Linear Circuits Analysis. Superposition, Thevenin /Norton Equivalent circuits

Transcription

1 Lnear Crcuts Analyss. Superposton, Theenn /Norton Equalent crcuts So far we hae explored tmendependent (resste) elements that are also lnear. A tmendependent elements s one for whch we can plot an / cure. The current s only a functon of the oltage, t does not depend on the rate of change of the oltage. We wll see latter that capactors and nductors are not tmendependent elements. Tmendependent elements are often called resste elements. Note that we often hae a tme dependent sgnal appled to tme ndependent elements. Ths s fne, we only need to analyze the crcut characterstcs at each nstance n tme. We wll explore ths further n a few classes from now. Lnearty A functon f s lnear f for any two nputs x 1 and x 2 f( x 1 x 2 )= f( x 1 ) f( x 2 ) Resste crcuts are lnear. That s f we take the set {x } as the nputs to a crcut and f({x }) as the response of the crcut, then the aboe lnear relatonshp holds. The response may be for example the oltage at any node of the crcut or the current through any element. Let s explore the followng example. 1 R 2 KVL for ths crcut ges Or 1 2 R = 0 (1.1) 1 2 = (1.2) R 6.071/ Sprng Chanotaks and Cory 1

2 And as we see the response of the crcut depends lnearly on the oltages 1 and 2. A useful way of ewng lnearty s to consder suppressng sources. A oltage source s suppressed by settng the oltage to zero: that s by short crcutng the oltage source. Consder agan the smple crcut aboe. We could ew t as the lnear superposton of two crcuts, each of whch has only one oltage source R R 2 The total current s the sum of the currents n each crcut. = = R R 1 2 = R Whch s the same result obtaned by the applcaton of KVL around of the orgnal crcut. (1.3) If the crcut we are nterested n s lnear, then we can use superposton to smplfy the analyss. For a lnear crcut wth multple sources, suppress all but one source and analyze the crcut. Repeat for all sources and add the results to fnd the total response for the full crcut / Sprng Chanotaks and Cory 2

3 Independent sources may be suppressed as follows: Voltage sources: = suppress short =0 Current sources: Is =Is suppress open = / Sprng Chanotaks and Cory 3

4 An example: Consder the followng example of a lnear crcut wth two sources. Let s analyze the crcut usng superposton. Is 1 2 Frst let s suppress the current source and analyze the crcut wth the oltage source actng alone. 1 2 So, based on just the oltage source the currents through the resstors are: 1 = 0 (1.4) 2 = (1.5) Next we calculate the contrbuton of the current source actng alone Is Notce that R 2 s shorted out (there s no oltage across R 2 ), and therefore there s no current through t. The current through s Is, and so the oltage drop across s, 6.071/ Sprng Chanotaks and Cory 4

5 1= Is (1.6) And so 1 = Is (1.7) 2 = (1.8) How much current s gong through the oltage source? Another example: For the followng crcut let s calculate the node oltage. Is Nodal analyss ges: or Is = 0 (1.9) = Is (1.10) We notce that the answer gen by Eq. (1.10) s the sum of two terms: one due to the oltage and the other due to the current. Now we wll sole the same problem usng superposton The oltage wll hae a contrbuton 1 from the oltage source and a contrbuton 2 from the current source Is / Sprng Chanotaks and Cory 5

6 1 2 Is And 1 = R 1 R 2 R 2 2 = Is R 1 R 2 (1.11) (1.12) Addng oltages 1 and 2 we obtan the result gen by Eq. (1.10). More on the characterstcs of crcuts. As dscussed durng the last lecture, the characterstc cure s a ery good way to represent a gen crcut. A crcut may contan a large number of elements and n many cases knowng the characterstcs of the crcut s suffcent n order to understand ts behaor and be able to nterconnect t wth other crcuts. The followng fgure llustrates the general concept where a crcut s represented by the box as ndcated. Our communcaton wth the crcut s a the port AB. Ths s a sngle port network regardless of ts nternal complexty. In Vn A B If we apply a oltage across the termnals AB as ndcated we can n turn measure the resultng current. If we do ths for a number of dfferent oltages and then plot them on the space we obtan the characterstc cure of the crcut. For a general lnear network the characterstc cure s a lnear functon = m b (1.13) 6.071/ Sprng Chanotaks and Cory 6

7 Here are some examples of characterstcs R In general the characterstc does not pass through the orgn. Ths s shown by the next crcut for whch the current and the oltage are related by or R = 0 (1.14) = R (1.15) R /R Smlarly, when a current source s connected n parallel wth a resstor the relatonshp s = Is (1.16) R open crcut oltage Is R Is RIs short crcut current 6.071/ Sprng Chanotaks and Cory 7

8 Theenn Equalent Crcuts. For lnear systems the cure s a straght lne. In order to defne t we need to dentfy only two pnts on t. Any two ponts would do, but perhaps the smplest are where the lne crosses the and axes. These two ponts may be obtaned by performng two smple measurements (or make two smple calculatons). Wth these two measurements we are able to replace the complex network by a smple equalent crcut. Ths crcut s known as the Theenn Equalent Crcut. Snce we are dealng wth lnear crcuts, applcaton of the prncple of superposton results n the followng expresson for the current and oltage relaton. (1.17) = m mv bji j 0 j j j Where V j and I j are oltage and current sources n the crcut under nestgaton and the coeffcents m j and b j are functons of other crcut parameters such as resstances. j And so for a general network we can wrte Where And j = m b (1.18) m= m 0 (1.19) (1.20) b= mv b I j j j j j Theenn s Theorem s stated as follows: A lnear one port network can be replaced by an equalent crcut consstng of a oltage source VTh n seres wth a resstor Rth. The oltage VTh s equal to the open crcut oltage across the termnals of the port and the resstance RTh s equal to the open crcut oltage VTh dded by the short crcut current Isc The procedure to calculate the Theenn Equalent Crcut s as follows: 1. Calculate the equalent resstance of the crcut (RTh) by settng all oltage and current sources to zero 2. Calculate the open crcut oltage Voc also called the Theenn oltage VTh 6.071/ Sprng Chanotaks and Cory 8

9 The equalent crcut s now In Vn Orgnal crcut A B RTh Voc Equalent crcut A B If we short termnals AB, the short crcut current Isc s Example: VTh Isc = (1.21) RTh Fnd o usng Theenn s theorem 6k Ω 2k Ω 12 V 6k Ω 1k Ω o The 1kΩ resstor s the load. Remoe t and compute the open crcut oltage Voc or VTh. 6k Ω 2k Ω 12 V 6k Ω Voc Voc s 6V. Do you see why? Now let s fnd the Theenn equalent resstance RTh / Sprng Chanotaks and Cory 9

10 6k Ω 2k Ω 6k Ω RTh And the Theenn crcut s RTh = 6 kω//6kω 2kΩ= 5kΩ 5k Ω 6 V 1k Ω 5k Ω 6 V 1k Ω o And o=1 Volt. Another example: Determne the Theenn equalent crcut seen by the resstor RL. RL Resstor RL s the load resstor and the balance of the system s nterface wth t. Therefore n order to characterze the network we must look the network characterstcs n the absence of RL / Sprng Chanotaks and Cory 10

11 A B Frst lets calculate the equalent resstance RTh. To do ths we short the oltage source resultng n the crcut. A B A B The resstance seen by lookng nto port AB s the parallel combnaton of 3 = In seres wth the parallel combnaton 4 = (1.22) (1.23) The open crcut oltage across termnals AB s equal to RTh = 3 4 (1.24) 6.071/ Sprng Chanotaks and Cory 11

12 A A B B VTh = A B (1.25) = R 1 R 3 R 2 R 4 And we hae obtaned the equalent crcut wth the Theenn resstance gen by Eq. (1.24) and the Theenn oltage gen by Eq. (1.25) / Sprng Chanotaks and Cory 12

13 The Wheatstone Brdge Crcut as a measurng nstrument. Measurng small changes n large quanttes s one of the most common challenges n measurement. If the quantty you are measurng has a maxmum alue, V max, and the measurement dece s set to hae a dynamc range that coers 0 V max, then the errors wll be a fracton of V max. Howeer, many measurable quanttes only ary slghtly, and so t would be adantageous to make a dfference measurement oer the lmted range, V max V mn. The Wheatstone brdge crcut accomplshes ths. A B u Ru The Wheatstone brdge s composed of three known resstors and one unknown, Ru, by measurng ether the oltage or the current across the center of the brdge the unknown resstor can be determned. We wll focus on the measurement of the oltage u as ndcated n the aboe crcut. The analyss can proceed by consderng the two oltage dders formed by resstor pars, and,. A A Ru B B The oltage u s gen by u = A B (1.26) Where, 6.071/ Sprng Chanotaks and Cory 13

14 And And u becomes: A = R 1 R 3 B Ru = R 2 Ru (1.27) (1.28) Ru u = R 1 R 3 R 2 Ru (1.29) A typcal use of the Wheatstone brdge s to hae = and ~ Ru. So let s take Under these smplfcatons, Ru = ε (1.30) Ru u = R 1 R 3 R 2 Ru ε = R 1 R 3 R 1 R 3 ε (1.31) As dscussed aboe we are nterested n the case where the araton n Ru s small, that s n the case where ε. Then the aboe expresson may be approxmated as, u ε (1.32) R 1 R / Sprng Chanotaks and Cory 14

15 The Norton equalent crcut A lnear one port network can be replaced by an equalent crcut consstng of a current source In n parallel wth a resstor Rn. The current In s equal to the short crcut current through the termnals of the port and the resstance Rn s equal to the open crcut oltage Voc dded by the short crcut current In. The Norton equalent crcut model s shown below: In Rn By usng KCL we dere the relatonshp for ths crcut. or For = 0 (open crcut) the open crcut oltage s In = 0 (1.33) Rn = In (1.34) Rn Voc = InRn (1.35) And the short crcut current s Isc = In (1.36) If we choose Rn = RTh and Voc In = the Theenn and Norton crcuts are equalent RTh 6.071/ Sprng Chanotaks and Cory 15

16 Voc RTh A B In RTh Theenn Crcut Norton Crcut We may use ths equalence to analyze crcuts by performng the so called source transformatons (oltage to current or current to oltage). For example let s consder the followng crcut for whch we would lke to calculate the current as ndcated by usng the source transformaton method. 6 Ω 3 V 6 Ω 3 Ω 3 Ω 2 A By performng the source transformatons we wll be able to obtan the soluton by smplfyng the crcut. Frst, let s perform the transformaton of the part of the crcut contaned wthn the dotted rectangle ndcated below: 6 Ω 3 V 3 Ω 6 Ω 3 Ω 2 A The transformaton from the Theenn crcut ndcated aboe to ts Norton equalent ges 0.5 A 6 Ω 3 Ω 6 Ω 3 Ω 2 A 6.071/ Sprng Chanotaks and Cory 16

17 Next let s consder the Norton equalent on the rght sde as ndcated below: 0.5 A 6 Ω 3 Ω 3 Ω 6 Ω 2 A The transformaton from the Norton crcut ndcated aboe to a Theenn equalent ges 0.5 A 6 Ω 6 Ω 3 Ω 3 Ω 6 V Whch s the same as 0.5 A 6 Ω 6 Ω 6 Ω 6 V By transformng the Theenn crcut on the rght wth ts Norton equalent we hae 0.5 A 6 Ω 6 Ω 6 Ω 1 A And so from current dson we obtan = = A (1.37) / Sprng Chanotaks and Cory 17

18 Another example: Fnd the Norton equalent crcut at termnals XY. X Is Frst we calculate the equalent resstance across termnals XY by settng all sources to zero. The correspondng crcut s Y X Rn And Rn s Y Rn = ( ) (1.38) Next we calculate the short crcut current X Is Isc Y 6.071/ Sprng Chanotaks and Cory 18

19 Resstor does not affect the calculaton and so the correspondng crcut s X Is By applyng the mesh method we hae Isc Y Is Isc = = In (1.39) Wth the alues for Rn and Isc gen by Equatons (1.38) and (1.39) the Norton equalent crcut s defned X In Rn Y 6.071/ Sprng Chanotaks and Cory 19

20 Power Transfer. In many cases an electronc system s desgned to prode power to a load. The general problem s depcted on Fgure 1 where the load s represented by resstor RL. lnear electronc system RL Fgure 1. By consderng the Theenn equalent crcut of the system seen by the load resstor we can represent the problem by the crcut shown on Fgure 2. RTh VTh L RL Fgure 2 The power delered to the load resstor RL s The current s gen by And the power becomes P= 2 RL VTh = RTh RL (1.40) (1.41) VTh P= RTh RL 2 RL (1.42) For our electronc system, the oltage VTh and resstance RTh are known. Therefore f we ary RL and plot the power delered to t as a functon of RL we obtan the general behaor shown on the plot of Fgure / Sprng Chanotaks and Cory 20

21 Fgure 3. The cure has a maxmum whch occurs at RL=RTh. In order to show that the maxmum occurs at RL=RTh we dfferentate Eq. (1.42) wth respect to RL and then set the result equal to zero. 2 dp 2 ( RTh RL) 2 RL( RTh RL) = VTh 4 drl ( RTh RL) (1.43) and dp = 0 RL RTh = 0 (1.44) drl and so the maxmum power occurs when the load resstance RL s equal to the Theenn equalent resstance RTh. 1 Condton for maxmum power transfer: RL= RTh (1.45) The maxmum power transferred from the source to the load s 2 VTh P max = (1.46) 4RTh 2 d P 1 By takng the second derate and settng RL=RTh we can easly show that 2 drl the pont RL=RTh corresponds to a maxmum. 2 d P 0 2 drl <, thereby 6.071/ Sprng Chanotaks and Cory 21

22 Example: For the Wheatstone brdge crcut below, calculate the maxmum power delered to resstor RL. RL Preously we calculated the Theenn equalent crcut seen by resstor RL. The Theenn resstance s gen by Equaton (1.24) and the Theenn oltage s gen by Equaton (1.25). Therefore the system reduces to the followng equalent crcut connected to resstor RL. RTh VTh L RL For conenence we repeat here the alues for RTh and VTh. VTh = R 1 R 3 R 2 R 4 (1.47) RTh = (1.48) The maxmum power delered to RL s 2 2 VTh P max = = 4RTh RR (1.49) 6.071/ Sprng Chanotaks and Cory 22

23 In arous applcatons we are nterested n decreasng the oltage across a load resstor by wthout changng the output resstance of the crcut seen by the load. In such a stuaton the power delered to the load contnues to hae a maxmum at the same resstance. Ths crcut s called an attenuator and we wll nestgate a smple example to llustrate the prncple. Consder the crcut shown of the followng Fgure. RTh Rs attenuator a VTh Rp o RL b The network contaned n the dotted rectangle s the attenuator crcut. The constrants are as follows: 1. The equalent resstance seen trough the port ab s RTh 2. The oltage o = kvth Determne the requrements on resstors Rs and Rp. Frst let s calculate the expresson of the equalent resstance seen across termnals ab. By shortng the oltage source the crcut for the calculaton of the equalent resstance s attenuator RTh Rs a Rp Reff The effecte resstance s the parallel combnaton of Rp wth RsRTh. b Reff = ( RTh Rs)// Rp ( RTh Rs) Rp = RTh Rs Rp (1.50) 6.071/ Sprng Chanotaks and Cory 23

24 Whch s constraned to be equal to RTh. The second constrant ges RTh = ( RTh Rs) Rp RTh Rs Rp (1.51) kvth VTh Rp = Rp RTh Rs (1.52) And so the constant k becomes: k = Rp Rp RTh Rs (1.53) By combnng Equatons (1.51) and (1.53) we obtan And 1 k Rs = RTh (1.54) k 1 Rp= RTh (1.55) 1 k The maxmum power delered to the load occurs at RTh and s equal to 2 2 kvth Pmax = (1.56) 4RTh 6.071/ Sprng Chanotaks and Cory 24

25 Representate Problems: P1. Fnd the oltage o usng superposton. (Ans Volts) o 2 Ω 3 Ω 4 Ω 1 Ω 6 V 2 V P2. Calculate o and o for the crcut below usng superposton (Ans. o=1.6 A, o=3.3 V) 2 A 4 Ω 2 Ω 3 Ω o 12 V 3 Ω 1 A 1 Ω 4 Ω P3. usng superposton calculate o and o as ndcated n the crcut below (Ans. o=1.35 A, o=10 V) o o 4 Ω 3 Ω 3 Ω 24 V 1 Ω 2 A 3 Ω 12 V 6.071/ Sprng Chanotaks and Cory 25

26 P4. Fnd the Norton and the Theenn equalent crcut across termnals AB of the crcut. (Ans. In= 1.25 A, Rn = 1.7Ω, VTh = 2.12V ) A 4 Ω 3 Ω 5 A 1 Ω 3 Ω B P5. Calculate the alue of the resstor R so that the maxmum power s transferred to the 5Ω resstor. (Ans. 10Ω) R 24 V 5 Ω 10 Ω 12 V P6. Determne the alue of resstor R so that maxmum power s delered to t from the crcut connected to t. R P7 The box n the followng crcut represents a general electronc element. Determne the relatonshp between the oltage across the element to the current flowng through t as ndcated / Sprng Chanotaks and Cory 26