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# Class Note for ECE 3336 at UH

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G Rizzoni Principles and Applications of Electrical Engineering Problem solutions Chapter 8 and by motivating the student to pursue further study in electronics or instrumentation In many respects this chapter is the centerpiece of the book Learning Objectives 1 Understand the properties of ideal amplifiers and the concepts of gain input impedance and outputimpedance Section 1 2 Understand the difference between openloop and closedloop opamp configuration and compute the gain or complete the design of simple inverting noninverting summing and differential amplifiers using ideal opamp analysis Analyze more advanced opamp circuits using ideal opamp analysis and identify important performance parameters in opamp data sheets Section 2 3 Analyze and design simple active filters Analyze and design ideal integrator and differentiator circuits Sections 3 and 4 4 Understand the structure and behavior of analog computers and design analog computer circuits to solve simple differential equations Section 5 5 Understand the principal physical limitations of an opamp Section 6 Section 8 1 Ideal Amplifiers Problem 81 Solution Known quantities For the circuit shown in Figure P8l Gz R06ko RL06kQ PS Ril 3kQ Rig 2 3kQ R01 2kg R02 2kg Avol 100 GM2 350111971 V2 V12 Vil Vll 2 Find The power gain G PDPs in dB Analysis Starting from the last stage and going backward we get 82 G Rizzoni Principles and Applications of Electrical Engineering Problem solutions Chapter 8 V2 V0 2 GM2V12M161 5 V12 gt Po VoIo 704349 R02RL RL R V12 AvOlWl 60 V11 gt Po 1 566 105 i2 01 R V1VS 0333 VS PO 10875105 V52 il S 15 VS 277810 4 VS PS VSIS 277810quot V52 il S P 5 G Pi 3915108 20dB L0g103915108 17185dB s Problem 82 Solution Known quantities The temperature sensor shown in Figure P82 produces a no load ie sensor current 2 O voltage V Vmcosat R 4009 VW 500mV a 62M S The temperature is monitored on a display the load with a vertical line of light emitting diodes Normal conditions are indicated when a string of the bottommost diodes 2 cm in length are on This requires that a voltage be supplied to the display input terminals where RL 12kQ V0 2 Vocoswt V0 2 6V The signal from the sensor must therefore be amplified Therefore a voltage amplifier is connected between the sensor and CRT with Ri 2 2kg R0 3 k9 Find The required no load gain of the amplifier Analysis The overall loaded voltage gain using the amplitudes of the sensor voltage and the specified CRT voltage must be AV 2 L 76 V 12 V 500 mV An expression for the overall voltage gain can also be obtained using two voltage divider relationships V0 vaio L AWN Ri L R0RL RiRs R0RL A Z L ARRRRL 120423 12 1 18 V vi mm mm The loss in gain due to the two voltage divisions or quotloadingquot is characteristic of all practical amplifiers Ideally there is no reduction in gain due to loading This would require an ideal signal source with a source resistance equal to zero and a load resistance equal to infinity 83 G Rizzoni Principles and Applications of Electrical Engineering Problem solutions Chapter 8 Problem 83 Solution Known quantities For the circuit shown in Figure P83 P 2 4 pisz RSO7kQ RL16Q Pi Ril Ri1 11m Ri2 19kg R01 29m R02 229 AW1 65 GM 1301118 G Find The power gain G PDPi in dB Analysis 0L V0 szmm 1204V2 R02 RL LO 6519 VD v0 Mimi 1204 1204 6791 Riz v11 29 19 2 via G 2 L2 if amp 67912amp 3171103 20dB L0g103l7l103 110021dB VA V11 RL 1 Ril Problem 84 Solution Known quantities For the circuit shown in Figure P84 RS 2 03kQ RL 2 2kg Ri1 Ri2 77kQ R01 R02 13kQ Am AM 17 Lo 1499 V11 Find a The power gain in dB b The overall voltage gain VOVs Analysis a 2 V70 G E 1 LO2amp 14992E 86509 20L0g1086509 9874dB PI VA V11 RL 2 Ril 84 G Rizzoni Principles and Applications of Electrical Engineering Problem solutions Chapter 8 b The input voltage of the first stage can be determined in terms of the source voltage using voltage division R11 VS 7 7 A Lquot mm Lquot Rll RS 1499 7 1443 20 Log101443 4318 dB Vs 111 vs 1211 vs 0 Problem 85 Solution Known quantities Figure P85 Find What approximations are usually made about the voltages and currents shown for the ideal operational amplifier opamp model Analysis 139sz iNzO szO Problem 86 Solution Known quantities Figure P86 Find What approximations are usually made about the circuit components and parameters shown for the ideal operational amplifier opamp model Analysis n z Do u z oo r0 z 85 G Rizzoni Principles and Applications of Electrical Engineering Problem solutions Chapter 8 Section 82 The Operational Amplifier Problem 87 Solution Known quantities The circuit shown in Figure P87 Find The resistor RAv that will accomplish the nominal gain requirement and state what the maximum and minimum values of RE can be Will a standard 5 percent tolerance resistor be adequate to satisfy this requirement Analysis For a noninverting amplifier the voltage gain is given by V R R R R R AV7wr7f Si1 Amm16if1 R57f 1kQ v R R R 16 1 To find the maximum and minimum RS we note that RAv 0lt 4 so to find the maximum RS we consider v the minimum AV R ax i l02kQ 161 002 1 lSkQ Conversely to find the minimum RS we consider the maximum AV R 9809 m 161002 1 Since a standard 5 tolerance lkQ resistor has resistance 950 lt R lt 1050 a standard resistor will suffice in this application Problem 88 Solution Known quantities The values of the two 10 percent tolerance resistors used in an inverting amplifier RF 33kQ RS 12kQ Find a The nominal gain of the amplifier b The maximum value of AV c The minimum value of Analysis Rf 33 a The gain of the inverting amplifier is A R7 W 275 S b First we note that the gain of the amplifier is proportional to Rfand inversely proportional to RS This tells us that to find the maximum gain of the amplifier we consider the maximum Rfand the minimum Rs Rfmax 330133 336 WW R 12 0112 S min 86 G Rizzoni Principles and Applications of Electrical Engineering Problem solutions Chapter 8 R 39 33 01 33 c To find Avlmin we consider the opposite case lAvlmin fmm 225 RSmaX 12 0112 Problem 89 Solution Known quantities For the circuit shown in Figure P89 let V1l10103 sinatV RF 10ko Vm 20v Find a The value of RS such that no DC voltage appears at the output b The corresponding value of Vowt Analysis a The circuit may be modeled as shown Applying the principle of superposition For the 20V source 10k 2 20 R S Vol20 For the 10V source 10kg110 V010 R S 10000 10000 The total DC output is VOIDC v020 v010 20 110 0 S S 10000 Solving forRS 1020 10 gt RS 21012 S b Since we have already determined R S such that the DC component of the output will be zero we can simply treat the amplifier as if the AC source were the only source present Therefore R v0 r 0001sinat R f1 0001sinat 1 1 2 10 3 sinatV S Problem 810 Solution Known quantities For the circuit shown in Figure P810 let R 2 Mo A 200000 R0 509 V0L RS 1ko R11k 2 R2 100ko RLOAD 10ko 87 G Rizzoni Principles and Applications of Electrical Engineering Problem solutions Chapter 8 Find V The gain AV 70 V 1 Analysis The opamp has a Very large input resistance a Very large quotopen loop gainquot Ax00 and a Very small output resistance Therefore it 2n be modeled with small error as an ideal opamp The amplifier shown in Figure 13810 is a noninVerting ampli er so we have R2 100103 1717101 AV Rl 1103 Problem 811 Solution Known quantities We 2s1nwlt4sinagt2tSSinw3t165mw40V RF 5 k9 Find Design an inverting summing amplifier to obtain VW I and determine the require source resistors Analysis The inverting summing amplifier is shown in the following figure J RF t By superposition and by selecting RSI 7 VS S111 0111 1 4 the output Voltage is 4 R 4 4 F z l VW 2Rivsl 22 V31 22 smwlt 11 11 11 3 that coincides with the desired output Voltage So the required source resistors are R31 25m R3 1251lt 2RS3 62552 R ii 31259 Problem 812 Solution Known quantities For the circuit shown in Figure P812 88 G Rizzoni Principles and Applications of Electrical Engineering Problem solutions Chapmr 8 a 2M9 Ll 200000 r 2552 RS 22m R11k 2 RF 87k52 RL 2052 Find a An expression for the input resistance Vi including the effects of the opramp b The Value of the input resistance in including the effects of the opramp c The Value of the input resistance With ideal opramp Analysis a The circuit in Figure P812 can be modeled as in the following gure VG vi uwaopi RL Substituting the expression for V4 in all the other equations We obtain ViRirigttima riliTVG RFrf lF 7 V0 1igt Hrih Hri r0gtlF39 RL By solving the second equation above for v0 and substituting in the third equation Vi Rdwii win 0 1 u i F r Hi RarigtHrira K RLJ andfinally 89 G Rizzoni Principles and Applications of Electrical Engineering Problem solutions Chapter 8 1rih 2 ltR1ngt L z 1RF4h4ro L b The value of the input resistance is 1I u 2 1752105 41012 R1ri L 2001106 25 quot 1LO RFrmro 17 20087106410 25 RL 39 39 20 10001kQ c In the ideal case Rl1ko 11 Problem 813 Solution Known quantities For the circuit shown in Figure P8l3 v5t00210 3coswtv RF220kQ R147kQ R218kQ Find a An expression for the output voltage b The corresponding value of Vot Analysis a The circuit is a noninverting amplifier then v0 111FIS 2 b v0 1ampjvs 246401232 cosat V 18 Problem 814 Solution Known quantities For the circuit shown in Figure P8l3 v5t 0053010 3coswtv R5 509 RL 2009 Find The output voltage V0 810 G Rizzoni Principles and Applications of Electrical Engineering Problem solutions Chapter 8 Analysis It is a particular case of noninverting amplifier where vav5 Problem 815 Solution Known quantities For the circuit shown in Figure P815 v51t 2910 3 cosatV R1 1ko R2 33m v52t3110 3coswtv R3 10ko R4 18ko39 Find The output voltage V0 and a numerical value Analysis By using superposition v0 ampv51 R4 1 521Q 18 3110 3cosat 2910 Zcosat R1 R2R4 R1 1 1833 18310 3 cosat V Problem 816 Solution Known quantities For the circuit shown in Figure P815 v515mV R11kQ R2 15ko v52 7mV R3 72kg R4 47kg Find The output voltage V0 analytically and numerically Analysis From the solution of Problem 815 vo ampv51 R4 15 52172 47 710 3 36010 12737mV R R2R4 re1 1547 l Problem 817 Solution Known quantities If in the circuit shown in Figure P8l7 VSlzvsz 7mV R1 8509 R2 15kQ RF 2 22kQ OpAmp MotorolaMCI74IC n 2M9 it 200000 r0 259 811 G Rizzoni Principles and Applications of Electrical Engineering Problem solutions Chapter 8 Find a The output voltage b The voltage gain for the two input signals Analysis a The op amp has a very large input resistance a very large quotopen loop gainquot u and a very small output resistance Therefore it can be modeled with small error as an ideal op amp with KVL VDVN 0 Wuw wxmwmmw mnxvm Q E A W KCL 39 V39V V39V V39V lNN S2N SlN 00 R2 R1 RF VNO iNO RF RF 22 22 2 7mV 7mV V0 R1 V51 R2 Vs2 085 15 b Using 25887mV l4677mV 2838mV the results above Aw 2588 AW 1467 Note The output voltage and gain are not dependent on either the op amp parameters or the load resistance This result is extremely important in the majority of applications where amplification of a signal is required Problem 818 Solution Known quantities For the circuit shown in Figure P815 V31kT1 R111k 2 R227k 2 Vszszz R333k 2 R468k 2 T135 C T2100 C k50mV C Find a The output voltage b The conditions required for the output voltage to depend only on the difference between the two temperatures Analysis a From the solution of Problem 815 VO kT1i1 WkT2 12 i5 2175 R1 R2R4k R1 11 6827 11 9065 V b For R3R4 R1R2 k R3 the output voltage is 812 G Rizzoni Principles and Applications of Electrical Engineering Problem solutions Chapter 8 VO ampk11i 1 T2lk39iL T2T2Ti R1 R2R4 R1 k R1R3 R1 Problem 819 Solution Find In a differential amplifier if Ad 2 20 sz 22 derive expressions for and then determine the value of the common and differential mode gains Analysis There are several ways to do this Using superposition V06 If VSVD 0 V0 V07CV07D V06 j AvrC 7 Vsc V00 If Vsrc 0 V0 V07CV07D V00 j AVVD 7 mo Assume that signal source 2 is connected to the noninverting input of the opamp The commonmode output voltage can be obtained using the gains given above but assuming the signal voltages have only a commonmode component 1 1 V0 2 Vs1 Av1 Vszsz Vs1 Vsrc 39 5 er Vs2 Vsrc 5 er Let 2er 0 V06 Vsrc Av1VSrC sz Vsc Av1Av2 Vsc AvrC j AweAv1Av220222 The differencemode output voltage can be obtained using the gains given but assuming the signal voltage have only a differencemode component Let vsrc 0 l l 1 v00 EVSD Avl EVSVDAVZ ver 39AvlA2 2er AVVD Aw Av2Av1 lt22120 21 Note If signal source 1 were connected to the noninverting input of the op amp then the difference mode gain would be the negative of that obtained above Problem 820 Solution Known quantities For the circuit shown in Figure P815 v51 13V R1 R2 47m v52 19v R3 R4 10kQ Find a The output voltage b The commonmode component of the output voltage 813 G Rizzoni Principles and Applications of Electrical Engineering c The differentialmode component of the output voltage Analysis a From the solution of Problem 815 R3 10 v 7 v v 7061277V 0 R1 52 51 47 b The commonmode component is zero c The differentialmode component is v0 Problem solutions Chapter 8 Problem 821 Solution Known quantities For the circuit shown in Figure P815 V5172 ABPL2 A 03 B 07 Vpsi R1 R2 47kQ R3 R4 10kQ RL 18kQ P1 6kPa P2 5kPa Find a The commonmode input voltage b The differentialmode input voltage Analysis a The commonmode input voltage is V52 Vs1 71 PIPZ AB 0307145041 b The differentialmode input voltage is vm v52 v51 BP2 P1 071450410 103 01015v 0411103 2 0858V Problem 822 Solution Known quantities A linear potentiometer variable resistor RF is used to sense and give a signal voltage Vy proportional to the current 2 position of an xy plotter A reference signal VR is supplied by the software controlling the plotter The difference between these voltages must be amplified and supplied to a motor The motor turns and changes the position of the pen and the position of the pot until the signal voltage is equal to the reference voltage indicating the pen is in the desired position and the motor voltage 2 0 For proper operation the motor voltage must be 10 times the difference between the signal and reference voltage For rotation in the proper direction the motor voltage must be negative with respect to the signal voltage for the polarities shown An additional requirement is that ip 0 to avoid quotloadingquot the pot and causing an erroneous signal voltage Find a Design an op amp circuit which will achieve the specifications given Redraw the circuit shown in Figure P822 replacing the box drawn with dotted lines with your circuit Be sure to show how the signal voltage and output voltage are connected in your circuit b Determine the value of each component in your circuit The op amp is an MCl74l 814 G Rizzoni Principles and Applications of Electrical Engineering Problem solutions Chapter 8 Analysis a The output voltage to the motor must be dependent on the difference between two input voltages A difference amp is required The signal voltage must be connected to the inverting input of the amplifier However the feedback path is also connected to the inverting input and this will caused loading of the input circuit ie cause an input current This can be corrected by adding an isolation stage or voltage follower between the input circuit and the inverting input of the 2nd stage The isolation stage must have a gain 2 l and an input current 2 0 The difference amplifier the second stage must give an output voltage IOkQ S VM 10VR39Vy 10VR3910VY AV 2 10 AW 2 10 The circuit configuration shown will satisfy these specifications In this first approximation analysis assume the op amps can be modeled as ideal op amps VD z 0 iP iN z 0 Consider the first or isolation or voltage follower stage Ideal 1quotsz KVL VYVDv010 Vol 1 10 szO gt V01 2 VY KVL 39VN39VDVP0 VDzO j VNVP MM KCL VN39V01VN39VMiN 0 W x 0 j VN R1 R2 R1R2 R1 R2 i i RJRZ R1 R2 m KCL L39Vhw39o pzo ipzo VP7R3 R3R4 R3 R4 ii R3R4 R3 R4 The second stage V01 W W 2 VP j vo1R2 VMR VRR4 R2R1 R2R1 R4R3 VM vyampva VYAWVRAW R1 R1R4R3 AV amp 10 Choose R2 100kQ R1 10kQ R1 AV w 10 Choose R3 10kQ R4 100kQ R1R4R3 The resistances chosen are standard values and there is some commonality in the choices Moderately large values were chosen to reduce currents and the resistor power ratings Cost of the resistors will be determined primarily by power rating and tolerance 815 G Rizzoni Principles and Applications of Electrical Engineering Problem solutions Chapter 8 Problem 823 Solution Known quantities For the circuit shown in Figure P815 v5113mV R11kQ R213kQ v5219mV R381kQ R456kQ Find The output voltage V0 Analysis From the solution of Problem 815 v0 L1amp52 5v51181 56 1910 3 811310 30211V R2R4 R1 R1 5613 Problem 824 Solution Known quantities The circuit shown in Figure P824 Find Show that the current out through the lightemitting diode is proportional to the source voltage Vs as long as Vs gt Analysis V zV V l V7 V Assume the op amp is ideal S gt VS gt 0 out 7 is 1 z 0 R2 R2 Problem 825 Solution Known quantities The circuit shown in Figure P825 Find Show that the voltage V0 is proportional to the current generated by the CdS solar cell Show that the transimpedance of the circuit VowIs is 7R Analysis Assuming an ideal opamp VJr E V7 0 gt V RIS The transimpedance is given by R R mm I 816 G Rizzoni Principles and Applications of Electrical Engineering Problem solutions Chapter 8 Problem 826 Solution Known quantities The opamp voltmeter circuit shown in Figure P826 is required to measure a maximum input of E 20 mV The opamp input current is I B 02 MA and the meter circuit has Im 100 MA fullscale de ection and rm 2 10 k9 Find Determine suitable values for R 3 and R4 Analysis R E 9 V0 1m rm 100414 IOkQ 1v E E V0ut Vout R4 R3 R4 From KCL at the inverting input 0 or 2 1 R3 R 4 E R3 R3 R V 1 Then 4 0 1 3 1 49 Now choose R3 and R4 such that R3 E 20 X 10 I S 02u B Z R3ltR4 rmquot 20x10 3 At the limit 3 02gtlt106 Solving for R3 we have R3 z IOZkQ R3W49R3 10gtlt10 l Therefore R4 z 5M9 Problem 827 Solution Known quantities Circuit in Figure P827 Find The output voltage V0 Analysis The circuit is a cascade of a noninverting opamp with an inverting opamp Assuming ideal opamps the inputoutput voltage gain is equal to the product of the single gains therefore 817 G Rizzoni Principles and Applications of Electrical Engineering VOh 1h 5 R52 R51 Problem solutions Chapter 8 Problem 828 Solution Known quantities Circuit in Figure P827 Find Select appropriate components using standard 5 resistor values to obtain a gain of magnitude approximately equal to 1000 How closely can you approximate the gain Compute the error in the gain assuming that the resistors have the nominal value Analysis From the solution of Problem 827 the gain is given by Av R52 R51 From Table 22 if we select R5118kQ R52 1kQ R 82kQ R 180kQ weobtain szh h 18H 1000 R52 R51 1 18 So we can obtain a nominal error equal to zero Problem 829 Solution Known quantities Circuit in Figure P827 Find Same as in Problem 828 but use the i5 tolerance range to compute the possible range of gains for this amplifier Analysis From the solution of Problem 827 the gain is given by AV 2 Rm 1 Rm R52 R51 From Table 22 if we select Rm18k Rm1k Rm 82kQ RFquot 180kQ we obtain 818 G Rizzoni Principles and Applications of Electrical Engineering Problem solutions Chapter 8 AWRM 1Rm rg 1882 1000 Rm Rm 1 18 So we can obtain a nominal error equal to zero The maximum gain is given by AVRF2 1RF1 Rm1005 1Rm1005 1200 5 R51 Ema 005 aha 005 while the minimum gain is AV 2 R 1 R j RF2n1 005 1 RF1n1 005 j 834 R52 R51 Rm1005 Rm1005 Problem 830 Solution Known quantities For the circuit in Figure P830 RL 20kQ v0 010V imm 1mA Find The resistance R such that immx 1mA Analysis v V 10 1quot inLw 3 10ko R z 10 max Problem 831 Solution Known quantities Circuit in Figure P8l3 Find Select appropriate components using standard 5 resistor values to obtain a gain of magnitude approximately equal to 200 How closely can you approximate the gain Compute the error in the gain assuming that the resistors have the nominal value Analysis From the solution of Problem 813 the gain is given by RF A llwl From Table 22 if we select R2 33 RF 68kQ we obtain 819 G Rizzoni Principles and Applications of Electrical Engineering Problem solutions Chapter 8 R 6800 17F 17207 AVquot R j 33 2 The error in the gain is 8 zwzm 200 Problem 832 Solution Known quantities Circuit in Figure P8l3 Find Same as in Problem 831 but use the i5 tolerance range to compute the possible range of gains for this amplifier Analysis From the solution of Problem 831 the gain is given by RF A llwl From Table 22 if we select R2quot 33 RFquot 68kQ we obtain AVquot 1h 1w207 2quot 33 The maximum gain is given by AW 2 1 RF1005 21 6800105 2 228 75 R2quot 1 005 33 095 while the minimum gain is AV 2 1lean 005 21 6800095 218744 R2n1005 33105 Problem 833 Solution Known quantities For the circuit in Figure P815 R1 R2 R3 R4 Find Select appropriate components using standard 1 resistor values to obtain a differential amplifier gain of magnitude approximately equal to 100 How closely can you approximate the gain Compute the error in the gain assuming that the resistors the have nominal value 820 G Rizzoni Principles and Applications of Electrical Engineering Analysis From the solution of Problem 815 the gain is given by R3 AV R 1 From Table 22 if we select R11kQ R3100k9 we obtain R AV A 100 R1 and the error for the gain with nominal values is zero Problem solutions Chapter 8 Problem 834 Solution Known quantities For the circuit in Figure P815 R1 R2 R3 R4 Find Same as in Problem 833 but use the i1 tolerance range to compute the possible range of gains for this amplifier Analysis From the solution of Problem 833 AV 2ampjAvm 100lt1001102 R1 10 001 AV ampjAV 1001 001 R1 m 11001 821 G Rizzoni Principles and Applications of Electrical Engineering Problem solutions Chapter 8 Section 83 Active Filters Problem 835 Solution Known quantities For the circuit shown in Figure P835 C 1 HF R 10k 2 RL 1kQ Find a The gain in dB in the pass band b The cutoff frequency c If this is a low or high pass filter Analysis a Assume the opamp is ideal Determine the transfer function in the form V M 1 H w 0 H v J Vs w 0 1f0 KVL39Vn39VdO deO Vnzo w KCL M I V V O E L n R ij w Vsjw This is not in the standard form desired but is the best that can be done There are no cutoff frequencies and no clearly defined pass band The gain ie the magnitude of the transfer function and output voltage increases continuously with frequency at least until the output voltage tries to exceed the DC supply voltages and clipping occurs In a normal high pass filter the gain will increase with frequency until the cutoff frequency is reached above which the gain remains constant b There is no cutoff frequency c This filter is best called a high pass filter however since the output will be clipped and severely distorted above some frequency it is not a particularly good high pass filter It could even be called a terrible filter with few redeeming graces 822 G Rizzoni Principles and Applications of Electrical Engineering Problem 836 Solution Known quantities For the circuit shown in Figure P836 Find a Whether the circuit is a low or highpass filter b The gain VgVS in decibel in the passband c The cutoff frequency Analysis a From Figure 824 it results that the amplifier in Figure P836 is a highpass filter In fact the output voltage is ja CR2 V a V a 00 1 jw CR1 51 b unM 20L0g 20LOgQ1317 dB we VS 1a 8 R1 18 1 1 5555 rads 07 0 CR1 140443103 Problem solutions Chapter 8 C1uF R118kQ R82ko RL333Q Problem 837 Solution Known quantities For the circuit shown in Figure P836 Find a Whether the circuit is a low or highpass filter b The gain VgVS in decibel in the passband c The cutoff frequency C 200 pF R110kQ R2 220m Analysis a From Figure 824 it results that the amplifier in Figure P836 is a highpass filter In fact the output voltage is ja CR2 V a V a 00 1 jw CR1 51 b unM 20L0g 20LogE2684dB we VS 1a dB R1 10 c 1 1 5 105rads 07 0 CR1 200404240403 RL 1kQ 823 G Rizzoni Principles and Applications of Electrical Engineering Problem solutions Chapter 8 Problem 838 Solution Known quantities For the circuit shown in Figure P838 C 100 pF R1 47 kg R2 6813 RL 22013 Find Determine the cutoff frequencies and the magnitude of the voltage frequency response function at very low and at very high frequencies Analysis The output voltage in the frequency domain is 1 ja CR1R2Vlj R V a 1 2 a 01 J 1ijR1 1 R1 160 C The circuit is a lowpass filter The cutoff frequencies are ml 2 L 2127106rads are1 10010 4710 w 1 1 2 CR1R2 1004042727403 For high frequencies we have 60 1375105rads Aw 1mm 1 1 1546 Hw Vija R1 7 At low frequencies liIn V0060 mgt0 Problem 839 Solution Known quantities For the circuit shown in Figure P839 C 20 HF R1 le R2 47kQ R3 80kg Find V0 0 a An expression for Hu 39 Vi 1a b The cutoff frequencies c The passband gain 1 The Bode plot Analysis a The frequency response is 824 G Rizzoni Principles and Applications of Electrical Engineering Problem solutions Chapter 8 Rm 1jwc Hltjwgtvo1wgt1 jwcR2R3JwCR2R31 g R2R3 Vjw R2 R2 ijR2R3 k R2 1 jroCR3 b The cutoff frequencies are 625 rads CR3 2010 8010 7R2R37 847103 7 11263 rads CRZR3 2010 8010 4710 c The passband gain is obtained by evaluating the frequency response at low frequencies R 80 llmH w 1i31718 Ad Ml m gt R 4397 2 d The magnitude Bode plot for the given amplifier is as shown Bode Magnmme Diagram tor we 25 Magnnuds mat to to to Fre uentv ironsec to Problem 840 Solution Known quantities For the circuit shown in Figure P840 C 047 yF R1 91kQ R2 22kg RL 22kQ Find a Whether the circuit is a low or hi ghpass filter b An expression in standard form for the voltage transfer function c The gain in decibels in the passhand that is at the frequencies being passed by the filter and the cutoff frequency Analysis a From Figure 321 it results that the amplifier in Figure P840 is alowpass filter 825 G Rizzoni Principles and Applications of Electrical Engineering Problem solutions Chapter 8 In fact the output voltage is R 1 V quotD 2 V 390 01 R1 11 CR2 31 V0050 1 Vijco R1 ljCO CR2 c The gain in decibel is obtained by evaluating Hu at 00 ie Hugom 20L0g 20L0g 766 dB 1 b H0060 The cutoff frequency is 1 1 007 76 3 9671 rads CR2 04710 2210 Problem 841 Solution Known quantities For the circuit shown in Figure P840 C 047 11F R1 2213 R2 6813 RL le Find a An expression in standard form for the voltage frequency response function b The gain in decibels in the passband that is at the frequencies being passed by the filter and the cutoff frequency Analysis a From Figure 821 it results that the amplifier in Figure P840 is a lowpass filter The voltage frequency response function is a V co R 1 HowgtM 2 Via R1 1 1a CR2 b The gain in decibel is obtained by evaluating Hu at 00 ie R 68 H 390 20L0 i20L0 7298dB l UltJ lm g R1 g22 The cutoff frequency is 00 i 31289 rads CR2 047 10 6810 Problem 842 Solution Known quantities For the circuit shown in Figure P842 C1 C2 01 uF R1 R2 10kQ 826 G Rizzoni Principles and Applications of Electrical Engineering Problem solutions Chapter 8 Find a The passband gain b The resonant frequency c The cutoff frequencies d The circuit Q e The Bode plot Analysis a From Figure 826 we have A 11951112 7 MW 7 J39wC 22m 7 7 7 2 i 2 2 z1 1 1w C1R1X1 1w CZRZ 1 m clRl 1w CRgt 211 C1R11 The magnitude of the frequency response is ABP jaw 12 1 w2CR The passband gain is the maximum over 0 of the magnitude of the frequency response ie 1 b The resonant frequency is w 7 1000 rads i i c The cutoff frequencies are obtained by solving with respect to wthe equation DC R 1 A 39w 7w272 w ww2 Ow w il 22w gtt MIDWAY m t t tlt gt a1 2414 rads 12 414 rads 1 1 d In this case 1 which implies Q E E e The Bode plot of the frequency response is as shown Bade Magnmme Diagram Mag nude mat Fre uentv radSc 8 27 G Rizzoni Principles and Applications of Electrical Engineering Problem solutions Chapter 8 Problem 843 Solution Known quantities For the circuit shown in Figure P843 C 047 11F R1 2209 R2 6813 RL le Find a An expression in standard form for the voltage frequency response function b The gain in decibels in the passband that is at the frequencies being passed by the filter and the cutoff frequency Analysis a The voltage frequency response function is M l Vija R11ijR2 b The gain in decibel is obtained by evaluating Hu at 00 ie HDUYD lHuj0ldB 20 Log 20 Log 68000 2 498 dB 1 The cutoff frequency is we 71 1 31289 rads CR2 2 047 409 68103 Problem 844 Solution Known quantities For the circuit shown in Figure P844 C1 22 ul C2 1 11F R1 22kQ R2 100kQ Find Determine the passband gain Analysis The voltage frequency response is Z 39oC R 0C R ABPltJcogt i lABPmo Z1 H160 C1R11 160 CgRg 1co2C1R1 1co2 CR The cutoff frequencies are 1 1 a 2210 6 22103 1 1 C212 110 9100103 The passband gain can be calculated approximately by evaluating the magnitude of the frequency response at frequencies greater than 01 and smaller than 0 ie 2066 rads 01 10000 rads 2 828 Gt Rizzoni Principles and Applications of Electrical Engineering Problem solutions Chapter 3 wltltwltltw2 1w1CR2wZCRZ 1w1C1R121 wCR1 R 100 7 i 4545 sz CiRi 2 2 Problem 345 Ar Solution Known quanti es For the circuit shownin Figure p345 let C C1 C2 220 uF R R1 R2 R 2in Find Determine the frequency response Analysis With reference to the Figure shown below We have C1 1C1 V0 T i l i i l i l Ir2wRiVowVc2wf1r2w rm R Vow 2 J 2 I 2 2 I Mari wri vvm x1 R1 c2 I ijZRZRI o J v 1 1 i 12ij Rimw W I 12 2 2 1 uCV 12 J gt 1 MCsz 1 w gt jwcr l l i i C 1 1 In llquot Icrlw1n2lw 1nrlw 105 CI F myo I D 2 2 2 2 1 7 10647 12 R J jruCR2 01 2 if i 7 i 7 l i 7 i l i Vin07 10 RimIv iijRVAw 21wCRijRVow 2 7 2 uCR 12 gt J MGR vow 8229 G Rizzoni Principles and Applications of Electrical Engineering Problem solutions Chapter 8 v0 jw 1 jaCR Vmltjagtgt mow jw2CR22jaCR2 2 ijR i ijR Problem 846 Solution Known quantities The inverting amplifier shown in Figure P846 Find a The frequency response of the circuit b IfR R2 100 kg and C 01 HF compute the attenuation in dB at a 1000 rads c Compute gain and phase at a 2500 rads d Find range of frequencies over which the attenuation is less than 1 dB Analysis a Applying KCL at the inverting terminal 1 R2 7 VOUT JwC laR2C VIN R1 jaR1C b Gain 2 0043 dB Gain 2 0007 dB Phase 177710 d To find the desired frequency range we need to solve the equation 1 CORZC J39CORiC This yields a quadratic equation in a which can be solved to find a gt 1965 rads 0 lt 08913 since 2010g1008913 1dB Problem 847 Solution Known quantities For the circuit shown in Figure P847 let C C1 C2 100 uF R1 3kQ R2 2kg Find Determine an expression for the gain Analysis 830 G Rizzoni Principles and Applications of Electrical Engineering Problem solutions Chapter 8 1C2 VC3 39 C2 1511 R1 R3 c wv A w I i Vin LI VCI IQ V0 With reference to the Figure shown above we have VC1ja V0ja gt IC1ja ij1V0ja VAow Rzlclow V0 or 1 ij1R2VOltjwgt Imooo ij2VOjwgt VAjwgt jw2C102R2 Vooan Vinow VAltjwgt R1Iczltjwgt IC1ltjwgt 1jwclltR1 R2 1w2C1C2R1 R2V0ltjwgt And finally the expression for the gain is Avoan VOW 1 1 Vinjw 1 ijR1 R2ja2C2R1R2 1ijR1X1jaCR2 Problem 848 Solution Known quantities The circuit shown in Figure P848 Find Sketch the amplitude response of V2 V1 indicating the half power frequencies Assume the op amp is ideal Analysis G V y IN AAAA V From KCL at the inverting input x ZC 1 0 Vx jaJCi jaJCV1 V 1 1 V1 VxV1 R R JOJRC 1 Similarly from KCL at the output of the op amp 831 G Rizzoni Principles and Applications of Electrical Engineering Problem solutions Chapter 8 V V V V 2 720 V2 lij 4 VIV21jaRC R ZC R R V2 ijC Combining the above results we find i 2 V1 1 ijC V2 CORC or i 2 V1 1 wRC This function has the form of a bandpass filter with maximum value determined as follows g wRC dG 1aRC2RC aRC2aRC2 2 7 V1 1axRC dw 1ch22 Setting the derivative equal to zero and solving for the center frequency RCw2R3C3 2w2R3C30 1 w2R2C220 013 1 Then Gmax 7 and the halfpower frequencies are given by wRC 1 1 1 7G 77 w2R2C212 wRC RZCZwZ z RCw1o 1coRC2 5 m 52 w ZJERCixSRZCZ 4R2C2 ii R 2R2C2 C The curve is sketched below G 05 03 54 0414 1 2414 DRC Problem 849 Solution Known quantities The circuit shown in Figure P849 Find Determine an analytical expression for the output voltage for the circuit shown in Figure P849 What kind of filter does this circuit implement Analysis The resistance Rm does not in uence the output voltage so 832 G Rizzoni Principles and Applications of Electrical Engineering 1 ijF 1 V a V a 7V a 00 R51 510 ijR51 510 The circuit is an integrator lowpass filter Problem solutions Chapter 8 Problem 850 Solution Known quantities The circuit shown in Figure P850 Find Determine an analytical expression for the output voltage for the circuit shown in Figure P850 What kind of filter does this circuit implement Analysis Figure P850 shows a noninverting amplifier so the output voltage is given by Vsjw RF C Sja1C5jaCSRF F V0ja 1 ij5 The filter is clearly a highpass filter 833 G Rizzoni Principles and Applications of Electrical Engineering Problem solutions Chapter 8 Sections 84 85 Integrator and Differentiator Circuits Analog Computers Problem 851 Solution Known quantities Figure P850 Find If C 1 p F R 10k R L le determine an expression for and plot the output voltage as a function of time Analysis R O 39VN39VD0 VDzO VN M CdvNVS iC E S 3 E xxkxxg mws g gv c 1 d vN vs WV 120 r is K CL C 1N 0 dt R 3 3quot dt VNzO V039RC The derivative is the slope of the curve for the source voltage Which is zero for tlt25ms5mslttlt75msandtgt15ms For25 mslttlt5ms v0 10103l106 33 0 3 12V 510 2510 15 3 1540 3 7510 3 6V For75 mslttlt 15 ms V0 4010 3 Problem 852 Solution Known quantities Figure P852a and Figure P852b Find If C1uF R10ko RL 1k 2 834 G Rizzoni Principles and Applications of Electrical Engineering a An expression for the output voltage Problem solutions Chapter 8 b The value of the output voltage at t 5 75 125 15 and 20 ms and a plot of the output voltage as a function of time Analysis a As usual assume the op amp is ideal so 39 VN VD O VD z 0 gt VN z 0 C d VN 39 V0 i lC 1 d t dt VNzO dV039 Vsdt RC 1 f Integrating Wan mm IVS dr RC 1 b Integrating gives the area under a curve Recall area of triangle 2 12base X height and area of rectangle 2 base X height Integrating the source voltage when it is constant gives an output voltage which is a linear function of time Integrating the source voltage when it is a linear function of time gives an output voltage which is a quadratic function of time 1 1 rad RC 10103 110 6 s 2 1 V05ms ov 100E251033v1 375mV V075ms 37510393 10025103933 1125mV 1 V0125ms 4124103 100 51033 1875mV 1 V015ms 18751O3 1OO 25103 15 16875mV v020ms 1687510393 100510393 15 9375 mV Problem 853 Solution Known quantities In the circuit shown in Figure P853 the capacitor is initially uncharged and the source voltage is m r 10 10 3 sin20007rtV Find At t 0 the switch 51 is closed How long does it take before clipping occurs at the output if RS 10k 2 and CF 20008uF 835 b At what times does the integration of the DC input cause the op amp to saturate fully G Rizzoni Principles and Applications of Electrical Engineering Problem solutions Chapter 8 Analysis 1 1 T 7 001 39 2000 Vout RSCF JVm tdt RSCF Sln mpt a 1 r r 7 j 001dt sin2000mdt R F O s F O l l The peak amplitude of the AC portion of the output is VP 2000 1989 V s F 727 The output will begin to clip when VO DC Vp 15V so we need to find at what time the 7 condition J00ldt l3V is satisfied The answer is found below F 0 13R C 0011 13 j 7 104ms RSCF 001 15R C b Using the results obtained in part a L39 W 1201113 Problem 854 Solution Known quantities The circuit shown in Figure 821 Find a If RS 10kQ RF 2M9 CF 0008 H F and vst10sin2000mV find vm using phasor analysis b Repeat parta if RF 20013 and if RF 20kg Compare the time constants with the period of the waveform for part a and b What can you say about the time constant and the ability of the circuit to integrate Analysis a Replacing the circuit elements with the corresponding impedances Z f 0 ZS V ID V out R f ZS RS Z f v 1 Jwa C f For the signal component at a 20007 836 G Rizzoni Principles and Applications of Electrical Engineering Problem solutions Chapter 8 R Vout if in 200 in RS 1waCf 1 Vin i 11800 arctan2 5 19839190570 V 0 2 1 425 For the signal component at a 0 DC V0 200Vm 2000V Thus van t 2000 19839sin2000m 9057 V 2000 2cos2000mV b RF ZOOkQ For the signal component at a 20007 Rf 1 a your 2 7 39 m 21979719568 V RS 1 jCOR f C f For the signal component at a 0 DC V0 20Vm 200V Thus v t 200 19797 sin2000m 9568 V 200 200s2000mV RF 2 2013 For the signal component at a 20007 Rf 1 a V0 2 7 in 214111348 V RS 1 jCOR f C f For the signal component at a 0 DC V 2vm 20V Thus out v t 20 14lsin20007rt 135 V 20 14lcos20007rt 45 V Rf C T 2 M9 16 ms 1 ms 200 kg 16 ms 1 ms 20 kg 016 ms 1 ms In order to have an ideal integrator it is desirable to have 6 gtgt T Problem 855 Solution Known quantities For the circuit of Figure 826 assume an ideal opamp With VS t 10 1073 sin20007rtV CS 100uF CF 0008HF RF 2MQ and RS 10kQ Find v0 a The frequency response i060 Vs b Use superposition to find the actual output voltage remember that DC 0 HZ 837 G Rizzoni Principles and Applications of Electrical Engineering Problem solutions Chapter 8 Analysis 3 Zf vslttgt Vot Rf ZS RS 39 Zf 39 jCOCS lRfojCO Rf Viz Zi1ijfo 2 ijsz vs ZS RS 1 ijSCS 1ijRfo 1 ijs ja v0 3 7Jw51 0m S ja1 71 1 625 v0 H jOvs b V0 VS i060 By superposition Volwmv Vs 0V Vs 39 5 0 7 204041 8943 V MOO 1 j62831 1100 v0 t 2010 3 sin2000m 8943 V We can say that the practical differentiator is a good approximation of the ideal differentiator V Problem 856 Solution Find Derive the differential equation corresponding to the analog computer simulation circuit of Figure P856 Analysis xt 200jzdt or z L Also Z 40y 200 dt 1 dx dy Thf7i Al 4txtdt i4txt y 400 so y l m 0 or dt m 0 1 dzx dzx Therefore my 4 ft xt or 4000xt 16000ft 0 838 G Rizzoni Principles and Applications of Electrical Engineering Problem solutions Chapter 8 Problem 857 Solution Find Construct the analog computer simulation corresponding to the following differential equation d 2x dx 72100710x 5ft dt dt Analysis r dx 5 t 2 hi W E D KG dx 1001t 10 839 G Rizzoni Principles and Applications of Electrical Engineering Problem solutions Chapter 8 Section 86 Physical Limitations of Operational Amplifiers Problem 858 Solution Known quantities For the circuit shown in Figure 88 R5 RF 2213 Find Find the error introduced in the output voltage if the opamp has an input offset voltage of 2 mV Assume zero bias currents and that the offset voltage appears as in Figure 848 Analysis By superposition the error is given by Ava 1 gym 4mV 5 Problem 859 Solution Known quantities For the circuit shown in Figure 88 R5 RF 2213 Find Repeat Problem 858 assuming that in addition to the input offset voltage the opamp has an input bias current of l uA Assume that the bias currents appear as in Figure 849 Analysis By superposition the effect of the bias currents on the outputis v RIB v AV V v l l 4071 E 18 gt AVOJB RF 18 RRF sthn If R RF RS then mm RF 1 13 RFIOS 2200110 6 22mV The effect of the bias voltage is 3 So the total error on the output voltage is Ava AVOYVB AVOYIB 18mV AVOYVB 2 1 FV0ma 4mV Problem 860 Solution Known quantities The circuit shown in Figure P860 The input bias currents are equal and the input bias voltage is zero 840 G Rizzoni Principles and Applications of Electrical Engineering Problem solutions Chapter 8 Find The value of R that eliminates the effect of the bias currents in the output voltage Analysis We can write V RxIB V7 18 Br AVOJB v v 1 l R EIBr gt AVOJBzRF IBRxRFR1 B F By selecting RX RF II R1 a zero output voltage error is obtained Problem 861 Solution Known quantities For the circuit shown in Figure P860 RF 33kQ R1 le VS 15 sinat V Find The highestfrequency input that can be used without exceeding the slew rate limit of lV s Analysis The maximum slope of a sinusoidal signal at the output of the amplifier is R V 1 SOlAlwFFwlo ijw 1063D3105radls S max 1 Problem 862 Solution Known quantities The Bode plot shown in Figure 845 A0 106 00 107T rads Find The approximate bandwidth of a circuit that uses the opamp with a closed loop gain of A1 75 and A2 350 Analysis The product of gain and bandwidth in any given opamp is constant so 107 m1 J 4186405 rads A1 75 107 m2 J 897110 rads A2 350 841 G Rizzoni Principles and Applications of Electrical Engineering Problem solutions Chapter 8 Problem 863 Solution Known quantities For the practical charge amplifier circuit shown in Figure P863 the user is provided with a choice of three flung RLCF Tm a switch Assume that RL 10MQ RM 1MQ RS 01MQ and CF 01 HF Find Analyze the frequency response of the practical charge amplifier for each case and determine the lowest input frequency that can be amplified without excessive distortion for each case Can this circuit amplify a DC signal Analysis time constants RM CF L39Shm RS CF which can be selected by means of edium Applying KCL at the inverting terminal 1 VOL 0 1a 160C V or out JwC i 1jaRC This response is clearly that of a highpass filter therefore the charge amplifier will never be able to amplify a DC signal The low end of the magnitude frequency response is plotted below for the three time constants The figure illustrates how as the time constant decreases the cutoff frequency moves to the right solid line R 10 M9 dashed line R 1 M9 dotted line R 01 M9 100 res onse of ractical char e am lifier m 160 dsmudd mdd m 1071 100 101 102 frequency rads From the frequency response plot one can approximate the minimum useful frequency for distortionless response to be nominally l HZ for the 10 M9 case 10 Hz for the 1 M9 case and 100 HZ for the 01 MQ case 842 G Rizzoni Principles and Applications of Electrical Engineering Problem solutions Chapter 8 Problem 864 Solution Find Consider a differential amplifier We would desire the commonmode output to be less than 1 of the differentialmode output Find the minimum dB commonmode rejection ratio CMRR that fulfills this requirement if the differential mode gainAdm 1000 Let v1 sin20007rt 01s39m 1 207v v2 sin 2000m 180 0lsin1207tV V0 Adm V1 V2 AemVlT 2 Analysis We first determine which is the common mode and which is the differential mode signal v1 v2 23in20007rt WT 0lsin1207rt Therefore v0 Adm2sin20007rtAcm2sin1207rt Since we desire the common mode output to be less than 1 of the differential mode output we require Am01s 0012 or A lt 02 cm 1000 A CMRR So CMRRmin 5000 74 dB cm Problem 865 Solution Known quantities As indicated in Figure P865 the rise time tr of the output waveform is defined as the time it takes for the waveform to increase from 10 to 90 of its final value ie tr E tb ta L39ln 01 1n09 221 where Tis the circuit time constant Find Estimate the slew rate for the opamp Analysis dVW 15X 0398 X g 2731 Therefore the slew rate is approximately d m 145 101gtlt106 44 13 pg V 2737 us Problem 866 Solution Find Consider an inverting amplifier with openloop gain 105 With reference to Equation 818 843 G Rizzoni Principles and Applications of Electrical Engineering If RS 10kQ and RF 1MQ find the voltage gain Awa Problem solutions Chapter 8 b Repeatpartaif RS 10kQ and RF 10MQ c Repeatpartaif RS 10kQ and RF 100MQ d Using the resistors values of part 0 find AWCLgt if AWOL ltgt Analysis R l ACL iF Rs HM RSAOL a ACL 99899 b A 990 CL c ACL 9091 d As AOL ltgt ACL 10000 Problem 867 Solution Known quantities Figure P867 Find If the opamp shown in Figure P867 has an openloop gain of 45 X 105 find the closedloop gain for a RS RF 75kQ b Repeatpartaif RF 5RS 375kQ Analysis a all RS V39 v l A V OLV V Vout Vin v 7 O Vin V and vO A0Lvm v v v v7 0 v v0 Writing KCL at V7 RS RF 0 844 G Rizzoni Principles and Applications of Electrical Engineering Problem solutions Chapter 8 Substituting 10 mm 1 1 1 1 1 AOL A0L7LO j V0 777 Vm 77 Rs RF RF AOLRS AOLRF RF Rs RF 1 1 R R 1 1 V0 s F A0LRSRF i l39i in 1 1 i 1 K2 AOLRS AOLRF F 2 where K1RFRSRS RsAOLRF 9 V0 RJ K2RFRSRF2RSA0LRF Vi RS iRSl RF 1 iRSl RF 1 AOLRS AOLRS For the conditions of part a we obtain 1 1 ACL l999 2 5 2 5 A5x10 1 45gtlt10 1 b A 5 1 1 5 999 39 CL 6 5 6 5 39 A5x10 1 A5x10 1 Problem 868 Solution Find Given the unitygain bandwidth for an ideal opamp equal to 50 MHZ find the voltage gain at frequency of f 500 kHz Analysis A0500 K A1w1 K 1gtlt27r gtlt50MHZ107 gtlt106 6 A1 5 10 X10 10000 01 Zn x 500 Problem 869 Solution Find Determine the relationship between a finite and frequencydependent openloop gain AV0L 0 and the closedloop gain AVCL 0 of an inverting amplifier as a function of frequency Plot AVCL versus 0 Analysis As shown in Equation 884 if we consider a real opamp 845 G Rizzoni Principles and Applications of Electrical Engineering Problem solutions Chapter 8 A0 ja AV0Lw 1 O WhereAC RFRS andAD is the lowfrequency openloop gain If we choose A0 106 and COD 10 TC and since A060O Accol a 106107r 1 RF 39 Rs Therefore A 20LOgi0 lAVCL I 20LOgi0 AC V 01 Lng a Problem 870 Solution Find A sinusoidal sound pressure wave impinges upon a condenser microphone of sensitivity S mV Pa The voltage output of the microphone V3 is amplified by two cascaded inverting amplifiers to produce an amplified signal V0 Determine the peak amplitude of the sound wave in dB if V0 5 VRMS Estimate the maximum peak magnitude of the sound wave in order that V0 not contain any saturation effects of the opamps Analysis pt PO sinat Pa vs t S ptmV A A A A v0t1102vstV 1503311 srnatV A1A2 s10O 5000 mV v0 5 VRMS 5000 P 201 7 j 0 OgIO A1A2S Msizv PO 12000A1A2S Pa 1000 846 G Rizzoni Principles and Applications of Electrical Engineering Problem solutions Chapter 8 Problem 871 Solution Known quantities If in the circuit shown in Figure P87l v31 28 00lcosatV V 35 0007 cosatV Avl13 Av210 Find w4krad s a The common and difference mode input signals b The common and differential mode gains c The common and difference mode components of the output voltage d The total output voltage e The common mode rejection ratio Analysis a vsc vs1vs2 28V10chosat 35V 7chosat 315V15chosat VSD VSlVsz 28V10chosat 35V 7choswt 07 V 17 mV coswt Note that the expression for the differential input voltage and the differential gain below depends on which of the two sources in this case V31 is connected to the noninverting input 1 b AV Av1sz 3 Avd EAv139Av2 115 Vac st AVC 315V15chosat 3 9450V 45chosat VSDAM 07V17chosat 115 8050V 1955chosat VOVD d m v0cV0D 14V 2000mV coswt 39AWII 20dBLog10 5 1167dB e CMRR 20dB L0g10 Ivd Problem 872 Solution Known quantities If in the circuit shown in Figure P87l V31 2 35001 cosaxV V32 2 35 001 coswt krad Avc10dB Am 20dB a 47 S Find a The common and differential mode input voltages b The voltage gains for v51 and v52 847 G Rizzoni Principles and Applications of Electrical Engineering Problem solutions Chapter 8 c The common mode component and differential mode components of the output voltage d The common mode rejection ratio CMRR in dB Analysis a st V31Vsz 35V10chosat 35V 10chosat 2 35V VSD VSlVsz 35V10choswt 35V 10chosat 20choswt Note the expression for the difference mode voltage depends on which signal in this case v31 is connected to the noninverting input This is also true for the expression for the differential gain below b w AW 10ZOdB 3162 Aw 10MB 10 A Av1Av2 Av1 glowed 3162110 11581 1 1 1 AW 2 EAWAVZ j M EAwAvd E31621 10 8419 0 Vote MAW 35V3162 1107v 39 V01 2 VSDAM 20mV coswt10 200mV coswt d CMRR 20dBL0g10 AW 20dBL0g10 10dB Avd Problem 873 Solution Known quantities If in the circuit shown in Figure P873 the two voltage sources are temperature sensors with T 2 Temperature Kelvin and Vs1 kTJ Vsz sz Where k2120u R1R3R45k9 R23kQ RL6OOQ Find a The voltage gains for the two input voltages b The common mode and differential mode input voltage c The common mode and difference mode gains d The common mode component and the differential mode component of the output voltage e The common mode rejection ratio CMRR in dB Analysis a 848 G Rizzoni Principles and Applications of Electrical Engineering Problem solutions Chapter 8 KVL 39VN39VDVP 0 Assume the op amp is ideal VDzO gt VN VP MV0 KCL iNVNV51VNV0 0 W zO gt W R1 R3 R1R3 R1 R3 ii R1R3 R1 R3 v0 KCL ipVPV52VPV0 0 isz gt VP R2 R4 R2R4 R2 R4 i i R2134 R2 R4 Vs1R3V0R1 V52R4V0R2 Equating R3R1 R4R2 R R V52 4Vs139 3 V R4R2 R3R1 R4R2R3R1 0 R1 R2 R4R2R3R1 R3R1 R4R2 R4R3R1 R3R4R2 V52 Vs139 R1R4R2R2R3R1 R1R4R2R2R3R1 A R4R3R1 Slllol 2 V R1R4R2R2R3R1 58 310 39R3R4R2 5 ll8l AVI R1R4R2R2R3R1 5 8 3 10 V51 kT 120u310K 3720mV V52 kT2 120u335K 4020mV VSC V51V52 3720mV4020mV 3870mV VSLD 12521251 Note that the expression for the differential mode voltage and the differential mode gain below depends on Source 2 being connected to the non inverting input 849 G Rizzoni Principles and Applications of Electrical Engineering Problem solutions Chapter 8 l 1 v0 VSZAV2 Vs1As1 vsc EvsD Av2 vsc Ev Avl l VSVCAv2Av1VSVDEA239Av1 Vsrc AvrcVSVDAvd Avc Av2Avl 54 1 l l Avd EAv239Avl El539394l 4 5 v05 VSVCAW3870mV1 3870mV V01 2 VSDAM 3mV45 135mV e An ideal difference amplifier would eliminate all common mode output This did not happen here A figure of merit for a differential amplifier is the Common Mode Rejection Ratio CMRR CMRR 20dB L0g10 AV 20dB L0g10 1306dB A Problem 874 Solution Known quantities If for the differential amplifier shown in Figure P873 V31 13 mV V32 2 9mV v0 VOCVOD VOC 33mV VOD18V Find a The common mode gain b The differential mode gain c The common mode rejection ratio in dB Analysis a m V31Vs2 13mV9mV 11mV 33 V j AVC M i 3 m llmV VSD VSZv312 4mV b jAVDzmiDz 4500 VSD 4mV c CMRR 20dB L0g10 39AVcl 6352dB IAVDI 850 G Rizzoni Principles and Applications of Electrical Engineering Problem solutions Chapter 8 Chapter 8 Instructor Notes Chapter 8 introduces the notion of integrated circuit electronics through the most common building block of electronic instrumentation the operational amplifier This is in practice the area of modern electronics that is most likely to be encountered by a practicing nonelectrical engineer Thus the aim of the chapter is to present a fairly complete functional description of the operational amplifier including a discussion of the principal limitations of the opamp and of the effects of these limitations on the performance of opamp circuits employed in measuring instruments and in signal conditioning circuits The material presented in this chapter lends itself particularly well to a series of laboratory experiments see for examplel which can be tied to the lecture material quite readily After a brief introduction in which ideal amplifier characteristics are discussed open and closed loop models of the opamp are presented in section 82 the use of these models is illustrated by application of the basic circuit analysis methods of Chapters 2 and 3 Thus the Instructor who deems it appropriate can cover the first two sections in conjunction with the circuit analysis material A brief intuitive discussion of feedback is also presented to explain some of the properties of the opamp in a closedloop configuration The closedloop models include a fairly detailed introduction to the inverting noninverting and differential amplifier circuits however the ultimate aim of this section is to ensure that the student is capable of recognizing each of these three configurations so as to be able to quickly determine the closed loop gain of practical amplifier circuits summarized in Table 81 p 402 The section is sprinkled with various practical examples introducing practical opamp circuits that are actually used in practical instruments including the summing amplifier p 938 the voltage follower p 400 a differential amplifier Focus on Measurements Electrocardiogram EKG Amplifier pp 402404 the instrumentation amplifier pp 404 405 the level shifter p406 and a transducer calibration circuit Focus on Measurements Sensor calibration circuit pp 407409 Two features new in the third edition will assist the instructor in introducing practical design considerations the box Practical OpAmp Design Considerations p 410 illustrates some standard design procedures providing an introduction to a later section on opamp limitations the box Focus on Methodology Using Opamp Data Sheets pp 410412 illustrates the use of device data sheets for two common opamps The use of Device Data Sheets is introduced in this chapter for the first time In a survey course the first two sections might be sufficient to introduce the device Section 83 presents the idea of active filters this material can also be covered quite effectively together with the frequency response material of Chapter 6 to reinforce these concepts Section 84 discusses integrator and differentiator circuits and presents a practical application of the opamp integrator in the charge amplifier Focus on Measurements Charge Ampli ers pp 420421 The latter example is of particular relevance to the nonelectrical engineer since charge amplifiers are used to amplify the output of piezoelectric transducers in the measurement of strain force torque and pressure for additional material on piezoelectric transducers see for example A brief section 85 is also provided on analog computers since these devices are still used in control system design and evaluation Coverage of sections 84 and 85 is not required to complete section 86 The last section of the chapter 86 is devoted to a discussion of the principal performance limits of the operational amplifier Since the student will not be prepared to fully comprehend the reason for the saturation limited bandwidth limited slew rate and other shortcomings of practical opamps the section focuses on describing the effects of these limitations and on identifying the relevant parameters on the data sheets of typical opamps Thus the student is trained to recognize these limits and to include them in the design of practical amplifier circuits Since some of these limitations are critical even in low frequency applications it is easy and extremely useful to supplement this material with laboratory exercises The box Focus on Methodology Using Opamp Data Sheets 7 Comparison ofLM 741 and LMC 6061 pp 441 446 further reinforces the value of data sheets in realizing viable designs The homework problems present a variety of interesting problems at varying levels of difficulty many of these problems extend the ideas presented in the text and present practical extensions of the circuits discussed in the examples In a onesemester course Chapter 8 can serve as a very effective capstone to a first course in circuit analysis and electronics by stimulating curiosity towards integrated circuit electronics 1 Rizzoni G A Practical Introduction to Electronic Instrumentation 3 d Ed KendallHunt 1998 2 Doebelin E 0 Measurement Systems McGrawHill Fourth Edition 1987 81

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