Semiconductor Electronics: Materials, Devices and

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V
Vishal Baghel

Contributor-Level 10

Active region of the CE transistor is linear region and is best suited for its use as an amplifier.

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5 months ago

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A
alok kumar singh

Contributor-Level 10

Current through zener diode,

l z = 2 4 1 0 1 1 0 5 m A = 1 2 m A              

Power across zener diode,

P z = V z . l z = 1 0 * 1 2 = 1 2 0 m W          

New answer posted

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V
Vishal Baghel

Contributor-Level 10

Pentavalent materials have more electrons and so electron density increase. But overall semiconductor is neutral

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V
Vishal Baghel

Contributor-Level 10

A            B            X            Y            Z

1            0            0           

1            0            1     

...more

New answer posted

5 months ago

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A
alok kumar singh

Contributor-Level 10

    Y = ( A + B ¯ ¯ + A ) + ( A + B ¯ ¯ + B ) ¯ + ( A + B ¯ + A ) . ( A + B ¯ + B )           

              A            B            Y

              0            1

              0            1            0&nbs

...more

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Vishal Baghel

Contributor-Level 10

Very small change in minority charge carriers produces high value of reverse bias current.

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A
alok kumar singh

Contributor-Level 10

Y ->And gate

A

B

Y

0

0

0

1

1

1

0

1

0

1

0

0

 

New answer posted

5 months ago

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A
alok kumar singh

Contributor-Level 10

Voltage gain = I C R 0 I B R i  

R0 ->Output Resistance

Ri ->Input Resistance

IC ->Collector current

IB ->Base current

Voltage gain = I C I B R 0 R i = ( 5 * 1 0 3 ) ( 1 0 0 * 1 0 6 ) * 6 0 2 0 0 = 1 5  

New answer posted

5 months ago

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A
alok kumar singh

Contributor-Level 10

Kindly go through the solution

 

New answer posted

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V
Vishal Baghel

Contributor-Level 10

Truth table for Input & Output

A

B

y

1

1

0

0

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Now, Truth Table for option B.

             

A

B

y

1

1

0

1

0

1

0

1

1

0

0

1

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