The Time Constant

All Electrical or Electronic circuits or systems suffer from some form of "time-delay" between its input and output,

when a signal or voltage, either continuous, (DC) or alternating (AC) is firstly applied to it. This delay is generally

known as the time delay or Time Constant of the circuit and it is the time response of the circuit when a step voltage

or signal is firstly applied. The resultant time constant of any circuit or system will mainly depend upon the reactive

components either capacitive or inductive connected to it and is a measurement of the response time with units of,

Tau - τ

When an increasing DC voltage is applied to a discharged Capacitor the capacitor draws a charging current and

"charges up", and when the voltage is reduced, the capacitor discharges in the opposite direction. Because

capacitors are able to store electrical energy they act like small batteries and can store or release the energy as

required. The charge on the plates of the capacitor is given as: Q = CV. This charging (storage) and discharging

(release) of a capacitors energy is never instant but takes a certain amount of time to occur with the time taken for the

capacitor to charge or discharge to within a certain percentage of its maximum supply value being known as its Time

Constant (τ).

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 called the

transient response, required for this to occur is equivalent to about 5 time constants or 5T. This transient response

time T, is measured in terms of τ = R x C, in seconds, where R is the value of the resistor in ohms and C is the

value of the capacitor in Farads. This then forms the basis of an RC charging circuit were 5T can also be thought of

as "5 x RC".

RC Charging Circuit

The figure below shows a capacitor, (C) in series with a resistor, (R) forming a RC Charging Circuit connected

across a DC battery supply (Vs) via a mechanical switch. When the switch is closed, the capacitor will gradually

charge up through the resistor until the voltage across it reaches the supply voltage of the battery. The manner in

which the capacitor charges up is also shown below.

RC Charging Circuit

Let us assume above, that the capacitor, C is fully "discharged" and the switch (S) is fully open. These are the initial

conditions of the circuit, then t = 0, i = 0 and q = 0. When the switch is closed the time begins at t = 0 and current

begins to flow into the capacitor via the resistor. Since the initial voltage across the capacitor is zero, (Vc = 0) the

capacitor appears to be a short circuit and the maximum current flows through the circuit restricted only by the

resistor R. Then by using Kirchoff's voltage law (KVL), the voltage drops around the circuit are given as:

The current no flowing around the circuit is called the Charging Current and is found by using Ohms law as: i =

VR/R.

RC Charging Curves

The capacitor now starts to charge up as shown, with the rise in the RC charging curve steeper at the beginning

because the charging rate is fastest at the start and then tapers off as the capacitor takes on additional charge at a

slower rate. As the capacitor charges the potential difference across its plates increases with the actual time taken for

the charge on the capacitor to reach 63% of its maximum possible voltage, in our curve 0.63Vs being known as the

Time Constant, (T) of the circuit and is given the abbreviation of 1T. As the capacitor charges up, the voltage

difference between Vs and Vc reduces, so to does the circuit current, i. Then at the final condition, t = ∞, i = 0, q =

Q = CV. Then at infinity the current diminishes to zero, the capacitor acts like an open circuit condition therefore, the

voltage drop is entirely across the capacitor.

So mathematically we can say that the time required for a capacitor to charge up to one time constant is given as:

Where, R is in Ω's and C in Farads.

Since voltage V is related to charge on a capacitor given by the equation, Vc = Q/C, the voltage across the value of

the voltage across the capacitor, (Vc) at any instant in time during the charging period is given as:

Where:

Vc is the voltage across the capacitor

Vs is the supply voltage

t  is the elapsed time since the application of the supply voltage

RC is the time constant of the RC charging circuit

After a period equivalent to 4 time constants, (4T) the capacitor in this RC charging circuit is virtually fully charged

and the voltage across the capacitor is now approx 99% of its maximum value, 0.99Vs. The time period taken for the

capacitor to reach this 4T point is known as the Transient Period. After a time of 5T the capacitor is now fully

charged and the voltage across the capacitor, (VC) is equal to the supply voltage, (Vs). As the capacitor is fully

charged no more current flows in the circuit. The time period after this 5T point is known as the Steady State Period.

As the voltage across the capacitor Vc changes with time, and is a different value at each time constant up to 5T, we

can calculate this value of capacitor voltage, Vc at any given point, for example.

Example No1

Calculate the time constant of the following circuit.

The time constant τ is found using the formula T =

R x C in seconds.

Therefore the time constant τ is:

T = R x C = 47k x 1000uF = 47 Secs

a) What value will be the voltage across the capacitor at 0.7 time constants?

        At 0.7 time constants (0.7T) Vc = 0.5Vs. Therefore, Vc = 0.5 x 5V = 2.5V

b) What value will be the voltage across the capacitor at 1 time constant?

        At 1 time constant (1T) Vc = 0.63Vs. Therefore, Vc = 0.63 x 5V = 3.15V

c) How long will it take to "fully charge" the capacitor?

        The capacitor will be fully charged at 5 time constants.

        1 time constant (1T) = 47 seconds, (from above). Therefore, 5T = 5 x 47 = 235 secs

d) The voltage across the Capacitor after 100 seconds?

        The voltage formula is given as Vc = V(1 - e-t/RC)

        which equals: Vc = 5(1-e-100/47)    RC = 47 seconds from above, Therefore, Vc = 4.4 volts

We have seen that the charge on a capacitor is given by the expression: Q = CV and that when a voltage is firstly

applied to the plates of the capacitor it charges up at a rate determined by its time constant, τ. In the next tutorial we

will examine the current-voltage relationship of a discharging capacitor and look at the curves associated with it when

the capacitors plates are shorted together

RC Discharging Circuit

In the previous RC Charging Circuit tutorial, we saw how a Capacitor, C charges up through the resistor until it

reaches an amount of time equal to 5 time constants or 5T and then remains fully charged. If this fully charged

capacitor was now disconnected from its DC battery supply voltage it would store its energy built up during the

charging process indefinitely (assuming an ideal capacitor and ignoring any internal losses), keeping the voltage

across its terminals constant.

If the battery was now removed and replaced by a short circuit, when the switch was closed again the capacitor

would discharge itself back through the resistor, R as we now have a RC discharging circuit. As the capacitor

discharges its current through the series resistor the stored energy inside the capacitor is extracted with the voltage

Vc across the capacitor decaying to zero as shown below.

RC Discharging Circuit

In a RC Discharging Circuit, the time constant (τ) is still equal to 63%. Then for a RC discharging circuit that is

initially fully charged, the voltage across the capacitor after one time constant, 1T, has dropped to 63% of its initial

value which is 1 - 0.63 = 0.37 or 37% of its final value. So now this is given as the time taken for the capacitor to

discharge down to within 37% of its fully discharged value which will be zero volts (fully discharged), and in our curve

this is given as 0.37Vc.

As the capacitor discharges, it loses its charge at a declining rate. At the start of discharge the initial conditions of the

circuit, are t = 0, i = 0 and q = Q. The voltage across the capacitors plates is equal to the supply voltage and Vc =

Vs. As the voltage across the plates is at its highest value maximum discharge current flows around the circuit.

RC Discharging Curves

With the switch closed, the capacitor now starts to discharge as shown, with the decay in the RC discharging curve

steeper at the beginning because the discharging rate is fastest at the start and then tapers off as the capacitor

looses charge at a slower rate. As the discharge continues, Vc goes down and there is less discharge current. As

with the previous charging circuit the voltage across the capacitor, C is equal to 0.5Vc at 0.7T with the steady state

fully discharged value being finally reached at 5T.

For a RC discharging circuit, the voltage across the capacitor (Vc) as a function of time during the discharge period is

defined as:

Where:

Vc is the voltage across the capacitor

Vs is the supply voltage

t  is the elapsed time since the application of the supply voltage

RC is the time constant of the RC discharging circuit

Just like the RC Charging circuit, we can say that in a RC Discharging Circuit the time required for a capacitor to

discharge itself down to one time constant is given as:

Where, R is in Ω's and C in Farads.

So a RC circuit's time constant is a measure of how quickly it either charges or discharges.

Example No1

Calculate the time constant of the following RC discharging circuit.

The time constant, T of the circuit is found using the

following formula T = R x C given in seconds.

Therefore, the RC circuits time constant T is:

T = R x C = 100k x 22uF = 2.2 Seconds

a) What value will be the voltage across the capacitor at 0.7 time constants?

        At 0.7 time constants (0.7T) Vc = 0.5Vc. Therefore, Vc = 0.5 x 10V = 5V

b) What value will be the voltage across the capacitor after 1 time constant?

        At 1 time constant (1T) Vc = 0.37Vc. Therefore, Vc = 0.37 x 10V = 3.7V

c) How long will it take for the capacitor to "fully discharge" itself (5 time constants)?

        1 time constant (1T) = 2.2 seconds. Therefore, 5T = 5 x 2.2 = 11 Seconds

RC Waveforms

In the previous RC Charging and RC Discharging tutorials, we saw how a capacitor, C both charges up and

also discharges itself through a series resistor, R at an amount of time equal to 5 time constants or 5T when a

constant DC voltage is either applied or removed. But what would happen if we changed this constant DC supply to

an alternating AC waveform that constantly changes from a maximum value to a minimum value at a rate determined

by its frequency. How would this affect our RC time constant value and the output RC waveforms?.

We saw previously that the capacitor charges up to 5T when a voltage is applied and discharges down to 5T when it

is removed. This 5T time constant value always remains true as it is fixed by the resistor-capacitor combination. Then

the actual time required to fully charge or discharge the capacitor can only be changed by changing the value of

either the capacitor itself or the resistor in the circuit and this is shown below.

Typical RC Waveform

Square Wave Signal

Useful wave shapes can be obtained by using RC circuits with the required time constant. If we apply a continuous

square wave voltage waveform to the RC circuit whose frequency matches that exactly of the 5RC time constant

(5T) of the circuit, then the voltage waveform across the capacitor would look something like this:

A 5RC Input Waveform

The voltage drop across the capacitor alternates between charging up to Vc and discharging down to zero according

to the input voltage. Here in this example, the frequency (and therefore the time period, ƒ = 1/T) of the input square

wave voltage exactly matches that of the RC time constant, ƒ = 1/RC, and the capacitor is allowed to fully charge

and fully discharge on every cycle resulting in a perfectly matched RC waveform.

If the time period of the input waveform is made longer (lower frequency, ƒ< 1/RC) for example a time period

equivalent to say "8RC", the capacitor would then stay fully charged longer and also stay fully discharged longer as

shown.

An 8RC Input Waveform

If however we reduced the time period of the input waveform (higher frequency, ƒ> 1/RC), to "4RC" the capacitor

would not have sufficient time to either fully charge or discharge with the resultant voltage drop across the capacitor,

Vc being less than its maximum input voltage as shown below.

A 4RC Input Waveform

Frequency Response

The Integrator

The Integrator is a type of Low Pass Filter circuit that converts a square wave input signal into a triangular

waveform output. As seen above, if the RC time constant is long compared to the time period of the input RC

waveform the resultant output will be triangular in shape and the higher the input frequency the lower will be the

output amplitude compared to that of the input.

From which we derive an ideal voltage output for the integrator as:

The Differentiator

The Differentiator is a High Pass Filter type circuit that converts a square wave input signal into high frequency

spikes at its output. If the RC time constant is short compared to the time period of the input waveform the capacitor

will become fully charged quickly before the next change in the cycle. When the capacitor is fully charged the output

voltage across the resistor is zero. The arrival of the falling edge of the input waveform causes the capacitor to

reverse charge giving a negative output spike, then as the square wave input changes during each cycle the output

spike changes from a positive value to a negative value.

from which we have an ideal voltage output for the Differentiator as:

Sine Wave Input Signal

If we now change the input RC waveform of these RC circuits to that of a sinusoidal Sine Wave voltage signal the

resultant output waveform will remain unchanged and only its amplitude will be affected. By changing the positions of

the Resistor, R or the Capacitor, C a simple first order Low Pass or a High Pass filters can be made with the

frequency response of these two circuits dependant upon the input frequency value.

Low-frequency signals are passed from the input to the output with little or no attenuation, while high-frequency

signals are attenuated significantly to almost zero. The opposite is also true for a High Pass filter circuit. Normally, the

point at which the response has fallen 3dB (cut-off frequency, ƒc) is used to define the filters bandwidth and a loss of

3dB corresponds to a reduction in output voltage to 70.7 percent of the original value.

Cut-off Frequency

where RC is the time constant of the circuit previously defined and can be replaced by tau, T. This is another

example of how the Time Domain and the Frequency Domain concepts are related

Capacitance in AC Circuits

When capacitors are connected across a direct current DC supply voltage they become charged to the value of the

applied voltage, acting like temporary storage devices and maintain or hold this charge indefinitely as long as the

supply voltage is present. During this charging process, a charging current, ( i ) will flow into the capacitor opposing

any changes to the voltage at a rate that is equal to the rate of change of the electrical charge on the plates. This

charging current can be defined as: i = CdV/dt. Once the capacitor is "fully-charged" the capacitor blocks the flow of

any more electrons onto its plates as they have become saturated. However, if we apply an alternating current or AC

supply, the capacitor will alternately charge and discharge at a rate determined by the frequency of the supply. Then

the Capacitance in AC circuits varies with frequency as the capacitor is being constantly charged and discharged.

We know that the flow of electrons through the capacitor is directly proportional to the rate of change of the voltage

across the plates. Then, we can see that capacitors in AC circuits like to pass current when the voltage across its

plates is constantly changing with respect to time such as in AC signals, but it does not like to pass current when the

applied voltage is of a constant value such as in DC signals. Consider the circuit below.

AC Capacitor Circuit

In the purely capacitive circuit above, the capacitor is connected directly across the AC supply voltage. As the supply

voltage increases and decreases, the capacitor charges and discharges with respect to this change. We know that

the charging current is directly proportional to the rate of change of the voltage across the plates with this rate of

change at its greatest as the supply voltage crosses over from its positive half cycle to its negative half cycle or vice

versa at points, 0o and 180o along the sine wave. Consequently, the least voltage change occurs when the AC sine

wave crosses over at its maximum or minimum peak voltage level, (Vm). At these positions in the cycle the maximum

or minimum currents are flowing through the capacitor circuit and this is shown below.

AC Capacitor Phasor Diagram

At 0o the rate of change of the supply voltage is increasing in a positive direction resulting in a maximum charging

current at that instant in time. As the applied voltage reaches its maximum peak value at 90o for a very brief instant in

time the supply voltage is neither increasing or decreasing so there is no current flowing through the circuit. As the

applied voltage begins to decrease to zero at 180o, the slope of the voltage is negative so the capacitor discharges in

the negative direction. At the 180o point along the line the rate of change of the voltage is at its maximum again so

maximum current flows at that instant and so on. Then we can say that for capacitors in AC circuits the instantaneous

current is at its minimum or zero whenever the applied voltage is at its maximum and likewise the instantaneous

value of the current is at its maximum or peak value when the applied voltage is at its minimum or zero. From the

waveform above, we can see that the current is leading the voltage by 1/4 cycle or 90o as shown by the vector

diagram. Then we can say that in a purely capacitive circuit the alternating voltage lags the current by 90o.

We know that the current flowing through the capacitance in AC circuits is in opposition to the rate of change of the

applied voltage but just like resistors, capacitors also offer some form of resistance against the flow of current through

the circuit, but with capacitors in AC circuits this AC resistance is known as Reactance or more commonly in

capacitor circuits, Capacitive Reactance, so capacitance in AC circuits suffers from Capacitive Reactance.

Capacitive Reactance

Capacitive Reactance in a purely capacitive circuit is the opposition to current flow in AC circuits only. Like

resistance, reactance is also measured in Ohm's but is given the symbol X to distinguish it from a purely resistive

value. As reactance can also be applied to Inductors as well as Capacitors it is more commonly known as Capacitive

Reactance for capacitors in AC circuits and is given the symbol Xc so we can actually say that Capacitive

Reactance is Resistance that varies with frequency. Also, capacitive reactance depends on the value of the

capacitor in Farads as well as the frequency of the AC waveform and the formula used to define capacitive reactance

is given as:

Capacitive Reactance

 

Where:

  F is in Hertz and C is in Farads.

  2πF can also be expressed collectively as the Greek letter Omega, ω to denote an angular frequency.

From the capacitive reactance formula above, it can be seen that if either of the Frequency or Capacitance where to

be increased the overall capacitive reactance would decrease. As the frequency approaches infinity the capacitors

reactance would reduce to zero acting like a perfect conductor. However, as the frequency approaches zero or DC,

the capacitors reactance would increase up to infinity, acting like a very large resistance. This means then that

capacitive reactance is "Inversely proportional" to frequency for any given value of Capacitance and this shown

below:

Capacitive Reactance against Frequency

The capacitive reactance of the capacitor decreases as the frequency across it increases therefore capacitive reactance is inversely proportional to frequency. The opposition to current flow, the electrostatic charge on the plates (its AC capacitance value) remains constant as it becomes easier for the capacitor to fully absorb the change in charge on its plates during each half cycle. Also as the frequency increases the current flowing through the capacitor increases in value because the rate of voltage change across its plates increases.

Example No1.

Find the current flowing in a circuit when a 4uF capacitor is connected across a 880v, 60Hz supply.

So, the Capacitance in AC circuits varies with frequency as the capacitor is being constantly charged and

discharged with the AC resistance of a capacitor being known as Reactance or more commonly in capacitor circuits,

Capacitive Reactance. This capacitive reactance is inversely proportional to frequency and produces the opposition

to current flow around a capacitive AC circuit as we looked at in the AC Capacitance tutorial in the AC Theory

section.