question_answer

Draw the transfer characteristics of a common emitter n-p-n transistor. Point out the region in which the transistor operates as an amplifier.

Define the following terms used in transistor amplifiers.

(i) Input resistance

(ii) Output resistance

(iii) Current amplification factor

Answer:

The transfer characteristic curve of common emitter n-p-n transistor is drawn between \[{{I}_{C}}\] and \[{{I}_{B}},\] when \[{{V}_{CE}}\] is kept constant. Transfer characteristic of  CE \[n-p-n\]transistor A transistor works as amplifier when it is operated in an active region. In this region, emitter-base junction is forward biased, while collector-base is reverse biased. (i) Input resistance: It is defined as the ratio of change in base-emitter voltage \[(\Delta \,{{V}_{BE}})\] to the resulting change in the base current \[(\Delta \,{{I}_{B}})\] at constant collector-emitter voltage \[({{V}_{CE}}).\] It is reciprocal of slope of \[{{I}_{B}}-{{V}_{BE}}\] curve. Input resistance, \[{{R}_{i}}={{\left( \frac{\Delta \,{{V}_{BE}}}{\Delta \,{{I}_{B}}} \right)}_{{{V}_{CE}}=\,\,\text{constant}}}\] (ii) Output resistance: From the output characteristics, we define output resistance of transistor as the ratio of change in collector-emitter voltage to the resulting change in collector current at constant base current. Thus, Output resistance, \[{{R}_{0}}={{\left( \frac{\Delta \,{{V}_{CE}}}{\Delta \,{{I}_{C}}} \right)}_{{{I}_{B}}\,\,=\,\,\text{constant}}}\] = Reciprocal of slope of \[{{I}_{C}}-{{V}_{CE}}\]curve. (iii) Current amplification: The current amplification factor \[(\beta )\] of a transistor in CE configuration is defined as the ratio of change in collector current to the change in base current at a constant collector-emitter voltage when the transistor is in active state. +         \[\therefore \]                  \[{{\beta }_{AC}}={{\left( \frac{\Delta \,{{I}_{C}}}{\Delta \,{{I}_{B}}} \right)}_{{{V}_{CE}}\,\,=\,\,\text{constant}}}\] Its value is very large \[\left( {{\beta }_{AC}}>>1 \right).\]