lesson14

# lesson14 - 10 V 100 k Ω V OUT 150 k Ω 50V X 1 k Ω 2.2...

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Unformatted text preview: 10 V 100 k Ω V OUT 150 k Ω 50V X + _ 1 k Ω 2.2 k Ω + _ 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 Thévenin Equivalent circuit of the output (the 2.2 k Ω 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 μ , 50 in the example, determined by the manufacturer of the device. It would be ridiculous to sell circuits with every possible μ 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. Let’s 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 Ω (Short) 0 V 100 k Ω-6.63 V 127.5 k Ω-8.25 V 220 k Ω-13.83 V 255 k Ω-15.8 V 470 k Ω-27.46 V 510 k Ω-29.46 V 1.0 M Ω-50.8 V 3.06 M Ω-103.13 V 4.70 M Ω-124.9 V 6.12 M Ω-137.5 V ∞ Ω (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 Ω V OUT 150 k Ω 50V X 1 k Ω 2.2...

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