A Silicon PN junction diode under reverse bias has depletion region of width 10 -m-The relative permittivity of Silicon- - r- 11-7 and the permittivity of free space -0 -8-85 - 10-12 F-m- The depletion capacitance of the diode per square meter is

Ans : (b) C= ϵ _o ϵ _r Ad CA = 8.85× 10^ 12× 11.710× 10^ 6 = 10.35 μ F ...

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A Silicon PN ...

Question

A Silicon PN junction diode under reverse bias has depletion region of width 10 µm.The relative permittivity of Silicon, $ε_{r} = 11.7$ and the permittivity of free space $ε_{0} = 8.85 \times 10^{- 12} F / m$ . The depletion capacitance of the diode per square meter is

100 μ

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10 μ

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1 μ

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20 μ

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Solution

The correct option is B $10 μ$ F
Ans : (b)

$C = \frac{ϵ_{o} ϵ_{r} A}{d}$
$\frac{C}{A} = \frac{8.85 \times 10^{- 12} \times 11.7}{10 \times 10^{- 6}} = 10.35 μ F$

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A Silicon PN junction diode under reverse bias has depletion region of width 10 µm.The relative permittivity of Silicon, εr=11.7 and the permittivity of free space ε0=8.85×10−12F/m. The depletion capacitance of the diode per square meter is

The correct option is B 10 μFAns : (b) C=ϵoϵrAd CA=8.85×10−12×11.710×10−6=10.35 μF

A Silicon PN junction diode under reverse bias has depletion region of width 10 µm.The relative permittivity of Silicon, $ε_{r} = 11.7$ and the permittivity of free space $ε_{0} = 8.85 \times 10^{- 12} F / m$ . The depletion capacitance of the diode per square meter is

The correct option is B $10 μ$ F
Ans : (b)

$C = \frac{ϵ_{o} ϵ_{r} A}{d}$
$\frac{C}{A} = \frac{8.85 \times 10^{- 12} \times 11.7}{10 \times 10^{- 6}} = 10.35 μ F$