Ch_4prob_0005

Ch_4prob_0005 - v n leads to v n = AR i R 1 R o i s R s R L...

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Problem 4.5 For the op-amp circuit shown in Fig. P4.5: (a) Use the model given in Fig. 4-4 to develop an expression for the current gain G i = i L / i s . (b) Simplify the expression by applying the ideal op-amp model by (taking A , R i , and R o 0). + _ R L R 1 i L R s v p v n i s Figure P4.5: Circuit for Problem 4.5. Solution: (a) We start by replacing the op amp in Fig. P4.5 with its equivalent model, and to simplify the analysis, we will convert the parallel combination of ( i s , R s ) into a voltage source v s = i s R s and a series resistor R s . R i R o R 1 R L R s v p v n v n i 1 i n i L v o i s R s + + - ( v p - v n ) A ( v p - v n ) + - _ + _ At node v n , i 1 + i L + i n = 0 (1) i 1 = v n A ( v p v n ) R 1 + R o i L = v n R L i n = v n i s R s R s + R i Hence, v n A ( v p v n ) R 1 + R o + v n R L + v n i s R s R s + R i = 0 (2) All rights reserved. Do not reproduce or distribute. ©2009 National Technology and Science Press
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Additionally, v p v n = R i i n = R i p v n i s R s R s + R i P (3) Using Eq. (3) in Eq. (2) and then solving for
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Unformatted text preview: v n leads to: v n = ( AR i + R 1 + R o ) i s R s R L ( R s + R i )( R L + R 1 + R o )+ R L ( AR i + R 1 + R o ) . i L = v n R L , and G i = i L i s = v n R L i s = ( AR i + R 1 + R o ) R s ( R s + R i )( R L + R 1 + R o )+ R L ( AR i + R 1 + R o ) . (b) For the ideal op amp, A ≈ 10 6 and R i ≈ 10 6 Ω , so the product of the two is many orders of magnitude larger than all other products. Hence, G i ≃ AR i R s AR i R L = R s R L . 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_0005 - v n leads to v n = AR i R 1 R o i s R s R L...

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