In an npn transistor, 108 electrons enter the emitter in 10- 8 second. If 1 % electrons entering to emitter are lost in the base, the fraction of current that enters the collector and current amplification factor are respectively:

Question

In an npn transistor, 108 electrons enter the emitter in 10- 8 second. If 1 % electrons entering to emitter are lost in the base, the fraction of current that enters the collector and current amplification factor are respectively: (A) 0.8 and 49 (B) 0.9 and 90 (C) 0.88 and 88 (D) 0.99 and 99

Tardigrade Admin · Accepted Answer

108 electron enter the emitter in 10- 8 second ∴ IE=(108 × 1 . 6 × 10- 19/10- 8)=172.8× 10- 11A ∴ 1 % of IE is lost in base ,IB=(IE/100) ⇒ 99 % of IE i.e., (99/100)IE enters the collector ⇒ IC=0.99IE Now, β =(IC/IB)=(0 . 99 IE/0 . 01 IE)=99

Q. In an npn transistor, $108$ electrons enter the emitter in $1 0^{- 8}$ second. If $1%$ electrons entering to emitter are lost in the base, the fraction of current that enters the collector and current amplification factor are respectively:

$0.8$ and $49$

$0.9$ and $90$

$0.88$ and $88$

$0.99$ and $99$

Q. In an npn transistor, 108 electrons enter the emitter in 10−8 second. If 1% electrons entering to emitter are lost in the base, the fraction of current that enters the collector and current amplification factor are respectively:

0.8 and 49

0.9 and 90

0.88 and 88

0.99 and 99

108 electron enter the emitter in 10−8 second ∴IE​=10−8108×1.6×10−19​=172.8×10−11A ∴1% of IE​ is lost in base ,IB​=100IE​​ ⇒99% of IE​ i.e., 10099​IE​ enters the collector ⇒IC​=0.99IE​ Now, β=IB​IC​​=0.01IE​0.99IE​​=99

Q. In an npn transistor, $108$ electrons enter the emitter in $1 0^{- 8}$ second. If $1%$ electrons entering to emitter are lost in the base, the fraction of current that enters the collector and current amplification factor are respectively:

$0.8$ and $49$

$0.9$ and $90$

$0.88$ and $88$

$0.99$ and $99$