Draw a simple circuit of a CE transistor amplifier. Explain its working. Show that the voltage gain $A_V$ of the amplifier is given by $A_V=-\dfrac{\beta_{ac}R_L}{r_i}$, where $\beta_{ac}$ is the current gain, $R_L$ is the load resistance and $r_i$ is the input resistance. What is the significance of the negative sign in the expression for the voltage gain?

In the CE amplifier a small base-current change gives a $\beta_{ac}$-times larger collector-current change; the output across $R_L$ is $A_V=-\beta_{ac}R_L/r_i$. The negative sign means the output is $180^\circ$ out of phase with the input.

CBSE Class 12 · 2012 · Delhi · SemiconductorsLong Answer

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Draw a simple circuit of a CE transistor amplifier. Explain its working. Show that the voltage gain

A_{V}

of the amplifier is given by

A_{V} = - \frac{β _{a c} R _{L}}{r _{i}}

, where

β_{a c}

is the current gain,

R_{L}

is the load resistance and

r_{i}

is the input resistance. What is the significance of the negative sign in the expression for the voltage gain?

Solution

Circuit: n-p-n transistor in CE configuration, base-emitter junction forward biased, collector fed through the load

R_{L}

, signal applied at the base.
Working: the input signal changes

V_{B E}

and hence the base current; this produces a

β_{a c}

-times larger change in collector current.
For an input signal

v_{i}

, the base-current change is

Δ I_{B} = \frac{v _{i}}{r _{i}},

and the collector-current change is

Δ I_{C} = β_{a c} Δ I_{B}

.
The output voltage is the drop this produces across

R_{L}

v_{0} = - Δ I_{C} R_{L} = - β_{a c} \frac{v _{i}}{r _{i}} R_{L} .

Hence

A_{V} = \frac{v _{0}}{v _{i}} = - \frac{β _{a c} R _{L}}{r _{i}} .

The negative sign shows the output is

18 0^{\circ}

out of phase with the input.

A_{V} = - \frac{β _{a c} R _{L}}{r _{i}}; output 18 0^{\circ} out of phase.

CBSE Class 12 Board · 2012 · Delhi · Subjective

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CBSE Class 12 · 2012 · Delhi · SemiconductorsLong Answer

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