Semiconductor Electronics: Materials, Devices and

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3 months ago

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A
alok kumar singh

Contributor-Level 10

Maximum current through zener diode,

  i = 2 1 0 = 0 . 2 A             

When unregulated voltage is 14V,

Potential difference across Rs,

              VS = 14 – 10 = 4V

R S = V S i = 4 0 . 2 = 2 0 Ω

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3 months ago

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New answer posted

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alok kumar singh

Contributor-Level 10

β = I C I E = I C I E I C = I C / I E 1 l C / I E = α 1 α

New answer posted

3 months ago

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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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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          

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Vishal Baghel

Contributor-Level 10

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

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Vishal Baghel

Contributor-Level 10

A            B            X            Y            Z

1            0            0           

1            0            1     

...more

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3 months ago

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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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3 months ago

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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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alok kumar singh

Contributor-Level 10

Y ->And gate

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