The transistor in the circuit shown is operating in

F1 Shubham.B 21-01-21 Savita D6

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  1. cut off region
  2. active region
  3. saturation region
  4. either in the active or saturation region

Answer (Detailed Solution Below)

Option 1 : cut off region
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Detailed Solution

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Concept:

Transistor operation regions

Either forward or reverse biasing is done to the emitter and collector junctions of the transistor. These biasing methods make the transistor circuit to work in four kinds of regions as Active region, Saturation region, Cutoff region, and the Inverse active region.

The below table shows the different operation regions of the transistor.

Emitter junction

Collector junction

Region of operation

Forward Bias

Forward Bias

Saturation

Forward Bias

Reverse Bias

Active

Reverse Bias

Forward Bias

Inverse Active

Reverse Bias

Reverse Bias

Cutoff

 

NPN and PNP transistors schematic are shown below:

F1 ShubhamB Madhuri 12.03.2021 D16

Calculation:

Consider the given transistor and applying the KVL we get:

F1 ShubhamB Madhuri 12.03.2021 D17

- 5 + IE(5k) + 0.7 + IB(100K) + 5 = 0

We know that IE = (1 + β)IB

(1 + β)IB(5k) + IB(100) = - 0.7

IB < 0

The above equation implies that the base current is opposite to the traditional PNP transistor so the emitter junction is reverse biased.

The collector voltage is zero and this junction also reverse bias due to the grounding of the collector terminal.

Conclusion: Both the junctions are Reverse Bias so the transistor is in the cut-off.

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