Common emitter amplifier circuit: the base (b) of triode inputs small signal, and the emitter (c) outputs large signal.

The common emission amplifier circuit amplifies both current and voltage. （1） DC potential of each part

Calculate the DC potential to determine the appropriate static working point to avoid waveform distortion.

Vb = ［R2/（R1+R2）］*Vcc

Ve = Vb – 0.7

Ic = Ie = Ve/Re

Vo = Vcc – Ic*Rc

（2） AC voltage magnification

Δ Ie = VI / re (base voltage and emitter voltage AC equipotential ve = VI)

Δ Vc = Δ ic*Rc = Δ ie*Rc = ［Vi/Re］*Rc

Vo = Δ Vc = Vi/Re*Rc

Av = Vo/Vi = Rc/Re

Conclusion: the amplification factor AV is independent of the DC current amplification factor HFE of the transistor, but is determined by the ratio of RC to re

(since the base current is considered to be 0, it has nothing to do with HFE, but strictly speaking, it has something to do with it.)

Note: the maximum value of the input voltage corresponds to the minimum value of the voltage at point C (ensure that the triode works in the amplification area).

Design process of common emission amplifier circuit:

1. Select the power supply voltage according to the input and output signal amplitude

In order to output an output voltage of 5V, a power supply voltage of more than 5V must be used. The minimum voltage of 1 – 2V is added to the emitter resistance re, so in order to make the collector current flow, the minimum power supply voltage must be 6 – 7V ((5 + 1) – (5 + 2) V).

The most easily available 15V power supply is used in this experiment.

2. Select RC according to load RL

RL =100kΩ Rc = 100kΩ * 1/10 = 10kΩ

3. Set re according to the magnification AV.

Re = Rc / 5 = 2kΩ

4. Calculate the base bias voltage VB according to RC and re

Let VC = VCC * 1 / 2 = 7.5 v

Ic = Vc / Rc = 0.75mA

Ve = Ie * Re = Ic * Re = 1.5V

Vb = Ve + 0.7 = 2.2V

5. Calculate R1 and R2 according to VB and re

In order to make the base current IB small enough, the current flowing on R2 should be large, that is, the resistance value of R2 should be small enough.

Re‘ = β Re = 200kΩ

［ Ib = Ie/ β = Ve/ β Re VI = ve (base set and collector AC equipotential) re ‘= VI / IB can deduce re’= β Re ］

R2 = Re’/10 = 20kΩ

I（R2） = I（R1） = Vb / R2 = 2.2 / 20 mA = 0.11mA

R1 = （Vcc – Vb）/ I（R1） = 116kΩ

R1 value does not exist in the resistance of E24 system series, so R1 = 120K Ω and R2 = 20K Ω are taken.

5. Calculate C1 and C2 according to signal frequency f, RI and ro.

C1 and RI constitute a high pass filter; C2 and RL form a high pass filter.

The selection of electrolytic capacitor is related to the reference frequency. The impedance of electrolytic capacitor to signal frequency is 0.

As long as the cut-off frequency of the high pass filter is lower than 1 / 10 of the signal frequency, it can be considered that the impedance to the input signal is 0.

f = 1/ （2πRiC1） 《 1KHz*1/10

Ri = R1 // R2 // Re‘

C1 = 1/（200*πRi） = 102 nF

C2 = 1/（200*πRL） = 15.9 nF

Here, take C1 = C2 = 10 UF

6. Input / output waveform 