# How to bias a BJT using voltage divider biasing

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 R1, R2, R3 and R4 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 RC 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(IC). This is required for calculating the resistor values. Let say the collector current is 5mA(that is IC=5mA).

Using the following equation we can then determine RC,

$R_{C}=\frac{V_{CC}-V_{C}}{I_{C}}$

Using the online calculator, RC=400Î©.

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 VE is approximated here to be equal to approximately 1/4th of Vcc. We will use 1.2V in our example. So let VE= 1.2V. The emitter current Ie is approxmiately equal to collector current Ic, so Ie = 5mA. This then gives RE=240Î©.

The 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(VB). From the circuit we can see that the dc voltage at the base is given by,

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

We have VE=1.2V and with VBE=0.75V, substituting above, we get, VB=1.95V

The resistor divider equation is given by,

$V_{B}=\frac{R_2}{R_1+R_2} V_{CC}$

Let use R2=2.2KÎ© and with VB=1.95V and VCC=5V

Rearranging the above equation,

$R_{1}=\frac{V_{CC}}{V_{E}+V_{BE}}*R_{2}-R_{2}$ we get R1=3.44KÎ©

So from DC biasing we have,

To 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 R1, R2 and Zinb forms a parallel resistor whose equivalent resistance can be denote by Zinb. Similarly at the output the resistors RC and RL forms a parallel resistor whose equivalent we will denote by rc

We can redraw the above circuit as follows.

The equivalent impedance Zin 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 re = 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 rc 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 RC=400Î© and RC=100KÎ© we get rc=398Ohm

Having determined the input impedance of the amplifier(Zin) and the output impedance(rc) we can calculate the coupling capacitor C1, C3 and the bypass capacitor C2. 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 C1 and C3

The reactance due to capacitor C1 at the operating frequency should be much smaller than the Zin 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 C1=4.37uF

Similarly, the reactance due to C3 should be less than 0.1rc 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 C3=4uF

The value for the bypass capacitor C2 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, RE=240Î© we get C2=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 design emitter biased BJT amplifier