~

FIGURING IT OUTPART 13 - SECOND-ORDER MODELS

By Owen Bishop

This series is intended to help you with the quantitative aspects of electronic design:predicting currents, voltage, waveforms, and other aspects of the behaviour of circuits.

Our aim is to provide more than just a collection of rnle-of-thumb formulas.We will explain the underlying electronic theory and, whenever

appropriate, render some insights into the mathematics involved.

Last month we examined waysof using first-order differentialequations to build circuit mod-els. This month we extend themethods to include second-orderequations. These allow us to modelcircuits of greater complexity,such as that in Fig. 110.

R

"

--","c--.c

930010-13-12

Fig.110

'I'his includes resistive, inductiveand capacitive elements. By Kirch-hoff's Voltaga Law:

As we did last rnonth. we uselower-case letters for quantitieswhich are inherent1y variablein time and capitals forconstants.Substituting equivalent expres-sions for the voltages, based onEq. 26 ofPart 4, Ohm's law, andEq. 19 ofPart 4:

Ldi R' q-+ t+-=udt C

We will restriet the analysis tosituations in which u is constant,henee duldt = O.Then, differen-tiating both si des of this equa-tion:

Ld2i Rdi i_O-+ -+--dt2 dt C

In differentiatingthe third term,useq =it, and therefore dq/dt =i.The term d2i/dt2 makes this a sec-ond-order equation. Divide through

by L to make unity the coefficientof the first term:

d2i R di 1.-+-'-+-'l=Odt2 L dt LC

[Eq.97J

[Eq.94] If D is zero, the roots are bothm and the solution to Eq. 94 is:

Equation 94 models the circuit.It now remains to solve it andsubstitute actual values of R, Land C.

Auxiliary equationFor an equation of the form ofEq. 94, in which R, Land C areconstants, there is an auxiliaryequation ofthe form:

m2 +fm +s= 0, [Eq. 95]

in whichfis the coefficientof di/dtand g is the coefficient of i. InEq. 94, f = R JL and g = 1ILC.Solving Eq. 95 (a straightfor-ward quadratic equation) for In

is much easierthan solving Eq. 94for i. With values offandg sub-stituted in Eq. 95:

m2 + R m+_1_=0 [Eq.96]L LC

'I'his is solvable by applying thewell-known quadratic formulawhich, when applied to Eq. 95,is:

-f±~m2

The value ofthe expression f2 - 4g(known as the discriminant,D)determines what kind ofsolutionthe equation has:

lfD is positive, the equationhas two real roots.IfD is zero, the equation hastwo equal roots.If'D is negative, the equationhas two irnaginary roots.

IfD 1Spositive, the two real rootsare m1 and m2 and the solutionto Eq. 94 is:

[Eq.98]

If D is negative, we ca1culatek = .,j (-Dl2) and the solution is:

i =Ae-{l/2 cos Iü+ Be-ft/2 sin kt

[Eq. 99]All three equations are generalsolutions (see Part 12) and havetwo arbitrary constants, A andB. Last rnonth we had just onesuch constant and needed oneborder condition in order to findtheparticularsolution. Therearetwo constants, and we need twoborder conditions for second-orderequations.

Worked exampleGivenR=500n,L= 100 mH andC = 2 MF:f = R IL = 500/100xlO-3= 5000; andg = 11LC= 1I100x10-3x2x10-0= 5x1()6.From thesevaluesoffandg:

D = f2 - 4g = 5000L20x106= 5x106.

This is positive, so the equationhas two real solutions:

-f+JD2

-5000+22362

=-1382

-f-JD2

-5000-22362

=-3618

Substituting in Eq. 97:

i= Ae -1382t + Be--3618L

[Eq. 100J'I'his is the general solution. Wecan apply this to any set of bor-der conditions. We can imaginethe voltage ftuctuating, perhapsregularly, perhaps irregularly,causing a varying current in thecircuit. Then, when t = 0, thevoltage 1S suddenly held con-stant (Eq. 94). The models teilswh at happens after that. Forthis example, suppose that thecurrent is 2 rnA wen tirning be-gins, 01' i =2xl0-3 when t =O.Alsoassume a second border condi-tion that the tate ofchange ofcur-rent, dildt, is 0.05 As-t whent = 0. With t equal to zero, e haszero index in both terms and soequals unity. Substituting thevalues for the first border con-dition in Eq. 100:

2xlO-3 =A + B

.. B = 0.002-A.

Substitute this value for B inEq.100:

i=A(e-1382t _e-3618t)

+0.002e-36181

[Eq.101JTa incorporate the effect of thesecond border condition into theequation, we must first differ-entiate(seePart5)Eq.101 to ob-tain an equation for dildt:

di = A( -1382e -13821dt

+3618e-36181) _7.236e-36181

[Eq. 102]If dildt = 0.05 when t = 0:

ELEKTOR ELECTRONICS FEßRUARV 1994

0.05 = A(-1382 + 3618)-7.236.

A = (0.05 + 7.236) 12236=0.003258.

Substituting this in Eq. 101:

;=0.003258(e-1382t _e-3618/)

+0. 002e -36181

Rearranging terms gives:

i=0.003258e-13821

-0.001258e-36181

[Eq. 103]This is the particular solutionand Fig. 111 shows its graph. Itshows that at t = 0 the currentis 2 m.A, as specified. The in-crease of current when t =0 1Stoosmall to show on this graph. Thetangent to the curve at this pointwould slope up to the right byonly 3.6°, This upward slope i8almost instantly countered bythe relatively strang dampingef-fects of capacitance and induc-tance. By the end ofthe third mil-lisecond, the currenthas been al-most entirely damped out.

Figure 112 shows what hap-pens if we keep the first bordercondition unchanged, but havethe current increasing at 2 A s-t,instead of at only 0.005 A s-t.Substituting dildt in Eq. 102:

A = (2 + 7.236)12236 =0.00413l.

From Eq. 101:

;=0.004131e-13821

_0.002131e-3618t

We have another particular so-lution, matehing the new bor-der conditions. Its graph (Fig.112)shows current continuing to in-crease, to about 2.12 mA, beforedamping takes effect. From theequations we have established,we can calculate the current andthe rate of change of current atany instant from t = 0 onwards.With this information, we cango on to calculate the volta geacross thecomponents, and quan-tities such as the charge on thecapacitor and the magnetic link-age of the inductor at any time.

Specifying loter currentBorder conditions need not be re-stricted to events occurringwhent = o. We can specify i or di/dt atany other instant after timingbegins. It makes the equations1ightly more complicated, be-cause there is no simplificationofthe equations due to the indexofe beingzero. Let us trythe samecircuit, with the same first bor-

! (mAl2 _ j=0.002

wllent=O

o 2 3 t(ms)

Fig. 111

i (mA)

2

I", 0.004131e -'3621_0.002131e -36181

o 2 3 I(ms)

Fig. 112

i (mAl

5

4

3

2 "'j=O.OO2wllenl=O

O~--------'---------~2--------~~3 t (ma)

"""10_,..,.

Fig. 113

der condition, but the second to adefinite value in a specifiedborder condition being that time. Since we have specified i,i = 0.003 when t = 0.00l. We are not dildt, we go back to Eq. 101sayingthat the current increases and proceed from there. Substitute

FIGURING IT OUT - PART 13

the second border condition intothis, which involves multiply-ing the indices by 0.001:

0.003= A(e-1.382 _e-3·618)

+0.002e-3.618

=0.2242A+5.3673x 10-5A=0.01314

Substituting the new value of Ainto Eq. 101:

i= O. 01314(e-13821_e-3618t)

+0.002e-36181

=0.01314e-1382t

_0.01114e-36181

Yet another solution, with thegraph shown in Fig.113. Currentincreases even further beforedamping takes effect.

Critical dampingThe curves we have seen so farrepresent overdamping ofthe cur-rent. Current is reduced to zerofairly promptly. The model canbe used also to investigate thecircuit behaviour when it is crit-ical1y damped. We reduce thecapacitance slightly, to l.6 uF,leaving Rand L as before and,for comparison with Fig. 113,keep the same pair of borderconditions.

With the new value for C, fremains unchanged, but g be-comes 6.25xl06, and:

D = 50002 - 4x6.25x106 = o.

With a zero discriminant, thesolution ofthe auxiliary equationIS

m = -f12 = -500012 = -2500.

The general solution takes a dif-ferent form (Eq. 98):

i= Ae -2500t +Bte -2500t

[Eg. 104]Substitutingthe first bordercon-dition into this:

0.002 =A,

Eq. 104 becomes:

i = 0.002e-25001+ Bte-2500/.

Now substitute i = 0.003 andt = 0.001 into this:

0.003 = 0.002e-2.5 + 0.00lBe-2.5

B = 34.55.

The particular solution is:

ELEKTOR ELECTRONICS FEßRUARY 1994

GENERAL INTEREST

i ee 0.002e-25001+34.55te-25001.[Eq 105]

The graph in Fig. 114 showscurrent risingto a peak at 5.9 mAbefore beingdamped. The secondbord er condition occurs on theway down.

Under-dampingThis occurs when the discrim-nant has a negative value. Forexample, let UB reduce the ca-pacitance drastically, to 10 nF.The fremains at 5000. butg be-comes 109 and D becomes-3.975xlO9. With a negative dis-criminant, we need to calculatek:

k~~3. 975x 109 12~44581

The general solution is given byEq.99:

i c= Ae -2500t cos 44581t

+Be-2500t sin 445811

[Eq. 106]Keeping to the same border con-ditions, applying the first con-dition to Eq. 106, and using thefacts that cos 0 ~ 1 and sin 0 ~ 0:

0.002 ~ Aeo + 0

.. A ~ 0.002.

Substituting th.is in Eq. 106, to-gether with the values for t andi under the second border COTI-

dition:

0.003~0.002e-2.5 cos4.4581

+Be-2.5 sin 4.4581B~-0.03828

This leads UB to the particularsolution:

i~0.002e-25001 cos44581t

- O. 03828e -2500' sin 44581t

The graph of th.is equation hasan interesting form-Fig. 115.It shows the current reversin gmany times with gradually de-creasing magnitude. Oscillationsofth.is type are typical ofan under-damped circuit. The oscillationstake about 1.5 ms to die away.

Graphie ealeulatorA graph 1S an important aid tovisualizing the behaviour of amodel, and hence the behaviourofthe circuit 'it is modelling. Butplotting graphs is a tedious mat-ter, particularly when equationscontain severaJ exponential terms.With over-dampled and critically-

1 (mA)

6

4

----i '"0.002""he<! 1=0

i:= 0.OO2e-25001+ 34.55le-Z5OOt

o 2 3 I(ms)

1[mAl

i0ae "••~I

Fig. 114

I. O,0Q2e-2liOOI cos 445611 _ O.03626e-25001 sln445611

us V V;V V ~ "

V '[m~~

IDDIIf'_lUl_M

Fig. 115

u c L

I,

R

damped circuits, it is possibleto sketch the shape of the graphroughly after calculating half adozen points. But the oscilla-.tions of an under-damped circuitoften need 30 or more points toproduce a reasonably represen-tative curve.

A computer graphies pack-age may save a lot of time: theillustrations to this article were

930010·13·"

Fig. 116

produced in this way. Agraphiecalculator produees results evenmore quickly. We used one whenwe were planning the values touse in the examples. To makegraph plottingeven quicker, pro-gram the calculator to aeceptvariables, then to plot the graph.

The following 1S an exampleof a ahort program used on agraphic ealculator to plot graphs

of equations ofthe same type asEq_l03: 'EXP01':"A":?~A:"Ml":?-?C:"B":?-?B:''JM2'':?-?D:GRAPHy~Ae--CX+Be--DX:.Tbe programname is 'EXPOl'. The user isrequested to key in values forA, ml, Band m2' As 500n as thefinal value has been keyed in, thegraph is plotted according to theequation in the program.

A program can also includecommands to set the ranges of tand i for the displayed graph,so that the curve fills the screenreasonably well. Ifthe rangecom-mand follows the input of vari-ables, one or more ofthese vari-ables can be used in the rangecommands. For example, the rangefor i can be set to run from -2 Ato +2 A. It is also possible to in-clude inputs to set the range oft direetly from the program. Forthose who like to play areundwith models, adjusting the valuesto produce the required results,agraphie calculator is a valuabletoo!.

Parallel eireuitFigure 116shows a circuitwi thresistance, eapaeitance and in-ductance in parallel. Buildingthemodel follows very much the samesequence as building the modelof the series circuit. By KCL:

Replacingthe currents by equiv-alent expressions:

du u Ntp .C-+-+-~, [Eq.107]dt R L

The third term on the right 15obtained by noting thatL ~Ntpl i,as in Eq. 22, Part 5, where q'l isthe magnetic flux and N 1S thenumber of turne in the coil.Differentiating Eq. 107, and as-suming that i is eonstant:

d2u 1 du uC-+_·_+-~Odt2 R dt L

In obtaining the third term,dq>/dt ~ u IN (see Eq. 23, Part 5),so the term reduces to u I L.Dividing throughout by C givesthe model equation:

d2u 1 du u-+_·_+-~Odt2 RC dt LC

[Eq.108]The auxiliary equation is:

m2+mIRC+1/LC~0

in which f ~ lJRCand g ~ lJLC.With given values of R, C andL, the discrirninant may be pos-itive, zero or negative, yieldingequations fOT u having the same

ELEKTOR ELECTRONICS FEBRUARY 1994

FIGURING IT OUT - PART 13

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difficult to differentiate and passbeyond the scope of this series,However, there are ways aroundthis difficulty, as will be explainednext month.

form as Equations 97 to 99.Here isan exampleofan under-

damped circuit with R = 1kD.,C = 10 nF andL = 18 mH.

RC= 1x10-D,so{= l/RC= 105and f2 = 1/(RC)2 = 1xlOlO.

LC = 1.8x10-1O, so 4g = 4/LC= 2.222xlO!O.

From these we find thatD = -1.222x1010 and k = 78166.The general equation ts:

Fig. 117. It begins with a mas-sive 'kick' of up to -34 A. Aftera few swings of rapidly dirnin-ishing amplitude, the currentis fully darnped out in about0.15 ms ..

These examples show whatcan be done when we make cer-tain simplifying assumptionsabout the model. If these as-sumptions are not valid, in par-ticula r, if we can not assurnethat voltage 01' current is con-starrt, the equations are more

which models aseries cir-cuit (Fig. 110) in whichR = 1 kQ, L = 200 mH andC = 100 nF, given the borderconditions that i =-0.01 whent = 0, and i = +0.01 whent = 0.005. Is the circuit over-damped, critically damped 01'

under-damped? Find the cur-rent when t = 0.0008 s.

Acknowledgment. The authorwould like to thank the CasioComputer Company Ltd fortheirvaluable assistance.

2. In another series circuit,R = 5 kQ, C = 22 ~F andL = 0.5 H. When 1= 0, i = 0.1and di/dl = 0.5. Find the par-ticular solution for this cir-cuit. Find the current whenze ü.ö msand whenz e 100ms.How long does the currenttake to fall to 0.01 A? (Plot agraph 01' solve the equationofdi/dl by takingnaturallogs).

Testyourself1. Find the particular equation

u=Ae-50000t cos 781661

Be-SOOOOtsin 78166t

Nowforsome border conditions.Whenz e O;zz=-1; whent=O.OOOl,u = -0.5. Using tbe first eondi-tion, rememberingthatcos 0 = 1and sin 0 = 0:

i(A)

5

°t:-------too~--~~~;0~.1__----~.G-l <, t(ms)

whent",O 1=-0.5when t = 0.0001-1 =A. -5

For the second condition: -io Answers toTestyourself (Part 12)1. q = ge-1.418tx10-4;

(a): u = 1.45 V(h): i = 74.9 ~A.

-0.5 = -e-5 cos 7.8166+Be-5 sin 7.8166B = -74.22.

-15

-20

-25The particular solution is:2. 0.857 s.

u=_e-50000t cos78166t

-74. 22e-500001sin 781661

-30

3. 532 ~A.

The graph of this is shown in Fig. 117

ELEKTOR ELECTRONICS FEBRUARY 1994