IndiaBIX

Electronics and Communication Engineering - Electromagnetic Field Theory

Exercise : Electromagnetic Field Theory - Section 3
41.
In a bipolar transistor
recombination in base regions of both n-p-n and p-n-p transistor is low
recombination in base regions of both n-p-n and p-n-p transistors is high
recombination in base region of n-p-n transistor is low but that in p-n-p transistor is high
recombination in base region of p-n-p transistor is low but that in n-p-n transistor is high
Answer: Option
Explanation:

Base is very thin and therefore recombination is minimum in both p-n-p and n-p-n transistors.

42.
If for a silicon n-p-n transistor, the base to emitter voltage (VBE) is 0.7 V and the collector to base voltage VCB is 0.2 Volt, then the transistor is operating in the
normal active mode
saturation mode
inverse active mode
cut off mode
Answer: Option
Explanation:

Transistor will operate in active mode because

VBE = 0.7 volt, (Base emitter junction is forward biased)

VCB = - VBC = - 0.2 V (Base to collector junction is reverse biased).

43.
The number of doped regions in a bipolar junction transistor is
1
2
3
4
Answer: Option
Explanation:

p, n, p or n, p, n.

44.
Donor energy level is n type semiconductor is very near valence band.
True
False
Answer: Option
Explanation:

It is near conduction band.

45.
GaAs has an energy gap 1.43 eV the optical cut off wavelength of GaAs would lie in the
visible region of the spectrum
infrared region of the spectrum
ultraviolet region of the spectrum
for ultraviolet region of the spectrum
Answer: Option
Explanation:

GaAs has very large band gap and high carrier mobility.

Quick links
Quantitative Aptitude
Verbal (English)
Reasoning
Programming
Interview
Placement Papers