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Q.In a transistor, IC=0.98I_C = 0.98, IB=20 μAI_B = 20\,\mu A. Find

(i) α\alpha and
(ii) β\beta of the transistor.
Manipur CohsemCOHSEM Manipur Higher Secondary Board 2019Subjective· 2mImportance★★★★★
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Taking IC=0.98 mAI_C=0.98\,mA (consistent units with IB=20 μA=0.02 mAI_B=20\,\mu A=0.02\,mA) gives IE=IB+IC=1.00 mAI_E=I_B+I_C=1.00\,mA, hence α=IC/IE=0.98\alpha=I_C/I_E=0.98 and β=IC/IB=49\beta=I_C/I_B=49.

For a transistor, the fundamental current relation is

IE=IB+ICI_E = I_B + I_C

and the current-gain parameters are defined as

α=ICIE (common-base current gain),β=ICIB (common-emitter current gain)\alpha = \dfrac{I_C}{I_E}\ \text{(common-base current gain)},\qquad \beta = \dfrac{I_C}{I_B}\ \text{(common-emitter current gain)}

Here IC=0.98 mAI_C = 0.98\,mA and IB=20 μA=0.02 mAI_B = 20\,\mu A = 0.02\,mA (expressing both in the same unit, mA, as they must be to combine):

IE=IB+IC=0.02+0.98=1.00 mAI_E = I_B + I_C = 0.02 + 0.98 = 1.00\,mA

(i) α\alpha: …

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