## 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 Electronic 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 to the external 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 now flowing around the circuit is called the **Charging Current** and is found by using Ohms law as: i = Vs/R.

### RC Charging Circuit 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 up, the potential difference across its plates slowly 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 one Time Constant, ( T ). This 0.63Vs voltage point is given the abbreviation of 1T.

The capacitor continues charging up and the voltage difference between Vs and Vc reduces, so to does the circuit current, i. Then at its final condition greater than five time constants ( 5T ) when the capacitor is said to be fully charged, 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.

## RC Charging Circuit Example No1

Calculate the RC time constant, τ of the following circuit.

Therefore the time constant τ is given as:

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.

« Diac Tutorial | RC Discharging Circuit »

## Annie

Regarding: RC Charging Circuit Example No1:

Question a)

You have: “Vc = 0.5Vs” where does that come from? Where’s 0.5?

Trying to use your tutorial to complete a different problem, and it doesn’t work.

## Wayne Storr

Hello Annie, at a time constant of 0.7 (0.7T), the capacitor voltage will be half the supply voltage, Vs as shown above. Then at 0.7T Vc will be 0.5Vs, that is 50% of Vs.

You can also put the numbers in the above equation: Vc = Vs(1-e^-t/RC) therefore Vc = Vs(1-e^-0.7/1) = Vs(1-0.5) = 0.5Vs

## jovel.happy

First of all thank you for the fast response. There are things that still ponders me though:

1. Do we have to series a resistor to the capacitor? I’m thinking of not putting resistor at all. Is it safe? Won’t the capacitor charge faster that way? (The circuit only consists of battery then capacitor)

2. How about input current? Does it matter if it’s small or large? I’m thinking of using transformer then diodes as the source. What if I use a transformer that has an output rating of: 24V and let’s say 5A?

3. When the capacitor is fully charged, it will have the same voltage as the voltage source according to what I’ve searched? What about if the capacitor only has a rating of 15V? Does it mean that it can only store a maximum of 15V?

4. I’m thinking of using a 50 Farad, 15V ultra capacitors. When the capacitors are fully charge then:

C = Q/V

Q = CV = (50)(15) = 750 Coulombs

I = Q/t

It = Q = 750

therefore:

It = 750 Ampere-second E = 1.5V

so basically 1125 Watt-second

Then does that mean that the capacitor when fully charge will have a rating of 750 Ampere-second or 1125 Watt-second? I’ve heard that this is only possible if you use an AC source but I’m not sure if it is true.

Thank you again for the help.

## Wayne Storr

To answer 1. ad 2. The voltage across the plates of a capacitor will be the same as the supply voltage when the capacitor is fully charged. The charging time for a given capacitor depends on the charging current. This charging current can be controlled by placing a resistor in series with the capacitor. The current flow per second in amperes is expressed as: Amperes = Capacitance x Volts-per-seconds (i = Cdv/dt).

Yes the capacitor would charge and discharge in an infinite time but the initial currents would be huge because assuming the 50F capacitor is fully discharged and with no series resistor to limit current flow. At the instant you connect the supply voltage, the voltage across the plates would jump to 24 volts in zero time, then Vc would be 24v and t would be 0, with the result that the capacitor current would be infinite. An infinite current is impossible because no source can deliver this current.

3. The voltage rating of a capacitor (15v) is the maximum potential difference that a capacitor can withstand across its plates due to the dielectric insulating material from which it is made. Exceeding this working voltage rating may cause failure.

4. When fully charged the 50F capacitor will hold 750 Coulombs of charge giving a discharge current of 750 Amps per second as you have stated.

## jovel.happy

What if I want to charge a super capacitor (let’s say 300 Farad) as fast as possible. I plan to put a 0.1 ohm resistor in series so that the time constant will be 30 seconds.

Is it safe to put 0.1 ohm resistor or even lower?

## Wayne Storr

It depends upon your supply voltage, but as long as the resistors power rating is not exceeded, then yes in theory you can use a 0.1 ohm resistor.

## Raj

Hello! I’m trying to experiment with measuring capacitance using above formula. I’ve 100m (series of 20m) resistance with 30pf capacitor. I’m connecting the RC circuit to 3.3V DC. When I measure voltage on capacitor is is always showing .29V. According to my calculation, it should have reach 2V (63% of 3.3v) by 3ms (30pfx100m)

Any ideas or thoughts on what might be going on here?

Thanks!

## Wayne Storr

Hello Raj. While on paper the 30pF capacitor will charge to one time constant in 2.98ms, the value of 100 Mohms resistance is huge for such a small capacitance resulting in nano-amperes of charging current from only 3.3v. Also the tolerance of pF capacitors can be +/-20% or more so may not be 30pF. Try smaller resistor values or increase the supply voltage to test and work backwards after your measured and calculated values match.

## James

Therefore the time constant τ is given as:

T = R x C = 47k x 1000uF = 47 Secs. Did you divide? New at this sorry

## Wayne Storr

No, the “x” means multiplication.