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Modeling a RLC Circuit’s Current with Differential Equations Kenny Harwood May 17, 2011 Abstract The world of electricity and light have only within the past century been explained in mathematical terms yet still remain a mystery to the human race. R. Buckminster Fuller said; ”Up to the twentieth century, ”reality” was everything humans could touch, smell, see, and hear. Since the initial publication of the chart of the electromagnetic spectrum ... humans have learned that what they can touch, smell, see, and hear is less than one-millionth of reality.”[4] This paper gives an abbreviated description of the photovaltaic effect (solar power production process) and then a RLC circuit will be modeled that is powered by a photovaltaic panel that has its output voltage passed through an inverter to produce an AC output signal where voltage becomes a sinusoidal function of time. 1 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 Components 8 6 Solving the Ordinary Differential Equation 10 7 Matlab Method 12 8 Curtain Call 14 9 Appendix 15 2 1 A Means to Produce Power The Sun has showered this blue speck that has been called home for billions of years, showering free energy upon Earth’s surface unconditionally. However, only recently has solar power been given the spotlight across this globe. Solar panels currently are being produced and marketed in mass to counteract the dependency humans have on the less forgiving fossil fuels. In 2007, 18.8 trillion kilowatthours of electricity were produced globally[1].In comparison, the sunlight received on the Earth’s surface in one hour is enough to power the entire world for a year[2]. The question is, how do those radiant warm rays of light become electricity? The answer in short, recombination. 3 2 The S.R.H process In the early 1900’s, a mathematical model was published expressing the process in which light photons hand off valence electrons to latices (such as silicon photovaltaic, or PV cells) and create an electric current and therefore electricity. It was titled ’Shockley-Read-Hall Recombination’[7]. There are many complicated ways to descrive recombination and the photovaltaic effect, but with the help of Jessika Toothman and Scott Aldous’s How Stuff Works 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 that has electrons occupying it, and when the photons come in contact with a crystalline structure of silicon, the valence electrons are magnetically captured by ’electron holes’ in the silicon lattice. The electron holes are present in all the atoms in the silicon lattice so the captured electrons are passed through the 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 when the N-type (neutrally charged) and P-type (positively charged) silicon come into contact. Suddenly, the free electrons on the N side see all the openings on the P side, and there’s a mad rush to fill them. Eventually, equilibrium is reached, and an electric field is present separating the two sides. This electric field acts as a diode, allowing (and even pushing) electrons to flow from the P side to the N side, but not the other way around. It’s like a hill – electrons can easily go down the hill (to the N side), but can’t climb it to the P side. And if one were to go through the mathematics of this process starting with six spatial partial derivative equations, a solution to the differential equation for the current being made by recombination would be found to be I = Is (eVD /nVt − 1) which is known as the Shockley Diode equation 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 4 When light, in the form of photons, hits a solar cell, its energy breaks apart electron-hole pairs. Each photon with enough energy will normally free exactly one electron, resulting in a free hole as well. If this happens close enough to the electric field, or if free electron and free hole happen to wander into its range of influence, the field will send the electron to the N side and the hole to the P side. This causes further disruption of electrical neutrality, and if an external current path is provided, electrons will flow through the path to the P side to unite with holes that the electric field sent there. The electron flow provides the current, and the cell’s electric field causes a voltage. With both current and voltage, there is power, which is the product of the two. Yet, Silicon happens to be a very shiny material, which can send photons bouncing away before they’ve done their job, so an antireflective coating is applied to reduce those losses. The final step is to install something that will protect the cell from the elements – often a glass cover plate. PV modules are generally made by connecting several individual cells together to achieve useful levels of voltage and current, and putting them in a sturdy frame complete with positive and negative terminals. 3 Applying Free Electricity Now that the process of S.R.H recombination and the workings of a PV cell have been presented, this power source can be put into a circuit consisting of a resistor, capacitor, and inductor. Using the software Matlab, and skills learned in differential equations, the circuit will be modeled where as the circuit tunes to the frequency of the Flathead Valley’s own Kool 105.1 FM station. Prior to modeling the circuit some assumptions must be laid out, the first of which being that the voltage source coming from the solar panel is an alternating current signal. Solar panels produce direct current (DC) electricity. However, the electricity used in homes for lighting and power is 240 volt Alternating Current (AC) electricity. Therefore, an electronic component called an inverter is used in the transformation of DC electricity to AC electricity. An inverter achieves this by use of electronic switches to alternate the flow of the DC signal produced from solar panels.That is, switch one opens and switch 2 is closed and the current flows one way across the 5 circuit. Then switch 1 closes and switch 2 opens and the current runs the opposite way across a circuit. Thus the DC electricity is converted to AC electricity. For the remainder of this paper, the following assumptions will be made: • The Solar panel has its DC signal transformed to AC through an inverter before 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 to current. • Initial charge of capacitor in circuit is zero. For a simple example of how solar power can be used, an RLC circuit will be modeled with a driving voltage that is produced from PV cells(about 0.5 Volts per cell)[6] in a 12-celled solar panel. The RLC circuit being powered must have values for its components that let the frequency resonate at 105.1 kHz, which will in turn tune the circuit to pickup and resonates the AC signal that oscillates at the frequency of Kool 105.1 FM. In the AC signal equation where VAC (t) = Vpeak sin(ωt), the angular frequency ω can be expressed as ω = 2πf , and Vpeak can be found as the product of the amount of PV cells and the voltage output per cell. From this information, the Voltage function for the AC signal can be expressed as; VAC = 6 sin(105.1 × 103 t) where t is time measured in seconds. 6 4 In Search of an ODE With a basic understanding of how light is transformed into electricity, a mathematical model can be presented of the electric current in an RLC parallel circuit, also known as a ”tuning” circuit or band-pass filter. To reach the ordinary differential equation needed to model the RLC circuit, R dI V = L dt + RI(t) + 1/C ∗ ((Qo ) + I(t)dt[5] must be differentiated. Therefore, V has been been replaced with VAC found above, and Qo (initial charge of capacitor), assumed to equal zero, will make the voltage equation for the RLC circuit Z 1 dI Vpeak sin(ω ∗ t) = L + RI(t) + ∗ I(t)dt dt C taking the derivative of both sides of the equation with respect to t, the Second Order ordinary differential equation (ODE) is found to be; L dI 1 d2 I + R + I(t) = Vpeak ω cos(ω ∗ t) dt2 dt C Where; • I is the circuit current, • V is the Voltage output from a solar panel system which will be assumed 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 side like so d2 I/dt2 = −R/L ∗ dI/dt − 1/(C ∗ L) ∗ I(t) + 6ω cos(ω ∗ t) L (1) However, before solving the ODE, values must be assigned for the circuit components. 7 5 Analyzing Circuit for Numerical Values of Circuit Components Figure 1. RLC parallel circuit V - the voltage of the power source I - the current in the circuit R - the resistance of the resistor L - the inductance of the inductor C - the capacitance of the capacitor The components of this circuit are needing to resonate a frequency of 105.1 kHz. In order to find the values needed in a tuning circuit for the Kool 105 radio frequency, common inductance and resistance values found in older tuning circuits will be used. With the values of resistance (R = 10Ω), inductance (L = 130pF ) and frequency (f = 105.1kHz), the value of capacitance can be found. From the equation for forced damped harmonic oscillation x00 + 2cx0 + ωo2 x = A cos(ωt) a comparison can be made with equation (1) where ωo2 is equivalent to 1/(C ∗ L). Since this circuit is needing to tune to 105.1 kHz, the natural angular frequency ωo must be equal to the driving angular frequency ω and therefore the capacitance needed for this model can be calculated by setting ω2 = 8 1 CL (2) After doing some algebra and substitution, equation(2) becomes C = 1/(L ∗ (2π ∗ f )2 ) and capacitance is calculated to be 17.64 ∗ 10−6 F or 17.64 µF. To support this calculation, Multisim, a virtual circuit simulator has been used to show that the magnitude of frequency peaks at 105.1 kHz with the given and found values for this RLC circuit. Values of Components L=130 nH for VHF (FM) f =105.1 kHz C=17.64 µF R= 10 Ω 9 6 Solving the Ordinary Differential Equation In order to find a closed form solution to the RLC ODE, a general solution to the homogeneous and particular equations must be found and then solved for the initial conditions. At time t = 0 seconds, there is no current going through the circuit and therefore no initial rate of change of current either, also stated I(0) = 0 and dI/dt(0) = I 0 (0) = 0. To solve for the homogeneous equation, the ODE(1) must have no driving force, therefore I 00 + R/L ∗ I 0 + 1/CL ∗ I(t) = 0 the characteristic roots are found; q R ± λ=− 2L ( L)2 R − 4 CL 2 which equate to λ1 = −5.6694 × 103 and λ2 = −7.6917 × 107 . The homogeneous part to the general solution is now known as Ih = C1 eλ1 ∗t + C2 eλ2 ∗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 ”Differential Equations With Boundary Value Problems”[5] will be used in finding the particular solution Ip where I 00 + R/L ∗ I 0 + 1/CL ∗ I(t) = 6ω/L ∗ cos(ωt) will be put in the complex form of z 00 + R/L ∗ z 0 + 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 some algebra, 6( 1 − ωL − iR) a = 1Cω ( Cω − ωL)2 + R2 10 In the complex method of solving particular solutions, it is known that the final particular solution to this initial problem will be equal to the real part 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 let I(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) = C1 eλ1 t + C2 eλ2 t + 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 time interval produces the graph; 11 7 Matlab Method With the values for the RLC components having set values, the ODE is ready to be solved by MATlab as well. By inputting equation [3] into the function file cited in the appendix, Matlab’s ODE45 numerical solver estimated this graph 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 0 , a phase plane plot can be viewed. 12 The phase plane graph shows a circular path with which the current amplitude remains constant (radius of the circle) as the change in current moves from positive to negative repeatedly in a circular fashion. This confirms the ODE45 system is solved for a resonating system. To show that both the current 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 of 10 milliseconds to further back the previous statement. As is evident, the graph of the closed form solution matches up nicely to the graph that ODE45 produced for current versus time. Given all the 13 assumptions and procedures made are in tandem with current scientific theories and laws, this paper has outlined the process of sunlight being harnessed as electric current and has successfully provided and solved a second order ordinary differential equation regarding the electric current within a RLC circuit. 8 Curtain Call Though the mathematics of recombination are beyond the scope of ordinary differential equations, the smaller scale RLC system shows how little an amount of PV cells can be used for electronics that are used day to day in society, like the tuner circuit. The RLC second order differential equation allowed for an easier access to the power of mathematical tools, built into software as well as taught in textbooks abroad. As Man ventures into this new era of scientific enlightenment, the principles of harmony and the logic that math attests to must be made a prominent means to the motives of this and future generations’ progress. The Sun will be shedding free energy upon the Earth for billions of years so it is a reliable resource where as fossil fuels are running low. Electricity is a realm that holds up to mathematical models in the micro and macroscopic domains which is not common in mathematical modeling of the known universe. With the combination of these; light and electricity and the mathematics pertaining to them, there are guaranteed to be massive leaps in technology and the human understanding of the universe. 14 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; 15 % 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); 16 References [1] U.S. Energy Information Administration, International Energy Outlook 2010. http://www.eia.doe.gov/oiaf/ieo/electricity.html, Mar., 2011. [2] Cool Earth Solar, FAQ: Frequently http://www.coolearthsolar.com/faq, Apr., 2011. Asked Questions. [3] Jessika Toothman, Scott Aldous, How Solar Cells http://science.howstuffworks.com/environmental/energy/solarcell3.htm, Apr., 2011. Work. [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 with Boundary Value Problems. Pearson Education, Inc., Upper Saddle River, NJ 07458, 2nd Edition, 2006. [6] Colorado Solar Inc., Solar http://www.solarpanelstore.com/, Apr., 2011. Power Store. [7] Sallese J.-M., Krummenacher F., Fazan P., Derivation of Shockley-ReadHall recombination rates. Pearson Education, Inc., Solid-State Electronics, 48 (9) pp. 1539-1548, 2004. 17