Modeling a RLC Circuit’s Current withDifferential Equations

Kenny Harwood

May 17, 2011

Abstract

The world of electricity and light have only within the past cen-tury been explained in mathematical terms yet still remain a mysteryto the human race. R. Buckminster Fuller said; ”Up to the twenti-eth century, ”reality” was everything humans could touch, smell, see,and hear. Since the initial publication of the chart of the electromag-netic spectrum ... humans have learned that what they can touch,smell, see, and hear is less than one-millionth of reality.”[4] This pa-per gives an abbreviated description of the photovaltaic effect (solarpower production process) and then a RLC circuit will be modeledthat is powered by a photovaltaic panel that has its output voltagepassed through an inverter to produce an AC output signal wherevoltage becomes a sinusoidal function of time.

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Contents

1 A Means to Produce Power 3

2 The S.R.H process 4

3 Applying Free Electricity 5

4 In Search of an ODE 7

5 Analyzing Circuit for Numerical Values of Circuit Compo-nents 8

6 Solving the Ordinary Differential Equation 10

7 Matlab Method 12

8 Curtain Call 14

9 Appendix 15

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1 A Means to Produce Power

The Sun has showered this blue speck that has been called home forbillions of years, showering free energy upon Earth’s surface unconditionally.However, only recently has solar power been given the spotlight across thisglobe. Solar panels currently are being produced and marketed in mass tocounteract the dependency humans have on the less forgiving fossil fuels.In 2007, 18.8 trillion kilowatthours of electricity were produced globally[1].Incomparison, the sunlight received on the Earth’s surface in one hour is enoughto power the entire world for a year[2]. The question is, how do those radiantwarm rays of light become electricity? The answer in short, recombination.

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2 The S.R.H process

In the early 1900’s, a mathematical model was published expressing the pro-cess in which light photons hand off valence electrons to latices (such assilicon photovaltaic, or PV cells) and create an electric current and thereforeelectricity. It was titled ’Shockley-Read-Hall Recombination’[7]. There aremany complicated ways to descrive recombination and the photovaltaic ef-fect, but with the help of Jessika Toothman and Scott Aldous’s How StuffWorks article[3], this should go by quite painlessly.

The basic idea of recombination is that photons (light wave-particles)have electrons in their valence shell, or the outer-most orbit of an atom thathas electrons occupying it, and when the photons come in contact with a crys-talline structure of silicon, the valence electrons are magnetically capturedby ’electron holes’ in the silicon lattice. The electron holes are present in allthe atoms in the silicon lattice so the captured electrons are passed throughthe lattice to make more room at the silicon’s surface for more electrons.But without an electric field, the cell wouldn’t work; the field forms whenthe N-type (neutrally charged) and P-type (positively charged) silicon comeinto contact. Suddenly, the free electrons on the N side see all the openingson the P side, and there’s a mad rush to fill them. Eventually, equilibriumis reached, and an electric field is present separating the two sides.

This electric field acts as a diode, allowing (and even pushing) electronsto flow from the P side to the N side, but not the other way around. It’slike a hill – electrons can easily go down the hill (to the N side), but can’tclimb it to the P side. And if one were to go through the mathematics of thisprocess starting with six spatial partial derivative equations, a solution tothe differential equation for the current being made by recombination wouldbe found to be I = Is(e

VD/nVt − 1) which is known as the Shockley Diodeequation where;• I is the diode current,• Is is the reverse bias saturation current (or scale current),• VD is the voltage across the diode,• Vt is the thermal voltage, and

• n is the ideality factor

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When light, in the form of photons, hits a solar cell, its energy breaksapart electron-hole pairs. Each photon with enough energy will normallyfree exactly one electron, resulting in a free hole as well. If this happensclose enough to the electric field, or if free electron and free hole happen towander into its range of influence, the field will send the electron to the Nside and the hole to the P side. This causes further disruption of electricalneutrality, and if an external current path is provided, electrons will flowthrough the path to the P side to unite with holes that the electric field sentthere. The electron flow provides the current, and the cell’s electric fieldcauses a voltage. With both current and voltage, there is power, which isthe product of the two.

Yet, Silicon happens to be a very shiny material, which can send photonsbouncing away before they’ve done their job, so an antireflective coating isapplied to reduce those losses. The final step is to install something that willprotect the cell from the elements – often a glass cover plate. PV modulesare generally made by connecting several individual cells together to achieveuseful levels of voltage and current, and putting them in a sturdy framecomplete with positive and negative terminals.

3 Applying Free Electricity

Now that the process of S.R.H recombination and the workings of a PV cellhave been presented, this power source can be put into a circuit consistingof a resistor, capacitor, and inductor. Using the software Matlab, and skillslearned in differential equations, the circuit will be modeled where as thecircuit tunes to the frequency of the Flathead Valley’s own Kool 105.1 FMstation. Prior to modeling the circuit some assumptions must be laid out,the first of which being that the voltage source coming from the solar panelis an alternating current signal. Solar panels produce direct current (DC)electricity. However, the electricity used in homes for lighting and poweris 240 volt Alternating Current (AC) electricity. Therefore, an electroniccomponent called an inverter is used in the transformation of DC electricityto AC electricity. An inverter achieves this by use of electronic switches toalternate the flow of the DC signal produced from solar panels.That is, switchone opens and switch 2 is closed and the current flows one way across the

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circuit. Then switch 1 closes and switch 2 opens and the current runs theopposite way across a circuit. Thus the DC electricity is converted to ACelectricity. For the remainder of this paper, the following assumptions willbe made:

• The Solar panel has its DC signal transformed to AC through an inverterbefore being connected to the RLC circuit.• Voltage efficiency of the PV cells do not waver.(constant voltage peak for AC signal function)• Frequency of AC signal is set at 105.1 kHz• All wires in setup are ideal, therefore offering negligible resistance tocurrent.• Initial charge of capacitor in circuit is zero.

For a simple example of how solar power can be used, an RLC circuit willbe modeled with a driving voltage that is produced from PV cells(about 0.5Volts per cell)[6] in a 12-celled solar panel. The RLC circuit being poweredmust have values for its components that let the frequency resonate at 105.1kHz, which will in turn tune the circuit to pickup and resonates the AC signalthat oscillates at the frequency of Kool 105.1 FM. In the AC signal equationwhere VAC(t) = Vpeak sin(ωt), the angular frequency ω can be expressed asω = 2πf , and Vpeak can be found as the product of the amount of PV cellsand the voltage output per cell. From this information, the Voltage functionfor the AC signal can be expressed as; VAC = 6 sin(105.1 × 103t) where t istime measured in seconds.

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4 In Search of an ODE

With a basic understanding of how light is transformed into electricity, amathematical model can be presented of the electric current in an RLCparallel circuit, also known as a ”tuning” circuit or band-pass filter. Toreach the ordinary differential equation needed to model the RLC circuit,V = LdI

dt+ RI(t) + 1/C ∗ ((Qo) +

∫I(t)dt[5] must be differentiated. There-

fore, V has been been replaced with VAC found above, and Qo (initial chargeof capacitor), assumed to equal zero, will make the voltage equation for theRLC circuit

Vpeak sin(ω ∗ t) = LdI

dt+RI(t) +

1

C∗∫I(t)dt

taking the derivative of both sides of the equation with respect to t, theSecond Order ordinary differential equation (ODE) is found to be;

Ld2I

dt2+R

dI

dt+

1

CI(t) = Vpeakω cos(ω ∗ t)

Where;• I is the circuit current,• V is the Voltage output from a solar panel system which will beassumed to be constant,• L is the inductance of the inductor in the circuit measured in Henrys,

H,• R is the resistance of the resistor in the circuit measured in ohms, Ω,

and• ω is the angular frequency of the AC signal which is also expressed as

ω = 2πf

.In order to have this equation set up for being evaluated by Matlab’s

ode45 solver, the second derivative must be isolated on the left hand sidelike so

d2I/dt2 = −R/L ∗ dI/dt− 1/(C ∗ L) ∗ I(t) +6ω

Lcos(ω ∗ t) (1)

However, before solving the ODE, values must be assigned for the circuitcomponents.

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5 Analyzing Circuit for Numerical Values of

Circuit Components

Figure 1. RLC parallel circuitV - the voltage of the power source

I - the current in the circuitR - the resistance of the resistor

L - the inductance of the inductorC - the capacitance of the capacitor

The components of this circuit are needing to resonate a frequency of105.1 kHz. In order to find the values needed in a tuning circuit for theKool 105 radio frequency, common inductance and resistance values foundin older tuning circuits will be used. With the values of resistance (R =10Ω), inductance (L = 130pF ) and frequency (f = 105.1kHz), the value ofcapacitance can be found. From the equation for forced damped harmonicoscillation

x′′ + 2cx′ + ω2ox = A cos(ωt)

a comparison can be made with equation (1) where ω2o is equivalent to 1/(C ∗

L). Since this circuit is needing to tune to 105.1 kHz, the natural angularfrequency ωo must be equal to the driving angular frequency ω and thereforethe capacitance needed for this model can be calculated by setting

ω2 =1

CL(2)

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After doing some algebra and substitution, equation(2) becomes C =1/(L ∗ (2π ∗ f)2) and capacitance is calculated to be 17.64 ∗ 10−6F or 17.64µF. To support this calculation, Multisim, a virtual circuit simulator hasbeen used to show that the magnitude of frequency peaks at 105.1 kHz withthe given and found values for this RLC circuit.

Values of ComponentsL=130 nH for VHF (FM)

f=105.1 kHzC=17.64 µF

R= 10 Ω

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6 Solving the Ordinary Differential Equation

In order to find a closed form solution to the RLC ODE, a general solutionto the homogeneous and particular equations must be found and then solvedfor the initial conditions. At time t = 0 seconds, there is no current goingthrough the circuit and therefore no initial rate of change of current either,also stated I(0) = 0 and dI/dt(0) = I ′(0) = 0.

To solve for the homogeneous equation, the ODE(1) must have no drivingforce, therefore

I ′′ +R/L ∗ I ′ + 1/CL ∗ I(t) = 0

the characteristic roots are found;

λ = − R

2L±

√(RL)2 − 4

CL

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which equate to λ1 = −5.6694× 103 and λ2 = −7.6917× 107. The homoge-neous part to the general solution is now known as

Ih = C1eλ1∗t + C2e

λ2∗t

Where C1 and C2 are constants.

However, before solving for C1 and C2, a particular solution for I(t) = Ih+Ip must be found. The complex method laid out on page 167 of ”DifferentialEquations With Boundary Value Problems”[5] will be used in finding theparticular solution Ip where

I ′′ +R/L ∗ I ′ + 1/CL ∗ I(t) = 6ω/L ∗ cos(ωt)

will be put in the complex form of

z′′ +R/L ∗ z′ + 1/CL ∗ z(t) = 6ω/L ∗ eωit

Where i is the imaginary number√−1.

In order to solve for z(t), a guess must be made first. Let z(t) = aeωit

and substitute that guess in for z(t). After taking a few derivatives and somealgebra,

a =6( 1

Cω− ωL− iR)

( 1Cω− ωL)2 +R2

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In the complex method of solving particular solutions, it is known thatthe final particular solution to this initial problem will be equal to the realpart of the z(t) solution. Therefore,

Ip = Re(z(t)) = −8.4229× 10−8 sin(ωt)− 0.6 cos(ωt)

The final step to solving for the explicit solution to equation1 is to letI(t) = Ih + Ip and solve for the initial conditions C1 and C2. By doing so,the closed form solution to the second order ODE is equated as,

I(t) = C1eλ1t + C2e

λ2t + A ∗ cos(ωt) +B ∗ sin(ωt) (3)

C1 = 0.0051

C2 = −0.0051

λ1 = −5.6694× 103

λ2 = −7.6917× 107

A = −8.4229× 10−8

B = 0.6000

ω = 2π105.1× 103

Plotting the closed form solution in Matlab over a 0.1 millisecond timeinterval produces the graph;

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7 Matlab Method

With the values for the RLC components having set values, the ODE is readyto be solved by MATlab as well. By inputting equation [3] into the functionfile cited in the appendix, Matlab’s ODE45 numerical solver estimated thisgraph of I(t)

which is the solution to the current I equation in this RLC resonating circuit.By graphing I with respect to its derivative I ′, a phase plane plot can beviewed.

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The phase plane graph shows a circular path with which the current am-plitude remains constant (radius of the circle) as the change in current movesfrom positive to negative repeatedly in a circular fashion. This confirms theODE45 system is solved for a resonating system. To show that both the cur-rent and the rate at which the current changes do not decay as time passes,a composite graph has also been provided with Matlab’s tools.

Here also is the same composite graph yet with a longer time interval of10 milliseconds to further back the previous statement.

As is evident, the graph of the closed form solution matches up nicelyto the graph that ODE45 produced for current versus time. Given all the

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assumptions and procedures made are in tandem with current scientific theo-ries and laws, this paper has outlined the process of sunlight being harnessedas electric current and has successfully provided and solved a second orderordinary differential equation regarding the electric current within a RLCcircuit.

8 Curtain Call

Though the mathematics of recombination are beyond the scope of ordi-nary differential equations, the smaller scale RLC system shows how littlean amount of PV cells can be used for electronics that are used day to dayin society, like the tuner circuit. The RLC second order differential equationallowed for an easier access to the power of mathematical tools, built intosoftware as well as taught in textbooks abroad. As Man ventures into thisnew era of scientific enlightenment, the principles of harmony and the logicthat math attests to must be made a prominent means to the motives of thisand future generations’ progress. The Sun will be shedding free energy uponthe Earth for billions of years so it is a reliable resource where as fossil fuelsare running low. Electricity is a realm that holds up to mathematical modelsin the micro and macroscopic domains which is not common in mathematicalmodeling of the known universe. With the combination of these; light andelectricity and the mathematics pertaining to them, there are guaranteed tobe massive leaps in technology and the human understanding of the universe.

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9 Appendix

Matlab mfile

function KennysolveRLCwithode45

[t,y]=ode45(@RLC,[0,0.0001],[0;0]);

%I vs t ODE45 graph

plot(t,y(:,1))

xlabel(’time (seconds)’)

ylabel(’Current (Amperes)’)

title(’ODE45 approximation of RLC current second order differential equation’)

shg

pause

%Phase plane graph

figure

plot(y(:,1),y(:,2))

xlabel(’I (current)’)

ylabel(’Iprime’)

title(’Phase Plane for RLC Current and its Derivative’)

shg

pause

%Composite graph

figure

plot3(y(:,1),y(:,2),t)

xlabel(’I (current)’)

ylabel(’Iprime’)

zlabel(’time (seconds)’)

title(’Composite plot for RLC Current and its Derivative’)

shg

pause

%calculating values for Capacitance, eigenvalues, particular constants and

%closed form constants with Initial conditions of I(0)=0 and I’(0)=0

%w=2*pi*105.1*10^3; R=10; L=130*10^-9; C=17.64*10^-6;

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% lam1=-(R/(2*L))+((((R/L)^2)-(4/(C*L)))^(1/2))/2

%

lam1=-5.669352080553770e+003;

%

% lam2=-(R/(2*L))-((((R/L)^2)-(4/(C*L)))^(1/2))/2

%

lam2=-7.691740757099637e+007;

% a=((6/(C*w))-6*w*L)/((((1/(C*w))-w*L)^2)+R^2)

a=-8.422914076544442e-008;

%

% b=(6*R)/((((1/(C*w))-w*L)^2)+R^2)

b=0.599999999999988;

%

% c1=(-b*w-0.6*lam1)/(lam2-lam1)

c1=0.0051;

%

% c2=a-c1

c2=-0.0051;

%graph closed form

t=linspace(0,0.0001,1000);

W=2*pi*105.1*10^3;

I=(c1)*exp((lam1)*t)+(c2)*exp((lam2)*t)+a*cos(W*t)+b*sin(W*t);

plot(t,I)

xlabel(’time (seconds)’)

ylabel(’Current (Amperes)’)

title(’Closed form solution of RLC current second order differential equation’)

function yprime=RLC(t,y)

yprime=zeros(2,1);

w=2*pi*105.1*10^3; R=10; L=130*10^-9; C=1.7640e-005;

yprime(1)=y(2);

yprime(2)=(6*w/L)*cos(w*t)-(R/L)*y(2)-(1/(C*L))*y(1);

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References

[1] U.S. Energy Information Administration, International Energy Outlook2010. http://www.eia.doe.gov/oiaf/ieo/electricity.html, Mar., 2011.

[2] Cool Earth Solar, FAQ: Frequently Asked Questions.http://www.coolearthsolar.com/faq, Apr., 2011.

[3] Jessika Toothman, Scott Aldous, How Solar Cells Work.http://science.howstuffworks.com/environmental/energy/solar-cell3.htm, Apr., 2011.

[4] R. Buckminster Fuller, Cheap Thoughts.

http://www.angelo.edu/faculty/kboudrea/cheap/cheap1_f.htm

, Apr., 2011.

[5] John Polking, Albert Boggess, David Arnold, Differential Equations withBoundary Value Problems. Pearson Education, Inc., Upper Saddle River,NJ 07458, 2nd Edition, 2006.

[6] Colorado Solar Inc., Solar Power Store.http://www.solarpanelstore.com/, Apr., 2011.

[7] Sallese J.-M., Krummenacher F., Fazan P., Derivation of Shockley-Read-Hall recombination rates. Pearson Education, Inc., Solid-State Electron-ics, 48 (9) pp. 1539-1548, 2004.

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