In a common base configuration, \[{I_e} = 1mA\], \[ \propto = 0.95\], the value of base current is
A) \[1.95mA\]
B) \[0.05mA\]
C) \[1.05mA\]
D) \[0.95mA\]
Answer
642.3k+ views
Hint: Transistor is a semiconductor device which is made up of three terminals. It can act as an insulator or a conductor by applying small voltage across it. The component $\alpha$ of the transistor indicates the amount of emitter current that reaches the collector. The component $\beta$ is the ratio of the collector current to the base current.
Complete step by step answer:
Step I:
Three possible ways are known to connect a bipolar junction transistor. One of them is Common base configuration of the transistor. In this type of transistor, as the name suggests, the base is common between the emitter and the collector. It is connected to both. So it helps in the input as well as the output signal of the transistor.
Step II:
As it is known that the emitter current is the sum of collector current and the base current. Therefore,
\[{I_e} = {I_c} + {I_b}\] ---(i)
Step III:
Given emitter current, \[{I_e} = 1mA\]
\[ \propto = 0.95\]
Collector current, \[{I_c} = ?\]
For a Common Base bipolar junction transistor
\[ \propto = \dfrac{{{I_c}}}{{{I_e}}}\]---(ii)
Step IV:
Substituting the values in equation (ii) and evaluating value of collector current,
\[{I_c} = \propto {I_e} = (0.95)(1)\]
\[{I_c} = 0.95\]
Step V:
Substitute value of collector current and emitter current in equation (i) to get the value of base current
\[1 = 0.95 + {I_b}\]
\[{I_b} = 1 - 0.95\]
\[{I_b} = 0.05mA\]
Step VI:
In the common base transistor, the value of base current is \[0.05mA\]
Therefore, Option B is the right answer.
Note: It is to be noted that the common base configuration transistor is not very commonly used. This is because it has very high voltage gain characteristics. The term \[ \propto \](alpha) refers to the bipolar junction transistor that refers to the proportion of the emitter current that flows to the collector terminal as output signal. The value of alpha is usually very less and lies between \[0.95\]and \[1\].
Complete step by step answer:
Step I:
Three possible ways are known to connect a bipolar junction transistor. One of them is Common base configuration of the transistor. In this type of transistor, as the name suggests, the base is common between the emitter and the collector. It is connected to both. So it helps in the input as well as the output signal of the transistor.
Step II:
As it is known that the emitter current is the sum of collector current and the base current. Therefore,
\[{I_e} = {I_c} + {I_b}\] ---(i)
Step III:
Given emitter current, \[{I_e} = 1mA\]
\[ \propto = 0.95\]
Collector current, \[{I_c} = ?\]
For a Common Base bipolar junction transistor
\[ \propto = \dfrac{{{I_c}}}{{{I_e}}}\]---(ii)
Step IV:
Substituting the values in equation (ii) and evaluating value of collector current,
\[{I_c} = \propto {I_e} = (0.95)(1)\]
\[{I_c} = 0.95\]
Step V:
Substitute value of collector current and emitter current in equation (i) to get the value of base current
\[1 = 0.95 + {I_b}\]
\[{I_b} = 1 - 0.95\]
\[{I_b} = 0.05mA\]
Step VI:
In the common base transistor, the value of base current is \[0.05mA\]
Therefore, Option B is the right answer.
Note: It is to be noted that the common base configuration transistor is not very commonly used. This is because it has very high voltage gain characteristics. The term \[ \propto \](alpha) refers to the bipolar junction transistor that refers to the proportion of the emitter current that flows to the collector terminal as output signal. The value of alpha is usually very less and lies between \[0.95\]and \[1\].
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