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महाराष्ट्र राज्य शिक्षण मंडळएचएससी विज्ञान (सामान्य) इयत्ता १२ वी

Derive the relation between α and β.

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प्रश्न

Derive the relation between α and β.

Obtain the relation between the current gain αDC and βDC for a transistor.

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उत्तर

In common emitter configuration, the emitter current is

IE = IB + lC    ...(i)

Dividing the above equation by lC we get

`"I"_"E"/"I"_"C" = "I"_"B"/"I"_"C" + 1`

By definition of αDC and βDC

∴ `1/alpha_"DC" = 1/beta_"DC" + 1`    ...(ii)

`1/alpha_"DC" = (beta_"DC" + 1)/beta_"DC"`

∴ `alpha_"DC" = beta_"DC"/(1 + beta_"DC")`

∴ `alpha_"DC" = beta/(1 + beta)`    ...(iii)

Also, from equation (ii),

`1/alpha_"DC" - 1 = 1/beta_"DC"` 

`(1 - alpha_"DC")/alpha_"DC" = 1/beta_"DC"`

∴ `beta_"DC" = alpha_"DC"/(1 - alpha_"dc")`

∴ `beta_"DC" = alpha/(1 - alpha)`    ...(iv)

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पाठ 16: Semiconductor Devices - Exercises [पृष्ठ ३६४]

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बालभारती Physics [English] Standard 12 Maharashtra State Board
पाठ 16 Semiconductor Devices
Exercises | Q 17 iii | पृष्ठ ३६४

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