Ch_4prob_0008

Ch_4prob_0008 - 2009 National Technology and Science Press...

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Problem 4.8 The op-amp circuit shown in Fig. P4.8 has a constant dc voltage of 6 V at the noninverting input. The inverting input is the sum of two voltage sources, a 6-V dc source and a small time-varying signal v s . (a) Use the op-amp equivalent-circuit model given in Fig. 4-5 to develop an expression for v o . (b) Simplify the expression by applying the ideal op-amp model, which lets A , R i and R o 0. R L v o v s 6 V 6 V + _ + _ _ _ Figure P4.8: Circuit for Problem 4.8. Solution: (a) Replacing the op amp with its equivalent circuit: R L R o R i v o v n v p v s _ _ 6 V _ 6 V + _ A ( v p - v n ) For node v o , v o R L + v o A ( v p v n ) R o = 0 . Also, v p = 6 V , v n = 6 V + v s . Solving for v o , we have: v o = Av s R L R L + R o . (b) For R L R o , v o = Av s . All rights reserved. Do not reproduce or distribute.
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Unformatted text preview: 2009 National Technology and Science Press Hence, the output will be amplied by A , but inverted. If | Av s | V cc , then the op amp will saturate. Hence, v o = Av s , for | v s | V cc A V cc , if | v s | > V cc A and v s is positive + V cc , if | v s | > V cc A and v s is negative. All rights reserved. Do not reproduce or distribute. 2009 National Technology and Science Press...
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This note was uploaded on 04/19/2010 for the course EE 220 taught by Professor Rawat during the Spring '09 term at Nevada.

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Ch_4prob_0008 - 2009 National Technology and Science Press...

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