A PNP transistor is operating in active region. If the emitter voltage decreases such that the emitter becomes less positive than the base while the biasing condition of the collector-base junction remains unchanged, the transistor will enter _______.
Cut-off region
To solve this question, we need to understand the operating regions of a PNP transistor and how changes in voltages affect these regions.
PNP Transistor Basic Operation
A PNP transistor in active mode typically has the emitter-base junction forward-biased and the collector-base junction reverse-biased. These conditions facilitate amplification.
- Active Region: Emitter-Base junction is forward-biased, Collector-Base junction is reverse-biased. The transistor amplifies the input signal.
- Cut-off Region: Both the Emitter-Base and Collector-Base junctions are reverse-biased. The transistor is effectively off, with no current flowing through.
- Saturation Region: Both junctions are forward-biased. The transistor conducts maximum current, but there's no amplification.
- Reverse Active Region: Emitter-Base junction is reverse-biased, Collector-Base junction is forward-biased. This is an unusual mode, with reverse direction current flow.
Analysis of Given Situation
The question states: "If the emitter voltage decreases such that the emitter becomes less positive than the base while the biasing condition of the collector-base junction remains unchanged."
This suggests that the forward-bias condition of the emitter-base junction is lost. With the emitter becoming less positive relative to the base, the emitter-base junction can become reverse-biased. Since the collector-base junction is mentioned to remain unchanged, it stays reverse-biased.
Conclusion
With both the emitter-base and collector-base junctions reverse-biased, the PNP transistor enters the Cut-off Region.
Therefore, the correct answer is: Cut-off region.
Similar Questions
In a properly biased PNP transistor operating in active region, the emitter terminal is _______.
A PNP transistor is fabricated with an increased base doping concentration while keeping all other physical dimensions and doping levels constant. What is the effect on the transistor's performance?
Important Questions from Bipolar Junction Transistor
Find $I_C$, $V_{BE}$=0.7V.
- Voltage gain of common emitter amplifier with $R_E$ is given as :
Match the following :
List – I (Biasing of BJT) List – II (Functions) a. E-B junction forward bias and C-B junction reverse bias i. Very low gain amplifier b. Both E-B and C-B junctions forward bias ii. Saturation condition c. E-B junction reverse bias and C-B junction forward bias iii. High gain amplifier d. Both E-B and C-B junctions reverse bias iv. Cut-off condition Codes :
The doping concentration of n-p-n transistor are
i. 5 × 1018 / cm3
ii. 1017 / cm3
iii. 2 × 107 / cm3Identify the regions in the above order i, ii and iii.
For the BJT to be biased in its linear or active operating region :
(a) the base-emitter Junction must be forward biased
(b) the base-collector Junction must be reverse biased
(c) the base-emitter Junction must be reverse biased
(d) the base-collector Junction must be forward biased
