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  • (a) Draw a circuit diagram of an n-p-n transistor with its emitter-base junction forward biased and base-collector junction reverse biased. Briefly describe its working. Explain how a transistor in its active state exhibits a low resistance at its emit

(a) Draw a circuit diagram of an n-p-n transistor with its emitter-base junction forward biased and base-collector junction reverse biased. Briefly describe its working. Explain how a transistor in its active state exhibits a low resistance at its emitter-base junction and high resistance at its base-collector junction.

(b) Derive the expression for the voltage gain of a transistor amplifier in CE configuration in terms of the load resistance RL, current gain a and input resistance. Explain why input and output voltages are in opposite phase

 

 

 

 
 

Answers (1)

(a) A circuit diagram of the n-p-n transistor;-

When the emitter-base junction is forward biased, the negative (-ve) pole of the battery is connected with n-diode and positive (+ve) pole with p-diode, electrons are majority carriers, repelled towards base due to -ve potential on V_{EB} which combines with holes present in the base.

As here the base is thin, most of the electron will go the collector 

In the active mode of the transistor, emitter-base junction is forward biased and collector-base junction is reversed biased. When transistor operates in this mode, collector current increases linearly with base current. So in the active state it exhibits low resistance at emitter-base and high resistance at its base-collector.

(b) Applying Kirchoff's loop rule to input loop 

\Delta V_{BE}=\Delta I_{B}\left ( R_{B}+r_i \right ) ____(i)

Where, I_{B}=Base\; current

                r_i=internal \; resistance

                R_{B}=base \; resistance

Output loop

\Delta V_{CE}=-R_{L}\Delta I_{C}______(ii)

Voltage gain:

A_{V}=\frac{V_{0}}{Vi}=\frac{\Delta V_{CE}}{\Delta V_{BE}}

A_{V}=\frac{-R_{L}\Delta L_{C}}{\Delta I_{B}(R_{B}+r_i)}

A_{V}=-\beta_{ac}\frac{R_{L}}{r}

Where,

 \beta_{ac}=current \; gain\; in\; C.E=\frac{\Delta I_{C}}{\Delta I_{B}}

The negative sign in the voltage gain shows that the input voltage and output voltage has a phase difference of 180 degrees.

Posted by

Safeer PP

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