lesson14 - 10 V 100 k V OUT 150 k 50V X + _ 1 k 2.2 k + _ V...

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Unformatted text preview: 10 V 100 k V OUT 150 k 50V X + _ 1 k 2.2 k + _ V X + _ R IN I IN V IN R F A Lesson 14 Dependent Sources #2 (Section 4-2) (CLO 4-1) This is a challenging lesson to both teach and to learn. It is important because it sets the underlying concept for the operation of Op-Amps namely feedback. To reinforce what the students are to learn in this lesson there is an accompanying computer exercise that asks them to vary a feedback resistor and view how the gain changes. They must take away from this lesson that they can set, within limits, the gain of a circuit with a dependent source by properly selecting the feedback resistor. A second concept is the important effect that feedback in a dependent source circuit can have on the input (as viewed by the source) and output (as seen by the load) resistance. This concept is important to understanding loading or not by Op-Amps. I the following lesson on dependent sources, we will find the Thvenin Equivalent circuit of the output (the 2.2 k resistor in this example). The effect of feedback can significantly alter V T and R T . The following example can be used to convey the concept of feedback. This is similar to the Computer Exercise they are asked to do. Start with the example below (without R F ) to point out that the engineer has no control over the gain it is set by , 50 in the example, determined by the manufacturer of the device. It would be ridiculous to sell circuits with every possible to satisfy every application. How then can we empower the engineer to design a circuit with any desired gain? The answer is by using feedback. Lets add a feedback resistor R F as shown. Now things behave very differently. We start by doing a voltage divider at the output: X OUT 375 34 2 2 1 50 2 2 V k k V k V X - = +- = . . . Then we write a node equation at Node A. We note that V A = V X . 150 100 10 50 X X F X X = +- +-- k V k V R V V ) ( Simplifying and solving for V X , we get k k k R V 100 10 150 1 100 1 50 1 F X = + + + F F F F F F X 3060 1 6 15300 5 30 300 2 3 300 51 10 1 150 1 100 1 51 10 1 R k k R R k R R R k k k k R k V F + = + = + + = + + = / / Plugging V X back into our output equation, we obtain our final relationship of the output and the feedback resistor We can develop a table that shows how V OUT varies as we vary R F from a short to an open. We solved the open case last lesson. By choosing R F the design engineer can select the gain desired from 0 to a maximum of -206.25. A similar problem is found in the attached computer exercise that can be done in class or for homework. R F V OUT (Short) 0 V 100 k -6.63 V 127.5 k -8.25 V 220 k -13.83 V 255 k -15.8 V 470 k -27.46 V 510 k -29.46 V 1.0 M -50.8 V 3.06 M -103.13 V 4.70 M -124.9 V 6.12 M -137.5 V (Open)-206.25 V F X OUT 3060 1 25 206 375 34 R k V V +- =- = ....
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lesson14 - 10 V 100 k V OUT 150 k 50V X + _ 1 k 2.2 k + _ V...

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