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Questions
Derive the relation between α and β.
Obtain the relation between the current gain αDC and βDC for a transistor.
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Solution
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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