Answer
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Hint: here, firstly, we will calculate the resistance which is the ratio of load resistance ${R_L}$ and the input resistance, ${R_i}$ and then we will calculate the current gain using the formula ${A_v} = \beta \times \dfrac{{{R_L}}}{{{R_i}}}$, where $\beta $ is the current gain, ${R_L}$ is the load resistance, and ${R_i}$ is the input resistance.
Complete step by step answer:
The common base amplifier is a transistor in which the input is taken from the base and output is taken from the collector. The emitter is common for both the input and output terminals. The common base is called so because the signal and the load share the same base of the transistor.
According to the question,
The current gain of the common base amplifier, $\beta = 50$
Load resistance, ${R_L} = 4k\Omega $
${R_L} = 4000\Omega $
Input resistance, ${R_i} = 500\Omega $
Now, the resistance gain is defined as the ratio of the load resistance to the input resistance and is given by,
Resistance gain $ = \dfrac{{{R_L}}}{{{R_i}}}$
$ = \dfrac{{4000\Omega }}{{500\Omega }}$
$ = 8$
Now, the voltage gain of a common emitter amplifier is defined as the product of the current gain of the amplifier and the resistance gain.
Now, for the common base-emitter transistor, the voltage gain is given by
Voltage gain $ = $ current gain $ \times $ resistance gain
$\Rightarrow {A_v} = \beta \times \dfrac{{{R_L}}}{{{R_i}}}$
$\Rightarrow {A_v} = 50 \times 8$
$\therefore {A_v} = 400$
Hence, the voltage gain of the common base transistor amplifier is $400$.
Hence, option D is the correct option.
Note: A common base amplifier is a less commonly used amplifier than that of a common emitter amplifier. The common base amplifier is used to produce voltage gain without any current gain. In some high-frequency substances, the common base amplifier is used as a voltage amplifier.
The common base amplifier has a relatively high voltage gain. In this amplifier, the current gain less than or approximately equal to $1$ . Also, in this amplifier the input impedance is low and the output impedance is high.
Complete step by step answer:
The common base amplifier is a transistor in which the input is taken from the base and output is taken from the collector. The emitter is common for both the input and output terminals. The common base is called so because the signal and the load share the same base of the transistor.
According to the question,
The current gain of the common base amplifier, $\beta = 50$
Load resistance, ${R_L} = 4k\Omega $
${R_L} = 4000\Omega $
Input resistance, ${R_i} = 500\Omega $
Now, the resistance gain is defined as the ratio of the load resistance to the input resistance and is given by,
Resistance gain $ = \dfrac{{{R_L}}}{{{R_i}}}$
$ = \dfrac{{4000\Omega }}{{500\Omega }}$
$ = 8$
Now, the voltage gain of a common emitter amplifier is defined as the product of the current gain of the amplifier and the resistance gain.
Now, for the common base-emitter transistor, the voltage gain is given by
Voltage gain $ = $ current gain $ \times $ resistance gain
$\Rightarrow {A_v} = \beta \times \dfrac{{{R_L}}}{{{R_i}}}$
$\Rightarrow {A_v} = 50 \times 8$
$\therefore {A_v} = 400$
Hence, the voltage gain of the common base transistor amplifier is $400$.
Hence, option D is the correct option.
Note: A common base amplifier is a less commonly used amplifier than that of a common emitter amplifier. The common base amplifier is used to produce voltage gain without any current gain. In some high-frequency substances, the common base amplifier is used as a voltage amplifier.
The common base amplifier has a relatively high voltage gain. In this amplifier, the current gain less than or approximately equal to $1$ . Also, in this amplifier the input impedance is low and the output impedance is high.
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