The Transistor II, Basics

 Dr.-Ing. Peter Strassacker
E-mail: peter@lautsprechershop.de
 

The common emitter circuit

The common emitter circuit is the most commonly used circuit since here both current and voltage are amplified. Therefore, we are going to determine and calculate the operating point on the basis of an example. A detailed description of the transistor can be found in Wikipedia. Let's assume we want to design an amplifier with 20 V supply voltage. The first question to be answered is: what current value is needed at the collector (higher current values result in more noise, on the other hand the power output will be higher). 3mA represent already high current for a pre-amplifier, we'll base our calculation on this value.

If the operating point at the output (i.e. the voltage on the collector) should be just above of half the supply voltage, e.g. around 11 V, then the voltage at the resistor drops by 9 V (20V - 11V). Therefore, for a quiescent current of around 3 mA a collector resistor of

Emittorschaltung
Picture above: common emitter circuit
 

RC = 9V / 3 mA = 3 kΩ

is needed. Since a common value for the E12 resistor range is 3.3 kΩ, we'll choose this value.

Since we do not want to have more negative feedback, we have to choose an emitter resistor that results in a matching voltage amplification VU. For VU = 3.3 = RC / RE a resistor of

RE = RC / 3.3 = 1 kΩ has to be chosen.

Since almost the same current flows through the emitter resistor than through the collector resistor (IE = IC + IB and IB << IC) the voltage should drop by 2.7 Volt by using the emitter resistor ((9V / 3.3 kΩ) * 1 kΩ = 2.7). Over the base - emitter distance the voltage usually drops by 0.7 V, resulting in a voltage of 2.7 + 0.7 = 3.4 V.

The base voltage is adjusted by a resistance divider. For this the arbitrary values of 150 kΩ and 33 kΩ are chosen. This results in a base voltage of:

20 V * (33 kΩ/(150kΩ+33kΩ)) = 3.61 V

Both resistors are in parallel resulting in total resistance of Rg = 1/((1/33kΩ) + (1/150kΩ)) = 27 kΩ.

There is a further voltage drop at both resistors, since some current flows into the base. Considering a base current of IB = IC / β = 3 mA / 300 = 0.01 mA
the additional voltage drop is 0.27 V (0.01 mA * 27 kΩ = 0.27 V).

Therefore, the base voltage is not 3.61 V but 0.27 Volt less, resulting in 3.34 V. This is almost as we wished for (3.4 V).

We have finished designing the circuit and measured the operating points (voltages, marked red in the diagram). They are matching our calculations closely.

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