Physics Semiconductor Devices

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3 weeks ago

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R
Raj Pandey

Contributor-Level 9

n i 2 = n e * n h where n e = 5 * 10 28 10 6

n h = n i 2 n e = 1.5 * 10 16 1.5 * 10 16 * 10 6 5 * 10 28 n h = 4.5 * 10 9 / m 3

 

New answer posted

3 weeks ago

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R
Raj Pandey

Contributor-Level 9

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

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R
Raj Pandey

Contributor-Level 9

Factual (theory based)

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R
Raj Pandey

Contributor-Level 9

Theory based

Capacitor in parallel removes the AC ripple from the rectified output.

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4 weeks ago

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Raj Pandey

Contributor-Level 9

In half wave rectification f output   = 60 H z

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Raj Pandey

Contributor-Level 9

In (a) & (c) circuits, both the junctions are in same biasing conditions so offers equal resistances.

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Raj Pandey

Contributor-Level 9

Power gain = (i_c² R_c) / (i_b² R_B) = (i_c/i_b)² (R_c/R_B) = (10²)² (10? /10³) = 10?
i_c/i_b = 100 ⇒ β = i_c/i_b = 100

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A
alok kumar singh

Contributor-Level 10

Photodiode operate in reverse bias. The photocurrent increases initially and saturates fi

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Raj Pandey

Contributor-Level 9

First part of figure shown is or GATE and Second part of figure shown is NOT GATE So, Y P = O R + N O T =  NOR GATE Y = A + B ? = A B

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