Tutorial 14 - DC Transients in Capacitors

Learning Objective

To recognise the time constant in a capacitor;

To calculate the voltage or current at any time in a capacitor discharge;

To calculate the voltage or current at any time in a capacitor charge.

To recognise and describe an exponential fall or rise.

Key Questions

What is an RC Circuit?

How does a capacitor behave when in charges or discharges?

How do we work out the charge or voltage at any time when a capacitor discharges or charges?

### RC Networks

At its simplest the RC network is a series circuit consisting of a capacitor and a resistor connected to a source.

If we discharge a capacitor, we find that the charge decreases by the same fraction for each time interval. If it takes time t seconds for the charge to decay to 50 % of its original level, we find that the charge after another t seconds is 25 % of the original (50 % of 50 %). This time interval is called the half-life of the decay. The decay curve against time is called an exponential decay.

### Discharging a Capacitor

The voltage, current, and charge all decay exponentially during the capacitor discharge.

We can note the voltage and current at time intervals and plot the data, which gives us the exponential graph, using a circuit like this.  This sort of experiment lends itself well to harvesting data with a data logger.

The graph is like this:

This graph is called an exponential decay curve.  We will look at an exponential rise curve later.

This graph shows voltage against time.  The current - time graph and charge - time graph show exactly the same pattern.

Some bear traps:

•  DC transients are confused with reactive behaviour in AC circuits.  We cannot talk of reactance or phase difference with DC circuits.  Therefore do not think about CIVIL or any other model that is used with AC circuits.

• Not every curved graph is an exponential.  Exponential graphs only happen when a quantity changes by a constant fraction in each unit period of time.

We should note the following about the graph:

• Its shape is unaffected by the voltage.

• The half-life of the decay is independent of the voltage.

The product RC (capacitance × resistance) is called the time constant. The units for the time constant are seconds. We can go back to base units to show that ohms × farads are seconds.  Time constant is sometimes given the physics code t (“tau”, a Greek lower case letter‘t’).

t = RC

After RC seconds the voltage is 37 % of the original. To increase the time taken for a discharge we can:

• Increase the resistance.

• Increase the capacitance.

The half-life is 69 % of the time constant.

Electronic engineers use the time constant in preference to the half-life. In theory the exponential decay should never allow a capacitor to discharge completely, but in practice, a rule of thumb is that the capacitor is discharged completely after 5 RC seconds.

 Worked example A 5000 mF capacitor is charged to 12.0 V and discharged through a 2000 W resistor. (a) What is the time constant? (b) What is the half-life of the decay? Answer (a) Time constant = RC = 2000 W × 5000 × 10-6 F = 10 s. (b) t1/2 = 0.693 × RC = 0.693 × 10 = 6.93 s.

 Question 1 A capacitor of 470 mF is discharged through a 150 k resistor, with a starting voltage of 12 V? (a) What is the time constant? (b) What is the time needed to get to 6 V? (c) What is the time needed to get to 3 V? (d) What is the voltage across the capacitor after RC seconds? (e) How long does it take for the capacitor completely to discharge?

### Charging a Capacitor

When we charge up a capacitor, we get an exponential rise in charge and voltage. We get an exponential fall in the current. This is because when we start to charge up the capacitor, the current is a maximum and the voltage is zero. When the voltage is at a maximum, the current is zero, because no charge can flow on.

The voltage graph like this:

The current graph is like this:

This is because the current fall from an initially high value to a low value.

After RC seconds the capacitor has charged up to 63 % of its final voltage.

As in a discharge there is a half-life in the charging of a capacitor; we can relate it to the time constant by the relationship:

After 5 RC time constants, the capacitor is almost completely charged up, so the voltage is almost V0. In practice, the voltage is close to V0 before this.

 Question 2 A capacitor of 470 mF is charged up through a 150 k resistor, with a starting voltage of 12 V? (a) What is the time constant? (b) What is the time needed to get to 6 V? (c) What is the time needed to get to 9 V? (d) What is the voltage across the capacitor after RC seconds? (e) How long does it take for the capacitor completely to charge?

### Quantitative Treatment of Capacitor Discharge and Charge

Many electronic circuits use the charge and discharge of a capacitor.  If we discharge a capacitor, we find that the charge decreases by the same fraction for each time interval.  If it takes time t for the charge to decay to 50 % of its original level, we find that the charge after another t seconds is 25 % of the original (50 % of 50 %).  This time interval is called the half-life of the decay.  The decay curve against time is called an exponential decay.

The voltage, current, and charge all decay exponentially during the capacitor discharge.

We can plot a graph using a circuit like this:

The graph is described by the relationship:

[Q – charge (C); Q0 – charge at the start; e – exponential number (2.718…); t – time (s); C – capacitance (F); R – resistance (W).]

For voltage and current, the equation becomes:

·

·

Since t = RC, we can rewrite the voltage equation as:

And do the same for the charge and current equations.

The product RC (capacitance × resistance) which we see in the formula is called the time constant.  The units for the time constant are seconds.  We can go back to base units to show that ohms × farads are seconds.  So if we discharge the capacitor for RC seconds, we can easily find out the fraction of charge left:

So after RC seconds the voltage is 37 % of the original.  This is used widely by electronic engineers.  To increase the time taken for a discharge we can:

·         Increase the resistance.

·         Increase the capacitance.

We can link the half-life to the capacitance.  At the half life:

·         Q = Q0/2

·     t = t1/2

Therefore:

The half-life is 69 % of the time constant.

 Example A 5000 mF capacitor is charged to 12.0 V and discharged through a 2000 W resistor.   (a)                            What is the time constant? (b)                           What is the voltage after 13 s? (c)                            What is the half-life of the decay? (d)                           How long would it take the capacitor to discharge to 2.0 V?
 Answer (a)     Time constant = RC = 2000 W × 5000 × 10-6 F = 10 s.     (b)    Use V = V0 e –t/RC Þ V = 12.0 V × e – 13 s/10 s = 12.0 × e – 1.3 = 12.0 × 0.273                    Þ  V = 3.3 volts         (c)     t1/2  = 0.693 × RC = 0.693 × 10 = 6.93 s.     (d)    We need to rearrange the formula by taking natural logarithms.   V = V0 e –t/RC  Þ  V / V0 = e –t/RC Þ loge V - loge Vo = -t/RC [When you divide two numbers, you subtract their logs] Þ 0.693 – 2.485 = - t/10 Þ -t/10 = -1.792 Þ +t/10 = +1.792 Þ t = 1.792 × 10 = 17.9 s

 Question 3 A 470 mF capacitor, charged up to 12.0 V is connected to a 100 kW resistor. (a)    What is the time constant? (b)   What is the voltage after 10 s? (c)  How long does it take for the voltage to drop to 2.0 V?

### Exponential Rise in a capacitor

Earlier we saw the graph of the exponential rise of a capacitor voltage.  The same is true of the charge.  But the current falls exponentially.  This is because at low voltage and low charge, it is easy for the electrons to crowd onto the plates.  However, as the charge goes up (and the voltage, as Q = CV), it get more difficult for the electrons to crowd on, so the current falls.

The voltage graph looks like this:

It is an exponential rise.  The equation for this is:

In this case, the values of the capacitor are 4.7 mF, and the resistor is 330 kW, giving an RC time constant of 1.55 s.

The current falls exponentially:

 Question 4 Show that when t = 1.55 s, the voltage is at about 63 % of its final value.