EECE 210C Lecture, Fall 2005

5 The Operational Amplifier Assessment Problems AP 5.1 [a] This is an inverting amplifier, so vo = (−Rf/Ri)vs = (−80/16)vs, so vo = −5vs vs( V) 0.4 2.0 3.5 −0.6 −1.6 −2.4 vo( V) −2.0 −10.0 −15.0 3.0 8.0 10.0 Two of the vs 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 = −5vs, vs = 3 V Use the positive power supply value to determine the smallest input voltage: 10 = −5vs, vs = −2 V Therefore − 2 ≤ vs ≤ 3 V AP 5.2 From Assessment Problem 5.1 vo = (−Rf/Ri)vs = (−Rx/16,000)vs = (−Rx/16,000)(−0.640) = 0.64Rx/16,000 = 4×10−5Rx Use the negative power supply value to determine one limit on the value of Rx: 4×10−5Rx = −15 so Rx = −15/4×10−5 = −375 kΩ Since we cannot have negative resistor values, the lower limit for Rx is 0. Now use the positive power supply value to determine the upper limit on the value of Rx: 4×10−5Rx = 10 so Rx = 10/4×10−5 = 250 kΩ Therefore, 0 ≤ Rx ≤ 250 kΩ 5–1 5–2 CHAPTER 5. The Operational Amplifier AP 5.3 [a] This is an inverting summing amplifier so vo = (−Rf/Ra)va + (−Rf/Rb)vb = −(250/5)va − (250/25)vb = −50va − 10vb Substituting the values for va and vb: vo = −50(0.1) − 10(0.25) = −5 − 2.5 = −7.5 V [b] Substitute the value for vb into the equation for vo from part (a) and use the negative power supply value: vo = −50va − 10(0.25) = −50va − 2.5 = −10 V Therefore 50va = 7.5, so va = 0.15 V [c] Substitute the value for va into the equation for vo from part (a) and use the negative power supply value: vo = −50(0.10) − 10vb = −5 − 10vb = −10 V; Therefore 10vb = 5, so vb = 0.5 V [d] The effect of reversing polarity is to change the sign on the vb term in each equation from negative to positive. Repeat part (a): vo = −50va + 10vb = −5 + 2.5 = −2.5 V Repeat part (b): vo = −50va + 2.5 = −10 V; 50va = 12.5, va = 0.25 V Repeat part (c): vo = −5 + 10vb = 15 V; 10vb = 20; vb = 2.0 V AP 5.4 [a] Write a node voltage equation at vn; remember that for an ideal op amp, the current into the op amp at the inputs is zero: vn 4500 + vn − vo 63,000 = 0 Solve for vo in terms of vn by multiplying both sides by 63,000 and collecting terms: 14vn + vn − vo = 0 so vo = 15vn Now use voltage division to calculate vp. 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 kΩ resistor and the Rx resistor: vp = Rx 15,000 + Rx (0.400) Problems 5–3 Now substitute the value Rx = 60 kΩ: vp = 60,000 15,000 + 60,000(0.400) = 0.32 V Finally, remember that for an ideal op amp, vn = vp, so substitute the value of vp into the equation for v0 vo = 15vn = 15vp = 15(0.32) = 4.8 V [b] Substitute the expression for vp into the equation for vo and set the resulting equation equal to the positive power supply value: vo = 15 � 0.4Rx 15,000 + Rx � = 5 15(0.4Rx) = 5(15,000 + Rx) so Rx = 75 kΩ AP 5.5 [a] Since this is a difference amplifier, we can use the expression for the output voltage in terms of the input voltages and the resistor values given in Eq. 5.22: vo = 20(60) 10(24)vb − 50 10va Simplify this expression and substitute in the value for vb: vo = 5(vb − va) = 20 − 5va Set this expression for vo to the positive power supply value: 20 − 5va = 10 V so va = 2 V Now set the expression for vo to the negative power supply value: 20 − 5va = −10 V so va = 6 V Therefore 2 ≤ va ≤ 6 V [b] Begin as before by substituting the appropriate values into Eq. 5.22: vo = 8(60) 10(12)vb − 5va = 4vb − 5va Now substitute the value for vb: vo = 4(4) − 5va = 16 − 5va Set this expression for vo to the positive power supply value: 16 − 5va = 10 V so va = 1.2 V Now set the expression for vo to the negative power supply value: 16 − 5va = −10 V so va = 5.2 V Therefore 1.2 ≤ va ≤ 5.2 V 5–4 CHAPTER 5. The Operational Amplifier AP 5.6 [a] Replace the op amp with the more realistic model of the op amp from Fig. 5.15: Write the node voltage equation at the left hand node: vn 500,000 + vn − vg 5000 + vn − vo 100,000 = 0 Multiply both sides by 500,000 and simplify: vn + 100vn − 100vg + 5vn − 5v0 = 0 so 21.2vn − vo = 20vg Write the node voltage equation at the right hand node: vo − 300,000(−vn) 5000 + vo − vn 100,000 = 0 Multiply through by 100,000 and simplify: 20vo + 6 × 106vn + vo − vn = 0 so 6 × 106vn + 21vo = 0 Use Cramer’s method to solve for vo: ∆ = ������� 21.2 −1 6 × 106 21 ������� = 6,000,445.2 No = ������� 21.2 20vg 6 × 106 0 ������� = −120 × 106vg vo = No ∆ = −19.9985vg; so vo vg = −19.9985 [b] Use Cramer’s method again to solve for vn: N1 = ������� 20vg −1 0 21 ������� = 420vg vn = N1 ∆ = 6.9995 × 10−5vg vg = 1 V, vn = 69.995 µ V Problems 5–5 [c] The resistance seen at the input to the op amp is the ratio of the input voltage to the input current, so calculate the input current as a function of the input voltage: ig = vg − vn 5000 = vg − 6.9995 × 10−5vg 5000 Solve for the ratio of vg to ig to get the input resistance: Rg = vg ig = 5000 1 − 6.9995 × 10−5 = 5000.35 Ω [d] This is a simple inverting amplifier configuration, so the voltage gain is the ratio of the feedback resistance to the input resistance: vo vg = −100,000 5000 = −20 Since this is now an ideal op amp, the voltage difference between the two input terminals is zero; since vp = 0, vn = 0 Since there is no current into the inputs of an ideal op amp, the resistance seen by the input voltage source is the input resistance: Rg = 5000 Ω 5–6 CHAPTER 5. The Operational Amplifier Problems P 5.1 [a] The five terminals of the op amp are identified as follows: [b] The input resistance of an ideal op amp is infinite, which constrains the value of the input currents to 0. Thus, in = 0 A. [c] The open-loop voltage gain of an ideal op amp is infinite, which constrains the difference between the voltage at the two input terminals to 0. Thus, (vp − vn) = 0. [d] Write a node voltage equation at vn: vn − 2.5 10,000 + vn − vo 40,000 = 0 But vp = 0 and vn = vp = 0. Thus, −2.5 10,000 − vo 40,000 = 0 so vo = −10 V P 5.2 vb − va 20 + vb − vo 100 = 0, therefore vo = 6vb − 5va [a] va = 4 V, vb = 0 V, vo = −15 V (sat) [b] va = 2 V, vb = 0 V, vo = −10 V [c] va = 2 V, vb = 1 V, vo = −4 V [d] va = 1 V, vb = 2 V, vo = 7 V [e] If vb = 1.6 V, vo = 9.6 − 5va = ±15 ·. . −1.08 ≤ va ≤ 4.92 V P 5.3 vo = −(0.5 × 10−3)(10 × 103) = −5 V ·. . io = −5 5000 = −1 mA Problems 5–7 P 5.6 [a] The gain of an inverting amplifier is the ratio of the feedback resistor to the input resistor. If the gain of the inverting amplifier is to be 6, the feedback resistor must be 6 times as large as the input resistor. There are many possible designs that use only 20 kΩ resistors. We present two here. Use a single 20 kΩ resistor as the input resistor, and use six 20 kΩ resistors in series as the feedback resistor to give a total of 120 kΩ. Alternately, Use a single 20 kΩ resistor as the feedback resistor and use six 20 kΩ resistors in parallel as the input resistor to give a total of 3.33 kΩ. [b] To amplify a 3 V signal without saturating the op amp, the power supply voltages must be greater than or equal to the product of the input voltage and the amplifier gain. Thus, the power supplies should have a magnitude of (3)(6) = 18 V. P 5.7 [a] The circuit shown is a non-inverting amplifier. [b] We assume the op amp to be ideal, so vn = vp = 3V. Write a KCL equation at vn: 3 40,000 + 3 − vo 80,000 = 0 Solving, vo = 9 V. 5–8 CHAPTER 5. The Operational Amplifier P 5.16 [a] This circuit is an example of an inverting summing amplifier. [b] vo = −220 33 va − 220 22 vb − 220 80 vc = −8 + 15 − 11 = −4 V [c] vo = −19 − 10vb = ±6 ·. . vb = −1.3 V when vo = −6 V; vb = −2.5 V when vo = 6 V ·. . −2.5 V ≤ vb ≤ −1.3 V P 5.17 [a] Write a KCL equation at the inverting input to the op amp: vd − va 40,000 + vd − vb 22,000 + vd − vc 100,000 + vd 352,000 + vd − vo 220,000 = 0 Multiply through by 220,000, plug in the values of input voltages, and rearrange to solve for vo: vo = 220,000 � 4 40,000 + −1 22,000 + −5 100,000 + 8 352,000 + 8 220,000 � = 14 V [b] Write a KCL equation at the inverting input to the op amp. Use the given values of input voltages in the equation: 8 − va 40,000 + 8 − 9 22,000 + 8 − 13 100,000 + 8 352,000 + 8 − vo 220,000 = 0 Simplify and solve for vo: 44 − 5.5va − 10 − 11 + 5 + 8 − vo = 0 so vo = 36 − 5.5va Set vo to the positive power supply voltage and solve for va: 36 − 5.5va = 15 ·. . va = 3.818 V Set vo to the negative power supply voltage and solve for va: 36 − 5.5va = −15 ·. . va = 9.273 V Therefore, 3.818 V ≤ va ≤ 9.273 V Problems 5–9 P 5.19 We want the following expression for the output voltage: vo = −(2va + 4vb + 6vc + 8vd) This is an inverting summing amplifier, so each input voltage is amplified by a gain that is the ratio of the feedback resistance to the resistance in the forward path for the input voltage: vo = − � 48 Ra va + 48 Rb vb + 48 Rc vc + 48 Rd vd � Solve for each input resistance value to yield the desired gain: ·. . Ra = 48,000/2 = 24 kΩ Rc = 48,000/6 = 8 kΩ Rb = 48,000/4 = 12 kΩ Rd = 48,000/8 = 6 kΩ The final circuit is shown here: P 5.22 [a] This circuit is an example of the non-inverting amplifier. [b] Use voltage division to calculate vp: vp = 10,000 10,000 + 30,000vs = vs 4 Write a KCL equation at vn = vp = vs/4: vs/4 4000 + vs/4 − vo 28,000 = 0 Solving, vo = 7vs/4 + vs/4 = 2vs [c] 2vs = 8 so vs = 4 V 2vs = −12 so vs = −6 V Thus, −6 V ≤ vs ≤ 4 V. 5–10 CHAPTER 5. The Operational Amplifier P 5.23 [a] vp = vn = 68,000 80,000vg = 0.85vg ·. . 0.85vg 30,000 + 0.85vg − vo 63,000 = 0; ·. . vo = 2.635vg = 2.635(4), vo = 10.54 V [b] vo = 2.635vg = ±12 vg = ±4.554 V, −4.554 ≤ vg ≤ 4.554 V [c] 0.85vg 30,000 + 0.85vg − vo Rf = 0 �0.85Rf 30,000 + 0.85 � vg = vo = ±12 ·. . 1.7×10−3Rf + 51 = ±360; 1.7×10−3Rf = 360 − 51; Rf = 181.76 kΩ P 5.29 Use voltage division to find vp: vp = 2000 2000 + 8000(5) = 1 V Write a KCL equation at vn and solve it for vo: vn − va 5000 + vn − vo Rf = 0 so � Rf 5000 + 1 � vn − Rf 5000va = vo Since the op amp is ideal, vn = vp = 1V, so vo = � Rf 5000 + 1 � − Rf 5000va To satisfy the equation, � Rf 5000 + 1 � = 5 and Rf 5000 = 4 Thus, Rf = 20 kΩ. Problems 5–11 P 5.30 [a] vp 72,000 + vp − vc 9,000 + vp − vd 24,000 = 0 ·. . vp = (2/3)vc + 0.25vd = vn vn − va 12,000 + vn − vb 18,000 + vn − vo 144,000 = 0 ·. . vo = 21vn − 12va − 8vb = 21[(2/3)vc + 0.25vd] − 12va − 8vb = 21(0.4 + 0.2) − 12(0.5) − 8(0.3) = 4.2 V [b] vo = 14vc + 4.2 − 6 − 2.4 ±15 = 14vc − 4.2 ·. . 14vc = ±15 + 4.2 ·. . vc = 1.371 V and vc = −0.771 V ·. . −771 ≤ vc ≤ 1371 mV P 5.31 [a] vo = Rd(Ra + Rb) Ra(Rc + Rd)vb − Rb Ra va = 47(110) 10(80) (0.80) − 10(0.67) vo = 5.17 − 6.70 = −1.53 V [b] vn = vp = (800)(47) 80 = 470 mV ia = (670 − 470)10−3 10 × 103 = 20 µA Ra = va ia = 670 × 10−3 20 × 10−6 = 33.5 kΩ [c] Rin b = Rc + Rd = 80 kΩ 5–12 CHAPTER 5. The Operational Amplifier P 5.39 [a] Replace the op amp with the model from Fig. 5.15: Write two node voltage equations, one at the left node, the other at the right node: vn − vg 5000 + vn − vo 100,000 + vn 500,000 = 0 vo + 3 × 105vn 5000 + vo − vn 100,000 + vo 1000 = 0 Simplify and place in standard form: 106vn − 5vo = 100vg (6×106 − 1)vn + 121vo = 0 Let vg = 1 V and solve the two simultaneous equations: vo = −19.9915 V; vn = 403.2 µV [b] From the solution in part (a), vn = 403.2 µV. [c] ig = vg − vn 5000 = vg − 403.2 × 10−6vg 5000 Rg = vg ig = 5000 1 − 403.2 × 10−6 = 5002.02 Ω [d] For an ideal op amp, the voltage gain is the ratio between the feedback resistor and the input resistor: vo vg = −100,000 5000 = −20 For an ideal op amp, the difference between the voltages at the input terminals is zero, and the input resistance of the op amp is infinite. Therefore, vn = vp = 0 V; Rg = 5000 Ω