Massachusetts Insttute of Technology Department of Electrcal Engneerng and Computer Scence 6.002 í Electronc Crcuts Homework 2 Soluton Handout F98023 Exercse 21: Determne the conductance of each network shown below as ewed from ts port. Note that the resstors n each network are specæed n terms of ther conductances. G 1 G2 G 1 G 2 G 3 G 1 G 2 G3 G1 G2 G3 G 3 Network (a) Network (b) Network (c) Network (d) Answer: We know that conductance s the dual concept to resstance. Conductors n parallel add, just as do resstors n seres. Smlarly, conductances n seres combne n the same way as resstors do n parallel. We hae a deæned operator k, such that a k b = ab a b networka The conductances G 1, G 2, and G 3 are all n seres, so G = G 1 k G 2 k G 3 = G 1 G 2 G 3 G 1 G 2 G 1 G 3 G 2 G 3 networkb G 1 s n seres wth the parallel combnaton of G 2, G 3. G = G 1 k G 2 G 3 = G 1G 2 G 1 G 3 G 1 G 2 G 3 networkc G 1 s n parallel wth the seres combnaton of G 2, G 3. networkd G 1, G 2, and G 3 are all n parallel. G = G 1 G 2 kg 3 =G 1 G 2G 3 G 2 G 3 G = G 1 G 2 G 3 1
Exercse 22: through 3. For both networks shown below, determne the oltage across and the current 2 I 3 2 3 Network (a) Network (b) Answer: These networks show the usefulness of transformng between Theenn and Norton equalent crcuts. In both networks, we can cut the crcut n half, so that the source and are on one sde, whle 2 and 3 are on the other. We then take the source and and conert them to ther other form. The other sde of the crcut and ths s the heart of Theenn and Norton equalents wll not know the dæerence. I=/ 2 3 2 3 networka We cut the crcut as shown aboe, left. Snce we know the relaton Th = eq I N, we can compute the Norton equalent crcut. Note that the resstance assocated wth the Norton equalent equals that of the Theenn resstance. The new crcut s shown aboe, rght. The oltage across and current through 3 s then 3 = 2 3 2 3 2 3 I = I 3 = 3 3 = 2 3 2 3 2 3 2 2 3 2 3 I 2 =I 2 3 1 3 networkb Agan, we cut the crcut as shown. Ths tme, we conert to a Theenn equalent. The three resstors are now n seres, as shown aboe, rght. The oltage across and current through 3 s then 2
I 3 = 2 3 = I 2 3 3 = 3 I 3 = 3 I 2 3 Problem 21: Fnd the Theenn and Norton equalents of the followng networks, and graph ther relatons as ewed at ther ports. I 2 I 2 2 Network (a) Network (b) Network (c) Network (d) Answer: Th eq I N eq Theenn equalent Norton equalent For each crcut, we want Th, I N, and eq, as represented n the crcuts aboe. Note that the current at each port s deæned to go nto the poste termnal. Ths s the opposte drecton from our deænton of sc, whch sthe current we ænd from the poste termnal to the negate when we load the termnals wth a short crcut. So, Th = oc the oltage at the port when t s opencrcut, I N = sc the current through the port when t s shortcrcut. Ths tells us further that the ntercept of the relaton s Th, whle the ntercept s,i N. Note that shuttng oæ the sources and ændng the equalent resstance seen at the port determnes the slope of the relaton. Note also that our relatons wll be of the form = eq Th whle another way to relate them s wth the nerse functon = 1 eq, I N a In ændng oc, there s no current through and thus no oltage across the resstor closest to the port. Thus oc s the oltage across the other resstor. Th = oc = I 3
When we short the termnals, the source current s splt equally between the two resstors, snce they are of equal alue and are n parallel. The equalent resstance follows. I N = sc = 2 I = I 2 eq = Th I N =2 b Ths s the dual problem to a. I N = sc = Th = oc = 2 = 2 eq = Th I N = 2 c In ths problem t s easest to ænd sc and eq, and determne Th from those. sc 0 2 I eq 2 Shortng the port as n the ægure aboe, we force the oltage across the 2 resstor to zero. Thus the current splts equally between the two paths no current æows through the 2 resstor. So, I N = sc = I 2 We now turn oæ the current source turns nto an open crcut and compute the equalent resstance, seen at the termnals. It s obous from the ægure aboe, rght that the resstance s Now, the Theenn oltage s eq =2k= Th = eq I N = I 2 d Ths s the most complcated network. It helps to deæne node oltages e 1 and e 2 as n the ægure left. Then oc = e 1, e 2. To ænd e 1 and e 2,we ærst notce that the left and rght pars of resstors are completely ndependent of one another. Ths s a result of the locaton of the oltage source, across both pars. We then hae two oltage dders. e 1 = 2 = 3 4
eq 2 2 e 1 e2 2 eq 2 2 2 e 2 = 2 2 = 2 3 Th = oc = e 1, e 2 =, 3 Lookng at the ægure aboe, center, we see the crcut wth the oltage source shut oæ. It s dæcult to see how to analyze ths crcut untl t s redrawn as aboe, rght. If you are unsure the rght drawng s the same as the mddle one, label the nodes and trace each path. We ænd that The Norton current s then eq =k22k= 4 3 I N = Th eq =, 4 e Plots The characterstcs are as follows. Note that the axes are labelled, as well as the ntercepts and slope. 2 I /2 /2 I/2 /4 I/2 4/3 /3 / I/2 (A) (B) (C) (D) Problem 22: Ths problem analyzes the network shown below by two methods. Note the deænton of the reference node. a Use the node method to analyze the network. Frst deæne the approprate node oltages and branch currents. Second wrte the approprate equatons and sole for the node oltages. b Use superposton to analyze the network. That s, superpose the two partal node oltages obtaned wth only sngle sources acte to ænd the total node oltages. emember that a zerooltage source s a short crcut, and a zerocurrent source s an open crcut. 5
I 3 2 4 c Compare the solutons to Parts A and B, and check ther unts. The two solutons should be the same. Answer: The node method for ths crcut s qute tedous, as we shall see. a Let us deæne two node oltages, e 1 and e 2 as shown, relate to ground. From ths we can determne 1;2;3;4, whch I shall deæne as the current through each correspondng resstor ;2;3;4, n the drecton from the top of the crcut to the bottom as ewed n the pcture. 1 e 1 I 3 2 e 2 4 3 4 2 Also deæned are nodes n 1 at e 1 and n 2 at e 2. We perform KCL at these nodes to get, 1 I, 3, 2 =0,I 3, 4 =0 We then determne these currents n terms of oltages. I wll use conductances G = ;=1;2;3;4, as ths s easer when usng the node method. 1 1 = G 1 e 1, 3 = G 3 e 1, e 2 2 = G 2 e 1, 0 4 = G 4 e 2, 0 substtutng nto the KCL equatons and collectng terms, we get!! G 1 G 2 G 3,G 3 e 1 =,G 3 G 3 G 4 e 2,I I G 1! Our nodal equaton s of the form Ge = I, so our soluton should be e = G,1 I To nert the matrx, let us ærst calculate the determnant: æ = detg =G 1 G 2 G 3 G 3 G 4,,G 3,G 3 =G 1 G 3 G 2 G 3 G 1 G 4 G 2 G 4 G 3 G 4 6
now we know that for a 2 æ 2 matrx, so, a c b d G,1 = 1 æ det a c!,1 = b d! 1 ad, bc = ad, bc d,b,c a!! G 3 G 4 G 3 G 3 G 1 G 2 G 3 so, multplyng and usng a bt of shorthand, we get the node oltages. e 1 = G 1G 3 G 4 G 4 æ æ I e 2 = G 1G 3 æ, G 1 G 2 æ b The crcut analyss gets easer to deal wth when we use superposton. The ægure shows the crcut wth the current source oæ left and wth the oltage source oæ rght. Node oltages are further ndexed A or B, then the total alues wll be I e 1 = e 1A e 1B e 2 = e 2A e 2B e 1A e 1B 3 2 I 3 e 2 2A e 2B 4 4 We wll stll use conductances and nodal analyss to sole ths. The node equatons for crcut A are G 1 e 1A, G 2 e 1A G 3 e 1A,e 2A =0,G 3 e 1A,e 2A G 4 e 2A =0 Ths becomes the followng:! G 1 G 2 G 3,G 3,G 3 G 3 G 4! e 1 = e 2 G 1 0! Wehae the same G matrx as before! Ths s really no surprse, because the resstors hae not moed wth respect to the nodes. The soluton s therefore e A = G,1 I A or 7
e 1A = G 1G 3 G 4 æ e 2A = G 1G 3 æ We follow the same procedure for the B crcut. The node equatons are G 1 e 1B, 0, I G 2 e 1B G 3 e 1B, e 2B =0 I,G 3 e 1B,e 2B G 4 e 2B =0 Ths becomes the followng:! G 1 G 2 G 3,G 3,G 3 G 3 G 4! e 1 = e 2 I,I! and the soluton s e 1B = G 4 æ I e 2B =, G 1 G 2 æ I c and the components of e 1 and e 2 n fact add up to the analyss when both sources are on. Problem 23: Gen the network shown below, ænd 0 as a functon of 1, 2 and 3 assumng 0 =0. Hnt: use superposton. Also, ænd the Theenn equalent of the network as ewed from ts port. Fnally, assume that the oltage sources can each take on only the alues of 0 or 3, and determne 0 for the 8 possble combnatons of 1, 2 and 3. Based on your analyss, of what electronc crcut mght the network be a part? 2 2 2 2 3 2 2 1 o o Answer: Ths problem s ery dæcult f approached the wrong way. Followng the hnts, we approach t from an easy drecton. Frst, we shut oæ the sources 2 and 3. Ths leaes us wth only 1 as shown n the ærst crcut below. Lookng at the leftmost crcut, we see we can combne many of the resstors to the left of the source 1. Hang done ths, we cut the crcut as shown mddle crcut, and conert the source sde to ts Theenn equalent. On the rght n the ægure, we see the reconnected crcut. It s clear that the oltage 0 s 0 = 2 2 Th = 1 4 8
2 2 2 2 2 1 o o 2 1 2 2 o o 0.5 1 2 o o 2 2 2 2 2 2 o o 2 2 2 2 2 o o A B Now we examne the contrbuton by 2. Lookng at the ægure, we see the orgnal crcut wth 1 and 3 oæ. To the rght, the resstors to the left of 2 are combned, and two cuts are shown. It s obous that the crcut to the left of cut A yelds the same Theenn equalent asfor the cut n the 1 crcut. The crcut to the left of cut B has the same equalent resstance, but the Theenn oltage s cut by half. So, 0 = 1 2 2 2 2 = 2 8 2 2 3 2 2 2 o o 3 /8 2 o o Lookng at the ægure aboe, left, we see the crcut when 1 and 2 are shut oæ. If we cut the crcut as shown, the porton on the left looks exactly lke the 2 only crcut. Ths means we can replace ths by ts Theenn equalent and reattach as shown aboe, rght. We then further reduce ths crcut by notcng the oltage dder hales the source oltage, or oc = 1 2 s = 1 16 3. By shuttng oæ the source, the equalent resstance s eq =2k=. Note that ths s the equalent resstance of the entre crcut. Assumng 0 = 0, the relaton between 3 and 0 s 0 = 1 2 3 3 2 = 3 16 Hang the three components of 0 when 0 = 0, and hang the equalent resstance, we can draw the Theenn equalent crcut: 9
(4 1 2 2 3 )/16 o o The last part of the problem asks for the behaor of 0 wth respect to the source oltages. 0 = 1 4 1 1 8 2 1 16 3 To understand the meanng of ths equaton when 1 ; 2 ; 3 2 f0;3 g, suppose we deæne b = 16 3 ;=1;2;3. Then we ænd that 0 = 3 16 22 b 1 2 1 b 2 2 0 b 3 = 3 16 b 1b 2 b 3 where b 1 b 2 b 3 s the bnary representaton of ntegers from 0 to 7. We hae a dgtaltoanalog conerter! Problem 24: Two networks, N1 and N2, are descrbed n terms of ther relatons, and connected together through a sngle resstor, as shown below. a Fnd the Theenn and Norton equalents of N1 and N2. b Fnd analytc expressons for the currents 1 and 2 that result from the nterconnecton of N1 and N2. N1 1 2 Network #1 1 1 2 N2 I 1 I 1 2 Network #2 Answer: a At ths pont, we can smply read the Theenn and Norton parameters rght from the graph. For clarty, I wll superscrpt the parameters by ther network labels. Network 1 n1 Th = 1 I n1 N = I 1 n1 eq = 1 I 1 Network 2 n2 Th =, 2 I n2 N =,I 1 n2 eq =, 2,I 1 = 2 I 1 Ths s a good tme to note that the equalent resstance s poste obous but reassurng. 10
node 1 2 I I 1 1 1 2 b It seems Norton equalents are the rght way to go for ths part. Lookng at node 1, we get the equaton I 1, 1,, 2, I 1 =0 1 2 =0 Ths s awkward, but no current æows through the resstors. All the current goes nto the node from the left current source, and s drawn away by that on the rght. The currents through the resstors must all go n the same drecton, and are therefore zero. So, 1 =, 2 =,I 1 Now, the reasonng stated for the answer aboe s perhaps less than obous, and s not ery appealng. So, let us determne the alues 1,, and 2 assumng the Norton current source for Network 2 s I 2 rather than,i 1, then usng superposton. 2 2 I 1 1A A 2A 1B B 2B I 2 A B We shut oæ I 2, as shown on the left n the ægure. Usng current dders, we ænd that 1A = 2 2 2 I 1 A = 2A = 2 2 2 I 1 2 2 I 1 Now shuttng oæ I 1, as shown on the rght n the ægure, and agan usng the current dder, we get 1B = 2 2 2 I 2 11
B = 2 2 2 I 2 Summng parts A and B, 2B = 2 2 I 2 1 = 2 2 2 I 1 I 2 = 2 = 2 2 2 I 1 I 2 2 2 I 1 I 2 Further, we know 1 =,I 1 1, and 2 =,I 2 2. If I 2 =,I 1, then 1 = = 2 = 0, and we get 1 =, 2 =,I 1 12