
In an NPN transistor, the collector current is \[24\,mA\]. If 80% of electrons reach the collector its base current in \[mA\] is?
A. 36
B. 26
C. 16
D. 6
Answer
162k+ views
Hint:In NPN transistors, the emitter, as well as collector, are made of n-type semiconductor and the base is made of p-type semiconductor. When a battery is connected, current flows into the base of the NPN transistor which is opposite to the direction of electron flow. As conventional current flows from high potential to low potential but electrons flow from low potential to high potential. The collector current flows towards the n-type of NPN transistor and the emitter current flows opposite from it. The collector current \[{I_c}\] split into emitter current \[{I_e}\] and base current\[{I_b}\]. Hence the sum of base current and collector current gives the emitter current in an NPN transistor.
Formula used
\[{I_e} = {I_c} + {I_b}\]
Here, \[{I_e}\] represents emitter current, \[{I_c}\] represents collector current and \[{I_b}\] represents base current.
Complete step by step solution:
Given, Collector current \[{I_c}\] =24mA
80% of \[{I_e} = {I_c} = 24mA\]
Now by using the given value, we get
\[80\% of{\rm{ }}{I_e} = 24\]
\[\Rightarrow {I_e} = \dfrac{{24}}{{80}} \times 100\]
Emitter current,\[{I_e} = 30mA\]
As we know that,
\[{I_e} = {I_c} + {I_b}\]
\[\Rightarrow {I_b} = {I_e} - {I_c}\]
\[\Rightarrow {I_b} = 30 - 24\]
\[\therefore {I_b} = 6\,mA\]
Therefore the base current is 6mA.
Hence option D is correct.
Note: Transistors are of two types: PNP transistors and NPN transistors. Both the transistors are formed with three layers of semiconductor material. An NPN transistor consists of two n-type semiconductors which are in between a p-type semiconductor. Hence in this type of transistor, the majority of carriers are electrons. These electrons are passed from the emitter to the collector side. So we say that the direction of current is from the collector side to the emitter side. It forms a device of three layers that is a negative-positive-negative transistor.
Formula used
\[{I_e} = {I_c} + {I_b}\]
Here, \[{I_e}\] represents emitter current, \[{I_c}\] represents collector current and \[{I_b}\] represents base current.
Complete step by step solution:
Given, Collector current \[{I_c}\] =24mA
80% of \[{I_e} = {I_c} = 24mA\]
Now by using the given value, we get
\[80\% of{\rm{ }}{I_e} = 24\]
\[\Rightarrow {I_e} = \dfrac{{24}}{{80}} \times 100\]
Emitter current,\[{I_e} = 30mA\]
As we know that,
\[{I_e} = {I_c} + {I_b}\]
\[\Rightarrow {I_b} = {I_e} - {I_c}\]
\[\Rightarrow {I_b} = 30 - 24\]
\[\therefore {I_b} = 6\,mA\]
Therefore the base current is 6mA.
Hence option D is correct.
Note: Transistors are of two types: PNP transistors and NPN transistors. Both the transistors are formed with three layers of semiconductor material. An NPN transistor consists of two n-type semiconductors which are in between a p-type semiconductor. Hence in this type of transistor, the majority of carriers are electrons. These electrons are passed from the emitter to the collector side. So we say that the direction of current is from the collector side to the emitter side. It forms a device of three layers that is a negative-positive-negative transistor.
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