Examoogle
ExamsTest SeriesCBATRank CheckPrevious Year PapersPassBook StoreMy BooksAI Tutor
ЁЯЫТ0
рдЕA
Examoogle

India's most trusted platform for competitive exam PDF books. Expert-authored, watermark-protected, instant access.

Exams & Practice
All Exams & SyllabusMock Test SeriesPrevious Year PapersPractice Questions (MCQs)Recruitment Notifications
Quick Links
Examoogle AI TutorExam NewsBook StoreMy BooksLogin / Sign Up
Support
About UsRefund PolicyPrivacy PolicyTerms of UseContact Us
┬й 2026 Examoogle. India's #1 competitive exam AI tutor.
ЁЯФТ SSL SecuredЁЯУ▒ UPI AcceptedЁЯз╛ GST Invoice
Examoogle

Join 60,000+ competitive exam aspirants

or with email
By continuing, you agree to ourTerms of Service&Privacy Policy
Your Cart
SubtotalтВ╣0
TotalтВ╣0
Examoogle тАв User тАв info@examoogle.com тАв EE-2024-8821
Chapter 1 of 12 тАв Page 1 of 248ЁЯФТ Protected PDF тАв Watermarked
Back to Practice Questions
ElectricalElectronics
PrevNext

The base of a transistor is _____ doped.

A

Heavily

B

Moderately

C

Lightly

D

None

Correct Answer

тЪЩя╕П TE тАв Technical Concept & PrincipleElectricalElectronics
Option C

Lightly

Quick Summary:

The base of a Bipolar Junction Transistor (BJT) is intentionally made very thin and lightly doped compared to the emitter and collector regions. This specific design ensures that the majority of charge carriers injected from the emitter pass through the base to the collector with minimal recombination.

тЪЩя╕ПTETechnical SolutionConcept & Principle
ЁЯТб Explanation

The base of a Bipolar Junction Transistor (BJT) is intentionally made very thin and lightly doped compared to the emitter and collector regions. This specific design ensures that the majority of charge carriers injected from the emitter pass through the base to the collector with minimal recombination.

ЁЯФв Key Formulas

IE=IB+ICI_E = I_B + I_CIEтАЛ=IBтАЛ+ICтАЛ

╬▒=ICIE\alpha = \frac{I_C}{I_E}╬▒=IEтАЛICтАЛтАЛ

тЪЩя╕П Working Principle

In an NPN transistor, the base is P-type. Because it is lightly doped, the density of holes (the minority carriers in the base) is very low. This reduces the probability of electrons (injected from the emitter) recombining with holes in the base, which allows approximately 95% to 99% of the emitter current to reach the collector as the collector current (ICI_CICтАЛ).

ЁЯУМ Key Points
  • тЦ╕

    Base width is kept very thin to minimize recombination.

  • тЦ╕

    Emitter is the most heavily doped region to maximize charge injection.

  • тЦ╕

    Collector has the largest physical area to dissipate heat.

  • тЦ╕

    Light doping in the base ensures high current gain (╬▓\beta╬▓).

тЬЕ Advantages
  • тЦ╕

    High current amplification factor (╬▓\beta╬▓)

  • тЦ╕

    Efficient charge carrier transport

тЭМ Disadvantages / Limitations
  • тЦ╕

    Susceptibility to punch-through if base is too thin

  • тЦ╕

    High temperature sensitivity

ЁЯЫая╕П Applications / Uses
  • тЦ╕

    Switching circuits

  • тЦ╕

    Signal amplification

ЁЯФД Comparison Table
FeatureEmitterBase

Doping Level

Heavily

Lightly

ЁЯУД Additional Information
  • тЦ╕

    The emitter is heavily doped to provide a large number of majority carriers.

  • тЦ╕

    The collector is moderately doped and physically larger than the emitter to improve power handling capacity.

  • тЦ╕

    Option B (Moderately) is incorrect as it describes the collector region, not the base.

ЁЯУК Diagram / Illustration
BJT Doping ProfileEmitterHeavily DopedBaseLightly DopedCollectorModerately Doped
тЬЕ

C is correct тАФ The base of a BJT is lightly doped to minimize the recombination of charge carriers injected from the emitter.

Core Concepts Used
Click any tag to open in AI Tutor
BJT Structure Doping Concentration Charge Carrier Recombination
ЁЯТб EXAM TIP

Always remember the mnemonic: 'Emitter-Heavy, Base-Light, Collector-Moderate/Large' for BJT design parameters in exams.

Related Questions

ElectricalElectronics
Calculate the critical current through a long thin superconducting wire of radius 0.5 mm. The critical magnetic field is 7.2 kA/m
ElectricalElectronics
A superconductor tin has a critical temperature of 3.7 K in zero magnetic field and a critical field of 0.0306 T at 0 K. Find the critical field at 2
ElectricalElectronics
The critical temperature for a metal with isotopic mass 199.5 is 4.185 K. Calculate the isotopic mass if the critical temperature falls to 4.133 K
ElectricalElectronics
For mercury of mass number 202, Tc is 4.2 K. Find the transition temperature for its isotope of mass number 200.
ElectricalElectronics
The penetration depth is the __________ where the current drops to 1/e times of its value at the surface.

Discussion (0)

Loading discussion...
PrevNext