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ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc. Lecture 7 Amplifiers.

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Presentation on theme: "ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc. Lecture 7 Amplifiers."— Presentation transcript:

1 ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc. Lecture 7 Amplifiers

2 ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc. 1.Use various amplifier models to calculate amplifier performance for given sources and loads. 2.Compute amplifier efficiency. 3.Understand the importance of input and output impedances of amplifiers. Goals

3 ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc.

4 BASIC AMPLIFIER CONCEPTS Ideally, an amplifier produces an output signal with identical waveshape as the input signal, but with a larger amplitude.

5 ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc.

6 Inverting versus Non-inverting Amplifiers Inverting amplifiers have negative voltage gain, and the output waveform is an inverted version of the input waveform. Non-inverting amplifiers have positive voltage gain.

7 ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc. Exercise 11.1 Non-inverting amplifier with a voltage gain A V of 50: Inverting amplifier with a voltage gain A V of -50:

8 ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc. Voltage-Amplifier Model The input resistance R i is the equivalent resistance see when looking into the input terminals of the amplifier. R o is the output resistance. It causes the output voltage to decrease as the load resistance becomes smaller. A voc is the open circuit voltage gain.

9 ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc. Thevenin equivalents Voltage-Amplifier Model

10 ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc. Current Gain

11 ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc. Power Gain Upper case V and I indicate Root Mean Square (RMS) values

12 ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc. Root Mean Square (RMS) Values The average value of cos(  t+  ) is zero. This is the squared version of the signal, and its mean value is ½.

13 ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc. Root Mean Square (RMS) Values

14 ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc. Example 11.1 Find the voltage gain, the current gain and the power gain:

15 ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc. Example 11.1 Note that the output signal V o is not 8000 times the input signal, V S, since there is a voltage drop across the input resistance:

16 ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc. The current gain is very large since the input resistance (2M  ) allows only a small input current to flow whereas the small output resistance (8  ) allows a relatively large output current to flow. Example 11.1 Find the current gain and the power gain:

17 ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc. Exercise 11.2 20 mV rms 500  2000  75  25  500V i Find the voltage gain, the current gain and the power gain:

18 ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc. Exercise 11.3 20 mV rms 500  2000  25  500V i What load resistance maximizes the power gain? From the Thevenin equivalent model we know that the maximum power delivered to a load is when the load resistance is equal to the Thevenin resistance of 25 .

19 ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc. CASCADED AMPLIFIERS

20 ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc. CASCADED AMPLIFIERS The overall voltage gain of cascaded amplifiers is the product of the gain of the individual stages. This is also true for the current and power gains.

21 ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc. Example 11.2 Find the voltage gain for each stage and for the overall cascade connection:

22 ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc. Example 11.2 Find the current gain for each stage and for the overall cascade connection:

23 ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc. Example 11.2 Find the power gain for each stage and for the overall cascade connection:

24 ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc. Simplified Models for Cascaded Amplifier Stages First, determine the voltage gain of the first stage accounting for loading by the second stage. The overall voltage gain is the product of the gains of the separate stages. The input impedance is that of the first stage, and the output impedance is that of the last stage.

25 ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc. Voltage gain of first stage accounting for loading of second stage:

26 ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc. Input resistance and output resistance of the cascade:

27 ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc. Simplified Model

28 ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc.

29 POWER SUPPLIES AND EFFICIENCY Both power supplies are supplying power to the amplifier

30 ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc. Heat P o >P i The additional power comes from the power supply. The efficiency is given by:

31 ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc. Find the input power, output power, supply power and the power dissipated in the amplifier. Also find the efficiency of the amplifier.

32 ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc. Input power:Output power: Supply power


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