
In an n-p-n transistor circuit, the collector current is 9 mA. If \[90\% \] of the electrons emitted reach the collector. Then the emitter current is
\[A)\;\;\;\;8.1{\text{ }}mA\]
\[B)\;\;\;\;8{\text{ }}mA\]
\[C)\;\;\;\;9{\text{ }}mA\]
\[D)\;\;\;\;10{\text{ }}mA\]
Answer
565.5k+ views
Hint:Remember the direction of the flow of current and electrons in an n-p-n transistor. The direction of the flow of electrons is from the emitter region to the collector region.
Complete step by step answer:
According to the question, it is given that it is an n-p-n circuit. NPN transistor is the transistor in which one p-type material is placed between two n-type materials.
The NPN transistor amplifies the weak signals which enter into the base and starts producing the strong amplified signals at the collector end.
In NPN transistors, the direction of flow of an electron is from the emitter to the collector region owing to which the current produces in the transistor.
These types of transistors are mostly used in the circuit because their majority charge carriers are electrons which have high mobility as compared to positively charged holes.
Also, the collector current is \[{I_C} = {\text{ }}9{\text{ }}mA\]
And it is also given that \[90\% \] of electrons from the emitter reaches the collector.
So we have \[{I_C} = {\text{ }}\dfrac{9}{{10}}{\text{ }}{I_E}\]
\[ \Rightarrow {I_E}\; = {\text{ }}\dfrac{{10}}{9}{\text{ }}{I_C}\]
Substitute the value of \[{I_C} = {\text{ }}9{\text{ }}mA\] and we get
\[ = \dfrac{{10}}{9}{\text{ }} \times {\text{ }}9{\text{ }}\]
Cancel the terms we get,
\[ = {\text{ }}10{\text{ }}mA\]
Therefore option D is correct.
Additional information:
The NPN transistor contains the two diodes connected back to back. The diode present on the left side is called an emitter-base diode, and the diodes on the right side are called the collector-base diode.
Note: In the diagram, the arrow is representing the flow of current which is from collector to emitter but the direction of the flow of electrons is reverse of that of the flow of current i.e., emitter to collector.
Complete step by step answer:
According to the question, it is given that it is an n-p-n circuit. NPN transistor is the transistor in which one p-type material is placed between two n-type materials.
The NPN transistor amplifies the weak signals which enter into the base and starts producing the strong amplified signals at the collector end.
In NPN transistors, the direction of flow of an electron is from the emitter to the collector region owing to which the current produces in the transistor.
These types of transistors are mostly used in the circuit because their majority charge carriers are electrons which have high mobility as compared to positively charged holes.
Also, the collector current is \[{I_C} = {\text{ }}9{\text{ }}mA\]
And it is also given that \[90\% \] of electrons from the emitter reaches the collector.
So we have \[{I_C} = {\text{ }}\dfrac{9}{{10}}{\text{ }}{I_E}\]
\[ \Rightarrow {I_E}\; = {\text{ }}\dfrac{{10}}{9}{\text{ }}{I_C}\]
Substitute the value of \[{I_C} = {\text{ }}9{\text{ }}mA\] and we get
\[ = \dfrac{{10}}{9}{\text{ }} \times {\text{ }}9{\text{ }}\]
Cancel the terms we get,
\[ = {\text{ }}10{\text{ }}mA\]
Therefore option D is correct.
Additional information:
The NPN transistor contains the two diodes connected back to back. The diode present on the left side is called an emitter-base diode, and the diodes on the right side are called the collector-base diode.
Note: In the diagram, the arrow is representing the flow of current which is from collector to emitter but the direction of the flow of electrons is reverse of that of the flow of current i.e., emitter to collector.
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