# How to polarize BJT transistor? (Part 2)

This is the second part about BJT polarization. Are shown projects of circuits with fixed and emitter’s stable polarization.

If you aren’t from the field, I suggest to read the operation of BJT transistor and part 1 about this component’s polarization first.

### BJT fixed polarization

This is an example of fixed polarization circuit with NPN transistor. Emitter terminal is linked to ground.

Considering these information:

• Base current ($I_{b}$): 20 μA.
• Emitter current ($I_{e}$): 4 mA.
• Voltage between colector and emitter ($V_{CE}$): 7.2 V.

We have to find these information:

• Colector current ($I_{c}$).
• Beta (β).
• Supply voltage ($V_{CC}$).
• Base resistor ($R_{b}$).

Calculating $I_{c}$.

$I_{e}=I_{c}+I_{b}$

$I_{c}=4000\mu -20\mu = 3980\mu =3.98mA$

Finding out supply voltage $V_{CC}$.

$I_{c}=\frac{V_{CC}-V_{CE}}{R_{c}}$

$3.98m=\frac{V_{CC}-7.2}{2.2k}$

$V_{CC}=15.956 V$

Finding β and $R_{b}$.

$I_{c}=\beta \cdot I_{b}$

$\beta= \frac{3.98m}{20\mu }=199$

$R_{b}= \frac{V_{CC}-V_{BE}}{I_{b}}$

Considering $V_{BE}$ as 0.7 V.

$R_{b}= \frac{15.956-0.7}{20\mu }=762 k\Omega$

#### Designing current sources with fixed polarization

This is a current source project with a NPN BD139 transistor, whose datasheet is in this link. Voltage supply is 6 Volts, while current colector must be 100 mA. Transistor must be in saturation state.

Looking at datasheet, you can see that in saturation state, voltage $V_{CE}$ is 0.5 V.

Calculating $R_{c}$.

$R_{c}=\frac{V_{CC}-V_{CE}}{I_{c}}=\frac{6-0.5}{0.1}=55\Omega$

Collector’s resistor will dissipate sizeble power, therefore, must have a size bigger than 1/4 W resistors to avoid burn. The chosen commercial value is 47 Ω, because it was the closest I had with enough size to stand electric current.

To calculate base current $I_{b}$, current gain $h_{FE}$ or β was considered 25.

$I_{b}=\frac{I_{c}}{h_{FE}}=\frac{0.1}{25}=4mA$

Determining value of $R_{b}$.

$R_{b}=\frac{V_{CC}-V_{BE}}{I_{b}}=\frac{6-1}{4m}=1.25k\Omega$

Commercial value of $R_{b}$ is 1.2 kΩ.

What if use a PNP transistor like BD136? Just invert DC source’s poles, emitter must be connected to positive pole.

This is another fixed polarization project, collector current was arbitrarily chosen as 150 mA. While gain, $h_{FE}$ or $\beta$, is 40.

Calculating Rc.

$Rc=\frac{V_{CC}-V_{CE}sat}{I_{C}}=\frac{6-0.5}{150m}=36\Omega$

Finding Rb’s value. Parameters in the datasheet are negative, but since poles were exchanged, you can consider these values as positives.

$I_{b}=\frac{0,15}{40}=3,75mA$

$R_{b}=\frac{5}{3,75m}=1,33k\Omega$

The chosen commercial values for $R_{c}$ and $R_{b}$ are 30 Ω and 1,2 kΩ respectively. The value of $I_{c}$ can vary between 149 and 155 mA.

### Stable emitter polarization

A resistor between emitter terminal and ground improves circuit’s stability. Mathematically demonstrating using two following circuits as examples.

Calculating collector current $I_{c}$ and colector-emitter voltage $V_{CE}$ of T1 circuit, we have calculated values below. You can confirm considering base-emitter voltage $V_{BE}$ as 0.7 V.

$I_{c1}=2.925mA$

$V_{CE1}=8.089V$

If T1’s β is 135, values are:

$I_{c2}=4,38mA$

$V_{CE2}=4,174V$

Calculating the percentage of parameters’ variation using the following equations.

$%\Delta I_{c}=\left |\frac{I_{c}2-I_{c}1}{I_{c}1}\right |\cdot 100%$

$%\Delta V_{CE}=\left |\frac{V_{CE}2-V_{CE}1}{V_{CE}1}\right |\cdot 100%$

Results are respectively 49.4% and 48.3%. Now, calculating the same parameters in T2 circuit. This equation can be used to find base current in emitter’s stable polarization circuit. R3 is emitter’s resistor. While $V_{BE}$ is 0.7 V.

$I_{b}=\frac{V_{CC}-V_{BE}}{R_{b2}+(\beta +1)\cdot R3}$

Results are:

$I_{c}=2.9 mA$

$V_{CE}=8.6465V$

What if T2’s β is increased to 150?

$I_{c}=3.9mA$

$V_{CE}=4.75V$

Using percentage equations shown before.

$%\Delta I_{c}=\left | \frac{3.9m-2.9m}{2.9m}\right |\cdot 100=34.4%$

$%\Delta V_{CE}=\left | \frac{4.75-8.647}{8.647}\right |\cdot 100=45%$

T2 circuit is more stable, because collector current and collector-emitter voltage vary less when β increases 50%.

#### One more project

This is circuit project with a stabilizer resistor. Transistor used is BC547B, supply voltage $V_{CC}$ is 6 V and collector current must be 20 mA. Looking at this graphic from datasheet, you can see that current base can be 0.15 mA when $V_{CE}$ is 0.2 V.

Using equation to find out values of $R_{b}$ and $R_{e}$. β is:

$\beta =\frac{20m}{0.15m}=133.3$

Defining $R_{b}$ and $R_{e}$.

$I_{b}=\frac{V_{CC}-V_{BE}}{R_{b}+(\beta +1)\cdot R_{e}}$

$0.15m=\frac{6-0.7}{R_{b}+(133.3 +1)\cdot R_{e}}$

$0.15m\cdot R_{b}+20.145m\cdot R_{e}=5.3$

The value chosen arbitrarily for $R_{b}$ was 2.7 kΩ.

$0.405+20.145m\cdot R_{e}=5.3$

$R_{e}=242\Omega$

The commercial value of $R_{e}$ is 220 Ω. Since emitter current is the sum of $I_{b}$ and $I_{c}$, it’s value is 20.15 mA. Calculating emitter’s voltage.

$V_{e}=220\cdot 20.15m=4.433V$

Discovering voltage drop in $R_{c}$.

$V_{c}=6-(V_{CE}-V_{e})=6-(0,2+4,433)=1,367V$

$R_{c}=\frac{1,367}{20m}=0,06835k=68,35\Omega$

Fortunately, I have a 68 Ω resistor. Finally, assembling the circuit in a protoboard and making measurements.

Current value $I_{c}$ shown above aren’t exactly 20 mA. Because resistors have tolerance, causing a small variation in resistance.

Other parts about projects with BJT transistors will come. 