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

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5 The Operational Amplifer Assessment Problems AP 5.1 [a] This is an inverting amplifer, so v o =( R f /R i ) v s 80 / 16) v s , so v o = 5 v s v s ( V )0 . 42 . 03 . 5 0 . 6 1 . 6 2 . 4 v o ( V ) 2 . 0 10 . 0 15 . . 08 . 01 0 . 0 TwooFthe 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 s = 2 V ThereFore 2 v s 3 V AP 5.2 ±rom 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 / 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 amplifer 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 . 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 . 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 . 2 . 5 V Repeat part (b): v o = 50 v a +2 . 10 V ;5 0 v a =12 . 5 ,v a . 25 V Repeat part (c): v o = 5+10 v b =15 V ;1 0 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 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 so v o 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 v p = 15(0 .
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This note was uploaded on 04/20/2011 for the course EECE 222 taught by Professor Amsddasgf during the Spring '11 term at American University in Bulgaria.

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

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