There are number of methods to bias a bipolar junction transistor(BJT) to design an amplifier such as self bias, voltage divider bias, collector feedback bias etc. One of the biasing technique is the base biasing. The base biasing method forms the basis or foundation for designing and understanding other biasing methods. Base bias method creates a fixed base current. Here we illustrate base biasing method to bias a transistor for creating an amplifier.
For real implementation and testing of this see How to build base biased BJT amplifier on breadboard and test with PC soundcard based oscilloscope.
Base Biased BJT Amplifier
The following circuit diagram shows a base biased BJT amplifier in common emitter configuration.
In the circuit diagram above, the transistor is dc biased using the base resistor Rb and collector resistor Rc. Vcc supplies the DC voltage. The C1 and C2 are coupling capacitor. The capacitor C1 couples the input ac signal V1 into the base of the transistor and the capacitor C2 couples the output signal at the collector to the load RL.
Design Constraints
To bias the transistor and for amplifier design we have to calculate the resistors values Rb and Rc, the capacitor values C1 and C2. But to calculate these values we must know the characteristics of the input ac source(frequency and amplitude) and the desired output.
Let us assume that the ac voltage source has magnitude of 10mA and that the frequency is 1KHz. Let say we want to have collector voltage of 3V and collector current of 5mA. The transistor is assumed to be small signal transistor like 2N3904 or 2N2222A. Let the beta or dc gain(beta) be 100 and that the base-emitter junction voltage Vbe is 0.75V.
Design Steps
To design the amplifier we can use two design stages- dc bias circuit and ac bias circuit.
DC biasing of base bias circuit
The first step in amplifier design is creating fixed dc voltage at various point in the circuit. In the dc bias analysis, we remove the ac source and its effecting component such as the coupling capacitor. The load resistor is also removed. The following shows the dc biasing circuit diagram for base bias.
Calculating Collector Resistor(Rc)
We determine the collector resistor Rc using the following equation,
\(V_{CC}=I_{C} R_{C}+V_{C}\)
Rearranging,
\(R_{C}=\frac{V_{CC} -V_{C}}{I_{C}}\)
with our earlier assumption, Ic=5mA and V=3V and therefore,
\(R_{C}=\frac{5V -3V}{5mA}\)
that is, \(R_{C}=400\Omega\)
Calculating Base Resistor(Rb)
The following equation is used to calculate base resistor Rb,
\(V_{CC}=I_{B} R_{B}+V_{BE}\)
Rearranging,
\(R_{B}=\frac{V_{CC} -V_{BE}}{I_{B}}\)
Since, \(I_{C}=\beta I_{B}\) and with \(\beta=100\), \(R_{C}=400\Omega\) calculated above,
\(I_{B}=\frac{I_{C}}{\beta}=\frac{5mA}{100}=50\mu A\)
with our earlier assumption, Vbe=0.75V we get the value of Rb as follows,
\(R_{B}=\frac{5V -0.75V}{50\mu A}\)
that is, \(R_{B}=85k\Omega\)
Hence with these component value the DC biased base biased circuit is shown below.
AC biasing of base bias circuit
To amplify an signal we have to couple the ac signal source to the DC biased circuit above. Also we have to couple the output signal from the amplifier to a load.
The ac equivalent of base biased circuit is shown below.
The impedance due to the transistor Zinb is,
because, \(Z_{inb}=\beta r_{e}\)
and from solid state physics is,
\(r_{e}=\frac{25mV}{I_{E}}\)
thus,
\( Z_{inb}=\frac{\beta*25mV}{I_{E}} \)
with Ie=5mA and \(\beta=100\)
\( Z_{inb}=\frac{100*25mV}{5mA} =500\Omega\)
The resistors \(R_{B}\) and \(Z_{inb}\) forms a parallel resistor whose equivalent impedance can be calculated as follows,
\(\frac{1}{Z_{in}}=\frac{1}{R_{B}}+\frac{1}{Z_{inb}}\)
or, \(Z_{in}=\frac{R_{B} Z_{inb}}{R_{B}+Z_{inb}}\)
With Rb=85Kohm and Zin=500Ohm, we get,
\(Z_{in}=\frac{85k\Omega*500\Omega}{85k\Omega+500\Omega}\)
that is, \(Z_{in}=472\Omega\)
Similarly, the resistor Rc and RL forms a parallel resistor whose equivalent resistance rc is given by,
\(\frac{1}{r_{c}}=\frac{1}{R_{C}}+\frac{1}{R_{L}}\)
or, \(r_{c}=\frac{R_{C} R_{L}}{R_{C}+R_{L}}\)
with Rc=400ohm and RL=10kohm, we get
\(r_{c}=\frac{400\Omega*10k\Omega}{400\Omega+10k\Omega}\)
that is, \(r_{c}=384\Omega\)
If we insert the coupling capacitors we get the following circuit,
Calculating the coupling capacitors
The coupling capacitor C1 reactance of Xc1 which should be less than 0.1 of Zin, that is,
Xc1 < 0.1 Zin
as, \(X_{C1}=\frac{1}{2 \pi f C_{1}}\)
we get,
\(C_{1}=\frac{1}{2 \pi f 0.1 Z_{in}}\)
with f=1KHz, Zin=472Ohm
\(C_{1}=\frac{1}{2* \pi *1khz *0.1*472\Omega}\)
that is, \(C_{1}=3.37 \mu F\)
Similarly, the coupling capacitor C2 reactance Xc2 should be less than 0.1 of rc, that is,
Xc2 < 0.1 rc
as, \(X_{C2}=\frac{1}{2 \pi f C_{2}}\)
we get,
\(C_{2}=\frac{1}{2 \pi f 0.1 r_{c}}\)
with f=1KHz, rc=384Ohm
\(C_{2}=\frac{1}{2* \pi *1khz *0.1*384\Omega}\)
that is, \(C_{2}=4.14 \mu F\)
The final base biased BJT amplifier circuit diagram is shown below,
The following shows the transient analysis graph showing various signals at different nodes.
The base biased based BJT amplifier is easy to design than other circuit and also easy to understand the underlying theory of BJT amplifier analysis. The draw back of the base biased BJT amplifier is that the beta is not stable. That is under various enviroment conditions or due to transistor replacement or due to variation of beta among same transistor model, the amplifier may not have stable voltages and currrent(due to variation of beta).
A better approach to biasing a transistor is emitter biased circuit such as voltage divider circuit. See the tutorial how to bias a BJT using voltage divider biasing for this. Also during BJT amplifier design we want the output collector current and collector to emitter voltage lie in the active region of operation. For this it is helpful to plot characteristics curve of a transistor. The tutorial How to plot BJT characteristic curves in Multisim shows how to do this in multisim.
See BJT amplifier design using other biasing techniques.
- How to Design Collector-Emitter Feedback biased BJT Amplifier
