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Semiconductor Devices — Physics STD 12 Science — Question

Maharashtra BoardEnglish MediumSTD 12 SciencePhysicsSemiconductor Devices3 Marks
Question
In a common-base connection, the emitter current is $6.28\ mA$ and collector current is $6.20\ mA.$ Determine the common base $DC$ current gain.
✓

Answer

Data $: I _{ E }=6.28 mA , I _{ C }=6.20 mA$
$\alpha_{ dc }=\frac{I_C}{I_{ E }}$ and $\beta_{ dc }=\frac{I_{ C }}{I_{ B }}=\frac{\alpha_{ dc }}{1-\alpha_{ dc }}$
Common-emitter current gain, $\alpha_{d c}=\frac{6.20}{6.28}=0.9873$
Therefore, common-base current gain,
$\beta_{ dc }=\frac{0.9873}{1-0.9873}=\frac{0.9873}{0.0127}=77.74$
OR
$ I _E= I _B+ I _C$
$\therefore I _B= I _E- I _C=6.28-6.20=0.08 mA$
$\therefore \beta_{d c}=\frac{6.20}{0.08}=77.5$
[Note: The answer given in the textbook obviously refers to the common-emitter current gain.]

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