Capacitors Tutorial 2 - Exponential Discharge

 

Graphical Representation and Quantitative Treatment of Capacitor Discharge

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:

 

 

We can note the voltage and current at time intervals and plot the data, which gives us the exponential graph shown below.  We should note the following about the graph:

The graph is asymptotic, i.e. in theory the capacitor does not completely discharge.  In practice, it does.

The graph is described by the relationship:

 

Q = Q0 e –t/RC

 

[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:

 

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:

           

            V= V0 e –RC/RC = V0 e –1 = 0.37 V0

 

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:

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

Q0 ÷ 2 = Q0 e – t1/2/RC

Þ ½ = e – t1/2/RC

Þ 2-1 = e – t1/2/RC

Þ e + t1/2/RC = 2

Þ loge (2) = t1/2/RC  [In text books you may see the natural logarithm written as ‘ln’]

Þ t1/2  = loge (2) × RC = 0.693 × RC

 

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

 

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 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

(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 V0 = -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 1

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? 

Answer