# 2 common collector BJT amplifier projects

In this post, it is shown another type of amplifier configuration using BJT. This time, it is the common collector or emitter follower.

## Common collector amplifier with NPN transistor

The chosen transistor was the 2N3904, whose datasheet is on this link.

### Calculating resistors linked to transistor’s base

First, you must determine the voltage on transistor’s base. This value must be lower than supply voltage ($Vcc$) and higher than base-emitter voltage ($V_{BE}$). The 2N3904’s datasheet shows the possible range of V_{BE}. For this project, the chosen value is 0.65V. Source: 2N3904 datasheet.

The formula to obtain voltage base ($V_{b}$).

$V_{b}=V_{BE}+\frac{(V_{cc}-V_{be})}{2}$

$V_{cc}$ value was chosen arbitrarily as 8V.

$V_{b}=0.65+\frac{(8-0.65)}{2}=4.325V$

Using the voltage division equation.

$V_{b}=\frac{R_{b2}}{R_{b2}+R_{b1}}\cdot V_{cc}$

$4.325=\frac{R_{b2}}{R_{b2}+R_{b1}}\cdot 8$

$0.5406=\frac{R_{b2}}{R_{b1}+R_{b2}}$

$R_{b1}=0.8498\cdot R_{b2}$

The closest commercial values of $R_{b1}$ and $R_{b2}$ that meet this proportion are 1kΩ and 1.2kΩ, respectively.

### Calculating resistance on emitter

Calculating emitter voltage.

$V_{BE}=V_{b}-V_{e}$

$0.65=4.325-V_{e}$

$V_{e}=3.675V$

To design the resistor Re, it’s necessary define the emitter current ($I_{e}$). The arbitrarily defined value is 10mA, because it’s a safe value to prevent transistor overheating.

$V_{e}=R_{e}\cdot I_{e}$

$R_{e}=367.5\Omega$

The closest available commercial value is 470Ω, which produces an emitter current of 7.81mA.

### Designing capacitors

The formula to calculate the input capacitor Ci.

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

Where $Ri$ is the input resistance saw by Ci capacitor.

$Ri=R_{b1}||R_{b2}||\beta r_{e}$

In this case, it wasn’t possible to measure β of transistor on multimeter. On “ON CHARACTERISTICS” part of datasheet shown before, the gain range goes from 100 to 300, when collector current ($I_{C}$) is 10mA. The gain can be considered as 100. Calculating $r_{e}$ and $Ri$.

$r_{e}=\frac{26mV}{I_{E}}=\frac{26m}{7,81m}=3,329\Omega$

$R_{b1}||R_{b2}=\frac{R_{b1}\cdot R_{b2}}{R_{b1}+R_{b2}}=523\Omega$

$\beta\cdot r_{e}=332,9$

$Ri=\frac{523\cdot 332,9}{523+332,9}=203,42\Omega$

With cut frequency 20Hz, input capacitance is:

$Ci=39,119\cdot 10^{-6}F$

The closest commercial value is $47\cdot 10^{-6}F$ or 47μF. When higher Ci capacitance, lower will be the cut frequency. Below is the equation to calculate capacitance of Co.

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

The formula to calculate the output resistance saw by capacitor.

$Ro=\frac{R_{b1}||R_{b2}}{\beta}$

$Ro=5,23\Omega$

$Co=1521,5\mu F$

The closest commercial value is $2200\mu F$.

## Common collector amplifier with PNP transistor

In this project, it’s used the PNP transistor 2N3906, whose datasheet is on this link.

### Calculating the resistors linked to transistor base

Using the same procedure on previous project, considering negative voltages. The chosen supply voltage is -9V. The base-emitter voltage ($V_{BE}$) is -0.65V.

$V_{B}=V_{BE}+\frac{V_{CC}-V_{BE}}{2}$

$V_{B}=-0.65+\frac{-9+0.65}{2}=-4.825V$

$V_{B}=\frac{R_{b2}\cdot V_{CC}}{R_{b2}+R_{b1}}$

$-4.825=\frac{-9\cdot R_{b2}}{R_{b2}+R_{b1}}$

$1.15R_{b1}=R_{b2}$

The commercial values of $R_{b1}$ and $R_{b2}$ are 1kΩ e 1.2kΩ, respectively. Calculating Re resistance, whose emitter current ($I_{E}$) is 15mA, a value arbitrarily chosen.

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

$V_{E}=-4,175V$

$V_{E}=R_{E}\cdot I_{E}$

$R_{E}=278\Omega$

The commercial value is 330Ω, resulting in a $I_{E}$ of -12.6mA. Calculating Ci and Co values.

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

I choose the cut frequency $f$ as 5000Hz.

$Ri=R_{b1}||R_{b2}||\beta r_{e}$

$\beta$ can also be considered as 100.

$r_{e}=\frac{26m}{\left| I_{E}\right|}=2,63\Omega$

$\beta r_{e}=263$

$R_{b1}//R_{b2}=\frac{1k\cdot 1,2k}{2,2k}=545\Omega$

$Ri=\frac{545\cdot 263}{545+263}=177,39\Omega$

$Ci=\frac{1}{2\pi 5000\cdot 177,39}=179,44\cdot 10^{-9} F$

Ci commercial value is 100nF. Designing Co.

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

$Ro=\frac{R_{b1}//R_{b2}}{\beta}=5,45\Omega$

$Co=\frac{1}{2\pi 5000\cdot 5,45}=5,84\cdot 10^{-6}F$

The closest commercial value is 10μF.

## Common collector amplifiers video

In the video, blue signal is the input and yellow is output. The type of amplifier gain is 1, therefore, output signal isn’t amplified. 