ch05 - Attia, John Okyere. Transient Analysis. Electronics...

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Attia, John Okyere. “Transient Analysis.” Electronics and Circuit Analysis using MATLAB. Ed. John Okyere Attia Boca Raton: CRC Press LLC, 1999 © 1999 by CRC PRESS LLC
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CHAPTER FIVE TRANSIENT ANALYSIS 5.1 RC NETWORK Considering the RC Network shown in Figure 5.1 , we can use KCL to write Equation (5.1). RC V o (t) Figure 5.1 Source-free RC Network C dv t dt vt R oo () += 0 ( 5 . 1 ) i.e., dv t dt CR 0 If V m is the initial voltage across the capacitor, then the solution to Equation (5.1) is vt Ve m t CR 0 = ( 5 . 2 ) where CR is the time constant Equation (5.2) represents the voltage across a discharging capacitor. To obtain the voltage across a charging capacitor, let us consider Figure 5.2 . © 1999 CRC Press LLC
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V o (t) R C V s Figure 5.2 Charging of a Capacitor Using KCL, we get C dv t dt vt V R oos () + = 0 ( 5 . 3 ) If the capacitor is initially uncharged, that is vt 0 () = 0 at t = 0, the solution to Equation (5.3) is given as e S t CR 0 1 =− (5.4) Examples 5.1 and 5.2 illustrate the use of MATLAB for solving problems related to RC Network. Example 5.1 Assume that for Figure 5.2 C = 10 µ F, use MATLAB to plot the voltage across the capacitor if R is equal to (a) 1.0 k , (b) 10 k and (c) 0.1 k . Solution MATLAB Script % Charging of an RC circuit % c = 10e-6; r1 = 1e3; © 1999 CRC Press LLC
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tau1 = c*r1; t = 0:0.002:0.05; v1 = 10*(1-exp(-t/tau1)); r2 = 10e3; tau2 = c*r2; v2 = 10*(1-exp(-t/tau2)); r3 = .1e3; tau3 = c*r3; v3 = 10*(1-exp(-t/tau3)); plot(t,v1,'+',t,v2,'o', t,v3,'*') axis([0 0.06 0 12]) title('Charging of a capacitor with three time constants') xlabel('Time, s') ylabel('Voltage across capacitor') text(0.03, 5.0, '+ for R = 1 Kilohms') text(0.03, 6.0, 'o for R = 10 Kilohms') text(0.03, 7.0, '* for R = 0.1 Kilohms') Figure 5.3 shows the charging curves. Figure 5.3 Charging of Capacitor © 1999 CRC Press LLC
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From Figure 5.3 , it can be seen that as the time constant is small, it takes a short time for the capacitor to charge up. Example 5.2 For Figure 5.2 , the input voltage is a rectangular pulse with an amplitude of 5V and a width of 0.5s. If C = 10 µ F, plot the output voltage, vt 0 () , for resistance R equal to (a) 1000 , and (b) 10,000 . The plots should start from zero seconds and end at 1.5 seconds. Solution MATLAB Script % The problem will be solved using a function program rceval function [v, t] = rceval(r, c) % rceval is a function program for calculating % the output voltage given the values of % resistance and capacitance. % usage [v, t] = rceval(r, c) % r is the resistance value(ohms) % c is the capacitance value(Farads) % v is the output voltage % t is the time corresponding to voltage v tau = r*c; for i=1:50 t(i) = i/100; v(i) = 5*(1-exp(-t(i)/tau)); end vmax = v(50); for i = 51:100 t(i) = i/100; v(i) = vmax*exp(-t(i-50)/tau); end end % The problem will be solved using function program % rceval % The output is obtained for the various resistances c = 10.0e-6; r1 = 2500; © 1999 CRC Press LLC
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[v1,t1] = rceval(r1,c); r2 = 10000; [v2,t2] = rceval(r2,c); % plot the voltages plot(t1,v1,'*w', t2,v2,'+w') axis([0 1 0 6]) title('Response of an RC circuit to pulse input') xlabel('Time, s') ylabel('Voltage, V') text(0.55,5.5,'* is for 2500 Ohms') text(0.55,5.0, '+ is for 10000 Ohms') Figure 5.4 shows the charging and discharging curves.
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This note was uploaded on 12/19/2009 for the course ELECTRONIC elm 461 taught by Professor Oğuzkucur during the Spring '09 term at Gebze Institute of Technology.

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ch05 - Attia, John Okyere. Transient Analysis. Electronics...

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