Draw the typical input and output characteristics of a $n - p - n$ transistor in $CE$ configuration. Show how these characteristics can be used to determine
(a) the input resistance ( ${r_i}$ ) and
(b) current amplification factor $\left( \beta \right)$ .
Answer
Verified
115.8k+ views
Hint: The $n - p - n$ transistor has the input, output and the transfer characteristics. Draw the input characteristics between base current and base emitter voltage and the output characteristics between the collector current and the common emitter voltage.
Complete step by step solution:
(a) The input resistance is also known as input impedance. It is defined as the ratio of the changes in the base emitter voltage to the corresponding change in the base current at a given particular common emitter voltage in the transistor.
The formula of the input impedance is given by
${r_i} = \dfrac{{\Delta {V_{BE}}}}{{\Delta {I_B}}}$ ( at a given ${V_{CE}}$ )
(b) The term current amplification factor is also known as the current gain. It is obtained by dividing the small change in the collector current to that of the corresponding change in the base current of the transistor at constant common emitter voltage.
The formula of the current amplification factor is given by
$\beta = \dfrac{{\Delta {I_C}}}{{\Delta {I_B}}}$ ( at a given ${V_{CE}}$ )
The following graph represents the input characteristics of the $n - p - n$ transistor. It shows that to a certain point, when the base current increases the base emitter voltage remains the same.
The following graph shows the output characteristics of the $n - p - n$ transistor. It includes the cut off, active and the saturated regions of the base current where the common emitter voltage is constant.
Note: In the $n - p - n$ transistor, the $P$ - type of the semiconductor is made connected between the $N$ - type of the semiconductor. It is most commonly used and it looks like the $PN$ junction diode connected back to back. In this transistor, the electron transfers from the base to the collector.
Complete step by step solution:
(a) The input resistance is also known as input impedance. It is defined as the ratio of the changes in the base emitter voltage to the corresponding change in the base current at a given particular common emitter voltage in the transistor.
The formula of the input impedance is given by
${r_i} = \dfrac{{\Delta {V_{BE}}}}{{\Delta {I_B}}}$ ( at a given ${V_{CE}}$ )
(b) The term current amplification factor is also known as the current gain. It is obtained by dividing the small change in the collector current to that of the corresponding change in the base current of the transistor at constant common emitter voltage.
The formula of the current amplification factor is given by
$\beta = \dfrac{{\Delta {I_C}}}{{\Delta {I_B}}}$ ( at a given ${V_{CE}}$ )
The following graph represents the input characteristics of the $n - p - n$ transistor. It shows that to a certain point, when the base current increases the base emitter voltage remains the same.
The following graph shows the output characteristics of the $n - p - n$ transistor. It includes the cut off, active and the saturated regions of the base current where the common emitter voltage is constant.
Note: In the $n - p - n$ transistor, the $P$ - type of the semiconductor is made connected between the $N$ - type of the semiconductor. It is most commonly used and it looks like the $PN$ junction diode connected back to back. In this transistor, the electron transfers from the base to the collector.
Recently Updated Pages
How to find Oxidation Number - Important Concepts for JEE
How Electromagnetic Waves are Formed - Important Concepts for JEE
Electrical Resistance - Important Concepts and Tips for JEE
Average Atomic Mass - Important Concepts and Tips for JEE
Chemical Equation - Important Concepts and Tips for JEE
Concept of CP and CV of Gas - Important Concepts and Tips for JEE
Trending doubts
JEE Main Exam Marking Scheme: Detailed Breakdown of Marks and Negative Marking
Collision - Important Concepts and Tips for JEE
Ideal and Non-Ideal Solutions Raoult's Law - JEE
Young's Double Slit Experiment Derivation
Current Loop as Magnetic Dipole and Its Derivation for JEE
When Barium is irradiated by a light of lambda 4000oversetomathopA class 12 physics JEE_Main