For a p-n-p transistor in CB configuration- the emitter current IE is 1 mA and - - 0-95- The base current and collector current are

The correct option is C 0.05 mA , 0.95 mA Given that: I_e #160;= 1 mA α = 0.95 For transistor in CB mode, α = I_cI_e I_c = α I_e = (0. ...

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Standard XII

Physics

Toroid

For a p-n-p t...

Question

For a p-n-p transistor in CB configuration, the emitter current $I_{E}$ is $1 m A$ and $α$ $=$ $0.95$ . The base current and collector current are

0.95 m A, 0.05 m A

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0.05 m A, 0.95 m A

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9.5 m A, 0.5 m A

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0.5 m A, 9.5 m A

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Solution

The correct option is C $0.05 m A, 0.95 m A$
Given that:
$I_{e} = 1 m A$
$α = 0.95$
For transistor in CB mode,
$α = \frac{I_{c}}{I_{e}}$
$I_{c} = α I_{e} = (0.95) (1) = 0.95 m A$
Also, for transistors
$I_{e} = I_{b} + I_{c}$
$I_{b} = I_{e} - I_{c} = 1 - 0.95 = 0.05 m A$

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For a p-n-p transistor in CB configuration, the emitter current IE is 1 mA and α =0.95. The base current and collector current are

The correct option is C 0.05 mA,0.95 mAGiven that:Ie=1 mAα=0.95For transistor in CB mode,α=IcIeIc=αIe=(0.95)(1)=0.95 mAAlso, for transistorsIe=Ib+IcIb=Ie−Ic=1−0.95=0.05 mA

For a p-n-p transistor in CB configuration, the emitter current $I_{E}$ is $1 m A$ and $α$ $=$ $0.95$ . The base current and collector current are

The correct option is C $0.05 m A, 0.95 m A$
Given that:
$I_{e} = 1 m A$
$α = 0.95$
For transistor in CB mode,
$α = \frac{I_{c}}{I_{e}}$
$I_{c} = α I_{e} = (0.95) (1) = 0.95 m A$
Also, for transistors
$I_{e} = I_{b} + I_{c}$
$I_{b} = I_{e} - I_{c} = 1 - 0.95 = 0.05 m A$