  24 February 2009
Author: Giorgos Lazaridis
BJT Transistor theory

A Common Base amplifier example

We want to design a small signal common base amplifier. The input is 10mV at 1KHz. The voltage gain must be around 10 (Av=10). The transistor is a BC548 NPN with Î²=100. The DC supply (VDD) is 12V.

Here is the schematic diagram of the amplifier: We extract and calculate the DC equivalent: We start with an estimation of RE. RE = 10K VCE = VDD / 2 => VCE = 12 / 2 = 6V

VC = Av x 25 mV => VC = 10 x 25 mV = 250 mV

VE = VDD - VC - VCE = 12 - 0.25 - 6 = 5.75 V

IE = VB / RE = 5.75 / 10000 = 0.575 mA

VB = VBE + VE => VB = 0.7 + 5.75 => VB = 6.45 V

RB2 = (VB x 0.1 x Î² x RE) / VDD = (6.45 x 0.1 x 100 x 10000) / 12 => RB2 = 53750 Ohms

So we choose a 56K resistor for RB2.

RB1 = RB2 x (VDD - VB) / VB => RB1 = 56000 x (12 - 6.45) / 6.45 = 48186 Ohms

So we choose 47K for RB1.

RC = VC / IC = 0.25 / 0.575 = 434 Ohm

We choose 470 Ohms for RC.

Here is this circuit prototyped on a breadboard: And here is a closer look at the transistor and it's biasing components: The results can be seen in the oscilloscope. A 10 mVp-p signal is applied at the input. The first channel of the oscilloscope (yellow channel) is connected at the input, and the second channel (green channel) is connected at the output: It is obvious that the output is 10 times bigger than the input (Av=100 / 10 = 10). The amplifier operates as expected. With the same methodology, we can design a common base amplifier with voltage gain 100 (Av=100). From the calculations we get the part values:

RB1 = 65000 Ohms [ We will use 68 K ]

RB2 = 35000 Ohms [ We will use 33 K ]

RC = 7142 Ohms [ We will use 8.2 K ] And here are the results: The input is still 10 mV, but the output is now 2 divisions multiplied by 500 mV, which makes 1000 mV (1 Volt). The voltage gain is now 100.

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