JEE Main 12 Jan 2019 Evening Question 20
Question: In the figure, given that $ V _{BB} $ supply can vary from 0 to $ 5.0V, $ $ V _{CC}=5V,{\beta _{dc}}=200, $ $ R _{B}=100k\Omega , $ $ R _{C}=1,k\Omega $ and $ V _{BE}=1.0,V. $ The minimum base current and the input voltage at which the transistor will go to saturation, will be respectively
[JEE Main Online Paper Held On 12-Jan-2019 Evening]
Options:
A) $ 20\mu A $ and 3.5 V
B) $ 25\mu A $ and 3.5V
C) $ 20\mu A $ and 2.8 V
D) $ 25\mu A $ and 2.8 V
Show Answer
Answer:
Correct Answer: B
Solution:
- Applying KVL at output and input circuit $ V _{CB}=V _{CC}-I _{C}R _{C} $ $ V _{BB}=I _{B}R _{B}+V _{BE} $
At saturation, $ V _{CB}=0\Rightarrow V _{CC}=I _{C}R _{C} $
$ \Rightarrow $ $ I _{C}=\frac{V _{CC}}{R _{C}}=\frac{5v}{1k\Omega }=5mA $
Base current, $ I _{B}=\frac{I _{C}}{\beta }=\frac{5}{200}=25\mu A $
Using equation (i), $ V _{BB}=(25\times {10^{-6}})(100\times 10^{3})+1=2.5+1=3.5V $
