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GREYUIT [131]
3 years ago
11

Consider a step pn junction made of GaAs at T = 300 K. At zero bias, only 20% of the total depletion region width is in the p-si

de. The built-in potential ϕi​=1.20V.
acceptor density in the p-side = 5.68x10^16
donor density in the n-side = 1.42x10^16
Determine the depletion region width, xn​, in the n-side in unit of μm.
Engineering
1 answer:
Nat2105 [25]3 years ago
3 0

Answer:

0.31 μm

Explanation:

this question wants us to Determine the depletion region width, xn​, in the n-side in unit of μm. using the information below.

density in the p-side = 5.68x10^16

density in the n-side = 1.42x10^16

\sqrt{\frac{2*12.7*8.85E-10}{1.6E-14}(\frac{1}{5.68E16}+\frac{1}{1.42E16} )(1.2)  }

= √(1.42x10⁵)(1.76056335x10⁻¹⁷ + 7.042253521x10⁻¹⁷)(1.2)

= √150.74x10⁻¹¹

= 3.882x10⁻⁵

approximately 0.39μm

xn = 0.39 x 0.8

= 0.31μm

0.31 um is the depletion region width. thank you!

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I_L = 20 / (( 1 - 0.6 )² × 12.5)

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I_{Lmax = [20 / (( 1 - 0.6 )² × 12.5)] + [ (20 × 0.6 × 5) / (2 × 10) ]

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I_{Lmin = [20 / (( 1 - 0.6 )² × 12.5)] - [ (20 × 0.6 × 5) / (2 × 10) ]

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ΔV₀/V₀ = 0.6 / (12.5 × (40 × 10⁻⁶) × (2 × 10⁵) )

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under ideal components; diode current = output current

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I_D = V₀/R

as V₀ = 50 V and R = 12.5 Ω

we substitute

I_D = 50 / 12.5

I_D = 4 A

Therefore, the average current in the diode under ideal components is 4 A

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