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HW _9 Solutions

# HW _9 Solutions - Problem 4.20 Fig P4.20 Determine the...

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Problem 4.20 Determine the linear range of the source v s in the circuit of Fig. P4.20. v o v s 200 Ω 400 Ω 1.2 k Ω V cc = 12 V 2 V + _ + _ Figure P4.20: Circuit for Problem 4.20. Solution: The circuit is a standard summing amplifier. v o = R f R 1 v 1 R f R 2 v 2 = 1 . 2 × 10 3 200 v s 1 . 2 × 10 3 400 × 2 = 6 v s 6 (V) . For v o = V cc = 12 V, 12 = 6 v s 6 , or v s = 3 V . For v o = V cc = 12 V, 12 = 6 v s 6 , or v s = 1 V . Hence, the linear range of v s is 3 V v s 1 V . All rights reserved. Do not reproduce or distribute. ©2009 National Technology and Science Press

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Problem 4.22 The circuit in Fig. P4.22 uses a potentiometer whose total resistance is R = 10 k , with the upper section being β R and the bottom section ( 1 β ) R . The stylus can change β from 0 to 0.9. Obtain an expression for G = v o / v s in terms of β , and evaluate the range of G as β is varied over its own allowable range. v o v s R = 10 k 100 678 678 678 β R (1 - β) R + _ Figure P4.22: Circuit for Problem 4.22.
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HW _9 Solutions - Problem 4.20 Fig P4.20 Determine the...

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