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If for a silicon n-p-n transistor, the base-to-emitter voltage (VBE) is 0.7V and the collector-to-base voltage (VCB) is 0.2V. then the transistor is operating in the___
  • a)
    Normal active mode
  • b)
    Saturation mode
  • c)
    Inverse active mode
  • d)
    Cut-off mode
Correct answer is option 'A'. Can you explain this answer?
Verified Answer
If for a silicon n-p-n transistor, the base-to-emitter voltage (VBE) i...
VB = 0.7V [Forward-biased]
and VBC = -VCB = - 0.2V (Reverse-biased)
Hence, the transistor operates normal active mode.
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Most Upvoted Answer
If for a silicon n-p-n transistor, the base-to-emitter voltage (VBE) i...
Explanation:

To determine the operating mode of the transistor, we need to consider the voltages at the base, emitter, and collector.

Given:
- Base-to-emitter voltage (VBE) = 0.7V
- Collector-to-base voltage (VCB) = 0.2V

Operating Modes of a Transistor:
1. Normal Active Mode: In this mode, the transistor is used as an amplifier. The base-emitter junction is forward-biased, and the collector-base junction is reverse-biased. The transistor is in active mode when both junctions are biased.

2. Saturation Mode: In this mode, the transistor acts as a switch and is fully turned on. The base-emitter junction is forward-biased, and the collector-base junction is also forward-biased. The transistor is in saturation mode when both junctions are forward-biased.

3. Inverse Active Mode: In this mode, the transistor is used as an amplifier, but with reversed polarities. The base-emitter junction is reverse-biased, and the collector-base junction is forward-biased. The transistor is in inverse active mode when both junctions are biased.

4. Cut-off Mode: In this mode, the transistor is fully turned off and does not conduct current. Both the base-emitter junction and the collector-base junction are reverse-biased.

Determining the Operating Mode:
In our given scenario, the VBE (base-to-emitter voltage) is 0.7V, which indicates that the base-emitter junction is forward-biased. This eliminates the possibilities of inverse active mode and cutoff mode since both require a reverse-biased base-emitter junction.

The VCB (collector-to-base voltage) is 0.2V, which indicates that the collector-base junction is forward-biased. This eliminates the possibility of saturation mode since it requires a reverse-biased collector-base junction.

Therefore, the transistor is operating in the normal active mode because both the base-emitter junction and the collector-base junction are biased.

Conclusion:
The transistor is operating in the normal active mode because both the base-emitter voltage (VBE) and the collector-base voltage (VCB) indicate biased junctions.
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If for a silicon n-p-n transistor, the base-to-emitter voltage (VBE) is 0.7V and the collector-to-base voltage (VCB) is 0.2V. then the transistor is operating in the___a)Normal active modeb)Saturation modec)Inverse active moded)Cut-off modeCorrect answer is option 'A'. Can you explain this answer?
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