2 common base BJT amplifier projects

The common base is little known configuration for amplifiers using BJT. This type is used in high frequencies.

Common base amplifier with NPN transistor

First, the transistor must be defined, in this project was used the BC548, which has many datasheets that can be found in this link. The supply voltage (VCC) was arbitrarily chosen as 15V. While input impedance, which must have a low value, was considered as 50Ω.

Defining emitter and collector resistors

The value of resistance on emitter (Re) was chosen as 2.2kΩ. Below is the equation of input impedance for common base amplifiers.

$Zi=\frac{Re\cdot r_{e}}{Re+r_{e}}$

$50=\frac{2200\cdot r_{e}}{2200+r_{e}}$

$r_{e}=51.1\Omega$

Finding the current emitter value.

$r_{e}=\frac{V_{T}}{I_{E}}$

$I_{E}=\frac{26m}{51,1}=0.508mA$

Calculating emitter ($V_{E}$) and base ($V_{B}$) voltages. Knowing that base-emitter voltage ($V_{BE}$) is 0.7V.

$V_{E}=Re\cdot I_{E}=22k\cdot 0.508m=1.1176V$

$V_{BE}=V_{B}-V_{E}$

$V_{B}=1.876V$

Collector voltage value ($V_{C}$) is arbitrary. However, it must be higher than $V_{B}$, otherwise the circuit won’t amplify the signal. As was explained in the first part of BJT polarization. I choose $V_{C}$ as 7.5V, half of VCC. The collector current ($I_{C}$) and emitter current ($I_{E}$) can be considered as equal.

$I_{E}\simeq I_{C}=0,508mA$

$Rc=\frac{VCC-V_{C}}{I_{C}}=14.76k\Omega$

The closest commercial value is 15kΩ. In common base, output impedance ($Z_{o}$) has the same value of Rc.

Defining base resistors

Using voltage divider equation.

$V_{B}=\frac{Rb2\cdot VCC}{Rb2+Rb1}$

$1.876=\frac{Rb2\cdot 15}{Rb2+Rb1}$

$Rb1=6.996\cdot Rb2$

Considering Rb2 as 1kΩ, Rb2 is equal to 6.996kΩ. The closest commercial available value is 8.2kΩ.

Calculating capacitors

Equation to calculate input capacitor value.

$f=\frac{1}{2\pi \cdot Ci\cdot Zi}$

Input impedance $Zi$ is 50Ω and frequency arbitrary value is 100kHz.

$Ci=\frac{1}{2\pi fZi}=\frac{1}{2\pi \cdot 10^{5}\cdot 50}=31,83nF$

The closest commercial available value is 47nF, which gives a cutoff frequency of 67.7kHz. Can also use one of 33nF. Designing output capacitor, where output impedance ($Zo$) is 15kΩ.

$Co=\frac{1}{2\pi fZo}=\frac{1}{2\pi \cdot 10^{5}\cdot 15000}=106pF$

The commercial value is 100pF. Calculating bypass capacitor Cb, cutoff frequency was arbitrarily chosen as 500kHz.

$Cb=\frac{1}{2\pi fRb2}=\frac{1}{2\pi \cdot 5\cdot 10^{5}\cdot 10^{3}}=318pF$

The commercial value is 270pF, which gives a cut frequency of 589.4kHz.

Common base amplifier with PNP transistor

The used BJT is BC558, whose datasheets are on this link. Input impedance (Zi) is 50Ω again and supply voltage is -20V.

Calculating resistors

Re was chosen as 1.8kΩ.

$Zi=\frac{Re\cdot r_{e}}{Re+r_{e}}$

$50=\frac{1800\cdot r_{e}}{1800+r_{e}}$

$r_{e}=51.4\Omega$

Finding collector current. With PNP transistor, it has a negative value.

$r_{e}=\frac{V_{T}}{\left| I_{E}\right|}$

$\left| I_{E}\right|=\frac{26m}{51.4}=0.506mA$

Calculating voltages on BC558’s terminals. Since transistor is PNP, voltages must be negative.

$V_{E}=Re\cdot I_{E}=-0.9108V$

$V_{BE}=V_{B}-V_{E}$

$-0.7=V_{B}+0.9108$

$V_{B}=-1.6108V$ $V_{C}$ value is arbitrarily, but it must be lower than $V_{B}$, because it’s a PNP amplifier. $V_{C}$ is equal to -10V, half of supply voltage. Calculating collector resistance.

$Rc=\frac{VDC-V_{C}}{I_{C}}$

$I_{E}\simeq I_{C}=-0.506mA$

$Rc=\frac{-20+10}{-0.506m}=19.7k\Omega$

Commercial value of Rc is 22kΩ and it’s equal to output impedance Zo. Calculating base resistors.

$V_{B}=\frac{Rb2\cdot VCD}{Rb2+Rb1}$

$-1.6108=\frac{-20\cdot Rb2}{Rb2+Rb1}$

$-1.6108\cdot Rb1=-18.3892\cdot Rb2$

Arbitrary value: Rb2=1kΩ.

$Rb1=11.41k\Omega$

Rb1 commercial value is 12kΩ.

Calculating capacitors

Defining input capacitor (Ci) arbitrarily as 10nF and calculating cutoff frequency f.

$f=\frac{1}{2\pi \cdot Ci\cdot Zi}=\frac{1}{2\pi \cdot 10^{-8}\cdot 50}=318.3kHz$

Calculating output capacitor Co.

$f=\frac{1}{2\pi \cdot Co\cdot Zo}$

$Co=\frac{1}{2\pi \cdot 318.3\cdot 10^{3}\cdot 22\cdot 10^{3}}=22.7pF$

The commercial value is 22pF. Designing the bypass capacitor Cb, cutoff frequency is 1MHz.

$Cb=\frac{1}{2\pi f\cdot Rb2}=\frac{1}{2\pi \cdot 10^{6}\cdot 10^{3}}=159pF$

Cb commercial value is 150pF. The complete amplifier below.

Common base amplifiers’ video

The common base amplifier must receive a signal with very low amplitude, otherwise the output signal will be saturated and will have a different shape from input signal. Blue=input and yellow=output.