For a common emitter amplifier, dc (direct current) current gain is 100. If the base current is `20 muA`, calculate the collector and emitter current.

Explanation

`beta=100, I_(b)= 20 mu A= 20 xx 10^(-6)A, I_(c)`= ? `I_(e)=` ?
`beta= (I_(c))/(I_(b))`
`I_(c)= beta I_(b) = 100 xx 20 xx 10^(-6)= 2 xx 10^(-3) A= 2mA`
We know `I_(e)= I_(b) + I_(c)`
`2 xx 10^(-3) = 20 xx 10^(-6) + I_(c)`
`I_(c)= 2 xx 10^(-3) - 20 xx 10^(-6)`
`=200 xx 10^(-5) -2 xx 10^(-5)`
`=(200-2) 10^(-5)`
`=198 xx 10^(-5)`
`= 1.98 xx 10^(-3)`
`I_(c)= 1.98 mA`