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In an NPN transistor circuit, the collector current is 10 mA. If 90% of the electrons emitted reach the collector, the emitter current (iE) and base current (iB) are given by 
(1) iE = –1 mA, iB = 9 mA

(2) iE = 9 mA, iB = – 1 mA

(3) iE = 1 mA, iB = 11 mA

(4) iE = 11 mA, iB = 1 mA

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In a common emitter transistor, the current gain is 80. What is the change in collector current, when the change in base current is 250 μ
(1) 80 × 250 μA

(2) (250 – 80) μA

(3) (250 + 80) μA

(4) 250/80 μA

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Level 1: 80%+
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Least doped region in a transistor 

(1) Either emitter or collector

(2) Base

(3) Emitter

(4) Collector

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The transistors provide good power amplification when they are used in 

(1) Common collector configuration

(2) Common emitter configuration

(3) Common base configuration

(4) None of these

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The transfer ratio of a transistor is \(50\). The input resistance of the transistor when used in the common-emitter configuration is\(1~\text{K}\Omega\). The peak value for an AC input voltage of \(0.01~\text{V}\) peak is:
1. \(100~\mu\text{A}\)
2. \(0.01~\text{mA}\)
3. \(0.25~\text{mA}\)
4. \(500~\mu\text{A}\)

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For a transistor the parameter β = 99. The value of the parameter α is

(1) 0.9                 

(2) 0.99

(3) 1                   

(4) 9

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In a PNP transistor the base is the N-region. Its width relative to the P-region is 

(1) Smaller           

(2) Larger

(3) Same             

(4) Not related

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A common emitter amplifier is designed with NPN transistor (α = 0.99). The input impedance is 1 KΩ and load is 10 KΩ. The voltage gain will be:

(1) 9.9                           

(2) 99

(3) 990                         

(4) 9900

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The symbol given in figure represents 

(1) NPN transistor

(2) PNP transistor

(3) Forward biased PN junction diode

(4) Reverse biased NP junction diode

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The most commonly used material for making transistor is

(1) Copper                 

(2) Silicon

(3) Ebonite                 

(4) Silver

 89%
Level 1: 80%+
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