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Introduction to Capacitors: Properties and Time Varying Behavior

Learn about the properties and behavior of capacitors, including the role of dielectric insulators, voltage and current relationships, power and energy calculations. Also, explore series and parallel configurations, and the basics of inductors and time-dependent signals.

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Introduction to Capacitors: Properties and Time Varying Behavior

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  1. Capacitors Inductors Time Constant TC Waveforms RC Circuits – Step Input RL Circuits – Step Input Chapter 4 and 5 Time Varying Circuits Grossman/ Melkonian

  2. CAPACITORS: Section 4.1 Dielectric Insulator  = permittivity  A A C = d Metal plates of area ‘A’ d • Where: ‘C’ is the capacitance in Farads. The farad is a large unit. Typical values of capacitance are in F or pF. ‘’ is the permittivity of the dielectric medium (8.854x10-12 F/m for air). ‘A’ is the cross-sectional area of the two parallel conducting plates. ‘d’ is the distance between the two plates. Grossman/ Melkonian

  3. CAPACITORS: + + Dielectric Insulator  = permittivity Vs Metal plates of area ‘A’ - - iC(t) iC(t) + + Standard Notations (passive sign convention) vC vC - - • When a voltage is applied to the plates, an electric field is produced that causes an electric charge q(t) to be produced on each plate. Grossman/ Melkonian

  4. CAPACITORS: q (t) = CvC(t) Charge of a capacitor since: i(t) = dq(t)/dt and: q(t) = CvC(t) Differentiating: dq(t) = C•dvC(t) iC(t) = dq(t)/dt = C(dvC(t)/dt ) i-v relationship for a capacitor iC(t) = C[dvC(t)/dt] Grossman/ Melkonian

  5. CAPACITORS: • Integrating the i-v relationship of the capacitor: t t  dvC(t) = 1/CiC(t)dt = vC(t) • Letting VC(0) = V0 represent the initial voltage across the capacitor at some time t = t0: t Voltage across a capacitor vC(t) = V0 + 1/C iC(t)dt t  0 t0 iC (t) = C[dvC(t)/dt] Current through a capacitor Grossman/ Melkonian

  6. CAPACITORS: • Capacitor Power: pC(t) = iC(t)•vC(t) = C[dvC(t)/dt]vC(t) pC(t) = d/dt[½CvC2(t)] Power associated with a capacitor • Capacitor Energy: Energy is the integral of power, therefore, wC(t) = ½CvC2(t) Energy associated with a capacitor Grossman/ Melkonian

  7. CAPACITORS: Example 1: Given iC (t) = Io[e-t/Tc] for t  0 and vC(0) = 0V, find the capacitor’s power and energy. Power: pc(t) = iC(t)•vC(t) t t vC(t) = V0 + 1/C  iC(x)dx = 0V + 1/C  Io[e-x/Tc]dx 0 0 vc (t) = [(IoTc/C)(1-e-t/Tc)]V pC(t) = [Ioe-t/Tc][(IoTC/C)(1-e-t/Tc)]W vC(t) iC(t) Power can be positive or negative pc(t) = Io2Tc/C(e-t/Tc -e-2t/Tc) Grossman/ Melkonian

  8. CAPACITORS: Example 1 cont.: Energy: wC(t) = ½CvC2(t) Energy is always positive wC(t) = [(IoTC)2/2C](1-e-t/Tc)2 Waveforms: iC(t) Io iC(t) = Io[e-t/Tc] t 0 Grossman/ Melkonian

  9. CAPACITORS: Example 1 cont.: vC(t) IoTc/C vc(t) = [(IoTc/C)(1-e-t/Tc)]V t 0 pC(t) Io2Tc/4C pc(t) = Io2Tc/C(e-t/Tc -e-2t/Tc) t Tcln2 0 Grossman/ Melkonian

  10. CAPACITORS: Example 1 cont.: wc (t) Io2Tc2/2C t 0 wC(t) = [(IoTC)2/2C](1-e-t/Tc)2 Grossman/ Melkonian

  11. CAPACITORS: Series and Parallel Capacitors: Series: Capacitors connected in series combine like resistors connected in parallel. 1/CEQ = 1/C1 + 1/C2 + 1/C3 + 1/C4 +….+ 1/CN C1 C3 C2 + i(t) vc(t) C4 - C5 C7 C6 Grossman/ Melkonian

  12. CAPACITORS: Series and Parallel Capacitors: Parallel: Capacitors connected in parallel add like resistors connected in series. CEQ = C1 + C2 + C3 + C4 + … + CN iC2(t) iC1(t) iC3(t) vc(t) . . . C1 C2 CN C3 C4 C5 Grossman/ Melkonian

  13. CAPACITORS: Properties of the Capacitor: • The voltage across a capacitor cannot change instantaneously. • Current through a capacitor can change instantaneously. • The capacitor stores energy in its electric field. • The current through a capacitor is zero when the voltage across the capacitor is constant. For example, when the capacitor is fully charged. Therefore, a capacitor acts like an open circuit when a DC voltage is applied. • Capacitors in parallel add. • Capacitors in series combine like resistors connected in parallel. Grossman/ Melkonian

  14. INDUCTORS: vL(t) + - iL(t) Magnetic Field Lines • Inductors are typically made by winding a coil of wire around an insulator or ferromagnetic material. • Current flowing through an inductor creates a magnetic field. i-v relationship for an inductor vL(t) = L[diL(t)/dt] • Where L is the inductance of the coil measured in henrys (H). 1 H = 1 V-s/A Grossman/ Melkonian

  15. INDUCTORS: • The basic results for the inductor can be derived in the same way as we have done for the capacitor: t iL(t) = iL(0) + 1/L  vL(t) dt Current through an inductor pL(t) = iL(t)•vL(t) = d/dt[WL(t)] Power associated with an inductor WL(t) = 1/2LiL2(t) Energy associated with an inductor Grossman/ Melkonian

  16. INDUCTORS: Properties of the Inductor: • Inductor current is continuous. Cannot change instantaneously. (Would require infinite power). • The voltage across an inductor can change instantaneously. • When the current through an inductor is constant, the voltage is zero. Therefore, the inductor acts like a short circuit when a DC source is applied. • Inductors connected in series add like resistors connected in series. • Inductors connected in parallel combine like resistors connected in parallel Grossman/ Melkonian

  17. TIME-DEPENDENT SIGNALS: Section 4.2 • Sources that produce currents or voltages that vary with time are called time-dependent signal sources. • The sinusoidal waveform is one of the most important time-dependent signals. + + V(t) i(t) ~ V(t), i(t) - - Sinusoidal source Generalized time-dependent signals Grossman/ Melkonian

  18. TIME-DEPENDENT SIGNALS: • An important class of time-dependent signals is a periodic signal. • A periodic signal satisfies the following equation: x(t) = x(t + nT) n = 1, 2, 3, … where T is the period of the signal • A generalized sinusiod is defined as follows: x(t) = Acos(t + ) where f = natural frequency = 1/T cycles/s or Hz  = radian frequency = 2f rad/s  = 2t/T = 360t/T rad Grossman/ Melkonian

  19. TIME-DEPENDENT SIGNALS: v(t) v(t) t t Damped sinusoid Square wave v(t) v(t)   Pulse width  t t Pulse train Sawtooth wave Grossman/ Melkonian

  20. TIME-DEPENDENT SIGNALS: Sinusoidal Waveform Radial frequency Amplitude Phase shift x(t) = Acos(t + ) Time A T Reference cosine t A T Arbitrary sinusoid t t Grossman/ Melkonian

  21. TIME-DEPENDENT SIGNALS: Root Mean Square • The average value of a sinusoidal signal is zero, independent of its amplitude and frequency. Therefore, another method must be used to quantify the strength of a time-varying signal. • The operation of computing the Root-Mean-Square of a waveform is a method for quantifying the strength of a time-varying signal. T Root-mean-square xrms =  1/T  x2 (t) dt 0 Grossman/ Melkonian

  22. TIME-DEPENDENT SIGNALS: Example 2: Calculate the rms value of the sinusoidal voltage v(t) = Vsin(t). T vrms =  1/T  v2 (t) dt 0 T = 1/f = 2/ 2/ vrms =  /2V2sin2(t)dt 0 2/ vrms =  V2/21/2 - 1/2cos(2t)dt 0 Grossman/ Melkonian

  23. TIME-DEPENDENT SIGNALS: Example 2 cont.: 2/ vrms =  V2/2[t/2 - 1/4sin(2t)] 0 vrms =  V2/2[(2/2 - /4sin(4/) – (0 – 0)] 0 vrms =  V2/2(/ - /4sin(4) – 0) V vrms = vrms = 0.707V  2 Grossman/ Melkonian

  24. RC CIRCUIT, STEP RESPONSE: Sections 4.3 (pgs 161 & 162), 5.1, 5.2, 5.3 • Assume we have the following series circuit containing a voltage source, a resistor, and a capacitor. How do we write an expression for the voltage across the capacitor? R Switch t = 0 iC(t) + + VT vC(t) C - - Grossman/ Melkonian

  25. RC CIRCUIT, STEP RESPONSE: • Writing KVL for the circuit: – VS + Ric(t) + vC(t) = 0 ic(t) = C(dvc(t)/dt ) RC(dvC(t)/dt) + vC(t) = VS First order linear differential equation. RC(dvC(t)/dt) + vC(t) = VS for t  0 R Switch t = 0 iC(t) + + VS vC(t) C - - Grossman/ Melkonian

  26. RC CIRCUIT, STEP RESPONSE: • Since the circuit is linear, we can separate the solution vC(t) into two components: vC(t) = vN(t) + vF(t) • Natural Response: 1. RC(dvN(t)/dt) + vN(t) = 0 t  0 Natural Response: VS = 0 • The classical approach to solving this type of an equation is to try a solution of the following form: vN(t) = Kest where K and s are constants Grossman/ Melkonian

  27. RC CIRCUIT, STEP RESPONSE: • Substituting vN(t) = Kest into the above equation: RCsKest + Kest = 0 Kest(RCs + 1) = 0 K = 0 is trivial solution RCs + 1 = 0 Characteristic Equation • Solving the Characteristic Equation: s = -1/RC • Therefore, the Natural Response has the form: vN(t) = Ke-t/RC t > 0 Grossman/ Melkonian

  28. RC CIRCUIT, STEP RESPONSE: • Forced Response: 2. RC(dvF(t)/dt) + vF(t) = VS t  0 Forced Response • Equation 2 requires the linear combination of vF(t) and its derivative equal a constant VS. Setting vF(t) = VS satisfies this condition. Combining the Natural and Forced Responses: vC(t) = vN(t) + vF(t) = Ke-t/RC + VSt > 0 (General Solution) • Evaluating K using initial conditions (at t=0, the voltage across the capacitor): v(0) = V0 = Ke0 + VS = K + VS This requires K = (V0 – VS) Grossman/ Melkonian

  29. RC CIRCUIT, STEP RESPONSE: • Substituting into the general solution: vC(t) = (V0 – VS)e-t/RC + VSt > 0 Decaying exponentialConstant Important Values: RC: Defined as the time constant, sometimes written as TC or . The time constant depends only on fixed circuit parameters. V0: The initial condition or initial voltage across the capacitor. Sometimes written as Vi or vC(0) VS: The Thevenin voltage seen by the capacitor as t. Sometimes written as Vf or vC(). Grossman/ Melkonian

  30. RC CIRCUIT, STEP RESPONSE: Important Values cont.: R: The Thevenin resistance or equivalent resistance seen by the capacitor. Sometimes written as RT. t(0-): The instant of time just prior to t=0 or the switch changing its position. t(0): The time t=0 or the instant the switch changes its position. t(0+): The instant of time just after t=0 or just after the switch changes its position. Grossman/ Melkonian

  31. RC CIRCUIT, STEP RESPONSE: Important Concepts: • The voltage across a capacitor cannot change instantaneously. Therefore, vC(0-) = vC(0) = vC(0+). • The current through a capacitor can change instantaneously. • A capacitor acts like a short circuit the instant a voltage is applied across its terminals. • When a DC source is applied to a capacitor, it becomes an open circuit as t. vC(t) = (Vi – Vf)e-t/TC + Vft > 0 Grossman/ Melkonian

  32. RL CIRCUIT, STEP RESPONSE: • Assume we have the following series circuit containing a voltage source, a resistor, and an inductor. We can write an expression for the current through the inductor using a similar procedure as that applied previously for finding the voltage across a capacitor. R Switch t = 0 iL(t) + + VT vL(t) L - - Grossman/ Melkonian

  33. RL CIRCUIT, STEP RESPONSE: • Applying a similar procedure as that applied to finding the voltage across a capacitor, we obtain the equation for the current through an inductor when the input is a step function. iL(t) = (I0 – IS)e-Rt/L + ISt > 0 Decaying exponentialConstant Important Values: L/R: Defined as the time constant, sometimes written as TC or . The time constant depends only on fixed circuit parameters. I0: The initial condition or initial current through the inductor. Sometimes written as Ii or iL(0) IS: The Norton current seen by the inductor as t. Sometimes written as If or iL(). Grossman/ Melkonian

  34. RL CIRCUIT, STEP RESPONSE: Important Values cont.: R: The Norton resistance or equivalent resistance seen by the inductor. Sometimes written as RN. t(0-): The instant of time just prior to t=0 or the switch changing its position. t(0): The time t=0 or the instant the switch changes its position. t(0+): The instant of time just after t=0 or just after the switch changes its position. Grossman/ Melkonian

  35. RL CIRCUIT, STEP RESPONSE: Important Concepts: • The current through an inductor cannot change instantaneously. Therefore, iL(0-) = iL(0) = iL(0+). • The voltage across an inductor can change instantaneously. • A inductor acts like an open circuit the instant a voltage or current is applied to its terminals. • When a DC source is applied to a inductor, it becomes a short circuit as t. iL(t) = (Ii – If)e-t/TC + Ift > 0 Grossman/ Melkonian

  36. RC CIRCUIT, STEP RESPONSE: Example 3: • Calculate the RC time constant (TC), Vi, Vf, the voltage across the capacitor, and the current through the capacitor for t > 0. The switch has been open for a long time before closing. 1.5K Switch t = 0 iC(t) + + 16V vC(t) 1.5F 1.5K - - Grossman/ Melkonian

  37. RC CIRCUIT, STEP RESPONSE: Example 3 cont.: • RC time constant (TC): TC = RTH•CEQ = 1.5K 1.5K•1.5F TC = 1.125ms 1.5K Switch t = 0 iC(t) + + 16V vC(t) 1.5F 1.5K - - Grossman/ Melkonian

  38. RC CIRCUIT, STEP RESPONSE: Example 3 cont.: Vi = 0V The capacitor has had time to discharge. 1.5K Vf = 16V Voltage divider 1.5K + 1.5K Vf = 8V Switch t = 0 1.5K iC(t) + + 16V vC(t) 1.5F 1.5K - - Grossman/ Melkonian

  39. RC CIRCUIT, STEP RESPONSE: Example 3 cont.: vC(t) = (Vi – Vf)e-t/TC + Vft > 0 • vC(t) t > 0 vC(t) = (0V - 8V)e-t/TC + 8V t > 0 vC(t) = 8 - 8e-t/1.125msV t > 0 1.5K Switch t = 0 iC(t) + + 16V vC(t) 1.5F 1.5K - - Grossman/ Melkonian

  40. RC CIRCUIT, STEP RESPONSE: Example 3 cont.: iC(t) = C[dvC(t)/dt] • iC(t) t > 0 iC(t) = 1.5F d/dt[8 - 8e-t/1.125ms]At > 0 iC(t) = 10.67e-t/1.125ms mA t > 0 Switch 1.5K t = 0 iC(t) + + 16V vC(t) 1.5F 1.5K - - Grossman/ Melkonian

  41. RL CIRCUIT, STEP RESPONSE: Example 4: • Calculate the L/R time constant (TC), Ii, If, the current through the inductor, and the voltage across the inductor for t > 0. The switch has been closed for a long time. 5K 1K Switch t = 0 iL(t) + + 14V 3k 5mH vL(t) - - Grossman/ Melkonian

  42. RL CIRCUIT, STEP RESPONSE: Example 4 cont.: • L/R time constant (TC): 5mH TC = LEQ/RN = 3k + 1K TC = 1.25s 1K Switch t = 0 5K iL(t) + + 14V 3k 5mH vL(t) - - Grossman/ Melkonian

  43. RL CIRCUIT, STEP RESPONSE: Example 4 cont.: 1/1K Source transformation and current divider 2.8mA Ii = 1/5K + 1/3k + 1/1K If = 0A Ii = 1.83mA 5K 1K Switch t = 0 iL(t) + + 14V 3k 5mH vL(t) - - Grossman/ Melkonian

  44. RL CIRCUIT, STEP RESPONSE: Example 4 cont.: iL(t) = (Ii – If)e-t/TC + Ift > 0 • iL(t): iL(t) = (1.83 - 0)e-t/1.25s+ 0 mAt > 0 iL(t) = 1.83e-t/1.25s mAt > 0 5K 1K Switch t = 0 iL(t) + + 14V 3k 5mH vL(t) - - Grossman/ Melkonian

  45. RL CIRCUIT, STEP RESPONSE: Example 4 cont.: • vL(t): vL(t) = L[diL(t)/dt] vL(t) = 5mH•d/dt[1.83x10-3•e-t/1.25s]V vL(t) = - 7.32e-t/1.25s V t > 0 5K 1K Switch t = 0 iL(t) + + 14V 3k 5mH vL(t) - - Grossman/ Melkonian

  46. RC CIRCUIT, STEP RESPONSE: Example 5 : • The switch has been in position ‘A’ for a long time. At t=0, the switch moves to position ‘B’. Calculate Vi, Vf, TC, and vC(t), iC(t), and iR(t) for t > 0. 800 t = 0 A iC(t) B iR(t) + + + vC(t) 2F 10V 6V 400 - - - Grossman/ Melkonian

  47. RC CIRCUIT, STEP RESPONSE: Example 5 cont.: 400 Vi = 10V Vi = 3.34V Voltage divider 800 + 400 400 Vf = 6V Vf = 2V Voltage divider 800 + 400 800 t = 0 A iC(t) B iR(t) + + + vC(t) 2F 10V 6V 400 - - - Grossman/ Melkonian

  48. RC CIRCUIT, STEP RESPONSE: Example 5 cont.: • TC: TC = RTH•CEQ = 800 400•2F TC = 533.4 s 800 t = 0 A iC(t) B iR(t) + + + vC(t) 2F 10V 6V 400 - - - Grossman/ Melkonian

  49. RC CIRCUIT, STEP RESPONSE: Example 5 cont.: • vC(t) t > 0 vC(t) = (Vi – Vf)e-t/TC+ Vft > 0 vC(t) = (3.34V - 2V)e-1875t + 2V t > 0 vC(t) = 2 + 1.34e-1875t V t > 0 800 t = 0 A iC(t) B iR(t) + + + vC(t) 2F 10V 6V 400 - - - Grossman/ Melkonian

  50. RC CIRCUIT, STEP RESPONSE: Example 5 cont.: iC(t) = C[dvC(t)/dt] • iC(t) t > 0 iC(t) = 2F d/dt[2 + 1.34e-1875t]At > 0 iC(t) = -5.01e-1875tmA t > 0 800 t = 0 A iC(t) B iR(t) + + + vC(t) 2F 10V 6V 400 - - - Grossman/ Melkonian

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