CHAPTER SIX SOLUTIONS

Transcription

1 CHAPTE SIX SOLUTIONS. The first step is to perform a simple source transformation, so that a 0.5-V source in series with a 50-Ω resistor is connected to the inerting pin of the ideal op amp Then, out ( 0.5) -. V Copyright 00 McGraw-Hill, Inc. All ights esered.

2 CHAPTE SIX SOLUTIONS. In order to delier 50 mw to the 0-kΩ resistor, we need out 3 (0.5)(0 0 ) V. Writing a nodal equation at the inerting input, we find 5 5 out Using out 38.73, we find that 48. Ω. Copyright 00 McGraw-Hill, Inc. All ights esered.

3 CHAPTE SIX SOLUTIONS 3. Since the 670-Ω switch requires 00 ma to actiate, the oltage deliered to it by our op amp circuit must be (670)(0.) 67 V. The microphone acts as the input to the circuit, and proides 0.5 V. Thus, an amplifier circuit haing a gain 67/ is required. One possible solution of many: a non-inerting op amp circuit with the microphone connected to the non-inerting input terminal, the switch connected between the op amp output pin and ground, a feedback resistor f 33 Ω, and a resistor Ω. Copyright 00 McGraw-Hill, Inc. All ights esered.

4 CHAPTE SIX SOLUTIONS 4. We begin by labeling the nodal oltages - and at the inerting and non-inerting input terminals, respectiely. Since no current can flow into the non-inerting input, no current flows through the 40-kΩ resistor; hence, 0. Therefore, we know that - 0 as well. Writing a single nodal equation at the non-inerting input then leads to or Soling, ( - S) 00 ( - - ) out 0 - S out 0 out -0 S Copyright 00 McGraw-Hill, Inc. All ights esered.

5 CHAPTE SIX SOLUTIONS 5. We first label the nodal oltage at the output pin V o. Then, writing a single nodal equation at the inerting input terminal of the op amp, V 7000 o 0 Soling, we find that V o V. Since no current can flow through the 300-kΩ resistor, V as well. Copyright 00 McGraw-Hill, Inc. All ights esered.

6 CHAPTE SIX SOLUTIONS 6. A source transformation and some series combinations are well worthwhile prior to launching into the analysis. With 5 kω 3 kω.875 kω and ( ma)(.875 kω).875 V, we may redraw the circuit as Ω Ω V This is now a simple inerting amplifier with gain f / / Thus, V -38.4(3.975) -5.6 V. Copyright 00 McGraw-Hill, Inc. All ights esered.

7 CHAPTE SIX SOLUTIONS 7. This is a simple inerting amplifier, so we may write out (t 3 s) V out 000 ( sin3t) - 4( sin 3t )V Copyright 00 McGraw-Hill, Inc. All ights esered.

8 CHAPTE SIX SOLUTIONS 8. We first combine the MΩ and 700 kω resistors into a 58.5 kω resistor. We are left with a simple non-inerting amplifier haing a gain of 58.5/ Thus, out (3.074) in 8 so in V. Copyright 00 McGraw-Hill, Inc. All ights esered.

9 CHAPTE SIX SOLUTIONS 9. This is a simple non-inerting amplifier circuit, and so it has a gain of f /. We want out 3.7 cos 500t V when the input is 0. cos 500t V, so a gain of 3.7/0. 37 is required. One possible solution of many: f 36 kω and kω. Copyright 00 McGraw-Hill, Inc. All ights esered.

10 CHAPTE SIX SOLUTIONS 0. Define a nodal oltage V - at the inerting input, and a nodal oltage V at the noninerting input. Then, At the non-inerting input: V [] Thus, V -4.5 V, and we therefore also know that V V. At the inerting input: V out [] 6 V - V 7 Soling and making use of the fact that V V, 7 7 out ( 4.5) V 6 6 Copyright 00 McGraw-Hill, Inc. All ights esered.

11 CHAPTE SIX SOLUTIONS. (a) B must be the non-inerting input: that yields a gain of 70/0 8 and an output of 8 V for a -V input. (b), A 0. We need a gain of 0/0, so choose B Ω. (c) A is the inerting input since it has the feedback connection to the output pin. Copyright 00 McGraw-Hill, Inc. All ights esered.

12 CHAPTE SIX SOLUTIONS. It is probably best to first perform a simple source transformation: ( ma)( kω) V. 3 kω kω V - 3 V kω V out V Since no current can flow into the non-inerting input pin, we know that V V, and therefore also that V - V. A single nodal equation at the inerting input yields: which yields out - V out Copyright 00 McGraw-Hill, Inc. All ights esered.

13 CHAPTE SIX SOLUTIONS 3. We begin by find the Théenin equialent to the left of the op amp: V th -3.3(3) π -9.9 π 000 S S 00 th 3.3 kω, so we can redraw the circuit as: 00 kω 3.3 kω out -9 S which is simply a classic inerting op amp circuit with gain of -00/ Thus, out (-30.3)( -9 S ) 7.7 S For S 5 sin 3t mv, out.364 sin 3t V, and out (0.5 s) V. Copyright 00 McGraw-Hill, Inc. All ights esered.

14 CHAPTE SIX SOLUTIONS 4. We first combine the 4.7 MΩ and.3 kω resistors: 4.7 MΩ.3 kω.30 kω. Next, a source transformation yields (3 0-6 )(300) mv which appears in series with the 0 mv source and the 500-Ω resistor. Thus, we may redraw the circuit as 37.7 kω 500 Ω out.8 kω 370 Ω mv -6 V Since no current flows through the.8 kω resistor, V mv and hence V mv as well. A single nodal equation at the inerting input terminal yields Soling, out.86 V out Copyright 00 McGraw-Hill, Inc. All ights esered.

15 CHAPTE SIX SOLUTIONS 5. We first combine the 4.7 MΩ and.3 kω resistors: 4.7 MΩ.3 kω.30 kω. Next, a source transformation yields (7 0-6 )(300) 35. mv which appears in series with the 0 mv source and the 500-Ω resistor. Thus, we may redraw the circuit as 37.7 kω 500 Ω out.8 kω 370 Ω 55. mv -6 V Since no current flows through the.8 kω resistor, V 55. mv and hence V mv as well. A single nodal equation at the inerting input terminal yields Soling, out 4. V out Copyright 00 McGraw-Hill, Inc. All ights esered.

16 CHAPTE SIX SOLUTIONS 6. The 3 ma source, kω resistor and 0 kω resistor may be replaced with a 3 V source ( reference up) in series with a kω resistor. No current flows through either MΩ resistor, so that the oltage at each of the four input terminals is identically zero. Considering each op amp circuit separately, 00 - (-3) 00 - (5) 0 out LEFTOPAMP out IGH OPAMP 4.9 V - 50 V x out LEFTOPAMP V. out IGH OPAMP Copyright 00 McGraw-Hill, Inc. All ights esered.

17 CHAPTE SIX SOLUTIONS Copyright 00 McGraw-Hill, Inc. All ights esered. 7. A general summing amplifier with N input sources:. a b 0. A single nodal equation at the inerting input leads to: N N a a a f a... 0 out Simplifying and making use of the fact that a 0, we may write this as N i N N N i N i N i f... i i i out i or simply N N f... out Thus, out N f - i i i N out N f a b

18 CHAPTE SIX SOLUTIONS 8. A general difference amplifier: 4 out 3 Writing a nodal equation at the inerting input, a - a - out 0 f Writing a nodal equation at the non-inerting input, b b Simplifying and collecting terms, we may write ( f ) a out f [] ( 3 ) b 3 [] 3 From Eqn. [], we hae b Since a b, we can now rewrite Eqn. [] as and hence out 3 ( ) f out f f 3 f Copyright 00 McGraw-Hill, Inc. All ights esered.

19 CHAPTE SIX SOLUTIONS 9. In total darkness, the CdS cell has a resistance of 00 kω, and at a light intensity L of 6 candela it has a resistance of 6 kω. Thus, we may compute the light-dependent resistance (assuming a linear response in the range between 0 and 6 candela) as CdS -5L 00 Ω. Our design requirement (using the standard inerting op amp circuit shown) is that the oltage across the load is.5 V at candela, and less than.5 V for intensities greater than candela. Thus, out ( candela) - CdS S / -70 V S /.5 ( in kω). Pick 0 kω. Then S V. Copyright 00 McGraw-Hill, Inc. All ights esered.

20 CHAPTE SIX SOLUTIONS 0. We want f / instrument K, and f / ocal K, where K is a constant not specified. Assuming K, one possible solution of many is: f Ω ocal Ω ocals microphone instruments Ω out instruments microphone Copyright 00 McGraw-Hill, Inc. All ights esered.

21 CHAPTE SIX SOLUTIONS. One possible solution of many: kω 99 kω out V S Copyright 00 McGraw-Hill, Inc. All ights esered.

22 CHAPTE SIX SOLUTIONS. out of stage is ()(-0/ ) -0 V. out of stage is (-0)(-000/ 0) 000 V Note: in reality, the output oltage will be limited to a alue less than that used to power the op amps. Copyright 00 McGraw-Hill, Inc. All ights esered.

23 CHAPTE SIX SOLUTIONS Copyright 00 McGraw-Hill, Inc. All ights esered. 3. We hae a difference amplifier as the first amplifier stage, and a simple oltage follower as the second stage. We therefore need only to find the output oltage of the first stage: out will track this oltage. Using oltage diision, then, we ind that the oltage at the non-inerting input pin of the first op amp is 3 3 V and this is the oltage at the inerting input terminal also. Thus, we may write a single nodal equation at the inerting input of the first op amp: out 3 3 f 3 3 V - V V - V 0 Stage which may be soled to obtain: f 3 3 f out out V - V V V Stage

24 CHAPTE SIX SOLUTIONS 4. The output of the first op amp stage may be found by realising that the oltage at the non-inerting input (and hence the oltage at the inerting input) is 0, and writing a ingle nodal equation at the inerting input: Vout stage which leads to V out -4. V steage 47 7 This oltage appears at the input of the second op amp stage, which has a gain of 3/ Thus, the output of the second op amp stage is 0(-4.) 4 V. This oltage appears at the input of the final op amp stage, which has a gain of 47/ Thus, the output of the circuit is 56.7(4) kv, which is completely and utterly ridiculous. Copyright 00 McGraw-Hill, Inc. All ights esered.

25 CHAPTE SIX SOLUTIONS 5. The output of the top left stage is (0/ ) -5 V. The output of the middle left stage is (0/ ) -0 V. The output of the bottom right stage is 3(0/ ) -5 V. These three oltages are the input to a summing amplifier such that 00 Soling, we find that Ω. V out ( 5 0 5) 0 Copyright 00 McGraw-Hill, Inc. All ights esered.

26 CHAPTE SIX SOLUTIONS 6. Stage is configured as a oltage follower: the output oltage will be equal to the input oltage. Using oltage diision, the oltage at the non-inerting input (and hence at the inerting input, as well), is V The second stage is wired as a oltage follower also, so out.667 V. Copyright 00 McGraw-Hill, Inc. All ights esered.

27 CHAPTE SIX SOLUTIONS 7. (a) a b nv d 0 and out 0. Thus, P 8Ω 0 W. (b) a 0, b nv d - nv out ( 0 5 )( ) -9.8 µv. Thus, P 8Ω 75 8 out pw. (c) a pv, b fv d.999 pv out ( 0 5 )( ) nv. Thus, P 8Ω 75 8 out aw. (c) a 50 µv, b -4 µv d 54 µv out ( 0 5 )( ).04 V. Thus, P 8Ω 75 8 out mw. Copyright 00 McGraw-Hill, Inc. All ights esered.

28 CHAPTE SIX SOLUTIONS 8. Writing a nodal equation at the - d node, 0 - d d S d out [] in - - V - - f or ( f in f in ) d in out - in f V S [] Writing a nodal equation at the out node, 0 - Eqn. [] can be rewritten as: so that Eqn. [] becomes: out - A - A out d out d [] in d - o ( ) - (- ) f o out [] o - Af in f ( A - ) in f f f o in VS o o in where for this circuit, A 0 6, in 0 TΩ, o 5 Ω, f 000 kω, 70 kω. (a) mv; (b) 7.78 mv; (c) V. Copyright 00 McGraw-Hill, Inc. All ights esered.

29 CHAPTE SIX SOLUTIONS 6 i 5 9. out A d A ( 80 0 ) sin t V i (a) A 0 5, i 00 MΩ, o alue irreleant. out 8 sin t nv (b) A 0 6, i TΩ, o alue irreleant. out 80 sin t nv Copyright 00 McGraw-Hill, Inc. All ights esered.

30 CHAPTE SIX SOLUTIONS 30. (a) Find out / in if i, o 0, and A is finite. The nodal equation at the inerting input is - d - in - d - [] 00 0 out At the output, with o 0 we may write out A d so d out / A. Thus, Eqn. [] becomes 0 out out out in A 00A 00 from which we find out -00A [] 0 A in (b) We want the alue of A such that out / in -99 (the ideal alue would be 00 if A were infinite). Substituting into Eqn. [], we find A 9999 Copyright 00 McGraw-Hill, Inc. All ights esered.

31 CHAPTE SIX SOLUTIONS 3. (a) δ 0 V d 0, and P 8Ω 0 W. (b) δ nv, so d 5 (5 0-9 ) -0-9 V Thus, out ( ) d -9.8 µv and P 8Ω ( out ) / pw (c) δ.5 µv, so d 5 ( ) V Thus, out ( ) d mv and P 8Ω ( out ) / µw Copyright 00 McGraw-Hill, Inc. All ights esered.

32 CHAPTE SIX SOLUTIONS 3. AD549 Writing a single nodal equation at the output, we find that 0 - out in out d [] i - A o Also, in out d, so Eqn. [] becomes and 0 ( out in ) o ( out A in A out ) i out ( A ) o o ( A ) i i in To within 4 significant figures (and more, actually), when in -6 mv, out -6 mv (this is, after all, a oltage follower circuit). Copyright 00 McGraw-Hill, Inc. All ights esered.

33 CHAPTE SIX SOLUTIONS 33. The ideal op amp model predicts a gain out / in -000/ 0-00, regardless of the alue of in. In other words, it predicts an input-output characteristic such as: out (V) -00 in (V) From the PSpice simulation result shown below, we see that the ideal op amp model is reasonably accurate for in 00 < 5 V (the supply oltage, assuming both hae the same magnitude), but the onset of saturation is at ±4.5 V, or in ~ 45 mv. Increasing in past this alue does not lead to an increase in out. Copyright 00 McGraw-Hill, Inc. All ights esered.

34 CHAPTE SIX SOLUTIONS 34. Positie oltage supply, negatie oltage supply, inerting input, ground, output pin. Copyright 00 McGraw-Hill, Inc. All ights esered.

35 CHAPTE SIX SOLUTIONS 35. This op amp circuit is an open-loop circuit; there is no external feedback path from the output terminal to either input. Thus, the output should be the open-loop gain times the differential input oltage, minus any resistie losses. From the simulation results below, we see that all three op amps saturate at a oltage magnitude of approximately 4 V, corresponding to a differential input oltage of 50 to 00 µv, except in the interest case of the LM 34, which may be showing some unexpected input offset behaior. op amp onset of negatie saturation negatie saturation oltage onset of positie saturation positie saturation oltage µa 74-9 µv -4.3 V 54.4 mv 4.34 V LM µv -4.7 V 337. mv 3.87 V LF µv -3.8 V mv 3.86 V Copyright 00 McGraw-Hill, Inc. All ights esered.

36 CHAPTE SIX SOLUTIONS 36. This is a non-inerting op amp circuit, so we expect a gain of 000/ With ±5 V DC supplies, we need to sweep the input just aboe and just below this alue diided by the gain to ensure that we see the saturation regions. Performing the indicated simulation and a DC sweep from 0. V to 0. V with 0.00 V steps, we obtain the following input-output characteristic: Using the cursor tool, we see that the linear region is in the range of 68. mv < V in < 68.5 mv. The simulation predicts a gain of 7.03 V/ 3.87 mv 6., which is reasonably close to the alue predicted using the ideal op amp model. Copyright 00 McGraw-Hill, Inc. All ights esered.

37 CHAPTE SIX SOLUTIONS 37. eferring to the detailed model of an op amp, shorting the input terminals together shorts the dependent source to ground. Therefore, any -V source connected to the output through a -Ω resistor should see Ω o. For the µa 74, we expect Ω. For the LF 4, we expect ~ Ω or ~ Ω. Simulating the µa74 circuit, we find: Supply oltages Current into output Total resistance Output resistance ±5 V -4.5 ma Ω -.53 Ω ±5 V ma Ω Ω ± V ma Ω Ω 0 V 579. ma.77 Ω 77 mω Conclusion: as we might expect from preious experience in determining Théenin equialent resistances, we must short out the oltage supplies to the op amp when performing such an experiment (hence the negatie resistance alues obtained aboe). Howeer, we obtained 0.77 Ω instead of the expected 75 Ω, which leads to two possible conclusions: () The PSpice model is not designed to represent the op amp behaior accurately in such a circuit configuration or () such an experimental connection is not adequate for measuring the output resistance. Simulating the LF4 circuit, we find: Supply oltages Current into output Total resistance Output resistance ±5 V 5.46 ma 39.8 Ω 38.8 Ω ±5 V 5.43 ma 39.3 Ω 38.3 Ω ± V 5.48 ma 39.4 Ω 8.4 Ω 0 V 000 ma Ω 0 Ω Conclusion: as we might expect from preious experience in determining Théenin equialent resistances, we must short out the oltage supplies to the op amp when performing such an experiment. Howeer, we obtained ~0 Ω instead of the expected Ω, which leads to two possible conclusions: () The PSpice model is not designed to represent the op amp behaior accurately in such a circuit configuration or () such an experimental connection is not adequate for measuring the output resistance. Howeer, it is interesting that PSpice did predict a much lower output resistance for the LF 4 than the µa 74, as we would expect. Copyright 00 McGraw-Hill, Inc. All ights esered.

38 CHAPTE SIX SOLUTIONS 38. Based on the detailed model of the LF 4 op amp, we can write the following nodal equation at the inerting input: 0 - d in x - 0 d 4 Ad - d 6 0 o Substituting alues for the LF 4 and simplifying, we make appropriate approximations and then sole for d in terms of x, finding that 6 0 d 6 x x With a gain of 000/0-00 and supply oltage magnitudes of 5 V, we are effectiely limited to alues of x < 50 mv. For x -0 mv, PSpice predicts d 6 µv, where the hand calculations based on the detailed model predict 50 µv, which is about one order of magnitude larger. For the same input oltage, PSpice predicts an input current of - µa, whereas the hand calculations predict 99.5 x ma -995 na (which is reasonably close). Copyright 00 McGraw-Hill, Inc. All ights esered.

39 CHAPTE SIX SOLUTIONS (a) The gain of the inerting amplifier is 000. At a sensor oltage of 30 mv, the predicted output oltage (assuming an ideal op amp) is 30 V. At a sensor oltage of 75 mv, the predicted output oltage (again assuming an ideal op amp) is 75 V. Since the op amp is being powered by dc sources with oltage magnitude equal to 5 V, the output oltage range will realistically be limited to the range 5 < V out < 5 V. (b) The peak input oltage is 75 mv. Therefore, 5/ , and we should set the resistance ratio f / < 99 to ensure the op amp does not saturate. Copyright 00 McGraw-Hill, Inc. All ights esered.

40 CHAPTE SIX SOLUTIONS 39. (a) We see from the simulation result that negatie saturation begins at V in 4.7 V, and positie saturation begins at V in 4.67 V. (b) Using a pω resistor between the output pin and ground, we obtain an output current of 40.6 ma, slightly larger than the expected 35 ma, but not too far off. Copyright 00 McGraw-Hill, Inc. All ights esered.

41 CHAPTE SIX SOLUTIONS 4. We assume that the strength of the separately-broadcast chaotic noise signal is receied at the appropriate intensity such that it may precisely cancel out the chaotic component of the total receied signal; otherwise, a ariable-gain stage would need to be added so that this could be adjusted by the user. We also assume that the signal frequency is separate from the carrier or broadcast frequency, and has already been separated out by an appropriate circuit (in a similar fashion, a radio station transmitting at 9 MHz is sending an audio signal of between 0 and 0 khz, which must be separated from the 9 MHz frequency.) One possible solution of many (all resistances in ohms): Copyright 00 McGraw-Hill, Inc. All ights esered.

42 CHAPTE SIX SOLUTIONS 4. One possible solution of many: This circuit produces an output equal to the aerage of V, V, and V 3, as shown in the simulation result: V aerage ( )/ V. Copyright 00 McGraw-Hill, Inc. All ights esered.