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ch05_ssm - The Operational Amplier 5 Assessment Problems AP...

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5 The Operational Amplifier Assessment Problems AP 5.1 [a] This is an inverting amplifier, so v o = ( R f /R i ) v s = ( 80 / 16) v s , so v o = 5 v s v s ( V ) 0 . 4 2 . 0 3 . 5 0 . 6 1 . 6 2 . 4 v o ( V ) 2 . 0 10 . 0 15 . 0 3 . 0 8 . 0 10 . 0 Two of the v s values, 3 . 5 V and 2 . 4 V, cause the op amp to saturate. [b] Use the negative power supply value to determine the largest input voltage: 15 = 5 v s , v s = 3 V Use the positive power supply value to determine the smallest input voltage: 10 = 5 v s , v s = 2 V Therefore 2 v s 3 V AP 5.2 From Assessment Problem 5.1 v o = ( R f /R i ) v s = ( R x / 16 , 000) v s = ( R x / 16 , 000)( 0 . 640) = 0 . 64 R x / 16 , 000 = 4 × 10 5 R x Use the negative power supply value to determine one limit on the value of R x : 4 × 10 5 R x = 15 so R x = 15 / 4 × 10 5 = 375 kΩ Since we cannot have negative resistor values, the lower limit for R x is 0 . Now use the positive power supply value to determine the upper limit on the value of R x : 4 × 10 5 R x = 10 so R x = 10 / 4 × 10 5 = 250 kΩ Therefore, 0 R x 250 k 5–1
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5–2 CHAPTER 5. The Operational Amplifier AP 5.3 [a] This is an inverting summing amplifier so v o = ( R f /R a ) v a + ( R f /R b ) v b = (250 / 5) v a (250 / 25) v b = 50 v a 10 v b Substituting the values for v a and v b : v o = 50(0 . 1) 10(0 . 25) = 5 2 . 5 = 7 . 5 V [b] Substitute the value for v b into the equation for v o from part (a) and use the negative power supply value: v o = 50 v a 10(0 . 25) = 50 v a 2 . 5 = 10 V Therefore 50 v a = 7 . 5 , so v a = 0 . 15 V [c] Substitute the value for v a into the equation for v o from part (a) and use the negative power supply value: v o = 50(0 . 10) 10 v b = 5 10 v b = 10 V ; Therefore 10 v b = 5 , so v b = 0 . 5 V [d] The effect of reversing polarity is to change the sign on the v b term in each equation from negative to positive. Repeat part (a): v o = 50 v a + 10 v b = 5 + 2 . 5 = 2 . 5 V Repeat part (b): v o = 50 v a + 2 . 5 = 10 V ; 50 v a = 12 . 5 , v a = 0 . 25 V Repeat part (c): v o = 5 + 10 v b = 15 V ; 10 v b = 20; v b = 2 . 0 V AP 5.4 [a] Write a node voltage equation at v n ; remember that for an ideal op amp, the current into the op amp at the inputs is zero: v n 4500 + v n v o 63 , 000 = 0 Solve for v o in terms of v n by multiplying both sides by 63 , 000 and collecting terms: 14 v n + v n v o = 0 so v o = 15 v n Now use voltage division to calculate v p . We can use voltage division because the op amp is ideal, so no current flows into the non-inverting input terminal and the 400 mV divides between the 15 k resistor and the R x resistor: v p = R x 15 , 000 + R x (0 . 400)
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Problems 5–3 Now substitute the value R x = 60 k : v p = 60 , 000 15 , 000 + 60 , 000 (0 . 400) = 0 . 32 V Finally, remember that for an ideal op amp, v n = v p , so substitute the value of v p into the equation for v 0 v o = 15 v n = 15 v p = 15(0 . 32) = 4 . 8 V [b] Substitute the expression for v p into the equation for v o and set the resulting equation equal to the positive power supply value: v o = 15 0 .
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