Tutorial: Mechanic -electrician Topic: Electronics II. class Transistors: Transistor CE Amplifier Prepared by: Ing. Jaroslav Bernkopf Projekt Anglicky.

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Principles & Applications Small-Signal Amplifiers

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Principles & Applications Small-Signal Amplifiers

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Tutorial: Mechanic -electrician Topic: Electronics II. class Transistors: Transistor CE Amplifier Prepared by: Ing. Jaroslav Bernkopf Projekt Anglicky v odborných předmětech, CZ.1.07/1.3.09/ je spolufinancován Evropským sociálním fondem a státním rozpočtem České republiky.

A transistor common emitter amplifier is a circuit where the input signal is applied between the base and the emitter while the output is taken from between the collector and the emitter. The emitter is grounded and is common to both the input and output. Hence the name: Common emitter amplifier. Transistor CE Amplifier Transistors 2

Transistor CE Amplifier Transistors 3

Transistor CE Amplifier Transistors 4

Transistor CE Amplifier Transistors 5 The collector current I C creates a voltage drop V RC across the collector resistor R C. The higher the base voltage, the bigger the voltage drop V RC, and the lower the collector voltage V out. An increase in base input voltage V in causes a decrease in collector output voltage V out. This is why we call this circuit an "Inverting Amplifier“. A small change of the base voltage makes a big change of the collector voltage. The input signal applied on the base appears, amplified, on the collector. V RC V CC V out

Transistor CE Amplifier Transistors 6 All bias components have been omitted to simplify the above explanations. In the next task, we will design a real amplifier stage including the setting and stabilization of the operating point. The operating point of a transistor, also known as bias point, quiescent point, or Q- point, is the steady-state operating condition of a transistor with no input signal applied. (Wikipedia) RERE

Transistor CE Amplifier Transistors 7 Given the following conditions, calculate the values of all resistors for the circuit below. I C = 1 mA V CE = 5 V V E = 2 V ß ≥ 100 Current I R1R2 through the voltage divider R 1, R 2 about 10 times the max I B I C =1mA I R1R2 V E =2V V CE =5V RERE

Transistor CE Amplifier Transistors 8 I C =1mA I R1R2 RERE I E =1mA 2k V E =2V V CE =5V

Transistor CE Amplifier Transistors 9 I C =1mA I R1R2 RERE I E =1mA V B = V R2 = 2.7 V 2k 27k V E =2V V CE =5V

Transistor CE Amplifier Transistors 10 I C =1mA I R1R2 RERE I E =1mA V B = V R2 = 2.7 V 2k 27k V R1 = 9.3 V 93k V E =2V V CE =5V

Transistor CE Amplifier Transistors 11 I C =1mA I R1R2 V E =2V V CE =5V RERE I E =1mA V B = V R2 = 2.7 V 2k 27k V R1 = 9.3 V 93k5k V RC =5V V CC =12V

Transistor CE Amplifier Transistors 12 I C =1mA I R1R2 V E =2V V CE =5V RERE I E =1mA V B = V R2 = 2.7 V 2k 27k V R1 = 9.3 V 93k5k V RC =5V V CC =12V When solving the task we have only needed three simple rules: 1)Ohm‘s law 2)Kirchhoff‘s voltage law 3)Kirchhoff‘s current law

Transistor CE Amplifier Transistors 13 I C =1mA I R1R2 V E =2V V CE =5V RERE I E =1mA V B = V R2 = 2.7 V 2k 27k V R1 = 9.3 V 93k5k V RC =5V V CC =12V

Transistor CE Amplifier Transistors 14 I C =1mA I R1R2 V E =2V V CE =5V RERE I E =1mA V B = V R2 = 2.7 V 2k 27k V R1 = 9.3 V 93k5k V RC =5V V CC =12V

Transistor CE Amplifier Transistors 15 I C =1mA I R1R2 V E =2V V CE =5V RERE I E =1mA V B = V R2 = 2.7 V 2k 27k V R1 = 9.3 V 93k5k V RC =5V V CC =12V

Transistor CE Amplifier Transistors 16      