There are many ways to amplify signal using transistor and one of the most common method is Common emitter amplifier.
It is one of the easiest way to amplify signal and it require only one transistor. A common emitter amplifier , amplifies the entire input signal. It doesn't matter whether the signal is in the positive or negative half of its cycle; the amplifier will boost both parts of the waveform.
It provide phase shift of 180 degree . It means if input is 1 , the output will be 0 and vice-versa.
The purpose of C1 is to block dc component and allow only ac component.
The purpose of C2 is to provide low resitance path to ac signal , it will helps to increase the gain of output.
Calculation:
Before doing calculation , you have to assume the quiescent current(IQ), Supply voltage, input signal and Amplification ratio.
Quiescent current(IQ): In a transistor amplifier, quiescent current is the DC current that flows through the transistor when there is no input signal. Lets 1mA be quiescent current.
Supply voltage: 4V
Input signal: Sine wave 20mVpeak to peak , 20khz.
Amplification ratio: 10 times the input voltage.
Now lets calcuate,
1) Collector voltage: It should be half of supply voltage so there will be enough room for output to swing up and down for postive and negative singal without distortion.
for this circuit collector voltage(Vc)= (1/2)*4= 2V
2) Calcuation of R3:
R3= (Supply voltage - collector voltage) / IQ
= (4-2) / 1m
=2kΩ
3) Calcuation of R4:
Amplification ratio = R3 / R4
or, 10 = 2k / R4
or, R4 = 200Ω
4) Voltage drop on R4:
V(R4) = (IQ) * R4
or, V(R4) = 1m * 200
or, V(R4) = 0.2V
5) Base Voltage ( Vb):
Vb = V(R4) + 0.7 (base voltage is alwasy 0.7V higher than emitter voltage for transistor to turn ON)
Vb = 0.9V.
6) Base current (Ib):
Ib = Ic / β ( β, beta is current in common-emitter mode and its value is found in datasheet, usually β= more than 100 and we will assume its 100)
= 1m / 100
= 10u
7) Calculation of R2:
according to 10% rule of thumb, The current through R2 shluld be 10 times the current through Ib. So,
Current through R2 (IR2) = 10 * Ib
= 10 * 10u
Current through R2 (IR2) = 100u
So, R2 = Vb / IR2
= 0.9 / 100u
or, R2 = 9kΩ
8) Calculation of R1:
Vb = [ R2 / (R1 + R2) ] * Vsupply
so, R1 = 31kΩ
9) Calculation of C1, C2 and C3:
C = 1 / (2 * pie * R * f )
or you can just use 10uf or higher. The higher the value , the less resistance it provide to ac signal.
For above circuit i choose 10uF.
Figure: Output without C2, the gain is 10 times only .
Figure: Output with C2, the gain is more than 10 times.
0 Comments