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# Analog Electronics ECE 3050

GPA 3.64

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This 0 page Class Notes was uploaded by Cassidy Effertz on Monday November 2, 2015. The Class Notes belongs to ECE 3050 at Georgia Institute of Technology - Main Campus taught by Staff in Fall. Since its upload, it has received 12 views. For similar materials see /class/233908/ece-3050-georgia-institute-of-technology-main-campus in ELECTRICAL AND COMPUTER ENGINEERING at Georgia Institute of Technology - Main Campus.

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Date Created: 11/02/15

OSCILLATOR DEMONSTRATION The condition for oscillation to occur in a single loop feedback system is simply that the loop gain must equal lO Both the magnitude and phase are important and allow us to solve for l the frequency at which oscillation occurs and 2 the value of loop gain at this frequency necessary for oscillation The The loop gain is defined as AB where A and B are shown in Fig 1 i0 Oufu No infw kw 016 Fig 1 minus is due to the inversion that Xf receives as it passes through the summer A is the amplifier39s forward transfer gain and B is the reverse transmission factor Since we want the condition that Loop gain AB 1g 1 then either the amplifier A or the feedback network B or both must contribute a phase shift of 180 so that condition of l is satisfied Let us consider an example of an oscillator shown in Fig 2 This oscillator is called a Wien bridge oscillator The loop gain can be calculated as vf v where the loop has been opened at the amplifiers input In opening loops one must be careful not to change the performance of the network Since the input impedance of the amplifier is infinite we have met this requirement One could also open the loop at the output of the amplifier The loop V6 gain is found as V V Id at a I I f L e W a WMquot 2 7quot Fig 2 V f szRC 2 V 2 2 f s 3mRCsmRC where L Rc RC 3 Equating 2 to 1L0 gives two equations They are V wK U ll RC 1 4 Vf 2 2 2 2 wRC u Buncm 3 w WEVi g m 0 5 NRC A From the phase shift we get the frequency of oscillation Thus from 5 we see that when wch that ArgVfquotVf 0 Therefore the frequency of oscillation no is given as l quot 0 I no quot E6 6 The necessary value of K for oscillation is found from the magnitude expression substi tuting the value of mo Thus from 4 and 6 we get K3 7 Consequently we have now found the operating condition of the oscillator Unfortunately if Fig 2 was connected with the above values it would probably not oscillate The reason is that to start the oscillator going requires K greater than 3 However once the oscillator is oscillating then K must return to the value of 3 In addition there is nothing in the problem that allows us to predict the amplitude of oscillation Therefore we must consider some further aspects of practical oscillators 00 The normal transfer function of the amplifier K in Fig 2 could be rep resented as shown in Fig 3 Power sup y Hugquotts However if we modify this transfer function through nonlinear methods we can obtain a transfer function such as that shown in Fig 4 Va Power supply 9 limits Power suf y HHS KIlt3 Fig 3 VA This transfer function is suitable for deter mining the amplitude of oscillation and func gtVf tions in the following manner At low signal levels the value of K is greater than 3 thus Kgt3 the oscillator can build up in oscillation However as the amplitude grows above V the effective gain of the amplifier begins 0 decrease because of the lesser value of K Poultf wfly limiis ie K at higher levels At some value of magnitude the effective or average value of K will be exactly 3 and the oscillator will produce a constant level sinusoid It is difficult to predict this value because of the nonlinearity of the transfer functions However it is easy to change the amplitude by simply changing K1 K2 or both The distortion of the sinusoid will be greatly influenced by the shape of Fig 4 For low distortion one prefers a very soft gradual limiting effect Thermistors light bulbs and other devices which change resistance as the power increases have been used to generate very low distortion sinusoids OSCILLATOR DEMONSTRATION The condition for oscillation to occur in a single loop feedback system is simply that the loop gain must equal lO Both the magnitude and phase are important and allow us to solve for l the frequency at which oscillation occurs and 2 the value of loop gain at this frequency necessary for oscillation The The loop gain is defined as AB where A and B are shown in Fig 1 i0 Oufu No infw kw 016 Fig 1 minus is due to the inversion that Xf receives as it passes through the summer A is the amplifier39s forward transfer gain and B is the reverse transmission factor Since we want the condition that Loop gain AB 1g 1 then either the amplifier A or the feedback network B or both must contribute a phase shift of 180 so that condition of l is satisfied Let us consider an example of an oscillator shown in Fig 2 This oscillator is called a Wien bridge oscillator The loop gain can be calculated as vf v where the loop has been opened at the amplifiers input In opening loops one must be careful not to change the performance of the network Since the input impedance of the amplifier is infinite we have met this requirement One could also open the loop at the output of the amplifier The loop V6 gain is found as V V Id at a I I f L e W a WMquot 2 7quot Fig 2 V f szRC 2 V 2 2 f s 3mRCsmRC where L Rc RC 3 Equating 2 to 1L0 gives two equations They are V wK U ll RC 1 4 Vf 2 2 2 2 wRC u Buncm 3 w WEVi g m 0 5 NRC A From the phase shift we get the frequency of oscillation Thus from 5 we see that when wch that ArgVfquotVf 0 Therefore the frequency of oscillation no is given as l quot 0 I no quot E6 6 The necessary value of K for oscillation is found from the magnitude expression substi tuting the value of mo Thus from 4 and 6 we get K3 7 Consequently we have now found the operating condition of the oscillator Unfortunately if Fig 2 was connected with the above values it would probably not oscillate The reason is that to start the oscillator going requires K greater than 3 However once the oscillator is oscillating then K must return to the value of 3 In addition there is nothing in the problem that allows us to predict the amplitude of oscillation Therefore we must consider some further aspects of practical oscillators 00 The normal transfer function of the amplifier K in Fig 2 could be rep resented as shown in Fig 3 Power sup y Hugquotts However if we modify this transfer function through nonlinear methods we can obtain a transfer function such as that shown in Fig 4 Va Power supply 9 limits Power suf y HHS KIlt3 Fig 3 VA This transfer function is suitable for deter mining the amplitude of oscillation and func gtVf tions in the following manner At low signal levels the value of K is greater than 3 thus Kgt3 the oscillator can build up in oscillation However as the amplitude grows above V the effective gain of the amplifier begins 0 decrease because of the lesser value of K Poultf wfly limiis ie K at higher levels At some value of magnitude the effective or average value of K will be exactly 3 and the oscillator will produce a constant level sinusoid It is difficult to predict this value because of the nonlinearity of the transfer functions However it is easy to change the amplitude by simply changing K1 K2 or both The distortion of the sinusoid will be greatly influenced by the shape of Fig 4 For low distortion one prefers a very soft gradual limiting effect Thermistors light bulbs and other devices which change resistance as the power increases have been used to generate very low distortion sinusoids h p 2 users ece gafech edupallen Ecs3osoFalzoo4 0er zz4o4 EIJci Jacaer meme VoI n again 04 ak I S fernm1 I Ilafag 34in 40 or AIM49 3 Mai I 0 Ar I MWer W fat aw 5 5 77 0 Common dram map150 WWI Vw zv and Kim 4W1quot KISQW Md Mat15 is iucn tquot 05mIz 3quot Mf39 Vmuv Asaozv VD SQ SWIMBiaud Modal 7 VM 04 JKnIpQ Mvom quot 50715 r0 1 223 l s I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I z 39 I g It 5 e I I I I I I I I I I I I I I I I I I I I I I I I I I I if 4 I Mm 6 L Id 1 4 git94 4 I if Ms MFM m anti14 zmnn f L 2 19613

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