Page 982 - Basic Electrical Engineering

An amplified output voltage is thus available across the load. The amplified
collector ac current is superimposed on the dc current, I CQ which will flow

through the collector when the ac input signal is not applied. It is the current
when the base current is I BQ .

Example 14.3 In an n–p–n transistor in the common emitter configuration,
an ac input signal of ± 40 mV is applied as shown in Fig. 14.22. The dc

current gain, β and ac current gain β are given as 80 and 100, respectively.
dc
ac
Calculate the voltage amplification, A of the amplifier. The I versus V BE
B
V
characteristic is such that for V = 0.7 V, I = 12 μA and for V = ± 40 mV,
i
B
B
I = ±4 μA. Also calculate the dc collector voltage.
b
Solution:

DC base current, I = 12 μA for dc voltage, V = 0.7 V and β = 80
dc
BB
B
I = β I = 80 × 12 μA = 0.96 mA
dc B
C

The collector voltage V is calculated as
CE

V CE = V – I R
C L
CC
–3
= 20 – 0.96 × 10 × 12 × 10 3
= 20 – 11.52

= 8.48 V

AC base current, I = ±4 μA for V = ±40 mV
i
b
I = β I = 100 × (±4 μA) = ±400 μA
ac b
C

AC output voltage across load resistance, V is calculated as
0

–6
3
V = I R = ±400 × 10 × 12 × 10 = ±4.8 V
C L
0