Explanation Calculation: The given diagram is shown in Figure 1 The conversion from 1 mV into V is given by, 1 mV = 10 − 3 V The conversion from 8 mV into V is given by, 8 mV = 8 × 10 − 3 V The conversion from 50 mV into V is given by, 50 mV = 50 × 10 − 3 V The expression for the output voltage is given by, v O = K v TR + V b …… (1) Substitute 8 × 10 − 3 V for v TR and 0 V for v O in the above equation. 0 V = K ( 8 × 10 − 3 V ) + V b …… (2) Substitute 50 × 10 − 3 V for v TR and 6 V for v O in the above equation. 6 V = K ( 50 × 10 − 3 V ) + V b Subtract above equation from equation (2). 6 V = K ( 50 × 10 − 3 V ) + V b − K ( 8 × 10 − 3 V ) + V b K = 6 42 × 10 − 3 V K = 142.9 Substitute 142.9 for K in equation (2). 0 V = ( 142.9 ) ( 8 × 10 − 3 V ) + V b V b = − 1143.2 × 10 − 3 V Substitute 142.9 for K and − 1.1432 V for V b in equation (1). v O = 142.9 v TR − 1.1432 V The second part of the above equation can be written as ( − 0.1429 ) ( 8 ) = 1.1432 V to get the required bias from the 8 V voltage source. The required equation is shown below. v O = 142

8th Edition

ISBN: 9781119235385

Author: Thomas

Publisher: WILEY

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Chapter 4, Problem 4.83P

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To design: An interface that will produce 0 to 6 V output.

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Chapter 4, Problem 4.82P

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Chapter 4 Solutions

ANALYSIS+DESIGN OF LINEAR CIRCUITS(LL)

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To design: An interface that will produce 0 to 6 V output.

Solution Summary: The author illustrates the required interface circuit in Figure 3. The conversion from 1mV into v

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Chapter 4 Solutions