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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.

How to polarize BJT transistor? (Part 1) Click here

How to polarize BJT transistor? (Part 2) Click here

Polarization of BJT transistor by voltage division

BJT polarization 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}.

Example of VD circuit

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.

Applying Thévenin to BJT polarization

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

Polarization of BJT transistor by 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.  

Explaining-currents-EN

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.

Polarizations of BJT transistor
(A) common base, (B) common emitter and (C) common colector. Source: Electrical4u.

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

Common base example

Considering this common base circuit.

Must find emitter’s current Ie, collector voltage V_{C} and the voltage between emitter and collector V_{CE}. R1 = emitter resistor and R2 = collector resistor.

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

common collector circuit
Must calculate I_{E} and V_{E}. R1 = base resistor and R2 = emitter resistor.

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

Common base with PNP
In this problem, must calculate emitter’s current Ie and voltage collector Vc.

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

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