies, isa)0.86 Vb)0.48 Vc)0.32 Vd)0.60 VCorrect answer is option 'B'. Can you explain this answer?.
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