Here we illustrate how to bias a bipolar hunction transistor(BJT) using voltage divider method. We will use the online voltage biasing calculator for BJT amplifier design. We will then show simulation of the designed amplifier in multisim.

A BJT small signal amplifier is designed to amplifier small signal of amplitude from micro volt to few hundreds of milli volts. These are useful in amplifying signals from sensors which are often very small in magnitude. They can also be used to amplify audio signal such as from microphone and are used in hobby FM and AM radios.

The **BJT amplifier** in voltage divider biasing configuration is shown below.

The voltage input source V1 is ac signal type which is fed into the C1 capacitor to remove any dc bias content in it. The resistors R1, R2 creates a fixed input bias voltage into the base. The name of the biasing circuit comes from using these two resistor because it is used in voltage divider configuration. These two resistors provides a fixed voltage and current into the base of the transistor which helps to stabilize the transistor from current gain fluctuations. The emitter resistor RE also helps to stabilize the transistor as it provides feedback back to the input if there is any changes in the collector current due to beta changes. RL represents the load or load circuit where the amplified signal is designed to be applied. The capacitor C3 is output coupling capacitor which couples the amplified signal into the load circuit and which helps to isolate the BJT amplifier from the load circuit. The capacitor C2 is the bypass capacitor which is used to ground the ac signal in the emitter to the ground.

The design steps for designing BJT amplifier with voltage divider biasing method can be grouped into two stages- the DC bias and AC bias.

**DC biasing**

The DC biasing refers to the technique to create DC voltages and current at different circuit points. In this, we only use the four resistors R_{1}, R_{2}, R_{3} and R_{4} along with the DC voltage source Vcc. The aim here is create DC voltage required at the collector(Vc). The following shows the circuit diagram for DC biasing of transistor using voltage divider biasing method.

In the following circuit the collector voltage R_{C} is shown which the designer has decide upon.

Consider that the applied DC voltage source is 5V, then we can choose the collector voltage of 3V. The designer must also choose the desired collector current(I_{C}). This is required for calculating the resistor values. Let say the collector current is 5mA(that is **I _{C}=5mA**).

Using the following equation we can then determine R_{C},

\[R_{C}=\frac{V_{CC}-V_{C}}{I_{C}}\]

Using the online calculator, **R _{C}=400Ω**.

The next step is to calculate the emitter resistor Re. The emitter resistance is given by the following equation,

\(R_{E}=\frac{V_{E}}{I_{E}}\)

The voltage V_{E} is approximated here to be equal to approximately 1/4th of Vcc. We will use 1.2V in our example. So let **V _{E}= 1.2V**. The emitter current Ie is approxmiately equal to collector current Ic, so Ie = 5mA. This then gives

**R**.

_{E}=240ΩThe next step is to calculate the resistors R1 and R2. This was explained in the tutorial How to bias a Bipolar Junction Transistor using Voltage Divider Biasing Technique. To determine the resistor values we must first determine the base
voltage(V_{B}). From the circuit we can see that the dc voltage at the base is
given by,

\[V_{B}=V_{E}+V_{BE}\]

We have V_{E}=1.2V and with V_{BE}=0.75V, substituting above, we get, **V _{B}=1.95V**

The resistor divider equation is given by,

\[V_{B}=\frac{R_2}{R_1+R_2} V_{CC}\]

Let use **R _{2}=2.2KΩ** and with

**V**and

_{B}=1.95V**V**.

_{CC}=5VRearranging the above equation,

\[R_{1}=\frac{V_{CC}}{V_{E}+V_{BE}}*R_{2}-R_{2}\] we get **R _{1}=3.44KΩ**

So from DC biasing we have,

**R _{1}=3.44KΩ**,

**R**,

_{2}=2.2KΩ**R**and

_{E}=240Ω**R**

_{C}=400ΩTo calculate the DC bias resistors we can use the BJT Bias Circuit Calculator Online.

### AC Bias

The next step is to bias the transistor in ac condition. In ac analysis we have to short circuit all capacitor and the DC source. The following circuit shows the ac equivalent circuit of the BJT amplifier circuit above.

The resistors R_{1}, R_{2} and Z_{inb} forms a parallel resistor whose equivalent resistance can be denote by Z_{inb}. Similarly at the output the resistors R_{C} and R_{L} forms a parallel resistor whose equivalent we will denote by r_{c}.

We can redraw the above circuit as follows.

The equivalent impedance **Z _{in}** is given by,

\( \frac{1}{Z_{in}}=\frac{1}{R_{1}}+\frac{1}{R_{2}}+\frac{1}{Z_{inb}}\)

because, \(Z_{inb}=\beta r_{e}\)

the above equation can be rewritten as,

\(\frac{1}{Z_{in}}=\frac{1}{R_{1}}+\frac{1}{R_{2}}+\frac{1}{\beta r_{e}} \)

then rearranging we get,

\(Z_{in}=\frac{R_{1}R_{2}\beta r_{e}}{R_{1}R_{2}+R_{1}\beta r_{e}+R_{2}\beta r_{e}}\)

From solid state physics we have,

\(r_{e}=\frac{25mV}{I_{E}}\)

with Ie = 5mV, we get **r _{e}** = 5Ohm

with re=5Ohm and beta=100 we get **Zinb=500Ohm**

with re=5Ohm, R1=3.44KOhm, R2=2.2KOhm, beta=100 we get **Zin=365Ohm**

The resistance **r _{c}** can be calculated as follows,

\(\frac{1}{r_{c}}=\frac{1}{R_{C}}+\frac{1}{R_{L}}\), that is,

\(r_{c}=\frac{R_{C}R_{L}}{R_{C}+R_{L}}\)

with R_{C}=400Ω and **R _{C}=100KΩ** we get

**=398Ohm****r**_{c} Having determined the input impedance of the amplifier(Z_{in}) and the output impedance(r_{c}) we can calculate the coupling capacitor C_{1}, C_{3} and the bypass capacitor C_{2}. If we insert the coupling capacitor back to the equivalent ac circuit we will get,

Now we proceed to calculate the values for coupling capacitor C_{1} and C_{3}.

The reactance due to capacitor** C _{1}** at the operating frequency should be much smaller than the Z

_{in}impedance. For this we can use,

\(X_{c1}<0.1 Z_{in}\)

Since,

\(X_{c1}=\frac{1}{2 \pi f C_{1}}\)

therefore,

\(C_{1}=\frac{1}{2 \pi f 0.1 Z_{in}}\)

with f=1KHz, Zin=365Ohm we get **C _{1}=4.37uF**

Similarly, the reactance due to** C _{3}** should be less than 0.1r

_{c}at the operating frequency that is,

\(X_{c3}<0.1 r_{c}\)

rearranging we get,

\(C_{3}=\frac{1}{2 \pi f 0.1 r_{c}}\)

with f=1KHz and rc=398 Ohm we have **C _{3}=4uF**

The value for the **bypass capacitor C _{2}** is calculated as follows,

\(X_{c2}<0.1 R_{E}\)

which on substituting \(X_{c2}=\frac{1}{2 \pi f C_{2}}\) and rearranging results in the following,

\(C_{2}=\frac{1}{2 \pi f 0.1 R_{E}}\)

with f=1KHz, R_{E}=240Ω we get **C _{2}=318.47uF**

The final designed BJT amplifier circuit with calculated values is below.

The above circuit was simulated with Multisim and following are the signal traces at the input and output of the BJT amplifier.

To understand which signal traces belongs which node in the circuit see below.

For implementation and real time testing seeHow to Build BJT amplifier and test with Soundcard based PC Oscilloscope using VDB.

For other biasing method for BJT amplifier design see the following:.

- How to bias a BJT using voltage divider biasing

- How to Design Collector-Emitter Feedback biased BJT Amplifier

- How to bias a BJT using base bias

- How to design emitter biased BJT amplifier