a = 3 × 10-4 cm और ND = 1015 electron/cm3 के साथ एक n-चैनल सिलिकॉन FET के लिए, पिंच-ऑफ वोल्टेज क्या है यदि सिलिकॉन का पारद्युतिक स्थिरांक ϵ = 12ϵ 0 और \(\rm \epsilon_0 = \frac{1}{36 \pi} \times 10^{-9} \) है) \(\frac{a^2qN_D}{2\epsilon}\) है?

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  1. 6.8 V
  2. 5.2 V
  3. 8.8 V
  4. 9.2 V

Answer (Detailed Solution Below)

Option 1 : 6.8 V
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Detailed Solution

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प्रयुक्त सूत्र:

पिंच ऑफ वोल्टेज (VP) = \(\frac{(3\times 10^{-4})^2\times 1.6\times 10^{-19}\times 10^{15}}{2\times 12 \epsilon_0}\)

अनुप्रयोग:

दिया गया,

a = 3 × 10-4 cm

ND = 1015 electron/cm3

ϵ = 12ϵ0

q = 1.6 × 10-19 C

उपरोक्त अवधारणा से,

VP = \(\rm \epsilon_0 = \frac{1}{36 \pi} \times 10^{-9} \)

जहां, \(\rm \epsilon_0 = \frac{1}{36 \pi} \times 10^{-9} \)

इसे हल करने के बाद,

VP = 6.8 V

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