Time Constant of RC Circuit

10 Key excerpts on "Time Constant of RC Circuit"

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  eBook - ePub

  Introductory Electrical Engineering With Math Explained in Accessible Language

  • Magno Urbano(Author)
  • 2019(Publication Date)
  • Wiley

    (Publisher)

  C is the capacitance, in Farads. RC is time constant.

  25.3 RC Time Constant

  The RC time constant is the time required to charge a capacitor through a resistor, from an initial charge of 0–63.2% of the value of an applied DC voltage or for a discharging capacitor to lose 63.2% of its initial charge.

  This value is derived from the mathematical formulas of charge and discharge for the capacitor. The RC time constant, also called tau (τ) and measured in seconds, is equal to the product of the circuit resistance in Ohms and the circuit capacitance in Farads.

  RC TIME CONSTANT

  • τ is the RC time constant, in seconds.
  • R is the resistance, in Ohms.
  • C is the capacitance, in Farads.

  25.3.1.1 Transient and Steady States

  The period of time in which an RC circuit exhibits variation of current and voltage, after a voltage pulse is applied to it, is called transient phase. After the transient phase is over, the circuit stabilizes and will not exhibit any other variations. The stable phase is called steady state.

  25.3.1.1.1 How Long Does the Transient Phase Last?

  When a capacitor charges, the amount of charge increases following an exponential curve. For this reason, the ratio of charge decreases with time. Hence, an infinite amount of time is required to reach the maximum charge.

  In practice, engineers cannot wait indefinitely for a capacitor to charge. In electrical engineering, a capacitor is considered charged after a time equal to five time constants.

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  Basic AC Circuits

  • Clay Rawlins(Author)
  • 2000(Publication Date)
  • Newnes

    (Publisher)

  R = IR. Therefore, in the circuit,

  The voltage across the resistor is equal to 10 volts when the switch is initially placed in position 2. The voltage across the resistor directly dependent on the circuit current flowing, decreases to zero exponentially as the current decreases to zero. This shown in Figure 10.13 . Note the shape is the same as the current waveform of Figure 10.10 .

  RC TIME CONSTANTS

  A summary of the current and voltage waveforms for dc RC circuits during the charge and discharge time of the capacitor is shown in Figure 10.14 . However, a question immediately comes to mind. How long does the charging and discharging take? The following discussion should answer that question.

  Figure 10.14 A Summary of the Current and Voltage Waveforms for DC RC Circuits During Capacitor Charge and Discharge Times

  RC Time Constant Defined

  The time required for a capacitor to fully charge is measured in terms of a quantity called the time constant of the circuit.

  The time constant of an RC circuit is defined as the amount of time necessary for the capacitor to charge to 63.2 percent of its final voltage.

  The time constant is symbolized by the Greek letter, tau, τ. For an RC circuit, tau is equal to the value of the resistance times the value of capacitance in the circuit. Expressed mathematically,

  (10–4)

  Where tau is measured in seconds, R in ohms, and C in farads.

  Calculation of an RC Time Constant

  The resistance of the example RC circuit being discussed (Figure 10.1 ) is 10 kilohms. Its capacitance is 1 microfarad. The time constant for the circuit is calculated:

  The time constant is 10 milliseconds.

  In the circuit, the capacitor is charging to 10 volts. 63.2 percent of 10 volts is 6.32 volts. Thus, as shown in Figure 10.15

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  eBook - ePub

  Further Electrical and Electronic Principles

  • C R Robertson, Christopher Robertson(Authors)
  • 2010(Publication Date)
  • Routledge

    (Publisher)

  o to zero. The equations for all the quantities will be as follows:

  The graphs for vC and i are shown in Fig. 8.8 .

  Fig. 8.8

  Note: The time constant for the C-R circuit was defined previously in terms of the capacitor charging. However, a time constant also applies to the discharge conditions. It is therefore better to define the time constant in a more general manner, as follows:

  The time constant of a circuit is the time that it would have taken for any transient variable to change, from one steady state to a new steady state, if it had maintained its original rate of change.

  Worked Example 8.2

  Q      A C-R charge/discharge circuit is shown in Fig. 8.9 . The switch has been in position ‘1’ for a sufficient time to allow the capacitor to become fully discharged. (a) If the switch is now moved to position ‘2’, calculate the capacitor p.d. after 0.3 s. (b) If after the 0.3 s the switch is now returned to position ‘1’, calculate the value of discharge current and capacitor p.d. after a further period of 0.3 s. Fig. 8.9 A

  C = 0.5 μF; R1 = 220 kΩ; R2 = 110 kΩ; E = 150 V

  (a)    When charging, only resistor R1 is connected in series with the capacitor, so R2 may be ignored.

  (b)    When discharging, both R1 and R2 are connected in series with the capacitor, so their combined resistance R = R1 + R2 , will determine the discharge time constant.

  Also, the capacitor will be discharging from a voltage of 140.2 V. Thus, the initial discharge current will be: Therefore after 0.3 s of the discharge cycle, the current will be:  

  Note: It should be obvious from equations (8.7) and (8.10) that, in the discharge circuit, the p.d. across the capacitor and the circuit resistance must be the same value. Thus the last part of (b) could have been obtained thus:

  Worked Example 8.3

  Q

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  eBook - ePub

  Digital Systems Design, Volume III

  Latch–Flip-Flop Circuits and Characteristics of Digital Circuits

  • Larry Massengale(Author)
  • 2019(Publication Date)
  • Momentum Press

    (Publisher)

  The charging and discharging rate of a capacitor in series with a resistor and a battery is determined by the resistance and capacitance within the circuit. This resistance (R) capacitance (C) factor is known as the time constant (TC) or tau (τ) of the system and is expressed as τ = R × C. The rate of current and voltage change follows the exponential function. During the charging phase, the transfer of electrons is very rapid at first, slowing down as the potential across the capacitor equals the battery voltage; the electron transfer will then cease and the plates will have a net charge determined by Q (charge potential across the plates). The current jumps to a value limited only by the resistance of the network and then decays to zero as the plates are charged. The charging process takes approximately five time constants, with the first time constant producing 63 percent of the charge while the time between the fifth and sixth time constant is less than 1 percent of the charge. During the discharging phase, the capacitor acts as a battery with a decreasing terminal voltage. The discharge rate is once again determined by the RC time constant and less than 1 percent during the firth to sixth time constant. The current will also decrease with the same exponential rate.

  Digital circuits or systems have delays between their inputs and outputs. These delays are known as time delays or time constants, which depend upon reactive components, either capacitive or inductive. The response time is measured in units of Tau ( τ ).

  The charge on the plates of the capacitor is given as charge (Q) = capacitance (C) times the voltage (V) or Q = CV. This charging (storage) and discharging (release) of a capacitor’s energy is never instant but takes a certain amount of time to occur with the time taken for the capacitor to charge or discharge within a certain percentage of its maximum supply value.

  2.2.1. CHARGING

  If a resistor is connected in series with the capacitor forming an RC circuit, the capacitor will charge up gradually through the resistor until the voltage across the capacitor reaches that of the supply voltage. The time, also called the transient response, required for the capacitor to fully charge is equivalent to about five time constants (5T).

  The initial conditions of a fully discharged circuit are shown in Figure 2.1 , where time t = 0, current i = 0, and charge q = 0. When the switch is closed at time t = 0, current begins to flow into the capacitor through the series resistor.

  Figure 2.1. Initial conditions of series RC circuit charging

  Vs = 10 v R1 = 25 kΩ C1 = 1,000 μf

  After four time constants, the capacitor in this series RC circuit is virtually fully charged and the voltage across the capacitor is at 98 percent of its maximum value and 98 percent of the supplied voltage (Vs). The time period taken for the capacitor to reach four time constants is known as the transient period, as shown in Figure 2.2 . After five time constants the capacitor is considered to be fully charged; the capacitor voltage is approximately equal to the supply voltage. At this point no further current will flow in the series RC circuit. The time period after four time constants is known as the steady state period, as shown in Figure 2.2

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  eBook - PDF

  Electrical and Electronic Principles

  Volume 3

  • S.A. Knight(Author)
  • 2014(Publication Date)
  • Newnes

    (Publisher)

  At any point on the charging cycle the voltage across C opposes the applied voltage, and this is a continuing process all the time the charge is taking place. Let the capacitor voltage be at some particular value v' at any instant during the charge, then the current R C. dv c and so dv c _ V at CR V/s This is clearly less than the initial rate of increase given by V/CR. Hence the curve of capacitor voltage against time becomes progres-sively less steep as time goes on and will constantly approach the limiting condition of a horizontal line as (V—v') approaches zero. Now, referring to Figure 2.8, at any particular instant. t x the capacitor has Figure 2.8 24 D.C. transients D.C. transients 29 still to be charged (V-v') volts. If this charging rate was then to be maintained, the time for the charge to be completed would be (t 2 -ίι) seconds; then tx = voltage available rate of increase of voltage V-v' = CRs (V-v')/CR Hence t 2 - t x = Γ, the time constant. This argument must be true for any selected point on the charging curve. Hence at any point on the curve the time remaining to complete the charge, if the charge then continued at a constant rate, is CR s. In theory the capacitor can never be completely charged, but after a time interval equal to SCR s the charge is within 1% of its final value. In Figure 2.9 we have used this information to obtain an approxi-mate graph of voltage against time. The time constant is the time it would take the capacitor to reach the value of the applied voltage V v X -0.8V + 0.6V + 0.4V + 0.2V + ΑΤ/ΐ,-^ΐ Figure 2.9 if the initial rate of increase could be continued. As we have seen, this statement applies to any instant in the charging cycle, but the final value of the voltage is V in each case. So we can draw a number of horizontal lines each of length equal to CR, at intervals corresponding to equal intervals of voltage.

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  Resistive, Capacitive, Inductive, and Magnetic Sensor Technologies

  • Winncy Y. Du(Author)
  • 2014(Publication Date)
  • CRC Press

    (Publisher)

  Instead, it takes some time to reach a value close to V DD . Similarly, when switching S back to Position 2, the volt-age does not immediately disappear, but it takes some time to reach a value close to zero. Thus, the time it takes for the voltage to reach or decay to a threshold V C , for example, 0.5 V DD , is a measure of the resistance R if C and V DD are fixed. The resis-tance R can directly relate to the time t by R t C V V RC DD C = ln( ) ( , ) / For an circuit discharging (6.65) The RC decay method provides an accurate and simple time measurement to determine the resistance value. It can be applied to both resistive and capacitive sensors and to convert resistance or capacitance into time. Its applications include thermistors, capacitive level sensors, and capacitive touch sensors. EXAMPLE 6.16 Derive Equation 6.65 based on the RC circuit shown in Figure 6.29. S OLUTION Applying Kirchhoff’s voltage law to the circuit with the switch S in Position 1 results in the charging circuit equation : V iR Q C DD --= 0 R R Other signal conditioning circuits V DD I SEN R SEN Outp ut FIGURE 6.28 A resistance-to-current conversion circuit. C V DD S V C 1 2 R FIGURE 6.29 An RC circuit converting resistance to time. 344 Resistive, Capacitive, Inductive, and Magnetic Sensor Technologies where Q is the charge of the capacitor; i is the current flowing through the circuit. With the switch S in Position 2, it yields the discharging circuit equation : --= iR Q C 0 Since i = dQ / dt , we have − ( dQ / dt ) R − ( Q / C ) = 0. Integrate this equation and use “ t = 0, Q = CV DD and t = t , Q = CV C ” as well as “ Q = CV ”, resulting in -= ⇒ -= ⇒ --[ ] = ∫ ∫ dQ Q RC dt Q t RC CV CV CV CV t CV CV C DD DD C DD C 1 0 ln ln( ) ln( ) t RC CV CV t RC V V t RC R t C V V V V e DD C DD C D D C C DD ln( ) ln( ) ln ln ( ) -= ⇒ = ⇒ = = / -t RC RC ( , ) circuit discharging where RC (often denoted as τ ) is the time constant . EXAMPLE 6.17 An RC circuit is used to monitor a RTD.

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  Electronic and Electrical Engineering

  Principles and Practice

  • Lionel Warnes(Author)
  • 2017(Publication Date)
  • Red Globe Press

    (Publisher)

  By KVL in Figure 4.1, for t 0 and by Ohm’s law, v R = iR . Now the current through a capacitor is given by i C = C d v C /d t , and Transients 86 (4.2) (4.3) (4.4) (4.5) (4.6) (4.7) Figure 4.2 The v ( t ) graph for a charging capacitor in this case i C = i , so we have Rearranging to separate the variables And then integrating which yields where is a constant of integration. Taking antilogs produces where is exp ( / RC ). By examining the initial conditions we can find . Initially, v C = 0 when t = 0 and Equation 4.6 reduces to v C = V = 0, as exp(0) = 1 and hence = V . The complete expression for v C ( t ) is when t 0. A graph of this equation is shown in Figure 4.2. The time scale is normalised to units of RC , known as the time constant , , of the circuit. After one time constant, Equation 4.7 shows that V C = V (1 e 1 ) = 0.63 V (the capacitor is 63% charged), and after five time constants have passed V C = 0.99 V the capacitor is effectively fully charged and the ‘transient’ is over. If the current-time graph for t 0, is needed it can be found from Equation 4.7 by differentiating: Chapter 4 87 (4.8) (4.9) Figure 4.3 The current decays exponentially as the capacitor discharges Figure 4.4 The series RL circuit excited by a direct voltage at t = 0 (4.10) (4.11) and multiplying the result by C Equation 4.9 shows that the current decays exponentially with time as in Figure 4.3. After five time constants have passed, the current has all but ceased and a steady state has been reached. 4.1.2 The series RL circuit Figure 4.4 shows a series RL circuit excited by a step voltage. The analysis takes the same path as for the series RC circuit. After the switch is closed, by KVL for t 0. As usual, v R = iR and v L = L d i /d t , giving Rearranging this:

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  The Fundamentals of Electrical Engineering

  for Mechatronics

  • Felix Hüning(Author)
  • 2014(Publication Date)
  • De Gruyter Oldenbourg

    (Publisher)

  Fig. 7.23 .

  Fig. 7.23 : A capacitor’s voltage as a function of time for an RLC series connection, voltage source U connected at t = 0 s : 1: overdamped; 2: underdamped; 3: critically damped.

  The underdamped case (periodic case): α < ω n

  If α < ωn than the roots of the characteristic polynomial are conjugate complex numbers:

  The general solution is:

  As shown in Fig. 7.23 the capacitor’s voltage oscillates with an angular frequency ωd around the steady-state value U. The peak value decreases with an exponential function with time constant 1/α.

   

  The critically damped case: α = ω n

  In case of α = ωn the roots of the characteristic polynomial are same:

  The solution is:

  In this case the capacitor’s voltage tends fastest towards the final steady-state value U as depicted in Fig. 7.23 .

    Example of an RLC series circuit

  The RLC circuit of Fig. 7.22 should be operated critically damped. The values U, L and C are given: U = 100 V, C = 80 nF, L = 40 mH. What about the value of the resistor for the critically damped case?

  In critically damped case α = ωn :

  For a resistor of R = 1.41 kΩ the circuit operates in critically damped mode and reaches the steady-state value of uC (∞) = U = 100 V in minimum time. If the value of the resistor is smaller the circuit starts to oscillate (underdamped case), if it is higher it takes more time to reach the steady-state value.

    Automotive application

  Switching is frequently required in automotive applications. Either single switching events, e.g. by the driver or frequent and continuous switching within the electronic system. In general the switching circuit consists of inductive, capacitive and resistive elements (taking parasitic effects into account even always). Depending on the size of these elements and the frequency a detailed analysis of the switching behavior has to be done to avoid an unwanted behavior, e.g. an oscillation or an overdamped case. Consider a switching event from 0 V to 5 V that has to be detected by a microcontroller (e.g. with an interrupt input pin or even with an ADC). In case of an oscillation the overshoot of the voltage (see curve 2 in Fig. 7.23

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  First and Second Order Circuits and Equations

  Technical Background and Insights

  • Robert O'Rourke(Author)
  • 2024(Publication Date)
  • Wiley-IEEE Press

    (Publisher)

  Figure 10.3 shows a series RC circuit driven by a 1 V step function starting at time equals 20 ns. The 25 nF capacitor has the initial voltage v C (0) set to 0. The resistor voltage is defined as V S minus VC, establishing a polarity of pos- itive current flowing clockwise around the loop. With no initial conditions in the schematic in Figure 10.3, this circuit shows step response, for comparison and contrast to the following two examples of complete response. Source voltage 1 0.5 Voltage 0 0 5e-07 1e-06 1.5e-06 Time 2e-06 3e-06 2.5e-06 Capacitor voltage Figure 10.4 Series RC step response simulation plot of capacitor voltage. Figure 10.4 shows the 1 V step function as a dashed line. The starting time of 20 ns is chosen to be able to see the signal not overlapped with the vertical axis. The capacitor voltage, VC in the schematic, builds up toward the source voltage V S . 0.06 Current –0.06 0 5e-07 1e-06 1.5e-06 Time 2e-06 3e-06 2.5e-06 –0.05 –0.04 –0.03 –0.02 –0.01 0 0.02 0.01 0.03 0.05 0.04 Resistor current Source current Figure 10.5 Series RC step response simulation plot of currents. Figure 10.5 shows the resistor current, defined by V 1 (V S ) minus VC, which is positive in the clockwise direction. It jumps up and then decays to zero exponentially over time. The negative source current indicates that the current polarity for the voltage source is positive coming out the bottom of the source and flowing in the counter- clockwise direction. 194 10 Complete Response of First-Order RC and RL Circuits 10.2.2 Second RC Example – Plus 0.5 V Initial Capacitor Voltage V S R1 R = 20 Ohm + – VC C1 C = 25 nF V = 0.5 V1 T2 = 20 us T1 = 20 ns U2 = 1 V U1 = 0 V Figure 10.6 Series RC circuit simulation schematic with 0.5 V initial condition. Figure 10.6 shows a series RC circuit driven by a 1 V step function start- ing at time equals 20 ns. The 25 nF capacitor has the initial voltage v C (0) set to 0.5 V.

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  Essentials of Advanced Circuit Analysis

  A Systems Approach

  • Djafar K. Mynbaev(Author)
  • 2024(Publication Date)
  • Wiley

    (Publisher)

  (4A.4) It’s important to remember that the voltage across a capacitor, v t C ( ) , cannot change instantly; therefore, its initial value, V C0 , is predetermined by the value which a capacitor collects at the moment before the switch is turned on. In other words, v v V V C C C 0 0 0 0 − + ( ) = ( ) = ≡ , (4A.5) Where t s ( ) = − 0 is the last instant when the switch in Figure 4A.1b was in position A, and t s ( ) = + 0 is the first instant when the switch turns to position B. Due to this capacitor’s property, determining its initial voltage is a simple process. This is why in analyzing an RC circuit, it’s always recommended to find the capacitor voltage first, as it’s done in our consideration here. Incidentally, the current through a capacitor can change instantaneously. Thus, at the last instant when the switch in Figure 4A.1b still was in position A, current i i R C ( ) ( ) 0 0 0 − − = = pro- vided that the switch was in position A long enough to fully charge the capacitor to V 0 . Just after the switch turns to position B, the current through a capacitor starts flowing, and its value becomes i i I V R R C 0 0 0 0 + + ( ) = ( ) = =− , (4A.6) as (4A.4) shows at t = 0. Appendix 4A More about Natural Response 231 We’ve mentioned several times that natural response is caused by the energy stored in a capacitor of a series RC circuit. What is the amount of this energy? Let’s start with power delivered to a resis- tor from a capacitor; it is equal to P t v t R V Re RC C t ( ) ( ) = ( ) = ( ) − 2 0 2 2α W . (4A.7) The energy accumulated by the RC circuit at t = 0 is precisely the amount of energy dissipated by the resistor during the circuit operation, t ≥ 0 ; it is given by E t P t dt CV e RC R t t ( ) ( ) . = = − ( ) ∫ − 0 0 2 2 1 2 1 α (J) (4A.8) To find the natural (source-free) response of an RL parallel circuit, consider Figure 4A.2a, where all values are taken from Example 4.3; that is, R L H I mA in = = = 1 0 5 12 Ω, .

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