Ch_4prob_0010

# Ch_4prob_0010 - v o v Th R f R Th + _ + _ v o = − R f R...

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Problem 4.10 In the circuit of Fig. P4.10, a bridge circuit is connected at the input side of an inverting op-amp circuit. (a) Obtain the Th´evenin equivalent at terminals ( a , b ) for the bridge circuit. (b) Use the result in (a) to obtain an expression for G = v o / v s . (c) Evaluate G for R 1 = R 4 = 100 , R 2 = R 3 = 101 , and R f = 100 k . v o v s R 2 b a R f R 1 R 4 R 3 + _ + _ Figure P4.10: Circuit for Problem 4.10. Solution: (a) The Th´evenin equivalent circuit at ( a , b ) : v s R 2 b a R 1 R 4 R 3 i 1 i 2 + _ v oc v s + i 1 ( R 1 + R 2 ) = 0 or i 1 = v s R 1 + R 2 . Also, v s + i 2 ( R 3 + R 4 ) = 0 and i 2 = v s R 3 + R 4 . v Th = v oc = i 1 R 2 + i 2 R 4 = v s R 2 R 1 + R 2 + v s R 4 R 3 + R 4 = [ R 4 ( R 1 + R 2 ) R 2 ( R 3 + R 4 )] v s ( R 1 + R 2 )( R 3 + R 4 ) . (1) Suppressing v s (by replacing it with a short circuit) leads to R Th = ( R 1 b R 2 )+ ( R 3 b R 4 ) = R 1 R 2 R 1 + R 2 + R 3 R 4 R 3 + R 4 = R 1 R 2 ( R 3 + R 4 )+ R 3 R 4 ( R 1 + R 2 ) ( R 1 + R 2 )( R 3 + R 4 ) . (b) For the new circuit: All rights reserved. Do not reproduce or distribute. ©2009 National Technology and Science Press

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Unformatted text preview: v o v Th R f R Th + _ + _ v o = − R f R Th v Th (inverting ampliﬁer) (3) Inserting Eqs. (1) and (2) into (3) leads to G = v o v s = − R f [ R 4 ( R 1 + R 2 ) − R 2 ( R 3 + R 4 )] R 1 R 2 ( R 3 + R 4 )+ R 3 R 4 ( R 1 + R 2 ) (c) For R 1 = R 4 = 100 Ω , R 2 = R 3 = 101 Ω , and R f = 10 5 Ω , G = − 10 5 [ 100 ( 100 + 101 ) − 101 ( 100 + 101 )] 100 × 101 ( 100 + 101 )+ 100 × 101 ( 100 + 101 ) = 4 . 9505 ≃ 5 . 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_0010 - v o v Th R f R Th + _ + _ v o = − R f R...

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