Draw a circuit diagram for the use of an NPN transistor as an amplifier in common emitter configuration. The input resistance of a transistor is $1000\ \Omega$. On changing its base current by $10\ \mu\mathrm{A}$, the collector current increases by $2\ \mathrm{mA}$. If a load resistance of $5\ \mathrm{k\Omega}$ is used in the circuit, calculate (i) the current gain and (ii) the voltage gain of the amplifier.

(i) Current gain $\beta=\Delta I_C/\Delta I_B=200$. (ii) Voltage gain $A_V=\beta\,(R_L/r_i)=200\times5000/1000=1000$.

CBSE Class 12 · 2006 · SemiconductorsShort Answer

Draw a circuit diagram for the use of an NPN transistor as an amplifier in common emitter configuration. The input resistance of a transistor is

1000 Ω

. On changing its base current by

10 μ A

, the collector current increases by

2 mA

. If a load resistance of

5 kΩ

is used in the circuit, calculate
(i) the current gain and
(ii) the voltage gain of the amplifier.

Answer: (i) Current gain $β = Δ I_{C} /Δ I_{B} = 200$ .
(ii) Voltage gain $A_{V} = β (R_{L} / r_{i}) = 200 \times 5000/1000 = 1000$ .

In the CE amplifier the base-emitter junction is forward biased and the collector-emitter output carries the load

R_{L}

.
(i) Current gain:

β = \frac{Δ I _{C}}{Δ I _{B}} = \frac{2 \times 1 0 ^{- 3}}{10 \times 1 0 ^{- 6}} = 200.

(ii) Voltage gain:

A_{V} = β \frac{R _{L}}{r _{i}} = 200 \times \frac{5000}{1000} = 1000.

β = 200, A_{V} = 1000.

CBSE Class 12 Board · 2006 · Subjective