  24 February 2009
Author: Giorgos Lazaridis
BJT Transistor theory

In the previous section we learned how can someone analyze a transistor circuit. But if someone wants to design an amplifier, the methodology is different. In this section we will discuss the design methodology for the three different transistor connections.

Common Base Transistor Design Methodology

A common base amplifier is usually used as an input amplification stage for high frequency applications (small signal amplifier). We will not discuss much about this type of connection, since it is very rarely used. Here is a typical Common Base transistor amplifier schematic with an NPN transistor: There will be a lot of mathematical functions and equations that have to be solved to analyze the circuit, and therefore i made an on-line calculator. You can use it to speed up your work. Here is the link:

• Common Base transistor amplifier - Analysis Calculator

• The resistor Rg is the resistance of the AC signal generator. The amplifier is biased with Voltage Divider (RB1-RB2). The output is taken from the collector and the input signal is applied at the emitter.

The designer must know the input signal size (amplitude and frequency range) and the required amplification factor. Also, the designer must know the DC voltage that will be used for the transistor bias (VDD). A common base amplifier has no current amplification, therefore we only need to calculate the voltage amplification. The design can begin with the DC equivalent:

First thing we can define with an educated guess is the RE resistor. We know that the AC input must be able to provide enough current for the emitter and the collector circuit. Therefore, we want to maintain the emitter current as low as possible. We can safely choose a large resistor such as 10 Kohms for RE.

Usually, we place the Q point in the middle of VCE. This means that:

VCEQ = VDD / 2

It can be proved that the voltage gain is calculate from this formula:

Av = RC / ri

To calculate ri we must first calculate the emitter resistor r'e. But we cannot calculate r'e since we do not know the emitter current. So we must find a way to calculate the emitter voltage first. Let's remember the formula for the internal emitter resistor:

r'e = 25mV / IC = 25mV / IE

We replace this formula to the gain amplification formula:

Av = RC / ( 25mV / IE ) => Av = IE x RC / 25mV = IC x RC / 25mV = VC / 25mV => VC = Av x 25mV

From the above formula we can directly calculate the collector voltage VC. Now we can use the Kirchhoff's law to calculate the emitter voltage:

VDD = VC + VCE + VE => VE = VDD - VC - VCE

Now we can calculate the base voltage;

VB = VBE + VE => VB = 0.7 + VE

So now we can calculate the emitter current:

IE = VE / RE

And the emitter AC resistance will be:

r'e = 25mV / IE

From the voltage divider formula, we can now calculate the resistor ratio needed to get this voltage:

VB = VDD x RB2 / ( RB1 + RB2 )

Let's assume that we choose to design a firm voltage divider. This means that the current through the voltage divider (IVD) must be at least 10 times greater than the base current. In other words:

IVD = 10 x IB => Rvd = 0.1 x β x RE => RB1 + RB2 = 0.1 x β x RE

From the two equations above we have:

VB = VDD x RB2 / (0.1 x β x RE)

And after some steps we calculate RB2:

RB2 = (VB x 0.1 x β x RE) / VDD

We can now calculate RB1 as follows:

RB1 = RB2 x (VDD - VB) / VB

Relative pages
• Basic transistor circuits
• 555 theory of operation
• Learn how to interface ICs
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• The TRIAC theory
• Learn about the most popular PC Cooling methods
• Dr.Calculus: Checking transistor functionality
• Op-Amp IC Pinouts

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