# How to polarize BJT transistor? (Part 3)

This is the third and last part of the series about BJT transistor polarization. Other circuit configurations are shown.

The other parts can be accessed by clicking on the following buttons.

### Polarization of BJT transistor by voltage division

This configuration is more stable than the emitter’s stable polarization. By that, variations of collector current variations ($I_{C}$) and collector-emitter voltage ($V_{CE}$) are very small when β changes. Exist two known methods for analyzing this circuit.

#### Accurate analysis

Using this circuit to apply accurate analysis, calculating values of $I_{C}$, $V_{CE}$, $I_{B}$, $V_{E}$ and $V_{B}$.

It’s necessary a Thévenin equivalent circuit of area inside the red quadrilateral, whose output, where there are the terminals, is pointed by the arrow.

Calculating resistance and Thévenin voltage respectively.

$R_{th}=R1||R2=\frac{R1\cdot R2}{R1+R2}=\frac{39k\cdot 8,2k}{39k+8,2k}=6,775k\Omega$

$E_{th}=\frac{R2\cdot V1}{R1+R2}=3,12V$

Applying the equation below to calculate base current $I_{B}$.

$I_{B}=\frac{E_{th}-V_{BE}}{R_{th}+(\beta +1)\cdot R_{E}}$

$I_{B}=\frac{3,12-0,7}{6,78k+(120 +1)\cdot 1k}=18,93\mu A\simeq 19\mu A$

$I_{C}=\beta \cdot I_{B}=120\cdot 19\mu =2,28mA$

Calculating collector-emitter voltage $V_{CE}$.

$V_{CE}=V_{CC}-I_{C}\cdot (R_{C}+R_{E})=18-2,28m\cdot (3,3k+1k)=$

Finding the remaining values.

$I_{E}=I_{B}+I_{C}=2,299mA\simeq 2,3mA$

$V_{E}=I_{E}\cdot R_{E}=2,3m\cdot 1k=2,3V$

$V_{BE}=V_{B}-V_{E}\rightarrow 0,7=V_{B}-2,3\rightarrow V_{B}=3V$

#### Approximate analysis

To know if you can apply approximate analysis in a circuit, you must make the following test. If the condition below is true, you can apply this analysis.

$\beta \cdot R_{E}\geq 10\cdot R_{2}$

$120 \cdot 1k\geq 10\cdot 8,2k$

$120k\geq 82k$

As the condition above is true.

$V_{B}=\frac{R_{2}\cdot V_{CC}}{R_{1}+R_{2}}$

$V_{B}=\frac{8,2k\cdot 18}{39k+8,2k}=\frac{147,6}{47,2}=3,12V$

$V_{BE}=V_{B}-V_{E}\rightarrow V_{E}=3,12-0,7=2,42V$

$I_{E}=\frac{V_{E}}{R_{E}}=\frac{2,42}{1k}=2,42mA$

In this analysis, can consider $I_{E}\cong I_{C}$. Calculating $V_{CE}$.

$V_{CE}=V_{CC}-I_{C}\cdot (R_{C}+R_{E})$

$V_{CE}=18-2,42m\cdot (3,3k+1k)=7,594V$

### DC polarization with voltage feedback

Another stable configuration, using Kirchhoff’s law for voltages in the mesh of emitter and collector’s terminals, can calculate current base as:

$I_{B}=\frac{V_{CC}-V_{BE}}{R_{B}+\beta (R_{C}+R_{E})}$

Applying the same Kirchhoff’s law in collector-emitter mesh can obtain collector-emitter voltage $V_{CE}$.

$V_{CE}=V_{CC}-I_{C}(R_{C}+R_{E})$

An important detail, current that passes by $R_{C}$ isn’t collector current $I_{C}$. But the one that enters the collector of BJT transistor.

The “Not Ic” current is the sum of base and collector currents. However, collector current and the one which passes through resistor $R_{C}$ are much higher than $I_{B}$. The equation shown before was demonstrated considering these current as approximately equal. With this proximity, saturation current $I_{Csat}$ can be calculated as:

$I_{Csat}=I_{Cmax}=\frac{V_{CC}}{R_{C}+R_{E}}$

### Other polarization configuration with BJT transistor

The BJT polarizations are common base, emitter, and collector. The circuits shown until now are common emitter because the circuit’s input is linked to base and output to collector.

Problems involving other configurations can be solved by applying Kirchhoff’s law for voltages.

#### Common base example

Considering this common base circuit.

Applying Kirchhoff’s law for voltages on the left mesh.

$-V_{EE}+R1\cdot I_{E}+V_{BE}=0$

$I_{E}=\frac{V_{EE}-V_{BE}}{R_{1}}$

$I_{E}=\frac{8-0,7}{2,2k}=3,32mA$

Calculating $V_{C}$ and considering emitter current proximately equal to collector’s ($I_{C}\cong I_{E}$).

$V_{CC}=R2\cdot I_{C}+V_{CB}$

$10=1,8k\cdot 3,32m+V_{CB}$

Since voltage in the base is zero.

$V_{CB}=V_{C}=4,024V$

And finally, finding the value of $V_{CE}$.

$V_{E}-V_{EE}=R_{1}\cdot I_{E}$

$V_{E}-(-8)=2,2k\cdot 3,32m$

$V_{E}=-8+7,304=-0,696V$

$V_{CE}=4,024-(-0,696)=4,72V$

#### Common collector example

Applying agains Kirchhoff’s law for voltages.

$V_{CC}=I_{B}R1+V_{BE}+I_{E}R2-V_{EE}$

$V_{CC}+V_{EE}-V_{BE}=I_{B}R1+I_{E}R2$

$V_{CC}+V_{EE}-V_{BE}=I_{B}R1+I_{B}(\beta +1)R2$

$I_{B}=\frac{V_{CC}+V_{EE}-V_{BE}}{R1+(\beta +1)R2}$

$I_{B}=\frac{6+6-0,7}{330k+(120 +1)1,2k}=0,02377m=23,77\mu A$

$I_{E}=(\beta +1)I_{B}=(120+1)\cdot 23,77\mu =2,88mA$

$V_{E}-V_{EE}=I_{E}R2$

$V_{E}-(-6)=2,88m\cdot 1,2k$

$V_{E}=-2,544V$

#### Common base with pnp transistor

$V_{EE}-V_{BE}=Re\cdot Ie$

$8-0,7=3,3k\cdot Ie$

$Ie=2,21mA$

Finding $Vc$.

$Vc-Vcc=Ic\cdot Rc$

$Vc-(-12)=2,21m\cdot 3,9k$

$Vc=-3,37V$