Alexander_2

4.8 Maximum Power Transfer 151This implies that (4.22) The source and load are said to be 0 ϭ (RTh ϩ RL Ϫ 2RL) ϭ (RTh Ϫ RL) matched when RL ϭ RTh.which yields RL ϭ RTh (4.23)showing that the maximum power transfer takes place when the loadresistance RL equals the Thevenin resistance RTh. We can readily confirm 2p 2that Eq. (4.23) gives the maximum power by showing that d ͞dR L 6 0. The maximum power transferred is obtained by substitutingEq. (4.23) into Eq. (4.21), for pmax ϭ V 2 (4.24) Th 4RThEquation (4.24) applies only when RL ϭ RTh. When RL RTh, wecompute the power delivered to the load using Eq. (4.21).Find the value of RL for maximum power transfer in the circuit of Example 4.13Fig. 4.50. Find the maximum power. 6Ω 3Ω 2Ω a 12 V +− 12 Ω 2 A RL b Figure 4.50 For Example 4.13.Solution:We need to find the Thevenin resistance RTh and the Thevenin voltageVTh across the terminals a-b. To get RTh, we use the circuit in Fig. 4.51(a)and obtain 6 ϫ 12 RTh ϭ 2 ϩ 3 ϩ 6 ʈ 12 ϭ 5 ϩ 18 ϭ 9 ⍀ 6Ω 3Ω 2Ω 6Ω 3Ω 2Ω 2A 12 Ω RTh 12 V +− 12 Ω i2 + i1 VTh − (a) (b) Figure 4.51 For Example 4.13: (a) finding RTh, (b) finding VTh.

152 Chapter 4 Circuit Theorems To get VTh, we consider the circuit in Fig. 4.51(b). Applying mesh analysis gives Ϫ12 ϩ 18i1 Ϫ 12i2 ϭ 0, i2 ϭ Ϫ2 A Solving for i1, we get i1 ϭ Ϫ2͞3. Applying KVL around the outer loop to get VTh across terminals a-b, we obtain Ϫ12 ϩ 6i1 ϩ 3i2 ϩ 2(0) ϩ VTh ϭ 0 1 VTh ϭ 22 V For maximum power transfer, RL ϭ RTh ϭ 9 ⍀ and the maximum power is pmax ϭ V 2 ϭ 222 ϭ 13.44 W Th 4ϫ9 4RLPractice Problem 4.13 Determine the value of RL that will draw the maximum power from the rest of the circuit in Fig. 4.52. Calculate the maximum power. 2Ω 4Ω + vx − Answer: 4.222 ⍀, 2.901 W. 1Ω 4.9 Verifying Circuit Theorems with PSpice9 V +− RL In this section, we learn how to use PSpice to verify the theorems cov- ered in this chapter. Specifically, we will consider using DC Sweep analy- + 3vx sis to find the Thevenin or Norton equivalent at any pair of nodes in a − circuit and the maximum power transfer to a load. The reader is advised to read Section D.3 of Appendix D in preparation for this section.Figure 4.52For Practice Prob. 4.13. To find the Thevenin equivalent of a circuit at a pair of open ter- minals using PSpice, we use the schematic editor to draw the circuit and insert an independent probing current source, say, Ip, at the termi- nals. The probing current source must have a part name ISRC. We then perform a DC Sweep on Ip, as discussed in Section D.3. Typically, we may let the current through Ip vary from 0 to 1 A in 0.1-A increments. After saving and simulating the circuit, we use Probe to display a plot of the voltage across Ip versus the current through Ip. The zero inter- cept of the plot gives us the Thevenin equivalent voltage, while the slope of the plot is equal to the Thevenin resistance. To find the Norton equivalent involves similar steps except that we insert a probing independent voltage source (with a part name VSRC), say, Vp, at the terminals. We perform a DC Sweep on Vp and let Vp vary from 0 to 1 V in 0.1-V increments. A plot of the current through Vp versus the voltage across Vp is obtained using the Probe menu after simulation. The zero intercept is equal to the Norton current, while the slope of the plot is equal to the Norton conductance. To find the maximum power transfer to a load using PSpice involves performing a DC parametric Sweep on the component value of RL in Fig. 4.48 and plotting the power delivered to the load as a function of RL. According to Fig. 4.49, the maximum power occurs

4.9 Verifying Circuit Theorems with PSpice 153when RL ϭ RTh. This is best illustrated with an example, and Ex-ample 4.15 provides one. We use VSRC and ISRC as part names for the independent volt-age and current sources, respectively.Consider the circuit in Fig. 4.31 (see Example 4.9). Use PSpice to find Example 4.14the Thevenin and Norton equivalent circuits.Solution:(a) To find the Thevenin resistance RTh and Thevenin voltage VTh atthe terminals a-b in the circuit in Fig. 4.31, we first use Schematics todraw the circuit as shown in Fig. 4.53(a). Notice that a probing currentsource I2 is inserted at the terminals. Under Analysis/Setput, we selectDC Sweep. In the DC Sweep dialog box, we select Linear for theSweep Type and Current Source for the Sweep Var. Type. We enter I2under the Name box, 0 as Start Value, 1 as End Value, and 0.1 asIncrement. After simulation, we add trace V(I2:–) from the PSpice A/Dwindow and obtain the plot shown in Fig. 4.53(b). From the plot, weobtainVTh ϭ Zero intercept ϭ 20 V, RTh ϭ Slope ϭ 26 Ϫ 20 ϭ 6 ⍀ 1These agree with what we got analytically in Example 4.9. R2 R4 26 V 24 V 22 22 V E1I1 R4 4 + +− R3 6 I2 − GAIN=2 20 V 0.4 A 0.6 A 0.8 A 1.0 A 0 A 0.2 A (b) 0 = V(I2:_) (a)Figure 4.53For Example 4.14: (a) schematic and (b) plot for finding RTh and VTh.(b) To find the Norton equivalent, we modify the schematic in Fig. 4.53(a)by replaying the probing current source with a probing voltage sourceV1. The result is the schematic in Fig. 4.54(a). Again, in the DC Sweepdialog box, we select Linear for the Sweep Type and Voltage Sourcefor the Sweep Var. Type. We enter V1 under Name box, 0 as Start Value,1 as End Value, and 0.1 as Increment. Under the PSpice A/D Window,we add trace I (V1) and obtain the plot in Fig. 4.54(b). From the plot,we obtain IN ϭ Zero intercept ϭ 3.335 A 3.335 Ϫ 3.165 GN ϭ Slope ϭ 1 ϭ 0.17 S

154 Chapter 4 Circuit Theorems 3.4 A R2 R1 2 2 3.3 A E1I1 R4 4 + +− R3 6 V1 +− 3.2 A − GAIN=2 3.1 A 0.4 V 0.6 V 0.8 V 1.0 V 0 V 0.2 V V_V1 0 I(V1) (b) (a)Figure 4.54For Example 4.14: (a) schematic and (b) plot for finding GN and IN. Practice Problem 4.14 Rework Practice Prob. 4.9 using PSpice. Answer: VTh ϭ 5.333 V, RTh ϭ 444.4 m⍀. Example 4.15 RL Refer to the circuit in Fig. 4.55. Use PSpice to find the maximum power transfer to RL. 1 kΩ Solution: 1 V +− We need to perform a DC Sweep on RL to determine when the power across it is maximum. We first draw the circuit using Schematics asFigure 4.55 shown in Fig. 4.56. Once the circuit is drawn, we take the followingFor Example 4.15. three steps to further prepare the circuit for a DC Sweep. PARAMETERS: The first step involves defining the value of RL as a parameter, RL 2k since we want to vary it. To do this: R1 1. DCLICKL the value 1k of R2 (representing RL) to open up the Set Attribute Value dialog box. 1k 2. Replace 1k with {RL} and click OK to accept the change. V1 Note that the curly brackets are necessary.DC=1 V + R2 {RL} The second step is to define parameter. To achieve this: − 1. Select Draw/Get New Part/Libraries p/special.slb. 0 2. Type PARAM in the PartName box and click OK. 3. DRAG the box to any position near the circuit.Figure 4.56 4. CLICKL to end placement mode. 5. DCLICKL to open up the PartName: PARAM dialog box.Schematic for the circuit in Fig. 4.55. 6. CLICKL on NAME1 ϭ and enter RL (with no curly brackets) in the Value box, and CLICKL Save Attr to accept change. 7. CLICKL on VALUE1 ϭ and enter 2k in the Value box, and CLICKL Save Attr to accept change. 8. Click OK.

4.10 Applications 155 The value 2k in item 7 is necessary for a bias point calculation; it 250 uWcannot be left blank. 200 uW The third step is to set up the DC Sweep to sweep the parameter.To do this: 150 uW 1. Select Analysis/Setput to bring up the DC Sweep dialog box. 100 uW 2. For the Sweep Type, select Linear (or Octave for a wide range 50 uW 2.0 K 4.0 K 6.0 K of RL). 0 3. For the Sweep Var. Type, select Global Parameter. 4. Under the Name box, enter RL. –V(R2:2)*I(R2) 5. In the Start Value box, enter 100. 6. In the End Value box, enter 5k. RL 7. In the Increment box, enter 100. 8. Click OK and Close to accept the parameters. Figure 4.57 For Example 4.15: the plot of power After taking these steps and saving the circuit, we are ready to across RL.simulate. Select Analysis/Simulate. If there are no errors, we selectAdd Trace in the PSpice A/D window and type ϪV(R2:2)*I(R2) inthe Trace Command box. [The negative sign is needed since I(R2) isnegative.] This gives the plot of the power delivered to RL as RL variesfrom 100 ⍀ to 5 k⍀. We can also obtain the power absorbed by RL bytyping V(R2:2)*V(R2:2)/RL in the Trace Command box. Either way,we obtain the plot in Fig. 4.57. It is evident from the plot that themaximum power is 250 mW. Notice that the maximum occurs whenRL ϭ 1 k⍀, as expected analytically.Find the maximum power transferred to RL if the 1-k⍀ resistor in Practice Problem 4.15Fig. 4.55 is replaced by a 2-k⍀ resistor.Answer: 125 mW. Rs4.10 Applications vs +− (a)In this section we will discuss two important practical applications ofthe concepts covered in this chapter: source modeling and resistancemeasurement.4.10.1 Source Modeling is RpSource modeling provides an example of the usefulness of the (b)Thevenin or the Norton equivalent. An active source such as a batteryis often characterized by its Thevenin or Norton equivalent circuit. An Figure 4.58ideal voltage source provides a constant voltage irrespective of the cur- (a) Practical voltage source, (b) practicalrent drawn by the load, while an ideal current source supplies a con- current source.stant current regardless of the load voltage. As Fig. 4.58 shows,practical voltage and current sources are not ideal, due to their inter-nal resistances or source resistances Rs and Rp. They become ideal asRs S 0 and Rp S ϱ. To show that this is the case, consider the effect

156 Chapter 4 Circuit Theorems of the load on voltage sources, as shown in Fig. 4.59(a). By the volt- age division principle, the load voltage is vL ϭ Rs RL RL vs (4.25) ϩ As RL increases, the load voltage approaches a source voltage vs, as illustrated in Fig. 4.59(b). From Eq. (4.25), we should note that: 1. The load voltage will be constant if the internal resistance Rs of the source is zero or, at least, Rs V RL. In other words, the smaller Rs is compared with RL, the closer the voltage source is to being ideal. Rs + vL Ideal source vs +− vL vs Practical source RL − (a) 0 (b) RL Figure 4.59 (a) Practical voltage source connected to a load RL, (b) load volt- age decreases as RL decreases. 2. When the load is disconnected (i.e., the source is open-circuited so that RL S ϱ), voc ϭ vs. Thus, vs may be regarded as the unloaded source voltage. The connection of the load causes the terminal volt- age to drop in magnitude; this is known as the loading effect. IL The same argument can be made for a practical current source when connected to a load as shown in Fig. 4.60(a). By the current division is Rp RL principle, (a) iL ϭ Rp Rp RL is (4.26) IL ϩ Ideal source Figure 4.60(b) shows the variation in the load current as the load resist- is ance increases. Again, we notice a drop in current due to the load (load- Practical source ing effect), and load current is constant (ideal current source) when the 0 RL internal resistance is very large (i.e., Rp S ϱ or, at least, Rp W RL). (b) Sometimes, we need to know the unloaded source voltage vs andFigure 4.60 the internal resistance Rs of a voltage source. To find vs and Rs, we fol-(a) Practical current source connected to a low the procedure illustrated in Fig. 4.61. First, we measure the open-load RL, (b) load current decreases as RL circuit voltage voc as in Fig. 4.61(a) and setincreases. vs ϭ voc (4.27) Then, we connect a variable load RL across the terminals as in Fig. 4.61(b). We adjust the resistance RL until we measure a load volt- age of exactly one-half of the open-circuit voltage, vL ϭ voc ͞2, because now RL ϭ RTh ϭ Rs. At that point, we disconnect RL and measure it. We set Rs ϭ RL (4.28) For example, a car battery may have vs ϭ 12 V and Rs ϭ 0.05 ⍀.

4.10 Applications 157 Signal + Signal + source voc source vL RL − − (a) (b) Figure 4.61 (a) Measuring voc, (b) measuring vL.The terminal voltage of a voltage source is 12 V when connected to a Example 4.162-W load. When the load is disconnected, the terminal voltage rises to12.4 V. (a) Calculate the source voltage vs and internal resistance Rs.(b) Determine the voltage when an 8-⍀ load is connected to the source.Solution:(a) We replace the source by its Thevenin equivalent. The terminalvoltage when the load is disconnected is the open-circuit voltage, vs ϭ voc ϭ 12.4 VWhen the load is connected, as shown in Fig. 4.62(a), vL ϭ 12 V and Rs iLpL ϭ 2 W. Hence, vs +− + pL ϭ v2L 1 RL ϭ vL2 ϭ 122 ϭ 72 ⍀ vL RL RL pL 2 −The load current is iL ϭ vL ϭ 12 ϭ 1 A (a) RL 72 6 2.4 Ω 12.4 V +−The voltage across Rs is the difference between the source voltage vsand the load voltage vL, or (b) 12.4 Ϫ 12 ϭ 0.4 ϭ RsiL, 0.4 Figure 4.62 + 8Ω Rs ϭ IL ϭ 2.4 ⍀ For Example 4.16. v −(b) Now that we have the Thevenin equivalent of the source, weconnect the 8-⍀ load across the Thevenin equivalent as shown inFig. 4.62(b). Using voltage division, we obtain v ϭ 8 8 ϭ 9.538 V ϩ 2.4 (12.4)The measured open-circuit voltage across a certain amplifier is 9 V. Practice Problem 4.16The voltage drops to 8 V when a 20-⍀ loudspeaker is connected to theamplifier. Calculate the voltage when a 10-⍀ loudspeaker is usedinstead.Answer: 7.2 V.

158 Chapter 4 Circuit Theorems 4.10.2 Resistance MeasurementHistorical note: The bridge was Although the ohmmeter method provides the simplest way to measureinvented by Charles Wheatstone resistance, more accurate measurement may be obtained using the(1802–1875), a British professor who Wheatstone bridge. While ohmmeters are designed to measure resist-also invented the telegraph, as Samuel ance in low, mid, or high range, a Wheatstone bridge is used to mea-Morse did independently in the sure resistance in the mid range, say, between 1 ⍀ and 1 M⍀. Very lowUnited States. values of resistances are measured with a milliohmmeter, while very high values are measured with a Megger tester. R1 R3 Galvanometer The Wheatstone bridge (or resistance bridge) circuit is used in a number of applications. Here we will use it to measure an unknown resistance. The unknown resistance Rx is connected to the bridge as shown in Fig. 4.63. The variable resistance is adjusted until no current flows through the galvanometer, which is essentially a d’Arsonval movement operating as a sensitive current-indicating device like an ammeter in the microamp range. Under this condition v1 ϭ v2, and the bridge is said to be balanced. Since no current flows through the gal- vanometer, R1 and R2 behave as though they were in series; so do R3 and Rx. The fact that no current flows through the galvanometer also implies that v1 ϭ v2. Applying the voltage division principle,v +− + v1 ϭ R2 v ϭ v2 ϭ Rx v (4.29) v2 Rx ϩ R2 ϩ Rx R2 + − R1 R3 v1 Hence, no current flows through the galvanometer when −Figure 4.63 R2 ϭ Rx 1 R2R3 ϭ R1Rx R1 ϩ R2 R3 ϩ RxThe Wheatstone bridge; Rx is theresistance to be measured. or Rx ϭ R3 R2 (4.30) R1 If R1 ϭ R3, and R2 is adjusted until no current flows through the gal- vanometer, then Rx ϭ R2. How do we find the current through the galvanometer when the Wheatstone bridge is unbalanced? We find the Thevenin equivalent (VTh and RTh) with respect to the galvanometer terminals. If Rm is the resistance of the galvanometer, the current through it under the unbal- anced condition is I ϭ VTh Rm (4.31) RTh ϩ Example 4.18 will illustrate this.Example 4.17 In Fig. 4.63, R1 ϭ 500 ⍀ and R3 ϭ 200 ⍀. The bridge is balanced when R2 is adjusted to be 125 ⍀. Determine the unknown resistance Rx. Solution: Using Eq. (4.30) gives Rx ϭ R3 R2 ϭ 200 ϭ 50 ⍀ R1 125 500

4.10 Applications 159A Wheatstone bridge has R1 ϭ R3 ϭ 1 k⍀. R2 is adjusted until no cur- Practice Problem 4.17rent flows through the galvanometer. At that point, R2 ϭ 3.2 k⍀. Whatis the value of the unknown resistance?Answer: 3.2 k⍀.The circuit in Fig. 4.64 represents an unbalanced bridge. If the gal- Example 4.18vanometer has a resistance of 40 ⍀, find the current through thegalvanometer.220 V +− 3 kΩ b 400 Ω a 40 Ω G 600 Ω 1 kΩFigure 4.64Unbalanced bridge of Example 4.18.Solution:We first need to replace the circuit by its Thevenin equivalent atterminals a and b. The Thevenin resistance is found using the circuitin Fig. 4.65(a). Notice that the 3-k⍀ and 1-k⍀ resistors are in parallel;so are the 400-⍀ and 600-⍀ resistors. The two parallel combinationsform a series combination with respect to terminals a and b. Hence,RTh ϭ 3000 ʈ 1000 ϩ 400 ʈ 600ϭ 3000 ϫ 1000 ϩ 400 ϫ 600 ϭ 750 ϩ 240 ϭ 990 ⍀ 3000 ϩ 1000 400 ϩ 600To find the Thevenin voltage, we consider the circuit in Fig. 4.65(b).Using the voltage division principle gives 1000 v2 ϭ 600 (220) ϭ 132 Vv1 ϭ 1000 ϩ 3000 (220) ϭ 55 V, 400 600 ϩApplying KVL around loop ab givesϪv1 ϩ VTh ϩ v2 ϭ 0 or VTh ϭ v1 Ϫ v2 ϭ 55 Ϫ 132 ϭ Ϫ77 VHaving determined the Thevenin equivalent, we find the currentthrough the galvanometer using Fig. 4.65(c). IG ϭ VTh ϭ Ϫ77 ϭ Ϫ74.76 mA RTh ϩ Rm 990 ϩ 40The negative sign indicates that the current flows in the directionopposite to the one assumed, that is, from terminal b to terminal a.

160 Chapter 4 Circuit Theorems 3 kΩ 400 Ω 3 kΩ 400 Ω 1 kΩ 600 Ω a RTh b 220 V +− + +− + (a) RTh a VTh b v2 600 Ω v1 1 kΩ − − (b) a IG 40 Ω VTh +− G b (c)Figure 4.65For Example 4.18: (a) Finding RTh, (b) finding VTh, (c) determining the current through the galvanometer.Practice Problem 4.18 Obtain the current through the galvanometer, having a resistance of 14 ⍀, in the Wheatstone bridge shown in Fig. 4.66.20 Ω 30 Ω Answer: 64 mA. G60 Ω 14 Ω 4.11 Summary 40 Ω 16 V 1. A linear network consists of linear elements, linear dependent sources, and linear independent sources.Figure 4.66For Practice Prob. 4.18. 2. Network theorems are used to reduce a complex circuit to a sim- pler one, thereby making circuit analysis much simpler. 3. The superposition principle states that for a circuit having multi- ple independent sources, the voltage across (or current through) an element is equal to the algebraic sum of all the individual voltages (or currents) due to each independent source acting one at a time. 4. Source transformation is a procedure for transforming a voltage source in series with a resistor to a current source in parallel with a resistor, or vice versa. 5. Thevenin’s and Norton’s theorems allow us to isolate a portion of a network while the remaining portion of the network is replaced by an equivalent network. The Thevenin equivalent consists of a voltage source VTh in series with a resistor RTh, while the Norton equivalent consists of a current source IN in parallel with a resis- tor RN. The two theorems are related by source transformation. RN ϭ RTh, IN ϭ VTh RTh

Review Questions 1616. For a given Thevenin equivalent circuit, maximum power transfer occurs when RL ϭ RTh; that is, when the load resistance is equal to the Thevenin resistance.7. The maximum power transfer theorem states that the maximum power is delivered by a source to the load RL when RL is equal to RTh, the Thevenin resistance at the terminals of the load.8. PSpice can be used to verify the circuit theorems covered in this chapter.9. Source modeling and resistance measurement using the Wheat- stone bridge provide applications for Thevenin’s theorem.Review Questions4.1 The current through a branch in a linear network is 4.7 The Norton resistance RN is exactly equal to the 2 A when the input source voltage is 10 V. If the Thevenin resistance RTh. voltage is reduced to 1 V and the polarity is reversed, the current through the branch is: (a) True (b) False(a) Ϫ2 A (b) Ϫ0.2 A (c) 0.2 A 4.8 Which pair of circuits in Fig. 4.68 are equivalent?(d) 2 A (e) 20 A (a) a and b (b) b and d4.2 For superposition, it is not required that only one (c) a and c (d) c and d independent source be considered at a time; any number of independent sources may be considered 5Ω 5Ω simultaneously. 20 V +− 4A(a) True (b) False (a) (b)4.3 The superposition principle applies to power calculation.(a) True (b) False4.4 Refer to Fig. 4.67. The Thevenin resistance at 4A 5 Ω 20 V +− 5Ω terminals a and b is:(a) 25 ⍀ (b) 20 ⍀ (c) (d)(c) 5 ⍀ (d) 4 ⍀ Figure 4.68 5Ω For Review Question 4.8.50 V +− a 20 Ω 4.9 A load is connected to a network. At the terminals to b which the load is connected, RTh ϭ 10 ⍀ andFigure 4.67 VTh ϭ 40 V. The maximum possible power suppliedFor Review Questions 4.4 to 4.6. to the load is:4.5 The Thevenin voltage across terminals a and b of the (a) 160 W (b) 80 W circuit in Fig. 4.67 is: (c) 40 W (d) 1 W(a) 50 V (b) 40 V 4.10 The source is supplying the maximum power to the(c) 20 V (d) 10 V load when the load resistance equals the source resistance.4.6 The Norton current at terminals a and b of the circuit (a) True (b) False in Fig. 4.67 is:(a) 10 A (b) 2.5 A Answers: 4.1b, 4.2a, 4.3b, 4.4d, 4.5b, 4.6a, 4.7a, 4.8c, 4.9c, 4.10a.(c) 2 A (d) 0 A

162 Chapter 4 Circuit TheoremsProblemsSection 4.2 Linearity Property 4.5 For the circuit in Fig. 4.73, assume vo ϭ 1 V, and use linearity to find the actual value of vo. 4.1 Calculate the current io in the circuit of Fig. 4.69. What value of input voltage is necessary to make io 2 Ω 3 Ω vo 2 Ω equal to 5 amps? 15 V +− 6Ω 6Ω 4Ω 5Ω 25 Ω io Figure 4.7330 V +− 40 Ω 15 Ω For Prob. 4.5.Figure 4.69 4.6 For the linear circuit shown in Fig. 4.74, use linearityFor Prob. 4.1. to complete the following table. 4.2 Using Fig. 4.70, design a problem to help other Experiment Vs Vo students better understand linearity. 1 12 V 4 V R2 R4 2 16 V 3 1V + 4 Ϫ2 V I R1 R3 R5 vo Vs + Linear + − − circuit VoFigure 4.70 –For Prob. 4.2.4.3 (a) In the circuit of Fig. 4.71, calculate vo and io Figure 4.74 when vs ϭ 1 V. For Prob. 4.6. (b) Find vo and io when vs ϭ 10 V. 4.7 Use linearity and the assumption that Vo ϭ 1 V to find the actual value of Vo in Fig. 4.75. (c) What are vo and io when each of the 1-⍀ resistors is replaced by a 10-⍀ resistor and 1Ω 4Ω vs ϭ 10 V? 1Ω + 3 Ω 2 Ω Vo 1Ω 1Ω 4 V −+vs +− – + 1 Ω vo io Figure 4.75 1Ω For Prob. 4.7. − Section 4.3 SuperpositionFigure 4.71 4.8 Using superposition, find Vo in the circuit of Fig. 4.76.For Prob. 4.3. Check with PSpice or MultiSim. 4 Ω Vo 1 Ω4.4 Use linearity to determine io in the circuit of Fig. 4.72. 3Ω 3Ω 2Ω 9A 5 Ω +− 3 V 6Ω io +− 9 V 4ΩFigure 4.72 Figure 4.76For Prob. 4.4. For Prob. 4.8.

Problems 1634.9 Given that I ϭ 4 amps when Vs ϭ 40 volts and Is ϭ 4 4.13 Use superposition to find vo in the circuit of Fig. 4.81. amps and I ϭ 1 amp when Vs ϭ 20 volts and Is ϭ 0, 4A use superposition and linearity to determine the value of I when Vs ϭ 60 volts and Is ϭ Ϫ2 amps. 8ΩVs +− −+ + vo I Is 12 V 5 Ω −Figure 4.77 2 A 10 ΩFor Prob. 4.9. Figure 4.81 For Prob. 4.13.4.10 Using Fig. 4.78, design a problem to help other 4.14 Apply the superposition principle to find vo in the students better understand superposition. Note, the circuit of Fig. 4.82. letter k is a gain you can specify to make the problem easier to solve but must not be zero. 6Ω 2A R kVab a 4Ω 2ΩV +− +− + 20 V +− + I Vab 1 A vo 3 Ω − Figure 4.82 For Prob. 4.14. − bFigure 4.78For Prob. 4.10.4.11 Use the superposition principle to find io and vo in 4.15 For the circuit in Fig. 4.83, use superposition to find i. the circuit of Fig. 4.79. Calculate the power delivered to the 3-⍀ resistor. io 10 Ω 20 Ω + vo − 20 V +− 1Ω 2A 4Ω 2Ω6 A 40 Ω 4io − 30 V i + 3Ω − + 16 VFigure 4.79 Figure 4.83For Prob. 4.11. For Probs. 4.15 and 4.56.4.12 Determine vo in the circuit of Fig. 4.80 using the 4.16 Given the circuit in Fig. 4.84, use superposition to superposition principle. obtain i0. 2A 4A 6Ω 5Ω 4Ω io 4 Ω 3Ω 2Ω12 V +− + vo − 12 Ω +− 19 V 3Ω 12 V +− 10 Ω 5Ω 2AFigure 4.80 Figure 4.84For Prob. 4.12. For Prob. 4.16.

164 Chapter 4 Circuit Theorems4.17 Use superposition to obtain vx in the circuit of 4.21 Using Fig. 4.89, design a problem to help other Fig. 4.85. Check your result using PSpice or students better understand source transformation. MultiSim. 30 Ω 10 Ω 6A 20 Ω io R1 I90 V +− + vx − 30 Ω +− 40 V V +− 60 Ω + R2 vo −Figure 4.85 Figure 4.89For Prob. 4.17. For Prob. 4.21.4.18 Use superposition to find Vo in the circuit of Fig. 4.86. 4.22 For the circuit in Fig. 4.90, use source transformation to find i. 2Ω 1Ω 5 Ω 10 Ω i10 V +− 2A 0.5Vo 2 A 5 Ω 4 Ω +− 20 V + 4 Ω Vo Figure 4.90 For Prob. 4.22. − 4.23 Referring to Fig. 4.91, use source transformation toFigure 4.86 determine the current and power absorbed by theFor Prob. 4.18. 8-⍀ resistor. 8Ω 3Ω4.19 Use superposition to solve for vx in the circuit of Fig. 4.87. ix 6A 4A + 3A 10 Ω 6 Ω −+ 15 V 2Ω −+ 8 Ω vxFigure 4.87 4ix −For Prob. 4.19. Figure 4.91Section 4.4 Source Transformation For Prob. 4.23. 4.20 Use source transformation to reduce the circuit in 4.24 Use source transformation to find the voltage Vx in Fig. 4.88 to a single voltage source in series with a the circuit of Fig. 4.92. single resistor. 3A 3 A 10 Ω 20 Ω 40 Ω 8Ω 10 Ω 12 V +− +− 16 V + Vx − 10 ΩFigure 4.88For Prob. 4.20. 40 V +− 2Vx Figure 4.92 For Prob. 4.24.

Problems 1654.25 Obtain vo in the circuit of Fig. 4.93 using source 4.29 Use source transformation to find vo in the circuit of transformation. Check your result using PSpice or Fig. 4.97. MultiSim. 4 kΩ 2A 9Ω 2 kΩ 3vo −+ 3A 4Ω 5Ω 6A 3 mA + + vo − 1 kΩ voFigure 4.93 +− Figure 4.97 −For Prob. 4.25. 2Ω 30 V For Prob. 4.29. 4.30 Use source transformation on the circuit shown in Fig 4.98 to find ix.4.26 Use source transformation to find io in the circuit of Fig. 4.94. ix 24 Ω 60 Ω 5Ω 12 V + 30 Ω 10 Ω 0.7ix − 3 A io 4 Ω6A 2Ω +− 20 V Figure 4.98 For Prob. 4.30.Figure 4.94 4.31 Determine vx in the circuit of Fig. 4.99 using sourceFor Prob. 4.26. transformation.4.27 Apply source transformation to find vx in the circuit 3Ω 6Ω of Fig. 4.95. + vx − 10 Ω a 12 Ω b 20 Ω 12 V +− 8Ω + 2vx50 V +− + vx − 8 A +− 40 V − 40 Ω Figure 4.99 For Prob. 4.31.Figure 4.95 4.32 Use source transformation to find ix in the circuit ofFor Probs. 4.27 and 4.40. Fig. 4.100.4.28 Use source transformation to find Io in Fig. 4.96. 1 Ω Io 4Ω 10 Ω 0.5ix + Vo − ix 15 Ω 60 V +− 50 Ω8V + 3Ω 31Vo 40 Ω −Figure 4.96 Figure 4.100For Prob. 4.28. For Prob. 4.32.

166 Chapter 4 Circuit TheoremsSections 4.5 and 4.6 Thevenin’s and Norton’s 4.37 Find the Norton equivalent with respect to terminals Theorems a-b in the circuit shown in Fig. 4.104.4.33 Determine the Thevenin equivalent circuit, shown in 2A Fig. 4.101, as seen by the 5-ohm resistor. 20 Ω Then calculate the current flowing through the 5-ohm resistor. a 120 V +− 40 Ω 12 Ω 10 Ω b4A 10 Ω 5Ω Figure 4.104 For Prob. 4.37.Figure 4.101 4.38 Apply Thevenin’s theorem to find Vo in the circuit ofFor Prob. 4.33. Fig. 4.105. 4Ω 1Ω4.34 Using Fig. 4.102, design a problem that will help 3A 16 Ω 5Ω + other students better understand Thevenin equivalent 10 Ω Vo circuits. – −+ 12 V I Figure 4.105 For Prob. 4.38. R1 R3V +− R2 4.39 Obtain the Thevenin equivalent at terminals a-b of the circuit shown in Fig. 4.106. a 3A b 10 Ω 16 ΩFigure 4.102 10 Ω aFor Probs. 4.34 and 4.49. b 24 V +− 5Ω4.35 Use Thevenin’s theorem to find vo in Prob. 4.12. Figure 4.106 For Prob. 4.39.4.36 Solve for the current i in the circuit of Fig. 4.103 using Thevenin’s theorem. (Hint: Find the Thevenin 4.40 Find the Thevenin equivalent at terminals a-b of the equivalent seen by the 12-⍀ resistor.) circuit in Fig. 4.107. i + Vo − 20 kΩ 12 Ω 10 kΩ a +− 30 V 10 Ω 40 Ω 70 V + b −+ 4Vo50 V +− −Figure 4.103 Figure 4.107For Prob. 4.36. For Prob. 4.40.

Problems 1674.41 Find the Thevenin and Norton equivalents at 4.45 Find the Thevenin equivalent of the circuit in terminals a-b of the circuit shown in Fig. 4.108. Fig. 4.112 as seen by looking into terminals a and b. 14 V 14 Ω a 6Ω a −+ 6Ω 4Ω1A 6Ω 3A 5Ω 4A b b Figure 4.112 For Prob. 4.45.Figure 4.108For Prob. 4.41.*4.42 For the circuit in Fig. 4.109, find the Thevenin 4.46 Using Fig. 4.113, design a problem to help other equivalent between terminals a and b. students better understand Norton equivalent circuits. 20 Ω R2 a 10 Ω 20 Ω −+ 20 V I R1 R3 5A b b a Figure 4.113 10 Ω 10 Ω 10 Ω For Prob. 4.46. 30 V +−Figure 4.109 4.47 Obtain the Thevenin and Norton equivalent circuitsFor Prob. 4.42. of the circuit in Fig. 4.114 with respect to terminals a and b. 4.43 Find the Thevenin equivalent looking into terminals a-b of the circuit in Fig. 4.110 and solve for ix. 12 Ω 10 Ω a 6 Ω b a ix 30 V −+ + 2Vx 10 Ω Vx 60 Ω 20 V +− 5Ω 2A – bFigure 4.110 Figure 4.114For Prob. 4.43. For Prob. 4.47.4.44 For the circuit in Fig. 4.111, obtain the Thevenin 4.48 Determine the Norton equivalent at terminals a-b for equivalent as seen from terminals: the circuit in Fig. 4.115. (a) a-b (b) b-c 3Ω 1Ω 10io 2Ω a +− a io 24 V +− 4Ω 2A 4Ω b 2Ω 5Ω 2A b c Figure 4.115 For Prob. 4.48.Figure 4.111 4.49 Find the Norton equivalent looking into terminalsFor Prob. 4.44. a-b of the circuit in Fig. 4.102. Let V ϭ 40 V, I ϭ 3 A, R1 ϭ 10 ⍀, R2 ϭ 40 ⍀, and R3 ϭ 20 ⍀.* An asterisk indicates a challenging problem.

168 Chapter 4 Circuit Theorems4.50 Obtain the Norton equivalent of the circuit in 4.54 Find the Thevenin equivalent between terminals a-b Fig. 4.116 to the left of terminals a-b. Use the of the circuit in Fig. 4.120. result to find current i. 6 Ω 12 V a 1 kΩ +− i a2A 4Ω 5Ω 4A 3 V −+ Io + 40Io + 50 Ω b 2Vx − Vx – b Figure 4.120 For Prob. 4.54.Figure 4.116For Prob. 4.50.4.51 Given the circuit in Fig. 4.117, obtain the Norton *4.55 Obtain the Norton equivalent at terminals a-b of the equivalent as viewed from terminals: circuit in Fig. 4.121.(a) a-b (b) c-d ab 8 kΩ I 6Ω 4Ω a c 2 V +− 0.001Vab + 80I 50 kΩ + b 2Ω − Vab120 V +− 3Ω 6A − d Figure 4.121Figure 4.117 For Prob. 4.55.For Prob. 4.51.4.52 For the transistor model in Fig. 4.118, obtain the 4.56 Use Norton’s theorem to find Vo in the circuit of Thevenin equivalent at terminals a-b. Fig. 4.122. 3 kΩ io a 12 kΩ 2 kΩ 10 kΩ6 V +− 20io 2 kΩ + b 36 V + 24 kΩ 3 mA 1 kΩ − Vo −Figure 4.118For Prob. 4.52. Figure 4.122 For Prob. 4.56.4.53 Find the Norton equivalent at terminals a-b of the circuit in Fig. 4.119. 0.25vo 4.57 Obtain the Thevenin and Norton equivalent circuits at terminals a-b for the circuit in Fig. 4.123.18 V +− 6Ω 2Ω a 50 V +− 3Ω 2Ω a b 10 Ω + + 3 Ω vo 6 Ω vx 0.5vx b − −Figure 4.119 Figure 4.123For Prob. 4.53. For Probs. 4.57 and 4.79.

Problems 1694.58 The network in Fig. 4.124 models a bipolar transistor *4.62 Find the Thevenin equivalent of the circuit in common-emitter amplifier connected to a load. Find Fig. 4.128. the Thevenin resistance seen by the load. ib R1 bib 0.1io vs +− R2 RL 10 Ω + a vo b −Figure 4.124 ioFor Prob. 4.58. 40 Ω 20 Ω 4.59 Determine the Thevenin and Norton equivalents at terminals a-b of the circuit in Fig. 4.125. Figure 4.128 +− For Prob. 4.62. 2vo 10 Ω 20 Ω8A a b 4.63 Find the Norton equivalent for the circuit in Fig. 4.129. 50 Ω 40 ΩFigure 4.125 10 Ω 0.5voFor Probs. 4.59 and 4.80. +*4.60 For the circuit in Fig. 4.126, find the Thevenin and vo 20 Ω Norton equivalent circuits at terminals a-b. − 2 A Figure 4.129 For Prob. 4.63. 18 V 4 Ω 6Ω 4.64 Obtain the Thevenin equivalent seen at terminals a-b of the circuit in Fig. 4.130. a +− b 3A 4Ω 1Ω ix 5Ω 2Ω a b +− 10ix + 10 V −Figure 4.126For Probs. 4.60 and 4.81.*4.61 Obtain the Thevenin and Norton equivalent circuits Figure 4.130 at terminals a-b of the circuit in Fig. 4.127. For Prob. 4.64. 2Ω 4.65 For the circuit shown in Fig. 4.131, determine the a relationship between Vo and I0.12 V +− 6Ω 6Ω +− 12 V 4Ω 2Ω Io 6Ω + Vo 2Ω − 12 V 2Ω 32 V −+ 12 Ω − + bFigure 4.127For Prob. 4.61. Figure 4.131 For Prob. 4.65.

170 Chapter 4 Circuit TheoremsSection 4.8 Maximum Power Transfer 4.70 Determine the maximum power delivered to the variable resistor R shown in the circuit of Fig. 4.136. 4.66 Find the maximum power that can be delivered to the resistor R in the circuit of Fig. 4.132. 3 Vx 2 Ω 10 V 5Ω 5Ω −+ 3Ω R20 V +− 5Ω 6A 4 V −+ 15 Ω RFigure 4.132 Figure 4.136 6ΩFor Prob. 4.66. For Prob. 4.70. +− 4.67 The variable resistor R in Fig. 4.133 is adjusted until Vx it absorbs the maximum power from the circuit. (a) Calculate the value of R for maximum power. (b) Determine the maximum power absorbed by R. 80 Ω 20 Ω 4.71 For the circuit in Fig. 4.137, what resistor connected across terminals a-b will absorb maximum power from the circuit? What is that power? 40 V R +− 3 kΩ 10 kΩ 10 Ω 90 Ω + a 40 kΩFigure 4.133 8 V +− vo 1 kΩ – 120voFor Prob. 4.67. − + b*4.68 Compute the value of R that results in maximum Figure 4.137 power transfer to the 10-⍀ resistor in Fig. 4.134. For Prob. 4.71. Find the maximum power. R 12 V +− 10 Ω 20 Ω 4.72 (a) For the circuit in Fig. 4.138, obtain the Thevenin +− 8 V equivalent at terminals a-b.Figure 4.134 (b) Calculate the current in RL ϭ 8 ⍀.For Prob. 4.68. (c) Find RL for maximum power deliverable to RL. (d) Determine that maximum power.4.69 Find the maximum power transferred to resistor R in the circuit of Fig. 4.135. 2A 10 kΩ 22 kΩ 4Ω 6Ω a + 4A 2Ω RL b100 V +− vo 40 kΩ 0.003vo 30 kΩ R +− − 20 VFigure 4.135 Figure 4.138For Prob. 4.69. For Prob. 4.72.

Problems 1714.73 Determine the maximum power that can be delivered 4.80 Use PSpice or MultiSim to find the Thevenin to the variable resistor R in the circuit of Fig. 4.139. equivalent circuit at terminals a-b of the circuit in Fig. 4.125. 10 Ω 25 Ω 4.81 For the circuit in Fig. 4.126, use PSpice or MultiSim60 V −+ R to find the Thevenin equivalent at terminals a-b. 20 Ω 5Ω Section 4.10 ApplicationsFigure 4.139 4.82 A battery has a short-circuit current of 20 A and anFor Prob. 4.73. open-circuit voltage of 12 V. If the battery is connected to an electric bulb of resistance 2 ⍀,4.74 For the bridge circuit shown in Fig. 4.140, find the calculate the power dissipated by the bulb. load RL for maximum power transfer and the maximum power absorbed by the load. 4.83 The following results were obtained from measurements taken between the two terminals of a resistive network. Terminal Voltage 12 V 0V Terminal Current 0 A 1.5 A vs +− R1 RL R3 Find the Thevenin equivalent of the network. R2 R4 4.84 When connected to a 4-⍀ resistor, a battery has aFigure 4.140 terminal voltage of 10.8 V but produces 12 V on anFor Prob. 4.74. open circuit. Determine the Thevenin equivalent circuit for the battery.*4.75 For the circuit in Fig. 4.141, determine the value of R such that the maximum power delivered to the 4.85 The Thevenin equivalent at terminals a-b of the load is 3 mW. linear network shown in Fig. 4.142 is to be determined by measurement. When a 10-k⍀ resistor is connected to terminals a-b, the voltage Vab is measured as 6 V. When a 30-k⍀ resistor is connected to the terminals, Vab is measured as 12 V. Determine: (a) the Thevenin equivalent at terminals a-b, (b) Vab when a 20-k⍀ resistor is connected to terminals a-b. R Linear a R network b R Figure 4.1421 V +− +− 2 V +− 3 V RL For Prob. 4.85.Figure 4.141 4.86 A black box with a circuit in it is connected to aFor Prob. 4.75. variable resistor. An ideal ammeter (with zero resistance) and an ideal voltmeter (with infiniteSection 4.9 Verifying Circuit Theorems resistance) are used to measure current and voltage with PSpice as shown in Fig. 4.143. The results are shown in the table on the next page. 4.76 Solve Prob. 4.34 using PSpice or MultiSim. Let V ϭ 40 V, I ϭ 3 A, R1 ϭ 10 ⍀, R2 ϭ 40 ⍀, and Black A i R3 ϭ 20 ⍀. box V R 4.77 Use PSpice or MultiSim to solve Prob. 4.44. Figure 4.143 For Prob. 4.86. 4.78 Use PSpice or MultiSim to solve Prob. 4.52. 4.79 Obtain the Thevenin equivalent of the circuit in Fig. 4.123 using PSpice or MultiSim.

172 Chapter 4 Circuit Theorems (a) Find i when R ϭ 4 ⍀. 4.90 The Wheatstone bridge circuit shown in Fig. 4.146 is used to measure the resistance of a strain gauge. The (b) Determine the maximum power from the box. adjustable resistor has a linear taper with a maximum value of 100 ⍀. If the resistance of the strain gauge R(⍀) V(V) i(A) is found to be 42.6 ⍀, what fraction of the full slider travel is the slider when the bridge is balanced? 2 3 1.5 8 8 1.0 Rs 4 kΩ 14 10.5 0.75 2 kΩ4.87 A transducer is modeled with a current source Is and vs +− G a parallel resistance Rs. The current at the terminals 100 Ω of the source is measured to be 9.975 mA when an ammeter with an internal resistance of 20 ⍀ is used. Rx (a) If adding a 2-k⍀ resistor across the source Figure 4.146 terminals causes the ammeter reading to fall to For Prob. 4.90. 9.876 mA, calculate Is and Rs. 4.91 (a) In the Wheatstone bridge circuit of Fig. 4.147, (b) What will the ammeter reading be if the resistance between the source terminals is select the values of R1 and R3 such that the bridge changed to 4 k⍀? can measure Rx in the range of 0–10 ⍀.4.88 Consider the circuit in Fig. 4.144. An ammeter with internal resistance Ri is inserted between A and B to measure Io. Determine the reading of the ammeter if: (a) Ri ϭ 500 ⍀, (b) Ri ϭ 0 ⍀. (Hint: Find the Thevenin equivalent circuit at terminals a-b.)30 kΩ a 2 kΩ b 5 kΩ −+ 60 V V +− R1 R3 20 kΩ G Rx Io 4 mA 50 Ω 10 kΩFigure 4.144 Figure 4.147For Prob. 4.88. For Prob. 4.91.4.89 Consider the circuit in Fig. 4.145. (a) Replace the (b) Repeat for the range of 0–100 ⍀. resistor RL by a zero resistance ammeter and determine the ammeter reading. (b) To verify the *4.92 Consider the bridge circuit of Fig. 4.148. Is the reciprocity theorem, interchange the ammeter and bridge balanced? If the 10-k⍀ resistor is replaced by the 12-V source and determine the ammeter reading an 18-k⍀ resistor, what resistor connected between again. terminals a-b absorbs the maximum power? What is this power? 2 kΩ 3 kΩ 6 kΩ 10 kΩ 20 kΩ RL 220 V +− ab 12 V −+ 12 kΩ 15 kΩ Figure 4.148 5 kΩ 10 kΩ For Prob. 4.92.Figure 4.145For Prob. 4.89.

Comprehensive Problems 173Comprehensive Problems4.93 The circuit in Fig. 4.149 models a common-emitter *4.96 A resistance array is connected to a load resistor R transistor amplifier. Find ix using source and a 9-V battery as shown in Fig. 4.151. transformation. (a) Find the value of R such that Vo ϭ 1.8 V. ix Rs (b) Calculate the value of R that will draw the maximum current. What is the maximum current?vs +− Ro ␤ix R 3 + Vo −Figure 4.149 10 ΩFor Prob. 4.93. 60 Ω4.94 An attenuator is an interface circuit that reduces the 8Ω 10 Ω 2 voltage level without changing the output resistance. 8Ω 40 Ω(a) By specifying Rs and Rp of the interface circuit in 4 Fig. 4.150, design an attenuator that will meet the 10 Ω following requirements:Vo ϭ 0.125, 1Vg Req ϭ RTh ϭ Rg ϭ 100 ⍀ + 9V−(b) Using the interface designed in part (a), calculate Figure 4.151 For Prob. 4.96. the current through a load of RL ϭ 50 ⍀ when Vg ϭ 12 V. Rg Rs 4.97 A common-emitter amplifier circuit is shown in Vg +− Rp Fig. 4.152. Obtain the Thevenin equivalent to the left of points B and E.Figure 4.150 AttenuatorFor Prob. 4.94. + RL Vo RL − 6 kΩ + Load B 12 V Req − 4 kΩ Rc E*4.95 A dc voltmeter with a sensitivity of 20 k⍀/V is used Figure 4.152 to find the Thevenin equivalent of a linear network. For Prob. 4.97. Readings on two scales are as follows: *4.98 For Practice Prob. 4.18, determine the current (a) 0–10 V scale: 4 V (b) 0–50 V scale: 5 V through the 40-⍀ resistor and the power dissipated by the resistor. Obtain the Thevenin voltage and the Thevenin resistance of the network.



Operational chapterAmplifiers 5He who will not reason is a bigot; he who cannot is a fool; and hewho dares not is a slave. —Lord ByronEnhancing Your Career Electronic Instrumentation used in medical research.Career in Electronic Instrumentation © Royalty-Free/CorbisEngineering involves applying physical principles to design devices forthe benefit of humanity. But physical principles cannot be understoodwithout measurement. In fact, physicists often say that physics is thescience that measures reality. Just as measurements are a tool for under-standing the physical world, instruments are tools for measurement.The operational amplifier introduced in this chapter is a building blockof modern electronic instrumentation. Therefore, mastery of operationalamplifier fundamentals is paramount to any practical application ofelectronic circuits. Electronic instruments are used in all fields of science and engi-neering. They have proliferated in science and technology to the extentthat it would be ridiculous to have a scientific or technical educationwithout exposure to electronic instruments. For example, physicists,physiologists, chemists, and biologists must learn to use electronicinstruments. For electrical engineering students in particular, the skillin operating digital and analog electronic instruments is crucial. Suchinstruments include ammeters, voltmeters, ohmmeters, oscilloscopes,spectrum analyzers, and signal generators. Beyond developing the skill for operating the instruments, someelectrical engineers specialize in designing and constructing electronicinstruments. These engineers derive pleasure in building their owninstruments. Most of them invent and patent their inventions. Special-ists in electronic instruments find employment in medical schools, hos-pitals, research laboratories, aircraft industries, and thousands of otherindustries where electronic instruments are routinely used. 175

176 Chapter 5 Operational AmplifiersThe term operational amplifier was in- 5.1 Introductiontroduced in 1947 by John Ragazziniand his colleagues, in their work on Having learned the basic laws and theorems for circuit analysis, we areanalog computers for the National now ready to study an active circuit element of paramount importance:Defense Research Council after World the operational amplifier, or op amp for short. The op amp is a versa-War II. The first op amps used vacuum tile circuit building block.tubes rather than transistors. The op amp is an electronic unit that behaves like a voltage-controlledAn op amp may also be regarded as a voltage source.voltage amplifier with very high gain. It can also be used in making a voltage- or current-controlled current source. An op amp can sum signals, amplify a signal, integrate it, or differentiate it. The ability of the op amp to perform these mathemat- ical operations is the reason it is called an operational amplifier. It is also the reason for the widespread use of op amps in analog design. Op amps are popular in practical circuit designs because they are ver- satile, inexpensive, easy to use, and fun to work with. We begin by discussing the ideal op amp and later consider the nonideal op amp. Using nodal analysis as a tool, we consider ideal op amp circuits such as the inverter, voltage follower, summer, and dif- ference amplifier. We will also analyze op amp circuits with PSpice. Finally, we learn how an op amp is used in digital-to-analog convert- ers and instrumentation amplifiers.Figure 5.1 5.2 Operational AmplifiersA typical operational amplifier. An operational amplifier is designed so that it performs some mathe-Courtesy of Tech America. matical operations when external components, such as resistors and capacitors, are connected to its terminals. Thus, The pin diagram in Fig. 5.2(a) corresponds to the 741 general- An op amp is an active circuit element designed to perform mathe- purpose op amp made by Fairchild matical operations of addition, subtraction, multiplication, division, dif- Semiconductor. ferentiation, and integration. The op amp is an electronic device consisting of a complex arrangement of resistors, transistors, capacitors, and diodes. A full dis- cussion of what is inside the op amp is beyond the scope of this book. It will suffice to treat the op amp as a circuit building block and sim- ply study what takes place at its terminals. Op amps are commercially available in integrated circuit packages in several forms. Figure 5.1 shows a typical op amp package. A typical one is the eight-pin dual in-line package (or DIP), shown in Fig. 5.2(a). Pin or terminal 8 is unused, and terminals 1 and 5 are of little concern to us. The five important terminals are: 1. The inverting input, pin 2. 2. The noninverting input, pin 3. 3. The output, pin 6. 4. The positive power supply Vϩ, pin 7. 5. The negative power supply V Ϫ, pin 4. The circuit symbol for the op amp is the triangle in Fig. 5.2(b); as shown, the op amp has two inputs and one output. The inputs are

5.2 Operational Amplifiers 177 Balance 1 8 No connection Inverting input 2 V+ 7 V+ Noninverting input 3 7 6 Output 6 Output 5 Balance − Inverting input 2 +Noninverting input 3 415 V− 4 V− (a) Offset NullFigure 5.2 (b)A typical op amp: (a) pin configuration, (b) circuit symbol.marked with minus (Ϫ) and plus (ϩ) to specify inverting and nonin- i1 i+ 6 +verting inputs, respectively. An input applied to the noninverting ter- 2 7 io VCCminal will appear with the same polarity at the output, while an input 3 −applied to the inverting terminal will appear inverted at the output. 4 i2 i− + As an active element, the op amp must be powered by a voltage VCCsupply as typically shown in Fig. 5.3. Although the power supplies are −often ignored in op amp circuit diagrams for the sake of simplicity, thepower supply currents must not be overlooked. By KCL, Figure 5.3 Powering the op amp. io ϭ i1 ϩ i2 ϩ iϩ ϩ iϪ (5.1) The equivalent circuit model of an op amp is shown in Fig. 5.4. v1The output section consists of a voltage-controlled source in series with − Ro vd Rithe output resistance Ro. It is evident from Fig. 5.4 that the input resis-tance Ri is the Thevenin equivalent resistance seen at the input termi- +nals, while the output resistance Ro is the Thevenin equivalent resistanceseen at the output. The differential input voltage vd is given by vd ϭ v2 Ϫ v1 (5.2) vowhere v1 is the voltage between the inverting terminal and ground and + Avdv2 is the voltage between the noninverting terminal and ground. The −op amp senses the difference between the two inputs, multiplies it by v2the gain A, and causes the resulting voltage to appear at the output.Thus, the output vo is given by Figure 5.4 The equivalent circuit of the nonideal vo ϭ Avd ϭ A(v2 Ϫ v1) (5.3) op amp.A is called the open-loop voltage gain because it is the gain of the op Sometimes, voltage gain is expressedamp without any external feedback from output to input. Table 5.1 in decibels (dB), as discussed in Chapter 14.TABLE 5.1 A dB ϭ 20 log10 ATypical ranges for op amp parameters.Parameter Typical range Ideal valuesOpen-loop gain, A 105 to 108 ϱInput resistance, Ri 105 to 1013 ⍀ ϱ⍀Output resistance, Ro 10 to 100 ⍀ 0⍀Supply voltage, VCC 5 to 24 V

178 Chapter 5 Operational Amplifiersvo shows typical values of voltage gain A, input resistance Ri, output resistance Ro, and supply voltage VCC. Positive saturationVCC The concept of feedback is crucial to our understanding of op amp circuits. A negative feedback is achieved when the output is fed back vd to the inverting terminal of the op amp. As Example 5.1 shows, when there is a feedback path from output to input, the ratio of the output voltage to the input voltage is called the closed-loop gain. As a result of the negative feedback, it can be shown that the closed-loop gain is almost insensitive to the open-loop gain A of the op amp. For this rea- son, op amps are used in circuits with feedback paths. A practical limitation of the op amp is that the magnitude of its output voltage cannot exceed |VCC |. In other words, the output voltage is dependent on and is limited by the power supply voltage. Figure 5.5 illustrates that the op amp can operate in three modes, depending on the differential input voltage vd:Negative saturation 0 1. Positive saturation, vo ϭ VCC. −VCC 2. Linear region, ϪVCC Յ vo ϭ Avd Յ VCC. 3. Negative saturation, vo ϭ ϪVCC.Figure 5.5 If we attempt to increase vd beyond the linear range, the op ampOp amp output voltage vo as a function of becomes saturated and yields vo ϭ VCC or vo ϭ ϪVCC. Throughoutthe differential input voltage vd. this book, we will assume that our op amps operate in the linear mode. Throughout this book, we assume that This means that the output voltage is restricted by an op amp operates in the linear range. Keep in mind the voltage constraint on ϪVCC Յ vo Յ VCC (5.4) the op amp in this mode. Although we shall always operate the op amp in the linear region, the possibility of saturation must be borne in mind when one designs with op amps, to avoid designing op amp circuits that will not work in the laboratory.Example 5.1 A 741 op amp has an open-loop voltage gain of 2 ϫ 105, input resis- tance of 2 M⍀, and output resistance of 50 ⍀. The op amp is used in the circuit of Fig. 5.6(a). Find the closed-loop gain vo͞vs. Determine current i when vs ϭ 2 V. 20 kΩ 20 kΩ i 10 kΩ 1 i 10 kΩ v1 Ro = 50 Ω vovs +− vs +− − O+ 1 O 741 vo − i + − vd Ri = 2 MΩ + Avd + − (a) (b)Figure 5.6For Example 5.1: (a) original circuit, (b) the equivalent circuit.

5.3 Ideal Op Amp 179Solution:Using the op amp model in Fig. 5.4, we obtain the equivalent circuitof Fig. 5.6(a) as shown in Fig. 5.6(b). We now solve the circuit inFig. 5.6(b) by using nodal analysis. At node 1, KCL gives vs Ϫ v1 ϭ v1 103 ϩ v1 Ϫ vo 10 ϫ 103 2000 ϫ 20 ϫ 103Multiplying through by 2000 ϫ 103, we obtain 200vs ϭ 301v1 Ϫ 100voor 2vs Ӎ 3v1 Ϫ vo 1 v1 ϭ 2vs ϩ vo (5.1.1) 3At node O, v1 Ϫ vo ϭ vo Ϫ Avd 20 ϫ 103 50But vd ϭ Ϫv1 and A ϭ 200,000. Then v1 Ϫ vo ϭ 400(vo ϩ 200,000v1) (5.1.2)Substituting v1 from Eq. (5.1.1) into Eq. (5.1.2) gives0 Ӎ 26,667,067vo ϩ 53,333,333vs 1 vo ϭ Ϫ1.9999699 vsThis is closed-loop gain, because the 20-k⍀ feedback resistor closesthe loop between the output and input terminals. When vs ϭ 2 V, vo ϭϪ3.9999398 V. From Eq. (5.1.1), we obtain v1 ϭ 20.066667 ␮V. Thus, i ϭ v1 Ϫ vo ϭ 0.19999 mA 20 ϫ 103It is evident that working with a nonideal op amp is tedious, as we aredealing with very large numbers.If the same 741 op amp in Example 5.1 is used in the circuit of Fig. 5.7, Practice Problem 5.1calculate the closed-loop gain vo͞vs. Find io when vs ϭ 1 V. + ioAnswer: 9.00041, 657 ␮A. 741 − vs +− 40 kΩ + 5 kΩ 20 kΩ vo −5.3 Ideal Op Amp Figure 5.7 For Practice Prob. 5.1.To facilitate the understanding of op amp circuits, we will assume idealop amps. An op amp is ideal if it has the following characteristics: 1. Infinite open-loop gain, A Ӎ ϱ. 2. Infinite input resistance, Ri Ӎ ϱ. 3. Zero output resistance, Ro Ӎ 0.

180 Chapter 5 Operational Amplifiers An ideal op amp is an amplifier with infinite open-loop gain, infinite input resistance, and zero output resistance. i1 = 0 −− Although assuming an ideal op amp provides only an approxi- vd mate analysis, most modern amplifiers have such large gains and+ input impedances that the approximate analysis is a good one. Unless i2 = 0 + stated otherwise, we will assume from now on that every op amp is + ideal.v1 + v2 = v1 For circuit analysis, the ideal op amp is illustrated in Fig. 5.8, which is derived from the nonideal model in Fig. 5.4. Two important−− characteristics of the ideal op amp are: + 1. The currents into both input terminals are zero: vo − i1 ϭ 0, i2 ϭ 0 (5.5)Figure 5.8 This is due to infinite input resistance. An infinite resistanceIdeal op amp model. between the input terminals implies that an open circuit exists there and current cannot enter the op amp. But the output current is not necessarily zero according to Eq. (5.1). 2. The voltage across the input terminals is equal to zero; i.e., vd ϭ v2 Ϫ v1 ϭ 0 (5.6) or v1 ϭ v2 (5.7)The two characteristics can be ex- Thus, an ideal op amp has zero current into its two input ter-ploited by noting that for voltage cal- minals and the voltage between the two input terminals is equalculations the input port behaves as a to zero. Equations (5.5) and (5.7) are extremely importantshort circuit, while for current calcula- and should be regarded as the key handles to analyzing op amptions the input port behaves as an circuits.open circuit.Example 5.2 Rework Practice Prob. 5.1 using the ideal op amp model. v2 i2 = 0 + i0 Solution: v1 − We may replace the op amp in Fig. 5.7 by its equivalent model in O Fig. 5.9 as we did in Example 5.1. But we do not really need to do i1 = 0 + this. We just need to keep Eqs. (5.5) and (5.7) in mind as we analyze vo 20 kΩ the circuit in Fig. 5.7. Thus, the Fig. 5.7 circuit is presented as invs +− 40 kΩ − Fig. 5.9. Notice that 5 kΩ v2 ϭ vs (5.2.1)Figure 5.9 Since i1 ϭ 0, the 40-k⍀ and 5-k⍀ resistors are in series; the sameFor Example 5.2. current flows through them. v1 is the voltage across the 5-k⍀ resistor. Hence, using the voltage division principle, v1 ϭ 5 5 40 vo ϭ vo (5.2.2) ϩ 9

5.4 Inverting Amplifier 181According to Eq. (5.7), v2 ϭ v1 (5.2.3)Substituting Eqs. (5.2.1) and (5.2.2) into Eq. (5.2.3) yields the closed-loop gain,vs ϭ vo 1 vo ϭ 9 (5.2.4) 9 vswhich is very close to the value of 9.00041 obtained with the nonidealmodel in Practice Prob. 5.1. This shows that negligibly small errorresults from assuming ideal op amp characteristics. At node O, io ϭ vo 5 ϩ vo mA (5.2.5) 40 ϩ 20From Eq. (5.2.4), when vs ϭ 1 V, vo ϭ 9 V. Substituting for vo ϭ 9 Vin Eq. (5.2.5) producesio ϭ 0.2 ϩ 0.45 ϭ 0.65 mAThis, again, is close to the value of 0.657 mA obtained in PracticeProb. 5.1 with the nonideal model.Repeat Example 5.1 using the ideal op amp model. Practice Problem 5.2Answer: Ϫ2, 200 ␮A.5.4 Inverting Amplifier i2 RfIn this and the following sections, we consider some useful op amp i1 R1 v1 0A − vi +− 1 −circuits that often serve as modules for designing more complex cir- 0Vcuits. The first of such op amp circuits is the inverting amplifier shownin Fig. 5.10. In this circuit, the noninverting input is grounded, vi is v2 + +connected to the inverting input through R1, and the feedback resistor +Rf is connected between the inverting input and output. Our goal is to voobtain the relationship between the input voltage vi and the output volt- −age vo. Applying KCL at node 1, Figure 5.10i1 ϭ i2 1 vi Ϫ v1 ϭ v1 Ϫ vo (5.8) The inverting amplifier. R1 Rf A key feature of the inverting amplifierBut v1 ϭ v2 ϭ 0 for an ideal op amp, since the noninverting terminal is that both the input signal and theis grounded. Hence, feedback are applied at the inverting terminal of the op amp. vi ϭ Ϫvo R1 Rf

182 Chapter 5 Operational Amplifiers or vo ϭ Ϫ Rf vi (5.9) R1Note there are two types of gains: The The voltage gain is Av ϭ vo͞vi ϭ ϪRf͞R1. The designation of the cir-one here is the closed-loop voltage cuit in Fig. 5.10 as an inverter arises from the negative sign. Thus,gain Av, while the op amp itself has anopen-loop voltage gain A.++ An inverting amplifier reverses the polarity of the input signal while Rf amplifying it.vi R1 – R1 vi vo− + − Notice that the gain is the feedback resistance divided by the input resistance which means that the gain depends only on theFigure 5.11 external elements connected to the op amp. In view of Eq. (5.9), anAn equivalent circuit for the inverter in equivalent circuit for the inverting amplifier is shown in Fig. 5.11.Fig. 5.10. The inverting amplifier is used, for example, in a current-to-voltage converter.Example 5.3 Refer to the op amp in Fig. 5.12. If vi ϭ 0.5 V, calculate: (a) the output voltage vo, and (b) the current in the 10-k⍀ resistor. 25 kΩ 10 kΩ Solution:vi +− − (a) Using Eq. (5.9), + + Ϫ Rf vo vo ϭ R1 ϭ Ϫ25 ϭ Ϫ2.5 − vi 10Figure 5.12 vo ϭ Ϫ2.5vi ϭ Ϫ2.5(0.5) ϭ Ϫ1.25 VFor Example 5.3. (b) The current through the 10-k⍀ resistor is i ϭ vi Ϫ 0 ϭ 0.5 Ϫ 0 ϭ 50 mA R1 10 ϫ 103Practice Problem 5.3 Find the output of the op amp circuit shown in Fig. 5.13. Calculate the current through the feedback resistor. 280 kΩ Answer: Ϫ3.15 V, 26.25 ␮A. 4 kΩ −45 mV +− ++ vo −Figure 5.13For Practice Prob. 5.3.

5.5 Noninverting Amplifier 183Determine vo in the op amp circuit shown in Fig. 5.14. Example 5.4Solution: 40 kΩApplying KCL at node a, 20 kΩ a − + va Ϫ vo ϭ 6 Ϫ va b + vo 40 k⍀ 20 k⍀ − va Ϫ vo ϭ 12 Ϫ 2va 1 vo ϭ 3va Ϫ 12 6 V +− 2 V +−But va ϭ vb ϭ 2 V for an ideal op amp, because of the zero voltagedrop across the input terminals of the op amp. Hence, Figure 5.14 For Example 5.4. vo ϭ 6 Ϫ 12 ϭ Ϫ6 VNotice that if vb ϭ 0 ϭ va, then vo ϭ Ϫ12, as expected from Eq. (5.9).Two kinds of current-to-voltage converters (also known as transresis- Practice Problem 5.4tance amplifiers) are shown in Fig. 5.15.(a) Show that for the converter in Fig. 5.15(a), vo ϭ ϪR is(b) Show that for the converter in Fig. 5.15(b), vo ϭ ϪR1a1 ϩ R3 ϩ R3 b is R1 R2Answer: Proof. R R1 R2 − R3 −is ++ ++ vo is vo −− (a) (b)Figure 5.15 i2 RfFor Practice Prob. 5.4. R1 i1 v1 − + + v2 vo5.5 Noninverting Amplifier vi +− −Another important application of the op amp is the noninverting ampli- Figure 5.16fier shown in Fig. 5.16. In this case, the input voltage vi is applied The noninverting amplifier.directly at the noninverting input terminal, and resistor R1 is connected

184 Chapter 5 Operational Amplifiers between the ground and the inverting terminal. We are interested in the output voltage and the voltage gain. Application of KCL at the invert- ing terminal gives i1 ϭ i2 1 0 Ϫ v1 ϭ v1 Ϫ vo (5.10) R1 Rf But v1 ϭ v2 ϭ vi. Equation (5.10) becomes Ϫvi ϭ vi Ϫ vo R1 Rf or vo ϭ a1 ϩ Rf b vi (5.11) R1 − + The voltage gain is Av ϭ vo͞vi ϭ 1 ϩ Rf͞R1, which does not have a + vo = vi negative sign. Thus, the output has the same polarity as the input. −vi +− A noninverting amplifier is an op amp circuit designed to provide a positive voltage gain.Figure 5.17The voltage follower. Again we notice that the gain depends only on the external resistors. Notice that if feedback resistor Rf ϭ 0 (short circuit) or R1 ϭ ϱ (open circuit) or both, the gain becomes 1. Under these conditions (Rf ϭ 0 and R1 ϭ ϱ), the circuit in Fig. 5.16 becomes that shown in Fig. 5.17, which is called a voltage follower (or unity gain amplifier) because the output follows the input. Thus, for a voltage follower − vo ϭ vi (5.12) +First + + Second Such a circuit has a very high input impedance and is therefore use-stage vi vo stage ful as an intermediate-stage (or buffer) amplifier to isolate one circuit from another, as portrayed in Fig. 5.18. The voltage follower mini- − − mizes interaction between the two stages and eliminates interstage loading.Figure 5.18A voltage follower used to isolate twocascaded stages of a circuit.Example 5.5 For the op amp circuit in Fig. 5.19, calculate the output voltage vo. Solution: We may solve this in two ways: using superposition and using nodal analysis. ■ METHOD 1 Using superposition, we let vo ϭ vo1 ϩ vo2

5.6 Summing Amplifier 185where vo1 is due to the 6-V voltage source, and vo2 is due to the 4-V 10 kΩ +input. To get vo1, we set the 4-V source equal to zero. Under this vocondition, the circuit becomes an inverter. Hence Eq. (5.9) gives 4 kΩ a − b 10 − Ϫ + vo1 ϭ 4 (6) ϭ Ϫ15 V 6 V +− 4 V +−To get vo2, we set the 6-V source equal to zero. The circuit becomesa noninverting amplifier so that Eq. (5.11) applies. vo2 ϭ a1 ϩ 10 ϭ 14 V Figure 5.19 b4 For Example 5.5. 4Thus, vo ϭ vo1 ϩ vo2 ϭ Ϫ15 ϩ 14 ϭ Ϫ1 V■ METHOD 2 Applying KCL at node a, 6 Ϫ va ϭ va Ϫ vo 4 10But va ϭ vb ϭ 4, and so 6 Ϫ 4 ϭ 4 Ϫ vo 1 5 ϭ 4 Ϫ vo 4 10or vo ϭ Ϫ1 V, as before.Calculate vo in the circuit of Fig. 5.20. Practice Problem 5.5Answer: 7 V. 4 kΩ + −+ 3 V +− 8 kΩ 5 kΩ vo 2 kΩ −5.6 Summing Amplifier Figure 5.20 i For Practice Prob. 5.5.Besides amplification, the op amp can perform addition and subtrac- +tion. The addition is performed by the summing amplifier covered in R1 i1 Rf vothis section; the subtraction is performed by the difference amplifier v1 −covered in the next section. R2 i2 i 0 A summing amplifier is an op amp circuit that combines several inputs and produces an output that is the weighted sum of the inputs. v2 − a The summing amplifier, shown in Fig. 5.21, is a variation of theinverting amplifier. It takes advantage of the fact that the inverting con- R3 i3 +figuration can handle many inputs at the same time. We keep in mind v3 0 Figure 5.21 The summing amplifier.

186 Chapter 5 Operational Amplifiers Example 5.6 that the current entering each op amp input is zero. Applying KCL at node a gives i ϭ i1 ϩ i2 ϩ i3 (5.13) But i1 ϭ v1 Ϫ va, i2 ϭ v2 Ϫ va R1 R2 (5.14) i3 ϭ v3 Ϫ va, i ϭ va Ϫ vo R3 Rf We note that va ϭ 0 and substitute Eq. (5.14) into Eq. (5.13). We get vo ϭ Ϫa Rf v1 ϩ Rf v2 ϩ Rf v3b (5.15) R1 R2 R3 indicating that the output voltage is a weighted sum of the inputs. For this reason, the circuit in Fig. 5.21 is called a summer. Needless to say, the summer can have more than three inputs. Calculate vo and io in the op amp circuit in Fig. 5.22. 5 kΩ 10 kΩ 2 V +− 2.5 kΩ a io +− 1 V − + + b 2 kΩ vo − Figure 5.22 For Example 5.6. Solution: This is a summer with two inputs. Using Eq. (5.15) gives vo ϭ Ϫ c 10 (2) ϩ 10 (1) d ϭ Ϫ(4 ϩ 4) ϭ Ϫ8 V 5 2.5 The current io is the sum of the currents through the 10-k⍀ and 2-k⍀ resistors. Both of these resistors have voltage vo ϭ Ϫ8 V across them, since va ϭ vb ϭ 0. Hence, io ϭ vo Ϫ 0 ϩ vo Ϫ 0 mA ϭ Ϫ0.8 Ϫ 4 ϭ Ϫ4.8 mA 10 2

5.7 Difference Amplifier 187Find vo and io in the op amp circuit shown in Fig. 5.23. Practice Problem 5.6 20 kΩ 8 kΩ 10 kΩ io 6 kΩ − + 1.2 V1.5 V − + +− vo 2V 4 kΩ − +− + Figure 5.23 For Practice Prob. 5.6.Answer: Ϫ3.8 V, Ϫ1.425 mA.5.7 Difference Amplifier The difference amplifier is also known as the subtractor, for reasons to beDifference (or differential) amplifiers are used in various applications shown later.where there is a need to amplify the difference between two input sig-nals. They are first cousins of the instrumentation amplifier, the mostuseful and popular amplifier, which we will discuss in Section 5.10.A difference amplifier is a device that amplifies the difference betweentwo inputs but rejects any signals common to the two inputs. Consider the op amp circuit shown in Fig. 5.24. Keep in mind thatzero currents enter the op amp terminals. Applying KCL to node a, v1 Ϫ va ϭ va Ϫ vo R1 R2or vo ϭ a R2 ϩ 1b va Ϫ R2 v1 (5.16) R1 R1 R2 R1 va 0 − R3 vb 0 +v1 +− +− v2 + vo R4 −Figure 5.24Difference amplifier.

188 Chapter 5 Operational Amplifiers Example 5.7 Applying KCL to node b, v2 Ϫ vb ϭ vb Ϫ 0 R3 R4 or vb ϭ R3 R4 R4 v2 (5.17) ϩ But va ϭ vb. Substituting Eq. (5.17) into Eq. (5.16) yields vo ϭ a R2 ϩ 1b R3 R4 R4 v2 Ϫ R2 v1 R1 ϩ R1 or vo ϭ R2(1 ϩ R1͞R2) v2 Ϫ R2 v1 (5.18) R1(1 ϩ R3͞R4) R1 Since a difference amplifier must reject a signal common to the two inputs, the amplifier must have the property that vo ϭ 0 when v1 ϭ v2. This property exists when R1 ϭ R3 (5.19) R2 R4 Thus, when the op amp circuit is a difference amplifier, Eq. (5.18) becomes vo ϭ R2 (v2 Ϫ v1) (5.20) R1 If R2 ϭ R1 and R3 ϭ R4, the difference amplifier becomes a subtractor, with the output vo ϭ v2 Ϫ v1 (5.21) Design an op amp circuit with inputs v1 and v2 such that vo ϭ Ϫ5v1 ϩ 3v2. Solution: (5.7.1) The circuit requires that vo ϭ 3v2 Ϫ 5v1 This circuit can be realized in two ways. Design 1 If we desire to use only one op amp, we can use the op amp circuit of Fig. 5.24. Comparing Eq. (5.7.1) with Eq. (5.18), we see R2 ϭ 5 1 R2 ϭ 5R1 (5.7.2) R1

5.7 Difference Amplifier 189Also, (1 ϩ R1͞R2) ϭ 3 1 6 ϭ 3or 5 5 5 (1 ϩ R3͞R4) 1 ϩ R3͞R4 2 ϭ 1 ϩ R3 1 R3 ϭ R4 (5.7.3) R4If we choose R1 ϭ 10 k⍀ and R3 ϭ 20 k⍀, then R2 ϭ 50 k⍀ andR4 ϭ 20 k⍀.Design 2 If we desire to use more than one op amp, we may cascade 3R3 5R1an inverting amplifier and a two-input inverting summer, as shown inFig. 5.25. For the summer, R3 v2 − vo ϭ Ϫva Ϫ 5v1 (5.7.4) 5R1 + va − vo +and for the inverter, va ϭ Ϫ3v2 (5.7.5) v1 R1Combining Eqs. (5.7.4) and (5.7.5) gives Figure 5.25 For Example 5.7. vo ϭ 3v2 Ϫ 5v1which is the desired result. In Fig. 5.25, we may select R1 ϭ 10 k⍀and R3 ϭ 20 k⍀ or R1 ϭ R3 ϭ 10 k⍀.Design a difference amplifier with gain 7.5. Practice Problem 5.7Answer: Typical: R1 ϭ R3 ϭ 20k⍀, R2 ϭ R4 ϭ 150 k⍀.An instrumentation amplifier shown in Fig. 5.26 is an amplifier of low- Example 5.8level signals used in process control or measurement applications andcommercially available in single-package units. Show that vo ϭ R2 a1 ϩ 2R3 b (v2 Ϫ v1) R1 R4Solution:We recognize that the amplifier A3 in Fig. 5.26 is a difference amplifier.Thus, from Eq. (5.20), vo ϭ R2 (vo2 Ϫ vo1) (5.8.1) R1Since the op amps A1 and A2 draw no current, current i flows throughthe three resistors as though they were in series. Hence, vo1 Ϫ vo2 ϭ i(R3 ϩ R4 ϩ R3) ϭ i(2R3 ϩ R4) (5.8.2)

190 Chapter 5 Operational Amplifiers + vo1 R1 R2 − A1 − v1 +− R3 + A3 v2 +− 0 va R2 R4 i vo 0 vb − R3 R1 A2 vo2 + Figure 5.26 Instrumentation amplifier; for Example 5.8. But (5.8.3) i ϭ va Ϫ vb R4 and va ϭ v1, vb ϭ v2. Therefore, i ϭ v1 Ϫ v2 R4 Inserting Eqs. (5.8.2) and (5.8.3) into Eq. (5.8.1) gives vo ϭ R2 a1 ϩ 2R3 b (v2 Ϫ v1) R1 R4 as required. We will discuss the instrumentation amplifier in detail in Section 5.10.Practice Problem 5.8 Obtain io in the instrumentation amplifier circuit of Fig. 5.27. 6.98 V + 40 kΩ − − − 20 kΩ + 7V + 20 kΩ io 40 kΩ 50 kΩ Figure 5.27 Instrumentation amplifier; for Practice Prob. 5.8. Answer: Ϫ800 ␮A.

5.8 Cascaded Op Amp Circuits 1915.8 Cascaded Op Amp CircuitsAs we know, op amp circuits are modules or building blocks fordesigning complex circuits. It is often necessary in practical applica-tions to connect op amp circuits in cascade (i.e., head to tail) to achievea large overall gain. In general, two circuits are cascaded when theyare connected in tandem, one behind another in a single file.A cascade connection is a head-to-tail arrangement of two or more opamp circuits such that the output of one is the input of the next. When op amp circuits are cascaded, each circuit in the string iscalled a stage; the original input signal is increased by the gain of theindividual stage. Op amp circuits have the advantage that they can becascaded without changing their input-output relationships. This is due tothe fact that each (ideal) op amp circuit has infinite input resistance andzero output resistance. Figure 5.28 displays a block diagram represen-tation of three op amp circuits in cascade. Since the output of one stageis the input to the next stage, the overall gain of the cascade connectionis the product of the gains of the individual op amp circuits, or A ϭ A1 A2 A3 (5.22)Although the cascade connection does not affect the op amp input-output relationships, care must be exercised in the design of an actualop amp circuit to ensure that the load due to the next stage in the cas-cade does not saturate the op amp.+ Stage 1 + Stage 2 + Stage 3 + A1 A2 A3v1 v2 = A1v1 v3 = A2v2 vo = A3v3− − − −Figure 5.28A three-stage cascaded connection.Find vo and io in the circuit in Fig. 5.29. Example 5.9Solution: +a + +This circuit consists of two noninverting amplifiers cascaded. At the − − iooutput of the first op amp, b vo va ϭ a1 ϩ 12b (20) ϭ 100 mV 20 mV +− 12 kΩ 10 kΩ 3 3 kΩ − 4 kΩAt the output of the second op amp, vo ϭ a1 ϩ 10 b va ϭ (1 ϩ 2.5)100 ϭ 350 mV Figure 5.29 4 For Example 5.9.The required current io is the current through the 10-k⍀ resistor. io ϭ vo Ϫ vb mA 10

192 Chapter 5 Operational Amplifiers But vb ϭ va ϭ 100 mV. Hence, (350 Ϫ 100) ϫ 10Ϫ3 ϭ 25 mA io ϭ 10 ϫ 103Practice Problem 5.9 Determine vo and io in the op amp circuit in Fig. 5.30. Answer: 6 V, 24 ␮A. ++ − −+1.2 V +− 200 kΩ vo io − 50 kΩFigure 5.30For Practice Prob. 5.9.Example 5.10 If v1 ϭ 1 V and v2 ϭ 2 V, find vo in the op amp circuit of Fig. 5.31. A 6 kΩ 2 kΩ − 5 kΩ C v1 +a 10 kΩ B − vo 8 kΩ + 4 kΩ − 15 kΩ v2 +b Figure 5.31 For Example 5.10. Solution: 1. Define. The problem is clearly defined. 2. Present. With an input of v1 of 1 V and of v2 of 2 V, determine the output voltage of the circuit shown in Figure 5.31. The op amp circuit is actually composed of three circuits. The first circuit acts as an amplifier of gain Ϫ3(Ϫ6 k⍀͞2 k⍀) for v1 and the second functions as an amplifier of gain Ϫ2(Ϫ8 k⍀͞4 k⍀) for v2. The last circuit serves as a summer of two different gains for the output of the other two circuits. 3. Alternative. There are different ways of working with this circuit. Since it involves ideal op amps, then a purely mathematical

5.8 Cascaded Op Amp Circuits 193 approach will work quite easily. A second approach would be to use PSpice as a confirmation of the math.4. Attempt. Let the output of the first op amp circuit be designated as v11 and the output of the second op amp circuit be designated as v22. Then we get v11 ϭ Ϫ3v1 ϭ Ϫ3 ϫ 1 ϭ Ϫ3 V, v22 ϭ Ϫ2v2 ϭ Ϫ2 ϫ 2 ϭ Ϫ4 V In the third circuit we have vo ϭ Ϫ(10 k⍀͞5 k⍀) v11 ϩ 3 Ϫ(10 k⍀͞15 k⍀) v22 4 ϭ Ϫ2(Ϫ3) Ϫ (2͞3)(Ϫ4) ϭ 6 ϩ 2.667 ϭ 8.667 V5. Evaluate. In order to properly evaluate our solution, we need to identify a reasonable check. Here we can easily use PSpice to provide that check. Now we can simulate this in PSpice. We see the results are shown in Fig. 5.32. R6 −3.000 R2 8.667 V R4 5 kΩ R1 2 kΩ 6 kΩ 10 kΩ + v11V OPAMP OPAMP − − − + U1 + U3 R7 − 4.000 R3 R5 15 kΩ 4 kΩ 8 kΩ + v22V OPAMP − − + U2 Figure 5.32 For Example 5.10. We note that we obtain the same results using two entirely different techniques (the first is to treat the op amp circuits as just gains and a summer and the second is to use circuit analysis with PSpice). This is a very good method of assuring that we have the correct answer.6. Satisfactory? We are satisfied we have obtained the asked for results. We can now present our work as a solution to the problem.

194 Chapter 5 Operational AmplifiersPractice Problem 5.10 If v1 ϭ 7 V and v2 ϭ 3.1 V, find vo in the op amp circuit of Fig. 5.33. − 20 kΩ 60 kΩ vo + 50 kΩ − v1 +− − + + 30 kΩ 10 kΩ v2 +− Figure 5.33 For Practice Prob. 5.10. Answer: 10 V. 5.9 Op Amp Circuit Analysis with PSpice PSpice for Windows does not have a model for an ideal op amp, although one may create one as a subcircuit using the Create Subcircuit line in the Tools menu. Rather than creating an ideal op amp, we will use one of the four nonideal, commercially available op amps supplied in the PSpice library eval.slb. The op amp models have the part names LF411, LM111, LM324, and uA741, as shown in Fig. 5.34. Each of them can be obtained from Draw/Get New Part/libraries . . . /eval.lib or by sim- ply selecting Draw/Get New Part and typing the part name in the PartName dialog box, as usual. Note that each of them requires dc sup- plies, without which the op amp will not work. The dc supplies should be connected as shown in Fig. 5.3.U4 U2 U33+ 7 5 6 2+ 85 6 7 3 + 4 U1A 3+ 7 5 6 V+ V+ V+ 1 V+ B2 BB /S 2 − V− 052 112− V− B1 3− V− G 2− V− 051 1 1 LM324 1 4 4 (c) Five– 4 connectionLF411 LM111 op amp subcircuit uA741(a) JFET–input op (b) Op amp (d) Five–connectionamp subcircuit subcircuit op amp subcircuitFigure 5.34Nonideal op amp model available in PSpice.

5.9 Op Amp Circuit Analysis with PSpice 195Use PSpice to solve the op amp circuit for Example 5.1. Example 5.11Solution:Using Schematics, we draw the circuit in Fig. 5.6(a) as shown inFig. 5.35. Notice that the positive terminal of the voltage source vs isconnected to the inverting terminal (pin 2) via the 10-k⍀ resistor, whilethe noninverting terminal (pin 3) is grounded as required in Fig. 5.6(a).Also, notice how the op amp is powered; the positive power supplyterminal Vϩ (pin 7) is connected to a 15-V dc voltage source, whilethe negative power supply terminal VϪ (pin 4) is connected to Ϫ15 V.Pins 1 and 5 are left floating because they are used for offset nulladjustment, which does not concern us in this chapter. Besides addingthe dc power supplies to the original circuit in Fig. 5.6(a), we have alsoadded pseudocomponents VIEWPOINT and IPROBE to respectivelymeasure the output voltage vo at pin 6 and the required current ithrough the 20-k⍀ resistor. 0 V2 + − U1VS + 2V 3 7 15 V + V+ 5 − –3.9983 6 052 R1 2− V− 051 + 0 10 K 1 − 15 V 4 uA741 V3 1.999E–04 R2 20 KFigure 5.35Schematic for Example 5.11. After saving the schematic, we simulate the circuit by selectingAnalysis/Simulate and have the results displayed on VIEWPOINT andIPROBE. From the results, the closed-loop gain is vo ϭ Ϫ3.9983 ϭ Ϫ1.99915 vs 2and i ϭ 0.1999 mA, in agreement with the results obtained analyticallyin Example 5.1.Rework Practice Prob. 5.1 using PSpice. Practice Problem 5.11Answer: 9.0027, 650.2 ␮A.

196 Chapter 5 Operational Amplifiers 5.10 Applications Digital Four-bit Analog The op amp is a fundamental building block in modern electronic input DAC output instrumentation. It is used extensively in many devices, along with(0000 –1111) resistors and other passive elements. Its numerous practical applications include instrumentation amplifiers, digital-to-analog converters, analog computers, level shifters, filters, calibration circuits, inverters, sum- mers, integrators, differentiators, subtractors, logarithmic amplifiers, comparators, gyrators, oscillators, rectifiers, regulators, voltage-to- current converters, current-to-voltage converters, and clippers. Some of these we have already considered. We will consider two more applica- tions here: the digital-to-analog converter and the instrumentation amplifier. (a) 5.10.1 Digital-to-Analog ConverterV1 V2 V3 V4 The digital-to-analog converter (DAC) transforms digital signals into R1 R2 R3 R4 Rf Vo analog form. A typical example of a four-bit DAC is illustrated inMSB LSB Fig. 5.36(a). The four-bit DAC can be realized in many ways. A sim- − ple realization is the binary weighted ladder, shown in Fig. 5.36(b). + The bits are weights according to the magnitude of their place value, by descending value of Rf͞Rn so that each lesser bit has half the (b) weight of the next higher. This is obviously an inverting summing amplifier. The output is related to the inputs as shown in Eq. (5.15).Figure 5.36 Thus,Four-bit DAC: (a) block diagram,(b) binary weighted ladder type. ϪVo ϭ Rf ϩ Rf ϩ Rf ϩ Rf (5.23) R1 V1 R2 V2 R3 V3 R4 V4In practice, the voltage levels may be Input V1 is called the most significant bit (MSB), while input V4 is thetypically 0 and ; 5 V. least significant bit (LSB). Each of the four binary inputs V1, . . . , V4 can assume only two voltage levels: 0 or 1 V. By using the proper input and feedback resistor values, the DAC provides a single output that is proportional to the inputs.Example 5.12 In the op amp circuit of Fig. 5.36(b), let Rf ϭ 10 k⍀, R1 ϭ 10 k⍀, R2 ϭ 20 k⍀, R3 ϭ 40 k⍀, and R4 ϭ 80 k⍀. Obtain the analog output for binary inputs [0000], [0001], [0010], . . . , [1111]. Solution: Substituting the given values of the input and feedback resistors in Eq. (5.23) gives ϪVo ϭ Rf V1 ϩ Rf V2 ϩ Rf V3 ϩ Rf V4 R1 R2 R3 R4 ϭ V1 ϩ 0.5V2 ϩ 0.25V3 ϩ 0.125V4 Using this equation, a digital input [V1V2V3V4] ϭ [0000] produces an ana- log output of ϪVo ϭ 0 V; [V1V2V3V4] ϭ [0001] gives ϪVo ϭ 0.125 V.

5.10 Applications 197Similarly, 3 V1 V2 V3 V4 4 ϭ 3 0010 4 1 ϪVo ϭ 0.25 V 3 V1 V2 V3 V4 4 ϭ 3 0011 4 1 ϪVo ϭ 0.25 ϩ 0.125 ϭ 0.375 V 3 V1 V2 V3 V4 4 ϭ 3 0100 4 1 ϪVo ϭ 0.5 V o 3 V1 V2 V3 V4 4 ϭ 3 1111 4 1 ϪVo ϭ 1 ϩ 0.5 ϩ 0.25 ϩ 0.125 ϭ 1.875 VTable 5.2 summarizes the result of the digital-to-analog conversion.Note that we have assumed that each bit has a value of 0.125 V. Thus,in this system, we cannot represent a voltage between 1.000 and 1.125,for example. This lack of resolution is a major limitation of digital-to-analog conversions. For greater accuracy, a word representation with agreater number of bits is required. Even then a digital representationof an analog voltage is never exact. In spite of this inexact represen-tation, digital representation has been used to accomplish remarkablethings such as audio CDs and digital photography. TABLE 5.2Input and output values of the four-bit DAC.Binary input Decimal value Output[V1V2V3V4] ϪVo 0 0000 1 0 0001 2 0.125 0010 3 0.25 0011 4 0.375 0100 5 0.5 0101 6 0.625 0110 7 0.75 0111 8 0.875 1000 9 1.0 1001 10 1.125 1010 11 1.25 1011 12 1.375 1100 13 1.5 1101 14 1.625 1110 15 1.75 1111 1.875A three-bit DAC is shown in Fig. 5.37. Practice Problem 5.12 (a) Determine |Vo| for [V1V2V3] ϭ [010]. 10 kΩ 10 kΩ (b) Find |Vo| if [V1V2V3] ϭ [110]. v1 (c) If |Vo| ϭ 1.25 V is desired, what should be [V1V2V3]? (d) To get |Vo| ϭ 1.75 V, what should be [V1V2V3]? 20 kΩ − vo v2 +Answer: 0.5 V, 1.5 V, [101], [111]. 40 kΩ v3 Figure 5.37 Three-bit DAC; for Practice Prob. 5.12.

198 Chapter 5 Operational Amplifiers 5.10.2 Instrumentation Amplifiers One of the most useful and versatile op amp circuits for precision measurement and process control is the instrumentation amplifier (IA), so called because of its widespread use in measurement systems. Typ- ical applications of IAs include isolation amplifiers, thermocouple amplifiers, and data acquisition systems. The instrumentation amplifier is an extension of the difference amplifier in that it amplifies the difference between its input signals. As shown in Fig. 5.26 (see Example 5.8), an instrumentation amplifier typically consists of three op amps and seven resistors. For conven- ience, the amplifier is shown again in Fig. 5.38(a), where the resistors are made equal except for the external gain-setting resistor RG, connected between the gain set terminals. Figure 5.38(b) shows its schematic symbol. Example 5.8 showed that vo ϭ Av(v2 Ϫ v1) (5.24)Inverting input v1 + RR Gain set − 1 RG R − + 3 vo Output R Gain set − 2 RNoninverting input v2 + R − + (a) (b)Figure 5.38(a) The instrumentation amplifier with an external resistance to adjust the gain, (b) schematic diagram. where the voltage gain is 2R (5.25) Av ϭ 1 ϩ RG As shown in Fig. 5.39, the instrumentation amplifier amplifies small differential signal voltages superimposed on larger common-mode − RG +Small differential signals riding on larger Instrumentation amplifier Amplified differential signal,common-mode signals no common-mode signalFigure 5.39The IA rejects common voltages but amplifies small signal voltages.Floyd, Thomas, L, Electronic Devices, 4th edition, © 1995, p. 795. Reprinted by permission of PearsonEducation, Inc., Upper Saddle River, NJ.

5.11 Summary 199voltages. Since the common-mode voltages are equal, they cancel eachother. The IA has three major characteristics: 1. The voltage gain is adjusted by one external resistor RG. 2. The input impedance of both inputs is very high and does not vary as the gain is adjusted. 3. The output vo depends on the difference between the inputs v1 and v2, not on the voltage common to them (common-mode voltage). Due to the widespread use of IAs, manufacturers have developedthese amplifiers on single-package units. A typical example is theLH0036, developed by National Semiconductor. The gain can be var-ied from 1 to 1,000 by an external resistor whose value may vary from100 ⍀ to 10 k⍀.In Fig. 5.38, let R ϭ 10 k⍀, v1 ϭ 2.011 V, and v2 ϭ 2.017 V. If RG Example 5.13is adjusted to 500 ⍀, determine: (a) the voltage gain, (b) the outputvoltage vo.Solution:(a) The voltage gain isAv ϭ 1 ϩ 2R ϭ 1 ϩ 2 ϫ 10,000 ϭ 41 RG 500(b) The output voltage is vo ϭ Av(v2 Ϫ v1) ϭ 41(2.017 Ϫ 2.011) ϭ 41(6) mV ϭ 246 mVDetermine the value of the external gain-setting resistor RG required Practice Problem 5.13for the IA in Fig. 5.38 to produce a gain of 142 when R ϭ 25 k⍀.Answer: 354.6 ⍀.5.11 Summary 1. The op amp is a high-gain amplifier that has high input resistance and low output resistance. 2. Table 5.3 summarizes the op amp circuits considered in this chap- ter. The expression for the gain of each amplifier circuit holds whether the inputs are dc, ac, or time-varying in general.

200 Chapter 5 Operational AmplifiersTABLE 5.3Summary of basic op amp circuits.Op amp circuit Name/output-input relationship R2 Inverting amplifier R1 − vo ϭ ϪR2 vivi + R1 vo R2 Noninverting amplifierR1 − vo ϭ a1 ϩ R2 b vi vi + R1 vo − vo Voltage followervi + vo ϭ vi R1 Rf Summerv1v2 R2 vo ϭ Ϫa Rf v1 ϩ Rf v2 ϩ Rf v3bv3 R3 R1 R2 R3 − vo R1 +v1 R2 Difference amplifier R1v2 vo ϭ R2 (v2 Ϫ v1) R1 − vo + R2 3. An ideal op amp has an infinite input resistance, a zero output resistance, and an infinite gain. 4. For an ideal op amp, the current into each of its two input termi- nals is zero, and the voltage across its input terminals is negligi- bly small. 5. In an inverting amplifier, the output voltage is a negative multiple of the input. 6. In a noninverting amplifier, the output is a positive multiple of the input. 7. In a voltage follower, the output follows the input. 8. In a summing amplifier, the output is the weighted sum of the inputs. 9. In a difference amplifier, the output is proportional to the differ- ence of the two inputs.10. Op amp circuits may be cascaded without changing their input- output relationships.11. PSpice can be used to analyze an op amp circuit.12. Typical applications of the op amp considered in this chapter include the digital-to-analog converter and the instrumentation amplifier.