
A CE transistor amplifies a weak current signal because collector current is:
A) $\beta $ times ${I_b}$.
B) $\beta $ times \[{I_c}\].
C) $\alpha $ times ${I_b}$.
D) $\alpha $ times ${I_c}$.
Answer
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Hint:A transistor comes in two forms n-p-n and p-n-p. Sometimes, it behaves as an amplifier when the emitter of the transistor is grounded and output is taken from the collector of the transistor. This is the same scenario where the transistor is behaving as an amplifier.
Formula used:The formula of the current gain in the collector emitter amplifier is given by,
$ \Rightarrow \dfrac{{{I_c}}}{{{I_b}}} = \beta $
Where,
${I_b}$ = current given in the base of the transistor.
${I_c}$ = current output in the collector.
$\beta $ = current gain.
Complete step by step answer:
We know that, in Collector Emitter amplifier, current gain of the collector is given by the expression$ \Rightarrow \dfrac{{{I_c}}}{{{I_b}}} = \beta $
Where,
${I_b}$ = current given in the base of the transistor
${I_c}$ = current output in the collector
$\beta $ = current gain
Therefore, collector current recorded at the output of the transistor is,
$ \Rightarrow \dfrac{{{I_c}}}{{{I_b}}} = \beta $
$ \Rightarrow {I_c} = \beta \times {I_b}$
The collector current is equal to ${I_c} = \beta \times {I_b}$ which means collector current is $\beta $ times ${I_b}$ .
Thus, option A is the correct answer.
Note:
-The value of the $\beta $ lies within the range of 40 to 100.
-If the base of the transistor is receiving a weak current signal, let say 1 Ampere, then the collector current will lie in the range of 40 to 100 amperes.
-An n-p-n transistor acts as an amplifier when the emitter is grounded and current input is fed in its base, then due to forward biasing, the output current is measured at the collector of the transistor.
-The current fed in the base of the transistor is generally pulled down by applying a resistor in the series in between base and current source.
-A emitter of the n-p-n transistor is generally accompanied by an arrow pointing outwards while the p-n-p transistor’s collector is marked with an arrow pointing inwards.
Formula used:The formula of the current gain in the collector emitter amplifier is given by,
$ \Rightarrow \dfrac{{{I_c}}}{{{I_b}}} = \beta $
Where,
${I_b}$ = current given in the base of the transistor.
${I_c}$ = current output in the collector.
$\beta $ = current gain.
Complete step by step answer:
We know that, in Collector Emitter amplifier, current gain of the collector is given by the expression$ \Rightarrow \dfrac{{{I_c}}}{{{I_b}}} = \beta $
Where,
${I_b}$ = current given in the base of the transistor
${I_c}$ = current output in the collector
$\beta $ = current gain
Therefore, collector current recorded at the output of the transistor is,
$ \Rightarrow \dfrac{{{I_c}}}{{{I_b}}} = \beta $
$ \Rightarrow {I_c} = \beta \times {I_b}$
The collector current is equal to ${I_c} = \beta \times {I_b}$ which means collector current is $\beta $ times ${I_b}$ .
Thus, option A is the correct answer.
Note:
-The value of the $\beta $ lies within the range of 40 to 100.
-If the base of the transistor is receiving a weak current signal, let say 1 Ampere, then the collector current will lie in the range of 40 to 100 amperes.
-An n-p-n transistor acts as an amplifier when the emitter is grounded and current input is fed in its base, then due to forward biasing, the output current is measured at the collector of the transistor.
-The current fed in the base of the transistor is generally pulled down by applying a resistor in the series in between base and current source.
-A emitter of the n-p-n transistor is generally accompanied by an arrow pointing outwards while the p-n-p transistor’s collector is marked with an arrow pointing inwards.
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