Response of RC Circuit Driven by a Square Wave, Second Order Circuits, Series RLC Circuit and the LC Circuit

Examples of Transient RC and RL Circuits. The Series RLC Circuit Impulse response of RC Circuit. Let’s examine the response of the circuit shown on Figure 1. The form of the source voltage Vs is shown on Figure 2. R C Vs + vc - Figure 1. RC circuit Vs Vp 0 tp t Figure 2. We will investigate the response vc(t ) as a function of the τ p and Vp . The general response is given by: −t ⎛ ⎞ vc(t ) = Vp ⎜1 − e RC ⎟ 0 ≤ t ≤ tp ⎝ ⎠ (1.1) If tp RC the capacitor voltage at t = tp is equal to Vp . Therefore for times t > tp the response becomes ⎛ − (t −tp ) ⎞ vc(t ) = Vp ⎜ e RC ⎟ tp ≤ t ⎝ ⎠ (1.2) A general plot of the response is shown on Figure 3 for RC = 1sec, tp = 6 sec, Vp = 10Volts Figure 3 If the pulse becomes narrower, the value of vc will not reach the maximum value. By expanding the exponential in Equation (1.1) we obtain, 2 3 ⎛ ⎡ ⎤⎞ t 1⎛ t ⎞ 1⎛ t ⎞ … vc(t ) = Vp ⎜1 − ⎢1 − + ⎜ − + ⎥ ⎟⎟ 0 ≤ t ≤ tp ⎜ ⎢ RC 2 ⎝ RC ⎟⎠ 6 ⎜⎝ RC ⎟⎠ ⎥⎦ ⎠ ⎝ ⎣ When RC (1.3) t the higher order terms may be neglected resulting in vc(t ) Vp t 0 ≤ t ≤ tp RC (1.4) At the end of the pulse (at t = tp ) the voltage becomes vc(t = tp ) Vptp RC (1.5) For t > tp the response becomes vc = t −tp ) ⎞ Vp tp ⎛ − (RC e ⎜ ⎟ RC ⎝ ⎠ (1.6) The product Vp tp is the area of the pulse and thus the response is proportional to that area. As the pulse becomes narrower (i.e. as tp → 0 ) equation (1.6) simplifies to vc −t ⎞ Vp tp ⎛ RC e ⎜ ⎟ RC ⎝ ⎠ (1.7) If we constrain the area of the impulse to a constant A = Vp tp , then as the pulse becomes narrower, the amplitude Vp increases, resulting in an impulse of strength A. Therefore the response of an impulse of strength A is −t A RC vc = e RC Figure 4. Impulse response of RC circuit (1.8) The spark plug in your car (a simplified model) Consider the circuit shown on Figure 5. The battery Vb corresponds to the 12 Volt car battery. The spark plug is connected actors the inductor and current may flow though it only if the voltage across the gap of the plug exceeds a very large value (about 20 kV). R + Vb - + vL L spark plug Figure 5 When the switch is closed, the current through the inductor reaches a maximum value of Vb / R . The equation that describes the evolution of the current with the switch closed is i(t ) = −t ⎞ Vb ⎛ L/R − 1 e ⎜ ⎟ R⎝ ⎠ (1.9) And the corresponding voltage across the inductor is given by vL(t ) = Vb e −t L/ R (1.10) When the switch is opened, the current path is effectively broken and thus the time rate of change of the current becomes arbitrarily large. Since the voltage is proportional to di / dt , the voltage developed across the inductor could become very large. As an example, let’s consider a system with a resistance of 5Ω, a solenoid with an inductance of 10mH connected to a 12 Volt battery. How long does it take for the solenoid to reach 99% of its maximum value? If the switch is opened in 1µs, what is the voltage developed across the solenoid? The time constant of the system is L 0.01 = = 0.002sec R 5 The maximum current that can flow in the system is of the maximum value is given by 12 A = 2.4 A . The time to reach 99% 5 −t 0.99 = 1 − e 0.002 The voltage across the coil when the switch is opened is v=L ∆i 2.4 = 0.01 = 24kV 1× 10−6 ∆t Response of RC circuit driven by a square wave. Let’s now consider the RC circuit shown on Figure 6(a) driven by a square wave signal of the form shown on Figure 6(b). + vR - R C Vs + vc - (a) Vs Vp T t -Vp (b) Figure 6 The response vc(t) is given by −t response = final value + [initial value - final value] e τ (1.11) By assuming that the initial value of the voltage across the capacitor is –Vp the response during the first half cycle of the square wave is vc(t ) = Vp + [ -Vp - Vp ] e −t ⎡ ⎤ = Vp ⎢1 - 2e RC ⎥ ⎣ ⎦ During the second half cycle the initial condition is −t RC (1.12) −T / 2 ⎡ ⎤ vc(T / 2) = Vp ⎢1 - 2e RC ⎥ ⎣ ⎦ (1.13) And the complete response during the second half of the first cycle becomes −T / 2 −t ⎡ ⎡ ⎤ RC ⎤ RC vc(t ) = -Vp + ⎢Vp ⎢1 - 2e ⎥ + Vp ⎥ e ⎢⎣ ⎣ ⎥⎦ ⎦ (1.14) Similarly the response during the first part of the second cycle starts with the value of vc at t=T and evolves towards the value Vp. If the time constant is small compared to the period of the square wave, the response will reach the maximum and minimum values of the square wave as shown on Figure 7, where RC = 1× 10 −4 sec and thus T/2=10RC. Figure 7 As the time constant RC increases, it takes longer for the response to reach the maximum value. Figure 8 shows a plot of the response for T/2=RC. Note that the response does not reach the maximum values of the input signal and the average value of the response is equal to the average value of the input signal. Figure 8 Figure 9(a) and Figure 9(b) show the system response for RC=5T/2 for a square wave with a duty factor of 50% that varies between 0 and 5 Volts. Notice that the average value is reached within a certain number of oscillations and that there is a variation of the response “ripple” about the average value. The magnitude of this ripple is inversely proportional to the time constant RC. This is the first step that one must take when an AC signal is converted to DC. Next week, when we learn about the diode, we will explore this circuit further. (a) (b) Figure 9 Second Order Circuits Series RLC circuit The circuit shown on Figure 10 is called the series RLC circuit. We will analyze this circuit in order to determine its transient characteristics once the switch S is closed. S + vR - + vL - L R Vs C + vc - Figure 10 The equation that describes the response of the system is obtained by applying KVL around the mesh vR + vL + vc = Vs (1.15) The current flowing in the circuit is i=C dvc dt (1.16) And thus the voltages vR and vL are given by vR = iR = RC vL = L dvc dt di d 2 vc = LC 2 dt dt (1.17) (1.18) Substituting Equations (1.17) and (1.18) into Equation (1.15) we obtain d 2 vc R dvc 1 1 + + vc = Vs 2 dt L dt LC LC (1.19) The solution to equation (1.19) is the linear combination of the homogeneous and the particular solution vc = vc p + vch The particular solution is vc p = Vs (1.20) And the homogeneous solution satisfies the equation d 2 vch R dvch 1 + + vch = 0 2 dt L dt LC (1.21) Assuming a homogeneous solution is of the form Ae st and by substituting into Equation (1.21) we obtain the characteristic equation s2 + R 1 =0 s+ L LC (1.22) R 2L (1.23) By defining α= And ωο = 1 LC (1.24) The characteristic equation becomes s 2 + 2α s + ωο2 = 0 (1.25) The roots of the characteristic equation are s1 = −α + α 2 − ωο2 (1.26) s 2 = −α − α 2 − ωο2 (1.27) And the homogeneous solution becomes vch = A1e s1t + A2 e s 2t (1.28) vc = Vs + A1e s1t + A2 e s 2t (1.29) The total solution now becomes The parameters A1 and A2 are constants and can be determined by the application of the dvc(t = 0) . initial conditions of the system vc (t = 0) and dt The value of the term α 2 − ωο2 determines the behavior of the response. Three types of responses are possible: 1. α = ωο then s1 and s2 are equal and real numbers: no oscillatory behavior Critically Damped System 2. α > ωο . Here s1 and s2 are real numbers but are unequal: no oscillatory behavior Over Damped System vc = Vs + A1e s1t + A2 e s 2 t 3. α < ωο . α 2 − ωο2 = j ωο2 − α 2 In this case the roots s1 and s2 are complex numbers: s1 = −α + j ωο2 − α 2 , s 2 = −α − j ωο2 − α 2 . System exhibits oscillatory behavior Under Damped System Important observations for the series RLC circuit. • • • As the resistance increases the value of α increases and the system is driven towards an over damped response. 1 The frequency ωο = (rad/sec) is called the natural frequency of the system LC or the resonant frequency. L The quantity has units of resistance C Figure 11 shows the response of the series RLC circuit with L=47mH, C=47nF and for three different values of R corresponding to the underdamped, critically damped and overdamped case. We will construct this circuit in the laboratory and examine its behavior in more detail. (a) Under Damped. R=500Ω (b) Critically Damped. R=2000 Ω (c) Over Damped. R=4000 Ω Figure 11 The LC circuit. In the limit R → 0 the RLC circuit reduces to the lossless LC circuit shown on Figure 12. S + vL - L C + vc - Figure 12 The equation that describes the response of this circuit is d 2 vc 1 vc = 0 + dt 2 LC (1.30) Assuming a solution of the form Ae st the characteristic equation is Where ωο = s 2 + ωο2 = 0 (1.31) s1 = + jωο (1.32) s 2 = − jωο (1.33) 1 LC The two roots are And the solution is a linear combination of A1e s1t and A2e s 2t vc(t ) = A1e jωot + A2e − jωο t (1.34) By using Euler’s relation Equation (1.34) may also be written as vc(t ) = B1cos(ωο t ) + B 2sin(ωο t ) (1.35) The constants A1, A2 or B1, B2 are determined from the initial conditions of the system. For vc(t = 0) = Vo and for dvc(t = 0) = 0 (no current flowing in the circuit initially) we dt have from Equation (1.34) A1 + A2 = Vo (1.36) jωo A1 − jωo A2 = 0 (1.37) And Which give A1 = A2 = Vo 2 (1.38) And the solution becomes Vo jωot − jωο t e +e 2 = Vo cos(ωot ) vc(t ) = ( ) (1.39) The current flowing in the circuit is dvc dt = −CVoωο sin(ωο t ) i=C (1.40) And the voltage across the inductor is easily determined from KVL or from the element di relation of the inductor vL = L dt vL = −vc = −Vo cos(ωot ) (1.41) Figure 13 shows the plots of vc (t ), vL(t ), and i (t ) . Note the 180 degree phase difference between vc(t) and vL(t) and the 90 degree phase difference between vL(t) and i(t). Figure 14 shows a plot of the energy in the capacitor and the inductor as a function of time. Note that the energy is exchanged between the capacitor and the inductor in this lossless system (a) Voltage across the capacitor (b) Voltage across the inductor (c)Current flowing in the ciruit Figure 13 (a) Energy stored in the capacitor (b) Energy stored in the inductor Figure 14