Tau is the time constant of an RC circuit that takes to change from one steady state condition to another steady state condition when subjected to a step change input condition.
Tau, symbol τ, is the greek letter used in electrical and electronic calculations to represent the time constant of a circuit as a function of time. But what do we mean by a circuits time constant and transient response.
Both electrical and electronic circuits may not always be in a stable or steady state condition, but can be subjected to sudden step changes in the form of changing voltage levels or input conditions. For example the opening or closing of an input switch or sensor.
However, whenever a voltage or state change occurs, the circuit may not respond instantaneously to the change, but may take an amount of time no matter how small if reactive components such as capacitors and inductors are present within the circuit.
The change of state from one stable condition to another generally occurs at a rate determined by the time constant of the circuit which itself will be an exponentially value. Then the time constant of a circuit defines how the transient response of the circuits currents and voltages are changing over a set period of time.
We have seen in these tutorials that when subjected to a steady state DC voltage, a capacitor will act as an open circuit, an inductor will act as a short circuit and a resistor will act as a current limiting device.
If the voltage across a capacitor as well as the current through an inductor cannot change instantly, then what will be their transient response when subjected to a step change condition.
But before we start applying some form of transient analysis to a capacitive circuit, let’s first remind ourselves of the V-I characteristics of an ordinary resistive circuit as shown below.
Resistive Circuit
With the switch in position S2, the 10Ω resistor is effectively shorted and therefore disconnected from the 10 volt supply voltage (V). As a result, zero current flows through the resistor, so IR = 0. However, when the switch is moved to position S1 at time t = 0, a step voltage of 10 volts is applied directly across the 10Ω resistor resulting in a current of 1 ampere (I = V/R) flowing around the closed circuit.
As the resistor is of a fixed non-inductive value, the current changes instantly from 0 to 1 ampere within a fraction of a second as soon as the switch is moved to position S1. Likewise, if the switch is returned back to position S2, the supply voltage (V) is removed so the circuit current will immediately drop to zero again as shown in the above graph.
Then for a resistive cicruit the change of electrical state from one to another is almost instant, as there is nothing to resist this change. Therefore, resistors only limit the flow of electrical current around a circuit to a value determined by Ohms Law, that is V/R and as such there is no time constant or transient response associated with them.
An RC circuit is an electrical circuit that contains resistances and capacitors so let’s look at the transient response of a resistor connected in series with a capacitor. What would be the V-I characteristics of this combination when subjected to an input step voltage change as before.
Time Constant of an RC Circuit
We have seen above that a resistance response instantly to any change in voltage applied to it. But a resistor is a passive linear device that does not store energy but instead dissipates energy in the form of heat. That is it gets hot.
A capacitor however, consists of two electrically conducting plates (electrodes) separated by a dielectric insulating material which has the ability to store electrical energy within itself in the form of an electrostatic charge (Q coulombs). In other words, capacitors store charge.
The result is that unlike the resistor, the capacitor cannot react instantly to quick or step changes in applied voltage so there will always be a short period of time immediately after the voltage is firstly applied for the circuit current and voltage across the capacitor to change state. In other words, there will be a certain amount of time required for the capacitor to change the amount of energy stored within its electric field, either increasing or decreasing in value.
The amount of time for the circuit to respond is expressed in multiples of R x C, that is the product of “Ohms x Farads” given in seconds (s). The current through the capacitor is given by: iC = C(dv/dt). Where: dv represents the change in voltage and dt represents the change in time.
Consider the simple resistor-capacitor RC circuit below.
Resistor-Capacitor (RC) Circuit
With the switch in position S2 for a while, the resistor-capacitor combination is shorted and therefore not connected to the supply voltage, VS. As a result, zero current flows around the circuit, so I = 0 and VC = 0.
When the switch is moved to position S1 at time t = 0, a step voltage (V) is applied to the RC circuit. At this instant in time, the fully discharged capacitor behaves like a short circuit due to the sudden dv/dt change of condition the exact moment the switch is closed to position S1.
This change causes the circuit current to increase to a value limited only by the resistance of the circuit, the same as before. Thus when the switch S1 is initially closed at t = 0, the current flowing around the closed circuit is approximately equal to VR/R amperes, as VR = I*R and VC = 0.
At the same instant the switch is moved to position S1, as well as current flow, the discharged capacitor starts to charge-up as it attempts to store electric charge onto its plates. The result is that the voltage, VC across the capacitors starts to gradually increase while the circuit current begins decreasing at a rate determined by time constant, tau, of the RC combination.
Thus we can define the voltage growth across the capacitors plates, (VC) starting from t = 0 as being:
Integrating both sides gives:
Thus the exponential natural growth of the voltage across the capacitor as it attempts to store charge onto its plates is given as:
Where:
- VC is the voltage across the capacitor
- V is the supply voltage
- e is the base of Natural Logs
- t is the time duration since the switch closed
- RC is the time constant tau of the RC circuit
We can show the exponential rate of growth of the voltage across the capacitor over time in the following table assuming normalised values for the supply voltage of 1 volt, and an RC time constant of one (1).
Capacitor Voltage Growth Over Time
| Time (s) | 0.5 | 0.7 | 1 | 2 | 3 | 4 | 5 | 6 |
| Voltage (VC) | 0.393V | 0.503V | 0.632V | 0.864V | 0.950V | 0.981V | 0.993V | 0.997V |
Clearly, we can see from the above table that the values of: VC = V(1 – e-t/τ ) increase over time from t = 0 to t = 6 seconds (6T) in our example, the voltage across the capacitor is an exponentially increasing function because as time (t) increases, the term e-t/τ gets smaller and smaller, so the voltage across the capacitor, VC gets larger towards that of the supply voltage driving the change.
Thus at time t = 0, the value of the function is zero, but as time (t) continues to grow towards ∞, the point at which t = RC when 1 – e-1 produces a value of 0.632 or 63.2% (0.632*100%) of its final steady state value.
Therefore, for an exponentially increasing function, the time constant, Tau (τ) is defined as the time taken for the function to reach 63.2% of its final steady state value at a rate starting from time, t = 0. Thus every time interval of tau, (τ) the voltage across the capacitor increases by e-1 of its previous value and the smaller the time constant tau, the faster is the rate of change.
We can show the variation of the voltage across the capacitor with respect to time graphically as follows:
Exponential Voltage Growth Over Time
Then we can see that the transient response of a capacitor to a step-input is not instant or linear, but increases exponentially at a rate determined by the time constant of the RC circuit and that one time constant is equal to a factor of 1 – e-1 = 0.6321.
Tau on its own does not describe how long it takes the capacitor to become fully charged and theoretically due to its exponentially increasing transient curve, a capacitor never becomes 100% fully charged.
However, after a time period equal too or greater than the equivalent of 5 time constants, that is ≥ 5τ or 5RC, from when the inital change in condition occured, the exponential growth has slowed to less than 1% of its maximum value so for most practical applications we can say that it has reached its final state or steady state condition with no more change taking place with time. That is, at 5T the capacitor is “fully charged”.
Tau Example No1
An RC series circuit has resistance of 50Ω and capacitance of 160µF. What is its time constant, tau of the circuit and how long does the capacitor take to become fully charged.
1. Time Constant, τ = RC. Therefore:
τ = RC = 50 x 160 x 10-6 = 8 ms
2. Time duration to fully charged:
5T = 5τ = 5RC = 5 x 50 x 160 x 10-6 = 40 ms, or 0.04s
Tau Example No2
A circuit consists of a resistance of 40Ω and a capacitance of 350uF connected together in series. If the capacitor is fully discharged, what will be the time taken for the voltage across the capacitors plates to reach 45% of its final steady state value once charging begins.
Data given: R = 40Ω, C = 350uF, t is the time at which the capacitor voltage becomes 45% of its final value, that is 0.45V
Then it takes 8.37 milli-seconds for the voltage across the capacitor to reach 45% of its 5T steady state condition when the time constant, tau is 14 ms and 5T is 70 ms.
Hopefully now we understand that the time constant of a series RC circuit is the time interval that equals 0.632V (usually taken as 63.2%) of its maximum value (V) at the end of one time constant, (1T) resulting from the product of R and C. Also, the symbol for time constant is a τ (Greek letter tau), and that τ = RC, where R is in ohms, C is in farads, and τ is in seconds.
But what about a capacitor that is already fully charged (VC >5T), what will be the V-I characteristics of the capacitor as it discharges back down to zero volts, and will the decay of the capacitor’s voltage follow the same exponential curve shape.
RC Transient Discharge Curve
The discharging of a fully-charged capacitor is similar to the charging process. The DC power supply used to charge the capacitor originally is disconnected and replaced by a short circuit as shown.
Assuming initial conditions in that switch (S) is “open” and the capacitor has been fully charged (VC >5T). When switch (S) is closed at time t = 0, the capacitor begins to discharged through the resistor with the amount of time required to discharge depending on the value of the resistor. Since initially VC = VR = V, the decay of voltage is given as:
Exponential Decay of Voltage Equation
Where; V(t) is the voltage across the capacitors plates, and VC is the initial capacitor voltage value before decay begins.
Previously the exponential function was for voltage growth. For an exponentially decaying function, the time required for the voltage to reach zero volts at a constant rate is still dependant on the RC time constant. Thus time constant is a measure of the “rate of decay”.
Therefore for an exponentially decaying function, the time constant, tau (τ) is also defined as the time required for the decaying voltage to reach approximately 36.8% of its final steady state value when the decay started at time t = 0. Thus if τ is one time constant, that is: τ = RC, and the RC circuit was at its fully charged steady state condition at t = 0, then:
Thus at time t = 0, the value of the function is at its maxium, but as time (t) moves towards ∞, the point at which t = RC when e-1 produces a value of 0.368 or 36.8% (0.368*100%) of its final steady state value, which is zero volts (fully discharged).
Again we can show the exponential rate of decay of the voltage across the capacitor over time in the following table with normalised values for the supply voltage of 1 volt, and an RC time constant of one (1).
Capacitor Voltage Decay Over Time
| Time (s) | 0.5 | 0.7 | 1 | 2 | 3 | 4 | 5 | 6 |
| Voltage (Vt) | 0.607VC | 0.497VC | 0.368VC | 0.135VC | 0.049VC | 0.018VC | 0.007VC | 0.002VC |
One point to notice here. The time constant, tau of a series RC circuit from its inital value at t = 0 to τ will always be 63.2% whether the capacitor is charging or discharging. For an exponential growth the inital condition is 0V, (zero volts) since the capacitor is fully discharged.
Thus the voltage grows exponentially upwards to 63.2% of VMAX at one time constant, 1T. But we could also think of the capacitor voltage at 1T as being 36.8% away from its final steady state value at 5T. That is fully charged.
The same idea is also true for an exponential decay. For a fully charged capacitor the inital steady state condition is VC(max), so the capacitor will discharge down to 36.8% of its final steady state condition of zero volts (0V) after 5T. But again, we can also think of the voltage across the capacitor at time 1T, as being 63.2% down from its inital starting when the capacitor was fully charged at t = 0.
Then the value of one time constant 1T, from the inital starting condition to 1T will always be 0.632V, or 63.2% of its final steady state condition. Likewise at 1T, the capacitor voltage will always be 0.368V, or 36.8% away from its final steady state condition after 5T as either fully charged at VC(max) or fully discharged at 0V.
We can show the decay of voltage with respect to time graphically as follows:
Exponential Voltage Decay Over Time
Again the rate of voltage decay over time relies greatly on the value of the RC time constant, tau.
Tau Example No3
An RC series circuit has a time constant, tau of 5ms. If the capacitor is fully charged to 100V, calculate: 1) the voltage across the capacitor at time: 2ms, 8ms and 20ms from when discharging started, 2) the elapsed time at which the capacitor voltage decays to 56V, 32V and 10V.
The voltage across a discharging capacitor is given as:
VC(t) = VC × e–t/RC Volts
RC time constant is given as 5ms, therefore 1/RC = 200. VC = 100V.
1a). Capacitor voltage after 2ms
VC(0.002) = 100 e–200t = 100 e–0.4 = 100 × 0.67 = 67.0 volts
1b). Capacitor voltage after 8ms
VC(0.008) = 100 e–200t = 100 e–1.6 = 100 × 0.202 = 20.2 volts
1c). Capacitor voltage after 20ms
VC(0.02) = 100 e–200t = 100 e–4 = 100 × 0.018 = 1.8 volts
Capacitor voltage (VC) at time duration from t = 0
2a). Elapsed time (t1) when VC(t) = 56 volts
56 = 100 e-200t, therefore: -200t1 = ln(56/100) = –0.5798
Thus: t1 = –0.5798 ÷ –200 = 2.9 ms
2b). Elapsed time (t2) when VC(t) = 32 volts
32 = 100 e-200t, therefore: -200t2 = ln(32/100) = –1.1394
Thus: t2 = –1.1394 ÷ –200 = 5.7 ms
2c). Elapsed time (t3) when VC(t) = 10 volts
10 = 100 e-200t, therefore: -200t3 = ln(10/100) = –2.3026
Thus: t3 = –2.3026 ÷ –200 = 11.5 ms
Tau the Time Constant Summary
We have seen here in this tutorial about Time Constant, Tau, symbol τ that the transient response of an RC circuit is the time it takes to change from one steady state condition to another steady state condition when subjected to a step change input condition.
When the capacitor is charges up from its initial zero voltage state to its final steady state voltage (V), the time duration is defined as: τ = RC. That is the product of R and C. This produces an exponentially growing function for VC with this RC time constant measured in seconds, and the smaller the time constant, the faster the rate of voltage change.
We have also seen that for an exponentially growing function, the value after one time constant, 1T will be equal to 63.2% of its final steady state value. That is for an exponentially increasing function, it is the time required for the voltage to reach its final steady state value which occurs after 5T.
The value of V(t) for an exponentially growing function at time t = τ is given as:
V(t) = V( 1 – e–1 ) = 0.632V
Likewise, for an exponentially decaying function, the value after one time constant, 1T is 36.8% of its final steady state value. That is for an exponentially decaying function it is time required for the voltage to reach zero value.
The value of V(t) for an exponentially decaying function at time t = τ is given as:
V(t) = V( e–1 ) = 0.368V
Either way, from t = 0 to τ will always be 63.2% of the time duartion, and from 1T to 5T will always be 36.8% of the time duration, exponentially increasing or decreasing.
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