Question

In an NPN transistor the collector current is $24mA$. If $80\%$ of electrons reach collector its base current in $mA$ is?

Answer 1

Step 1: Given data

Given,

$80\%$ of $I_e=I_c=24mA$

Step 2: Formula used

$I_e=I_c+I_b[where{I}_e=electroncurrent,I_c=collectorcurrent,I_b=basecurrent]$

Step 3: Analyzing the given data

From the given data

$$\begin{gathered} 80\%\hspace{0.17em}of{I}_e=I_c \\ \frac{80}{100}\times I_e=24mA \\ {I}_e=\frac{24\times 100}{80} \\ {I}_e=30mA \end{gathered}$$

$I_c$ is the current flowing into the collector terminal, $I_b$ is the current flowing into the base terminal and $I_e$ is the current flowing out of the emitter terminal.

$$\begin{gathered} I_e=I_c+I_b \\ {I}_b=I_e-I_c[where{I}_b=basecurrent] \\ {I}_b=30-24 \\ {I}_b=6mA \end{gathered}$$

Hence, its base current will be $6mA$.