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# Electronic Aids to Measurement PHYSICS 623

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This 48 page Class Notes was uploaded by Nichole Keebler on Thursday September 17, 2015. The Class Notes belongs to PHYSICS 623 at University of Wisconsin - Madison taught by Staff in Fall. Since its upload, it has received 32 views. For similar materials see /class/205231/physics-623-university-of-wisconsin-madison in Physics 2 at University of Wisconsin - Madison.

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Date Created: 09/17/15

Physics 623 Digital Circuits Aug 4 2006 1 Purpose 0 To introduce the basic principles of digital circuitry 0 To understand the small signal response of various gates and circuits typical of currently available TTL integrated circuitry 2 Discussion The circuits are TTL 7400 series of various types from various manufacturers Use the built in supply for V00 5 V DC The logic levels are 0 One 3 to 50 volts nominally 475 V 0 Zero 0 to 04 volts nominally ground 0 V 0 Switching Level 14 V The delay in passing a signal through one NAND gate is typically 13 ns for the SN7400 series and is about 7 ns for the 74L800 series The fanout capability is 107 ie ten inputs can be connected to each output The operating temperature range is from about 0 C to 70 C The general characteristics of TTL logic are described in the attached sheets together with the detailed speci cations of individual chips Every input of every logical unit you use must be connected to something Check each input pin around the symbol on the data sheet and decide where it should be connected Some unused inputs may not be shown on the simpli ed logic diagrams 3 Procedure H Verify the truth table for a JK flip flop7 using the 74LSll2 The Clock is CP The data sheet for the out dated 7473 ip op is included for its superior functional block diagram Study it and see how a JKFF works Here7 J and K refer to the respective input levels when the 71 clock is high7 and Qn1 refers to the output after clock returns from 1 to 0 These ip ops have a set and a reset feature Experiment with the and W inputs to determine their effect See the 74L8112 data sheet 2 9 7 J K QM1 0 0 Qn 0 1 0 1 0 1 1 1 Q Connect the 74L8112 JKFF7s to make an 4 bit counter 0 to 15 To display the sequence of bits use the lights on the board Dont forget to hook up all the inputs7 even if they arent shown on the functional logic diagram What do you want the J amp K inputs to be in this case What do you want on the preset and clear lines 74LS112 Make a ring oscillator Use an odd number of gates A signal is propagated around the loop with a speed depending on the delay of each gate and the number of gates For 7 gates the period is about 50 150 nsec7 depending on the type of gate Check the frequency by looking at the output of any gate with the scope Use your four bit counter to count the oscillations Determine the gate delay from your rneasured frequency Note that a full cycle requires delays going around the loop m Where must the unused input of each NAND gate be connected Study the behavior of the rnonostable rnultivibrator oneshot A dual oneshot has an id number 74LS221 Connect the circuit shown Check the spec sheet for valid resistor and capacitor valuesl Arrange the two outputs so that they have the appropriate relative tirning when triggered by a rising edge Determine the output pulse width relative to the RC time constant of your external circuit What changes would have to be made to trigger your 9 CT I circuit on the falling edge What would the result be if you used the rising edge trigger input B on the second one shot Use the data sheets to determine what to do with the unused inputs They are not all true I vcc vcc 39r 1 ms Output 1 L7 CEXT CEXT R1 C1 15 REXTCEXT REXTCEXT Output 2 1 13 13 Trlgger CHE 203 I372 9 o 393 m 3 30 cm 5 LE 3 30 an 6 LE 74L5221 74L5221 time Investigate the properties of the 8 bit serial input parallel output shift register lD 74HCT164 Connect the chip7 then clock various numbers into the register using the switches Use the eight LEDs on the breadboard to show the states of the output How does the Reset function work Keep the circuit set up for the next exercise The 7415283 is a four bit adder which performs the binary addition of one 4 bit number A1 to A4 to another B1 to B47 including a carry in 77 bit CO7 in order to produce a 4 bit sum 81 to S4 plus a carry out77 C4 Use the shift register to generate both the input binary numbers7 so that A1 to A4 corre spond to QE through QH and B1 to B4 to QA through QD For convenience it is useful to connect the input and output 4 bit numbers to a hex display using the TlL311 chip Verify the addition Optional 7 if time permits Replace the adder with a four bit magnitude comparator 74L885 Compare the two displayed numbers and verify all the outputs Now experiment with the carry inputs to determine what values are required to function correctly 3 A0 3 A1 s1 EI EDT A1 ET A2 82 TE EDT A2 EDT A3 33 TE 1 3 13 9 A3 3 A4 84 El EI 2 A 4 EI GT B0 6 E B QB 5 EI GT B1 393 2 B1 QC 5 9 IEI 1 B2 EDT B2 QD TE 13 2 B3 7 GT B3 3 GE TE I3 3 AltB e EI 3 B4 EI CLK QF TE 13 4 AB AB 5 EI 7 9 9 QG TE 3 AgtB AgtB El 3 00 04 El EI C CLR QH EI 74L585 74L5283 74HCT164 Transmission Lines Physics 623 Murray Thompson Sept 1 1999 Contents 1 Introduction 2 2 Equations for a lossless Transmission Line 2 3 The Voltage Solution 5 4 The Current Solution 5 5 The Characteristic Impedance Z077 6 6 Speed u of Signals 6 7 Impedances of Actual Cables 6 8 Eleven Examples 10 9 Capacitive Termination 16 10 Types of Transmission Lines 22 11 Imperfections of Transmission Lines 23 111 Remedies for Signal Loss 24 1 CONTENTS 2 12 Equations for a Lossy Transmission Line 25 13 Lossy Line With No Re ections 29 14 Attenuation if line is slightly lossy 29 15 Characteristic Impedance of a Lossy Transmission Line 30 16 Heavyside Distortionless Lines 30 17 Three Examples 31 18 Attenuation at DC and Low Frequencies 31 19 Attenuation at Higher Frequencies 33 191 Skin Effect Loss 33 192 Dielectric Loss 35 193 Radiation Loss 35 194 Actual Attenuation in Cables 35 1 INTRODUCTION 3 1 Introduction A Transmission line is a pair of conductors which have a cross which remains constant with distance For example7 a coaxial cable transmission line has a cross section of a central rod and an outer concentric cylinder Similarly a twisted pair transmission line has two conducting rods or wires which slowly wind around each other A cross section made at any distance along the line is the same as a cross section made at any other point on the line We want to understand the voltage Current relationships of transmission lines 2 Equations for a lossless Transmission Line A transmission line has a distributed inductance on each line and a distributed capacitance between the two conductors We will consider the line to have zero series resistance and the insulator to have in nite resistance a zero conductance or perfect insulator We will consider a Lossy line later in section 12 on page 25 De ne L to be the inductanceunit length and C to be the capacitanceunit length Consider a transmission line to be a pair of conductors diVided into a number of cells with each cell haVing a small inductance in one line and haVing small capacitance to the other line In the limit of these cells being very small7 they can represent a distributed inductance with distributed capacitance to the other conductor 5 TH llll V Consider one such cell corresponding to the components between position x and position x Ax along the transmission line 2 EQUATIONS FOR A LOSSLESS TRANSMISSION LINE 4 vAv V A 1quot Q 2 IAI AX 2Ax y A Ax w The small series inductance is LAx and the small parallel capacitance is CAx De ne the voltage and current to the right on the left side to be V and I De ne the voltage and current to the right on the right side to be V AV and I AI We now can get two equations 1 The current increment AI between the left and right ends of the cell is discharging the capacitance in the cell The charge on the cells capacitance capacitance X voltage CAxV and so the current leaving the capacitance to provide AI must be AI 7Charge 7CAmV The minus sign is due to the current leaving the capacitor av AI iCAm AI av E 03 Note the minus sign 2 The voltage increment AV between the left and right ends of the cell is due to the changing current through the cells inductance Lenzls Law BI BI AV iInductanceE iAxLE 7 AL Am 7 39 0t 39 Now take the limit of the cell being made very small so that the inductance and capacitance are uniformly distributed The two equations then become 70 Equation 1 AL Equation 2 Remember that L and C are the inductanceunit length measured7 in Henriesmeter and are the capacitanceunit length measured in Faradsmeter Differentiate equation 2 with respect to the distance m N EQUATIONS FOR A LOSSLESS TRANSMISSION LINE 5 as ass 32V 711131 312 31 at X and t are independent variables and so the order of the partials can be changed 32V 7 a 31 w L m Now substitute for from equation 1 above 32V 7 3 3V W L39 O39W 2 2 i LO Equation 3 This is usually called the Transmission Line Differential Equation Notes 0 L and C are NOT just the inductance and the capacitance They are both measured per unit length 0 The Transmission Line Differential Equation 3 above does NOT have a minus sign The Transmission Line Differential Equation 3 above is a normal 1 dimensional wave equa tion and is very similar to other wave equations in physics Erom experience with such wave equations7 we can try the normal solution of the form V Vs where s is a new variable 5 z ut Substituting this into the two sides of the Transmission Line Differential Equation 3 above we get the two sides being 2 2 agandi av m u239 3t2 Thus the form Vz ut can satisfy the Transmission Line Differential Equation 3 if and only if 7 LO Equation 4 Both roots of this satisfy the Equation 3 m 1 u iv The two roots give slightly different solutions and so7 since the equation 3 is linear7 any linear combination of the two solutions is a valid solution De ne u as the positive root u Jr Equation 5 3 THE VOLTAGE SOLUTION 6 3 The Voltage Solution Thus7 the general solution for the voltage is the linear combination V fz 7 wt g ut Equation 6 Where f0 and g are arbitrary single valued functions which can be very different 1 f 7 ut describes a wave propagating with no change in shape towards z 00 2 gz ut describes a wave propagating with no change in shape towards z 7oo 4 The Current Solution Consider one of the waves such as the forward wave77 propagating towards z 00 V Vz 7 wt From this we can show7 by differentiating7 that w 75 Equation 7 Also from equation 2 above 7L Equation 2 Equation 2 and equation 7 will have a common solution only if the two right hand sides are the same 1 3V 31 VuLI This can be rewritten using u from equation 5 Vg1 and and the current I of the forward wave is IVg and7 similarly for the backward wave 5 THE CHARACTERISTIC IMPEDANCE Z0 7 and the current I of the backward wave is I 7Vg Thus the general solution for both waves for the current I is I f7ut 79xut Equation 7 which can be compared with the earlier equation for the voltage V fz 7 ut g ut Equation 6 5 The Characteristic Impedance Z0 De ne the Characteristic lmpedance Z077 as the magnitude of the instantaneous ratio for either the forward wave or backward wave For the forward wave V 1 20705521 l l Ma For the backward wave 20 gsiiziii7i i7i i With this de nition of Z0 the voltage and current equations can be written V fz 7 ut g ut Equation 6 I 277 7 gig Out Equation 8 6 Speed u of Signals The Inductance per unit length L and Capacitance per unit length C can be calculated from Electromagnetic Theory The formulae depend upon the cross sectional shape ofthe conductors 7 Impedances of Actual Cables 0 Coaxial Cable EM theory says that a Coaxial Cable with inner rod having diameter a and outer tube having diameter b has 7 IMPEDANCES OF ACTUAL CABLES 8 7 27quot6 C 7 He and L 27f ln L le b From these7 u and Z0 can be obtained u L and Jue 7 A Z0 7 C For a vacuum7 of course7 Wile 60 Velocity of light and for polyethylene7 ET is about 4 and the speed is halved Velocit 0 li ht upolyethylene 7 47rze ln m glng u39uac Parallel Cylinders EM theory says that a pair of parallel conducting cylinders7 rods or wires7 With rod radius r and center to center separation D have C hale 13 L we 7 IMPEDANCES OF ACTUAL CABLES 9 From these u and Z0 can be obtained u and Z0 V 1712 6 ln m 240 ln 6M 1 xE Note that l in both cases the speed u is the same and depends only upon the medium In fact of course for all possible cross sectional shapes the speed is the same u 176 and if the c the speed of the voltagecurrent signal on the cable medium is a vacuum u is the speed of light N in both cases the speed and the Characteristic lmpedance depend upon logarithms of the ratio of two distances in the cross 0 thus a big transmission line can have the same impedance as a small transmission line if one is scaled in proportion from the other 0 For most lines it is not practical to vary the ratios 3 and much more than about 201 up to 101 Since the lug m 069 and ln1 10 m 23 the range of impedances is normally Within a moderate range about 20 ohm to 200 ohm Some typical values can be found for Z0 50 ohm cable With a polyethylene dielectric and speed u half of light 15 X 108 ms Use the speed equation from equation 5 of section 2 on page 4 and the characteristic impedance equation from section 5 on page 6 7 1 7 L u 7 7 and Z0 7 V By multiplying and dividing these equations we can get L and C L Q1 and C Z u on For the values assumed for Z0 and u 50 Ohm 3333 X 10 7 Henrymeter 15X108 m s L 333 X 10 9 Henrymeter 333 nanoHenrymeter 7 IMPEDANCES OF ACTUAL CABLES 10 OWlsxlowsm Faradm 1333 gtlt 1010 Faradm 1333 pFm Thus a foot of RG58 cable with Z50 ohm and uhalf of light has a capacitance of m 0305mft gtlt 1333pFm 40 pF 8 ELEVEN EXAMPLES 11 8 Eleven Examples To get a better feeling for the signi cance of the above consider 11 examples 1 Receiver Termination or Parallel Termination with R Z077 39 T Q7u Consider a transmission line which is terminated by a resistor to ground with resistance R which is equal to to characteristic impedance Z A signal is introduced by a module to the left at negative X The general solution for V and I from before is I 7 ut 7 91 utZ0 Equation 7 V fx 7 ut 91 ut Equation 6 At CE 0 the V and I must obey these 2 equations and ALSO obey the equation Ohm7s Law for the resistor V IR IZO These can only be consistent if the functions g 0 so that I 7 m 7 utgtgtzo V fx 7 ut Thus while there can be a forward wave there cannot be a backward wave Thus and forward wave cannot produce a backward wave There are NO REFLECTIONS This example with a resistor R Z0 is sometimes called using parallel termination 2 Any Resistance Value 8 ELEVEN EXAMPLES 9quot r 12 Now consider a similar signal introduced from the left but with a general value of R instead of requiring R Z0 Vfg Ie IfVIRthen VfgOR f g 1 25 7 92 f1 2 171 2 9 El Z Thus the forward wave causes a backward wave The backward wave is7 in general7 smaller and we call R the voltage re ection coef cient checks o fthe cable has Z0 50 Ohm and the resistor has R 100 Ohm then 71710075071 7a7f71005073 o If R Z07 the Voltage Re ection Coef cient R R ZO 0 as in the RZo preV1ous example i 2R Open Circuit Termination R 00 up coned If the cable is open circuited ie has no resistor then R 00 Ohm and using some sloppy algebra R f Q 1 0 00 The backward wave will have the same size and shape as the incident right forward wave Short Circuit Termination R 0 l r39j up If the cable is short circuited ie has zero resistance then R 0 Ohm and R 3 7 1 O O 8 ELEVEN EXAMPLES U 13 The backward wave will have the same size and shape as the incident right forward wave but will be inverted Joining one cable to another If one cable with characteristic impedance Z1 is connected to another long cable with characteristic impedance Z2 which is so long that signals have not had time to re ect from the far right7 then lets call the 2 functions on the left as xiut and 91 ut Since there is no re ection on the right7 the backward wave is absent Lets call the forward wave on the right as Fx 7 ut or F The equations for the left and right parts are Vfg VF L3 i 21 21 Zz We must have continuity of voltage at the boundary fgthenX0 Divide this trivial equation by Z1 f i F Z Z For continuity of current the currents on the two sides must be equal Add the last 2 equations to eliminate g f 7 1 1 7 Z 2 QZ HZ F d 22 F fZ2Zzi The factor 7 fozzl is called the Voltage Transmission Coef cient A challenge Prove that the re ected wave 9 here has the same formula as for the case of a simple resistor discussed before but with the notation changed slightly to have the cable with characteristic impedance Z1 being terminated with a resistor with resistance R Z2 6 Joining One Cable to three other cables 8 ELEVEN EXAMPLES consloler one Coaxlal oahle lolnrng 3 other long coaxial oahles with all lour oahles hayrng the same Charaotenstro lrnpeolanoe z Usrng the suhsorrpts f and y lor the orrgrnal anol a u u r r r the l t t three other cables the equatrons to be solved are Va aZIo V1 rVrZhZTrzh All a other long cablesquot wlll reoerye a transrnrtteol srgral whlch rs T tunes the orrgnal srgral V me e m 2 FiI The One Coaxlal cablequot wlll reoerye an rnyerteol xe ectlon whlch ls 72 trrnes the ongnal srgral 7231 E e N 7 When oloes a cable aot llke a Resrstorl r nsrnrssron hne aots exactly llke a reslstor to any orroult olrlyrng lt untll a re eotron the haokwarol wave y oan return As a ty boutUSo these t 150 mmns the srgnal oan propagate though a 15 meter oahle In about 10 ns an rts re eotron oan return 2 ns alter the srgnal was rst anected Thus suoh a transrnrssron hne am as a resistor to any external system drwing lt or about 20 ns How long must a oahle be to aot llke a resrstor lor 1 h 7 It a transrnrssron hne aots llke a resrstor to any external system olrryrng lt WHY DOES THE TRANSMISSION LINE NOT GET HOT Conslder chargmg up a 50 ohm Transmlsslon lme vla a 200 ohm reslstor from a battery lth voltage E Consldex the actlon when the swltch from the battery ls closed le connected Assume that the slgnals can propagate from one end to the other In a tlme T 8 ELEVEN EXAMPLES 15 2mm A z 5 DAM 12 41 a During the rst 2T7 the transmission line acts as a simple 50 ohm resistor to the battery7 switch and 200 ohm resistor Thus7 the junction point A rises as a step function77 to a voltage VA 20 350E 02 X E b During the rst T7 this signal a 02E step function propagates down the cable and meets the open far end B It re ects at the far end with a voltage re ection coe icient of R 10 and so a re ected 02E step function comes back Thus after time T7 the voltage at the far end rises to V3 forward wave re ected wave 2x02gtltE04E At the end of the rst 2T7 the re ected step function 02E meets the near end A Seen from the cable7 the resistance to ground for any voltage change is 200 ohm any resistance of the battery For all normal batteries7 the internal resistance is very small and much smaller than 200 ohm so we can neglect it Thus the signal meets 200 ohm to ground A O v Part of the signal 02E re ected signal is re ected again with a voltage re ection coe icient of H m o o l o l NIH mm 00 ll 0 o and so a re ection of a re ection7 now 06 X 02 X E 012 X E7 will be sent to towards the far end again A CL V At time 3T7 the re ection of a re ection7 012E7 will reach B and bounce there again with R 10 adding 2 X 012E 024E to the 04E already there and reaching 064E At time 4T7 the third re ection will reach A again and bounce with R 06 and so on A D V The result is a series of diminishing steps at both ends of the cable gradually rising up until both ends are nearly at voltage E of the battery 9 Source Termination or Series Termination with R Z077 12 6qu Dz v hr d 7 0 quot 8 ELEVEN EXAMPLES 16 The rst example with a resistor R Z0 is sometimes called using parallel termination While it has the advantage of causing re ections7 it has the disadvantage of requiring a steady current from the driving circuit to maintain any unchanging signal When it is important to minimize the power spent by the driver7 some folk use series resistors as shown above Note that we assume here that the driver has a low output impedance and the receiving module has a high input impedance Be careful7 the terms parallel termination and series termination are often mis used For example7 one can buy series terminations for SCSI busses on computers Although these little modules are plugged in series to the other modules on a SCSI cable7 the SCSI cable carries its own grounds and supplies and the module is really a parallel termination connecting each signal line via a suitable resistor to a voltage of about 3V The signal of voltage E injected on the left passes through a resistor with resistance R Z0 before it reaches the transmission line Thus for the rst few nanoseconds7 the signal on the cable is 05E This signal travels to the far end where it bounces with voltage re ection coef cient R 10 and the signal received by the receiving module is 2 gtlt 05E E The re ection travels back to the near end where it sees77 a total resistance to ground of R Z0 Here the voltage re ection coe icient R 0 and no further re ections occur Advantage of Source Termination 7 lower power load in the resistors and the drivers since there is no current for steady DC signals Disadvantage of Source Termination 7 if the transmission line has two or more receiving units7 then the unit near the end of the transmission line sees a single clean transition from 0 to E However7 a unit part way along the transmission line will see a voltage step from 0 to 05E then7 later7 a second voltage step from 05E to E The extra step will change the nal signal shape and may confuse a number of analog signal recorders Many digital circuits will misbehave if given an input which is half way between logical 0 and logical 1 Consider a sinusoidal wave form for both the forward wave and the backward wave with equal amplitudes and parameters V forward backward V asin27 Tm 7 wt asin2739Tm wt This can be changed using SinA Sin B 2 SinAB2CosA B2 to V a 2 sin2quotT coswt 9 CAPACITIVE TERMINATION 17 slglals are involved there slmply an alternatlve vo Srgnals llke these can be caus ry danc r end appear as stan ng w ves mplrtude ol the alternatmg voltage rs zero and at other posrtrons the amplrtude o t alternatlng voltage can be a appears to be no propagatlon Also the slgnal at any polnt rs age ed by srmply rellectlng a srnusoldal lnput lrom an open or me posltlons the 11 A puzzle Regard the Jumper cable you use to start a car wrth a llat battery lrom another car as a E rres perlect transmrssron lme wrth characterrstrc rmpedance z e 100 ohm wrth both batte havrng an rnternal resrstance ol about 3 D ohm How does the voltage at the at battery wry when you connect the Jumpers and how rs energy translerred to the llat battery 9 Capacitive Termination Consrder a transmrssron lme wrth a purely capacrtrve termmatron The transmrssron lme has an lmpedance Zn and the termrnatlon has a capacrtance CT to ground What wlll be the effect due to a step lunctlon wave In the cable meeting thls termlnatron h t l bl Assume We use the general solutron lunctrons we lound before m rm to descrrbe a mve travellmg towards posrtrve x and yeet to descrrbe a wave travellrng towards negatrve x The voltage lunctrons can be m a m m m and the correspondmg current lunctron rs then I 7 42 the step functlon f r m but we DO NOT YET KNOW ol the two functions we know the re ected lunctron got at For the step lunctrom the lunctron got a m ls say V0074 0when a0 tgt0 and In other words the step sr suddenly rlses m a step to the We label everythrng at the Termlnatron wrth the subscrrpt 1 At the capmtor wrth V007 Gwhen wrmlt gnal commg lrom lar away m the cable has the value v 0 then w a when 7 et 0 9 CAPA CITIVE TERMINATION 18 capacitance CT the current combination of the step fz 7 ut travelling towards X and its unknown re ection gXut travelling towards X will provide a current to charge the capacitor lf V0 and Io are the voltage and charging current at the capacitor7 then the two equations for the voltage and current of the two travelling waves are VTf9 IT fig 20 On the capacitor CT the charge is QT CTVT and the voltage rises due to the current TEA 7 9 d XE 7 ram 7 VT 7 f d EE 7 ram 7 VT Aftert0atx0fGandVVT andso 7Z010T VT 7 2G differential equation This has a solution for the voltage VT at the termination VT 7 2G Ire 233T 7 Proof Differentiate the tentative solution and get 2 7ZS 0K6 ZOCT Then sub 7 stitute the differential equation to eliminate K6 200T This gives 2 7 T 7 2G which satis es the differential equation Find the value of K At t 07 VT 7 2G60 K 0 7 2G K 1 The general solution after t 0 is VT 7 2G 72Ge VT 1 7 5235mm Since V f g and after t 07 we still have f G g VT 7 G g 1 7 26639 and in the general case with z lt 0 7 xut2 gz ut 1 7 26 OCT G 9 CAPACITIVE TERMINATION Usmg the srghal speed a and cable rmpedahce Zr where L rs th mductahce per umt length per umt length clta hce and wave traveumg towards the termmatrom and so 2 The general sohrtroh for gwm before t 0 is g 0 From the two rtems above the re ected waveform rs g0 m whexe 1 1f eet gt 0 thehgeet 17 225mg 2 1fwmlt 0 Lhengwut 0 The re ected wave due to a positwe step G starts wrth a h gative step to VG and is followed y ah expohehtral rrse to G Ah osc at any pomt wru see an addrtroh of the rmtral step wave m a m and re ected mv t The rmtral step wave of M at an arbrtrary e wru show the trace on the scope ruoscope e m a wru show the trace on the Some folk remember the shape of thrs re ected mve m Mt by saymg that 9 CAPACITIVE TERMINATION 20 1 the leadmg edge ol the step rs lormed lrom the hrgh lrequehores bemg m phase and these see the oapaortor as a short orrourt gwmg a leadmg hegatwe re eotroh 2 the trarlmg level ol the step rs lormed lrom the low lrequehores bemg m phase and these Se quot the oapaortor as an open orrourt gwmg a trarlmg posrtwe re eotroh The lollowlhg examples show how the lorward and bukwaxd wayelrohts may be seen 11 the scope rs used to look at the voltage at the termmatroh at w o a s p then the scope wlll show the sum ol the two preyrous traces but the two sharp trahsrehts at e or t 0 wlll cancel gwmg Comblned slglal m a ut ye 4 or scope 0 4 to a step then the two sharp trahsrehts wlll appear at dlf ferent trmes ahd wlll NOT cancel The scope may show Comblned slglal m a ut ye 4 or scope AmmJ39 quot39eup 39 Cohsrder a short pulse at 7 ut mstead ol a step At a long drstahoe belore the Lermlr hatrom the mput pulse may appear lohg belore ahy re eotloh can axnve as Short Pulse srghal at 7 ut or scope at large hegatwe e 9 CAPACITIVE TERMINATION in i i it i i Lethe in Li H at a negative due to f0 7 m being a positive going pulse then the scope may show Combined sigiai m 7 m 900 m on scope I the input pulse is lengthened so that long before any ie ections mun the tme looks h no u t I om i then the scope Win Show the Combined signal we 7 m 900 m 9 CAPACITIVE TERMINATION 10 TYPES OF TRANSMISSION LINES 23 10 Types of Transmission Lines Brief Notes Follow Any pair of parallel conductors which have a cross and shape which are constant indepen dent of distance and are far from other conductors or any pair of parallel conductors which have a cross and shape which do not change rapidly and retain the same ratios of dimensions independent of distance along the pair and are far from other conductors If one conductor does not fully enclose the other conductor eg twisted pair7 then some of the EM eld will gradually radiate away and the signal will show a steady attenuation and exponential decay which is frequency dependent Examples H An example could be a single cylindrical conductor wire above a conducting plane with the wire diameter changing slowly and the distance between the wire center and the conducting plane being kept proportional to the wire diameter E0 Twisted pair wire 7 often with Z0 120 ohm C40 60 Hertz High Voltage Power Lines across the country Strictly speaking these are usually a combination of 6 interacting transmission lines with the phase at 60 degree intervals The speed of power transmission is very close to that of light um Velocity of light Ethernet coaxial cables all around Sterling Hall and Chamberlin Hall Both have Z0 50i 1 ohm There are two types yellow thick ethernet77 cables have large conduc tors to have minimum resistance attenuation7 black RG58U cables thin ethernet77 are thinner for least cost 4 9 Electrical Transmission Lines are not Waveguides or Light Guides 03 The 60 Hertz power lines across the country are imperfect There is some leakage or radiation of the 60 Hertz eld since the conductors are open There is a nancial incentive to arrange the phases of the 6 conductors so that the minimum radiation occurs There is other attenuation due to slight breakdown on the insulators and corona in the air On the average about the USA7 there is about a 10 power loss 5 Besides the loss of power by radiation7 there is another problem of open conductors 7 unwanted coupling to other circuits This is not a problem at the lower frequencies but can be awful at high frequencies 11 IMPERFECTIONS OF TRANSMISSION LINES 11 1 E0 24 Imperfections of Transmission Lines Radiation from open conductors mentioned above Use Coaxial transmission lines if these are practical Skin Effect losses These are due to the actual currents being con ned by the skin effect to the surface layers of the conductors EM Theory Most current occurs within a distance skin thickness77 6 where 6 4 w ia where w 27139 X frequency Consider 3 examples 0 For copper which is pure and has been annealed at 20 C conductivity Ucu 580 gtlt 107 mhom M Mair 14 Hm w 27139 gtlt frequency giving 6 00661 722zii37tz Starting at the surface of the conductor the current density J drops to i at depth 6 and a total of 1 of the total current exists at depth beyond 56 For example in thick copper lines carrying power at 60 Hertz 660 Hm Cu 03 00085 meter 85 mm Thus at 60 Hertz only the outer centimeter of the 8 or 10 cm diameter lines carry most of the current 1 Resistivity The thick aluminum power lines used for high voltage distribution have a conduc tivity of HA 377 gtlt 107 mhometer 2 00105 meter 105 mmll wWAz 660 Hertz Al Why is aluminum rather than copper used for our National power sys tem Calculate the resistance of a 10000 km line or rod for both materials assuming that the effective cross sectional area used in each line is 27Tr6 and the radius is r 100 mm Also calculate the cost of each line or rod using Aluminum conductivity 0A1 377 X 107 mhom density 2708 legms costkg 770 kg Copper conductivity 00 580gtlt107 mhom density 2567 legm3 costkg 1540 kg Do the centers of the heavy power lines have much use 00661 meterhe39rtz m Since the skin depth is frequency dependent the effect is to attenuate the high frequencies of the signal more rapidly than the low frequencies of the signal Thus the signal CHANGES lTS SHAPE during the propagation due to the skin effect For copper lines at frequencies near 60 MHZ 6 85 microns 11 IMPERFECTIONS OF TRANSMISSION LINES 25 111 Remedies for Signal Loss On copper at high frequencies7 the skin depth 85 microns at 60 MHZ can be serious H The skin depth of 6 85 microns may have surface cracks7 surface oxide7 and other defects which increase its resistance and further attenuate the signal A good conductor has higher 039 and smaller 6 and although the depth 6 7 varies only as 7 it is still best to use high 039 to reduce the signal loss Therefore at high frequencies7 use gold or silver plating Why is gold used upon some critical circuits when copper costs less and copper is the better conductor A few conductivities are listed in decreasing order 0A9 107 UCu X on 107 mhometer tor of Al cables TV cables Avoid UAu 107 0A1 power lines E0 Use smooth polished solid conductors to minimize the distances for the currents near the surfaces Unfortunately7 to gain a exible cable7 often the central conductor is multistranded and the outer conductor is made of tinned braided thin copper 3 Use big conductors large 7quot ithe conducting cross sectional area A is about A m 27139 gtlt 6 although the resistance and power loss at a given current I7 frequency f and conductivity 039 is only proportional to The cost of the central conductor of a coaxial cable transmission line7 of course7 is proportional to er 4 The cross sectional conducting area is about A 27116 27mquot AM The resistance per i i 1 N 1 N 1 7 1 unit length R is about R a N 027mg N mug2L 7 dgg To reduce the power loss per unit length7 2R 21 7374 g5 use good conductors high 0 although 039 is in a square root and cannot help much 5 For high powered long distance systems7 use cheap metals 6 For high powered systems to deliver a particular power level P IV7 increase to a really high voltage V to reduce I and reduce the power loss 7 For high powered long distance systems7 use metals with a high tensile strength to span between the pylons and so need fewer pylons This leads to a compromise between 1 1 H EQUATIONS FOR A LOSSY TRANSMISSION LINE 26 conductivity 039 and tensile strength lf practical in each wire or rod carrying current use a high tensile inner wire surrounded by high conductivity but weaker outer wires Then worry about corrosion Beware of damaged coaxial cable causing unwanted re ections and reductions in the forward signal Crushed coax has a lower Z0 and the lumpy dielectric can change Z0 7 both cause re ections Speeds in some cables can be SLOW Even if no magnetic materials are used ie p no then u f If necessary use a foam dielectric with e m 1 giving u close to the speed of light 0 Minimize the resistance of the outer conductor 1 Minimize electrical breakdown Choose your connectors carefully Even the coaxial connectors often have a Z0 which is different from that of the cables and cause a pair of re ections which have the same shape and opposite sign but do not cancel because they occur at slightly different times Noncoaxial connectors often have short links with relatively high characteristic impedance and these impedance transitions can cause a number of upright and inverted re ections which are close together but not cancelling 3 For small signals being sent over long distance systems use optical bers 12 Equations for a Lossy Transmission Line In section 2 before on page 2 we began to consider Lossless77 transmission lines However sometimes we must use transmission lines with the series resistanceunit length being non zero and the insulation being non perfect with a parallel conductance of G being non zero A useful simple reference is Theory and Problems of Transmission Lines77 by Robert A Chipman Schaum Outline Series 1968 We will continue to use z for the distance along the transmission line whereas Chipman uses 2 for this Our x will avoid confusion between the distance and the characteristic impedance Z Consider a transmission line with distributed resistance R ohmunit length and parallel conductance G mhounit length 12 EQUATIONS FOR A LOSSY TRANSMISSION LINE J v 7F 2 As before consldex one such cell conespondlng to the componen39s between posmon at and poeltlon e Ae along the transmission hne VHW 4 The x slstance per umL ength R Includes both the resistance of the mug conductor and that contnhute d can algnl cantly to the total R In the theory o propagation we are only concerned W39th the total resistance per umL length Inner outer resistanceunit length and call the total as Rm 2 a 2 a 5E E39 n E E39 The equatlons for each element Aw of length ale oeAet 7 oet 73m aet 7 me 343 ae Jam 7 aet 7 am meat 7 cm 87537 ledmg by Aw and 1ettlng Aw 7 on we get slmllax equatlons to equatlons 3 3 and 3 4 of C 1 man 72w a L 1 lt9 a 7 aoet7clt j This can be wntten ln the most snnple ohm remembering that 1 and a are lunctlons of e and it an go a 7 7R2 7 L3 12 EQUATIONS FOR A LOSSY TRANSMISSION LINE 28 3139 31 E 7G0 7 05 These equations are similar of course to the equations 2 and 1 which we obtained upon page 3 but now include the non zero R and G As before we can substitute for 239 in one equation from the other to obtain a differential equation in 1 Similarly we can obtain a differential equation for 239 Log 7 LG RO RG2 L037 LG Bag RGZ39 Notes 1 These equations do not have the simple form and simple solution of the equations for a lossless line E0 Although we have assumed that the L C R and G are constants at high frequencies these can be frequency dependent For example they may be in uenced by the skin effect which is frequency dependent Never the less if we consider just one Fourier component of a signal we can obtain an understanding then by combining the Fourier components of any particular signal we can understand how a particular signal propagates 9 Although the equations above for 1 and 239 are identical they will usually have different boundary conditions and so will have different solutions We can replace ixt by R6Vxe7m and z39x t by R6I6 dt d2 R ijW 2 7G ijV From these we get two second order differential equations similar to Equations 315 and 316 of Whitman 7 R ijG wow 0 7 13 ijG Hum 0 We de ne y where 72 R ijG ij 1172 7 72V 0 i 72 0 The solutions for these are voltages and currents with an angular frequency w V Vle w V2 12 EQUATIONS FOR A LOSSY TRANSMISSION LINE 29 IQ 116 126 where V1 V2 1 and 2 are arbitrary constants and where 72 R ijG ij De ne Oz and 3 as the real and imaginary parts of 7 so that v a w ltR ijgtltG HMO Then the solution for V is Wyn eryleemzw w eje Zwmw m lz7 t ejeil e we j m j92l25Dm j m Using vzt Ra HM this becomes ML Reej tej91l e we j m 6792l3926mej w 11z7t Me Reejwquot m91 l3926D mReejW m92l corresponding to a forward X increasing wave a backward X decreasing wave The parts of this equation can be identi ed V1 is an arbitrary amplitude for a signal propagating in the X direction 0 6 is a real coef cient diminishing in the forward or X direction and describing the attenuation of the signal as its moves in that direction wt 7 x is a term describing a wave propagating in the X direction with phase speed u since ift is increased by At then the value is unchanged if the z is increased by Ax At The group speed is 91 is an arbitrary phase angle 0 e wt H is a sinusoidal wave of angular frequency w propagating in the X direction with phase speed and group speed Z V2 is an arbitrary amplitude for a signal propagating in the X direction 6 is a real coef cient diminishing in the backward or X direction and describing the attenuation of the signal as its moves in that direction wt x is a term describing a wave propagating in the X direction with phase speed u 7 since if t is increased by At then the value is unchanged if the z is decreased by Ax At The group speed is 7 02 is an arbitrary phase angle 0 57 t 92 is a sinusoidal wave of angular frequency w propagating in the X direction with phase speed u 7 and group speed 73 13 LOSSY LINE WITH NO REFLECTIONS 30 o vzt is the instantaneous voltage at position x and time t 13 Lossy Line with N0 Re ections If only the forward wave exists the equations become V Vle me j z I Ile we j z 14 Attenuation if line is slightly lossy Oz and B may be written a j R ijgtltG ij a j ijLoigwiL 17C 1 i 1 i R C Now use a Taylor expansion of each 2 as power series of m and m a j jw LO 5321WLL 57 1 5lt1 1gtjw 02 57 1 a j jmLO 7 ll 1 jc2 1 juJL L and C are usually log functions of ratios of cable radii or other cable dimensions and so cannot be very high or very small However if R and G are small but non zero OR the frequency f 1W is very high so that WiL ltlt 1 and ltlt 1 then these two Taylor series can be simpli ed by G jw C39 neglecting the higher powers of WiL and a m ijOilt 092 1 Equating the real parts 04 an LC a a Equating the imaginary parts j jWLO12Zl n w LO17 4550 15 CHARA CTERISTIC IMPEDANCE OF A LOSSY TRANSMISSION LINE 31 B x LO LO neglecting the product of WiL and E From 6 we can obtain the phase speed u and group speed 3 Make a check of these equations for the lossless transmission line If R G 0 then 04 0 0 and so as expected there is no attenuation of either the forward wave or backward wave w LG170 w LG The full equation can be written for the lossless case M957 t We amRe67 Wti m9l W5DrwR5ejwt z92 becomes vx t llReJLEjW HQO WReej t w92l 15 Characteristic Impedance of a Lossy Transmission Line Z 7 13ij we get I Gij This is the characteristic impedance and can be split into real and imaginary parts i R LIL ZO G u0 Zo R0 on If the transmission line is only slightly lossy then R and G are small and for most purposes we ignore X0 16 Heavyside Distortionless Lines lf then it can be shown that the speed of propagation is the same for all angular frequencies in and the shape of the signal with respect to position X remains constant although it gradually gets smaller with the attenuation Unfortunately this idea has little practical use because the R L G and G are suf ciently frequency dependent for the above theory to be insuf cient 17 THREE EXAMPLES 32 17 Three Examples Electrical parameters from the Alpha and Belden handbooks and dimensions from the CRC Handbook of CampP pg E117 L is obtained from L Z20 and L Z speed 39 Z ohm Ethernet Yellow Non Plenium 9059B neg negligible RG58U is a common coaxial cable used very frequently in all sciences In the next two sections7 we will consider the example of RG58U at frequencies of 07 10 MHZ and 1000 MHZ The speci cations of a given type of cable vary from manufacturer to manufacturer They may have slight differences multistranded or solid central conductor7 single or double outer conductor7 exible of rigid outer plastic covering7 etc Some have inconsistent values due to varying roundoff Some have metric measures and other have inches As a result7 comparisons are dif cult and the customer should be careful 18 Attenuation at DC and Low Frequencies Consider a coaxial cable and signal for which the frequency is suf ciently low that the skin depth is larger than the diameter d of the central wire 0 The resistance per meter at DC ie when frequency 0 is 18 ATTENUATION AT DC AND LOW FREQUENCIES 33 RDC 0821 mm2 580X107 mhomiOBZSleO G m2 039209X1071 Ohmm 0390209 Ohmm This agrees with the CRC handbook of 0amp1 pg F 117 of R2095 ohmkm for annealed copper at a temperature of 20 C o The inductance per unit length is L Z20 502 gtlt 853 gtlt 10 12 Hm L 2500 gtlt 853 X1042 Hm 0213 gtlt 10 6 Henrymeter o The signal speed is 078 gtlt c and so u 4 078 gtlt 3 gtlt108ms i i i L i o The impedance Z is approx1mately V i 50 ohm o The leakage as in most modern cables7 is negligible7 so set the leakage conductance G 0 We can use the speed7 the impedance and G 0 to obtain the attenuation The attenuation factor 04 is then i 2 E 2 Substitute Z g a by he a 04 Chipman pgs 497 55 amp 57 For RG58U with no skin effect ie at DC or low frequencies 04 meter 02095 gtlt 10 3 meter From 04 for any cable7 we can calculate the number of db loss of that cable over a length x by db loss 10 gtlt bylaw Power 1 km Vinput 2 Vat 1 km db loss 10 gtlt loglo thkm Vin m db loss 20 gtlt l0910 a p db loss 20 gtlt l0910e db loss 20 gtlt 101 095 db loss 20 gtlt This is often stated simply as db loss 86858 ow The attenuation of the voltage signal7 due to only the central conductor7 on RG58U over a distance of z 1 km 1000 m is 1 9 ATTEN UATI ON AT HIGHER FREQ UENCIES 34 i 02095x10 3 mxiooo m db loss 7 20 gtlt 23026 db db loss 1820 db for the 1 km length at zero and low frequencies 19 Attenuation at Higher Hequencies 191 Skin Effect Loss The loss rate calculated above is for low frequencies in which all of the central conductor conducts the signal At the higher frequencies7 the skin effect reduces the cross section of the copper which conducts and so the attenuation is increased a gel 7 1 Oz 7 2g 04 See Chiprnan pgs 49 amp 85 Since for all frequencies above some lower lirnit7 the skin effect restricts the current to approximately the skin depth of the copper 1 6 2 00661 meterbe39rtz tuba 1 F regency 7 we can estimate the approximate resistancerneter at these high frequencies For copper lines at frequencies near 10 MHZ7 6 W 209 microns 1f the di ameter ofthe central conductor is d then7 for most cables and frequencies above 1 MHZ7 6 ltlt d With this assumption then the conducting region is equivalent to that of a thin shell with a depth of 6 This relation can be proved and is called the Skin Effect Theorern See Chiprnan pgs 78 amp 85 From this7 we can estimate the effective resistanceunit length Rf at a particular frequency f Since the periphery is 7rd7 the cross section of 7rd X 6 Rfk l c7gtltcmducti0n cross sectional area N 1 Rf N ax7rdgtlt6 l c7gtlt7rdgtlt N 1 w RfN 39lEg Note that7 at the higher frequencies7 Rf cc V5 cc V7 Rf 2 Tm We estimate the attenuation due to only the central conductor of RG58U at 10 MHZ and 10 GHZ We do not have enough data to estimate the attenuation due to the outer conductor 1 9 ATTEN UATI ON AT HIGHER FREQ UENCIES 35 RG58U at f 10 MHz N 1 27rgtlt10gtlt106 47r R10 MHZ N 7rgtlt1024gtlt10 339 2gtlt580gtlt10739107 Ohmm N 1 10x106 A R10 MHZ N 1024x10339 580gtlt10739107 Ohmm N 1 40 R10 MHZ N 1024x10339 580x108 Ohmm R10 MHZ m W12626 gtlt 104 ohmm R10 MHZ 2564 gtlt10 1 ohmm R10 MHZ 02564 ohmm From this we can estimate a for RG58U at 10 MHZ iii 7 L a7 397rd39 20 NIH a 3 Igta a x 255604 0002564 meter For 10 MHZ signals in an RG58U cable 1 km long db loss 20 gtlt 2026 db 0002564 m 1000 m db loss 20 gtlt db db loss 223 db for the 1 km length at 10 MHZ RG58U at f 1000 MHz 10 Gsz N 1 27rgtlt10gtlt109 47r R1 GHZ N 7rgtlt1024gtlt10 339 2gtlt580gtlt10739107 Ohmm R1 GHZ m mama gtlt 103 ohmm R1 GHZ 2564 gtlt ohmm From this we can estimate a for RG58U at 10 GHZ R 04 a m 3523 002564 meter For 10 GHZ signals in an RG58U cable 1 km long db loss 20 gtlt db N 002564 mx1000 m db loss N 20 gtlt 23026 db 1 9 ATTEN UATI ON AT HIGHER FREQ UENCIES 36 db loss 223 db for the 1 km length at 10 GHZ 192 Dielectric Loss This attenuation is due to the dielectric absorbing energy as it is polarized in each direction This loss is usually small but becomes more signi cant at the higher frequencies The effect is that our conductanceunit length becomes non zero at the higher frequencies The difference of the phase of the current AI between the central conductor and the outer conductor and the phase of the voltage changes from being to g 7 e and data sheets can give a loss factor of Tame77 to give the relation Unfortunately7 we dont have convenient data sheets for the dielectric of RG58U The usual circuit boards made from epoxy glass have a signi cant attenuation due to the dielectric loss while polyethylene has a modest loss 193 Radiation Loss If the conductors form a tight electromagnetic system with the outer conductor having a thick ness greater than about 56 5 times the skin depth then external EM elds will be small and the radiation loss is negligible If the outer conductor is a loose braid7 the the external EM elds will cause radiation away from the cable and will cause attenuation If the transmission line has two open conductors with neither shielding the other7 then the external elds are minimized by twisting the two conductors forming a twisted pair77 line However7 even if the pitch of the twist is much smaller than the predominant wavelength of the electrical signal7 some radiation occurs and the signal is attenuated For this reason7 be careful when using twisted pair lines In addition such twisted pair lines can easy receive electrical noise from other circuits which have high current or voltage transients 194 Actual Attenuation in Cables We can compare the results of these approximate calculations with actual attenuation mea surements From the Alpha handbook7 o for RG58U pg 273 the loss at the relatively low frequency of 10 MHZ of RG58U is 12 db100 ft 394 db100 meters 394 dbkm 1 9 ATTEN UATI ON AT HIGHER FREQ UENCIES 37 The loss at the relatively high frequency of 1000 MHZ is 180 db100 ft 59 db100 meters 590 dbkm o for Thick Ethernet pg 233 the loss at 10 MHZ is 20 dbkm Obviously7 the estimate of the loss due to only the skin effect of the central conductor is about half of the db of the actual loss Probably the remaining loss is due to the imperfect outer conductor and to the dielectric loss7 o The central or inner conductor has a smaller periphery but usually has a smooth surface and high conducting metal 0 The outer conductor has a smaller periphery but is usually rough made of braided wires for exibility with only tinned surfaces and imperfect contacts The majority of the cable cost is due to the outer metal and so manufacturers may skimp a little on the conductivity of the outer conductor iNegatnmaFeedback 19hysks 623 Murray Thompson sept211999 Contents 1 Introduction 1 2 Negative Feedback With a Difference Ampli er 2 3 Negative Feedback With an Operational Ampli er 3 4 Summary 4 1 Introduction We will consider the use of a difference ampli er We will assume that this difference ampli er has some common attributes o relatively large negative voltage gain A o relatively high input impedances o relatively low output impedance An operational ampli er is similar but has extreme values An operational ampli er has N NEGATIVE FEEDBACK WITH A DIFFERENCE AMPLIFIER 2 o extremely large negative voltage gain A o extremely high input impedances 0 very low output impedance 2 Negative Feedback with a Difference Ampli er Place two resistors r and R about a difference ampli er and consider the overall characteristics of the combination Consider the current 239 in the feedback resistor R Because the input impedance ofthe difference is very high the current in the series resistor 7 will be the same as that in the feedback resistor namely 239 Give 239 the polarity of owing from input with U to the output with V Since the inverting input has an extremely high input impedance it can take no current Thus the combination of r and R acts as a potential divider generating a voltage Y from voltages U and V and the voltage Y at the inverting input is that of the join in the symmetric form r R EI O EI r UiV ZT R YV239R YVU VR TR OJ NEGATIVE FEEDBACK WITH AN OPERATIONAL AMPLIFIER 3 Y VrVIiigR7VR UR VT Y i TR TR But if the ampli er gives an output V due to an input Y and large gain A then V7AY70 Y Substituting this formula for Y into the earlier equation for Y z E Vr A 7 7 1 39r 7 UR V 7 71 7 TR 7V rRrA UAR V 7U 7I f39rA if R V WW Since we have assumed that the voltage gain A of the difference ampli er is relatively large this formula can be simpli ed V U By choosing a high gain difference ampli er we gain three clear advantages H We can make the ampli cation of the overall system be nearly fully dependent on the resistors and nearly independent of the ampli er gain For example if the gain A of the difference ampli er were A 2000 for small signals and A 1000 for large signals the choice of r 1k ohm and R 30k ohm would give an overall combination with a gain of 7 730 for signals of all sizes E0 The voltage gain ofthe combination is independent of production variations temperature variations and DC power levels in the transistors and is dependent only upon the ratio of two resistances It is easy to make integrated circuits with resistances if we are not concerned with precise resistances but want to control only the ratios of resistance values 00 The calculation of the voltage gain of the combination is made to be really simple 3 Negative Feedback with an Operational Ampli er Now consider the use of an Operational Ampli er in this mode with typically A 10 000 000 or more and an extremely high input impedance at the inverting and non inverting inputs 4 SUMMARY 4 typically Zm 109 ohms to 1012 ohms We can consider the limit of the formulae for V and Y as A a 00 V U f We can also consider the voltage at the inverting input of an Operational Ampli er From above R V Vii Ef e In the same limit asAaoo we get 7 R U Tl YirquRH R R iUR Y l Rfjw Rl Y0 Thus the negative feedback and grounded non inverting input causes the inverted input to act as a virtual ground 4 Summary For a combination using an Operational Ampli er the following effects including the 3 men tioned above are given by negative feedback Any difference ampli er with negative feedback gains these effects to a lesser degree H The ampli cation of the combination is less than one might expect from the raw transis tors E0 The voltage gain becomes a very stable 7 and is dependent only upon the ratio of two resistances The fabrication has been made simple 9 Both Operational Ampli ers and resistors are cheap 7 It does not matter if the operational ampli er or general difference ampli er has been designed with a number of approximations such as assume 6 is large77 or neglect the internal resistances re and 77 The voltage gain of the combination will still be 7 9 We can make the ampli cation of the overall system be nearly fully dependent on the resistors and nearly independent of the ampli er gain It is relatively easy to select or adjust resistors and so gain a desired ampli cation Thus the negative feedback overcomes an imperfection such as that due to B being de pendent upon the signal size ln other words negative feedback can make the amplifying combination amplify with good linearity 4 SUMMARY 5 6 5 00 p H H The voltage gain of the combination is independent of temperature variations in the transistors The voltage gain of the combination is independent of production variations in the tran sistors The voltage gain of the combination is independent of DC power levels in the transistors The voltage gain of the combination is independent of noise and other uctuations upon the DC power levels in the transistors The voltage gain of the combination is independent of variations in any voltage offsets in the transistors It is easy to integrate a ampli er with negative feedback since in such a circuit7 we are not concerned with precise resistances but want to control only the ratios of resistance values The calculation of the voltage gain of the combination is made to be really simple We can use the virtual ground77 for a number of other purposes in electronic systems It can be shown that this virtual ground77 has a relatively low impedance An Operational Ampli er combination can amplify DC signals with a precise gain In an AC or voltage transient system7 the full analysis with complex impedances gives the very simple form 7 It is practical to add capacitances and inductances so that the input signal is integrated or differentiated precisely making it possible to make analog computers The high gain operational ampli er must not use any internal coupling capacitors If the non inverting input is tied to a point with voltage lgmm then the inverting input becomes a virtual voltage level with voltage lgmm It can be shown that this virtual lgmm has a relatively low impedance It can be shown that the ouput impedance of the combination with negative feedback gives an output impedance which is lower than that of the operational ampli er by itself Beware of positive feedback Physics 623 Dlgltal to Analog and Analog to Dlgltal Convers1on Nov 191999 1 Purpose 0 To examine some of the techniques used for analog to digital and digital to analog conver sion 0 To illustrate one application of electronics used to interface instrumentation to digital com puters 2 Discussion The Integrated Circuits you need are MC1408 7 an eight bit R QR77 DAC which accepts TTL inputs 74193 7 a synchronous 4 bit UpDown TTL counter TlL 311 7 an LED Hexadecimal Hex display Integrated Circuit LM311 7 an Analog Comparator and the 7400 7 a quad NAND TTL gate In this experiment we will use an IC digital to analog converter DAC to create an analog voltage level proportional to a number held in a TTL digital register In order to obtain an Analog to Digital ADC function7 the DAC must be placed in a negative feedback loop The analog voltage generated by the DAC is compared to the unknown analog voltage and the number in the register adjusted7 depending on the difference7 until the difference is less than the voltage corresponding to one bit 3 Procedure H Wire a single 74193 TTL 4 bit counter with a TIL 311 7 segment display and toggle it with one of the debounced spring loaded switches on the bread board panel Verify that the counter is sequential over the 16 step range from zero to F and that it can count both up and down E0 Connect two 74193 units together to make an 8 bit counter with two display readouts The appropriate connections are shown in Fig 1 L 5V R High to clear7 Low otherwise CLK UP and Use one or the other to clock7 CLK DOWN the unused one must be Fig 1 8 bit Binary Ripple Counter 3 Now wire up the MC1408 DAC as shown in Figure 2 and connect the input data lines to the output of the 8 bit counter Run the entire system as a DAC7 driving the counter with the pulser The DAC will give approximately 0 4 Volts over the full 8 bit range 1 5V Vout VEE 1N4448 13 vcc 5V 2 GND 3 VEE 15V Open Fig 2 Connections to an MC1408 Digital to Analog Converter 4 If a comparator circuit is added7 as shown in Figure 37 you have a tracking analog to digital converter ADC The LM311 cornparator cornpares this with the DAC output and stops the clock when the comparator changes state Note that a NAND gate must be added Add two TlL 311 Hex displays to get a decimal two digit readout See Fig 4 Vin o gt 4V Fig 3 Analog to Digital Conversion Circuit 5 Determine the response characteristics of the circuit by measuring the counter output as a function of Vm Note that the and 7 input pin labels ofthe LM311 analog comparator are reversed from those ofthe non inverting and inverting 7 inputs of the 741 operational ampli er 03 This circuit will track only towards an increasing Negative Voltage Can you modify this circuit so that it will track in both directions 5 This modi cation introduces a problem called hunting where the count changes back and forth by one or two counts Can you think of a way to stabilize the display to make it easier to read I EN Law Tc Read If 0 us iiiqi to blank i Low otherwise 3 3 Fig 4 TlL 311 Hexadecimal Display 00

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