Capacitors Tutorial 2 - Exponential Discharge
Representation and Quantitative Treatment of Capacitor Discharge
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
seconds for the charge to
decay to 50 % of its original level, we find that the charge after another
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
voltage, current, and charge all decay exponentially during the capacitor
We can plot a graph using a circuit like this:
Why does a capacitor discharge exponentially?
Think about a capacitor that is fully charged. Let's suppose we remove 1 % of its charge. Its charge goes down to 99 %. The new charge is 99 % of the original
Now let's remove another 1 % of the charge that is left, i.e. the 99 % of the original. 1 % of 99 % = 0.99 %. New charge is now 98.01 %. And so on...
Now suppose we lose that fraction in each unit time. We can say that the capacitor discharges a constant fraction of what is left in each unit time. The charge lost in each unit of time can be worked out as:
Charge loss per unit time = change in charge ÷ time interval
At the moment of discharge, a current I flows which is the rate of change of charge. So we can write this as an equation:
Suppose we discharge the capacitor for a short amount of time dt. We know that:
So we can write:
And we can rearrange to:
The current is the rate of discharge. We can write the rate of discharge in calculus notation as:
We can rewrite the expression as:
The minus sign tells us that the charge is decreasing. From basic capacitance we know:
Q = CV
And we can substitute:
This is a differential equation of the form:
The term k is the fraction by which x decreases. Therefore:
1/CR is the fraction by which Q decreases.
The solution for this differential equation is:
Q – charge (C);
Q0 – original charge (C)
e – exponential number (2.718…)
t – time (s)
– time constant (s).
The mathematical justification for this is an extension at the end of the tutorial. You do not need to know this for the A-level exam.
Since Q = CV, we can also write:
Since V = IR, we can also write:
Graphical Representation of Discharge
We can note the voltage at time intervals and plot the data, which gives us the exponential graph shown below.
We should note the following about the graph:
shape is unaffected by the voltage.
half life of the decay is independent of the voltage.
current follows exactly the
same pattern as
I = V/R.
charge is represented by the voltage, as
Q = CV.
The graph is asymptotic, i.e. in theory the capacitor does not completely discharge. In practice, it does. For the current, the following graph is seen:
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. You may also find in some texts that the time constant is given the Physics code t (tau, a Greek lower case letter 't').
t = RC
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
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:
can link the half-life (a term from radioactivity) to the capacitance.
At the half life:
Q = Q0 ÷ 2
t = t1/2
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 ln (2) = t1/2/RC
ln (2) × RC
The half-life is 69 % of the time constant.
You must use the ln key on your calculator, NOT the log (or lg) key. The lg key will return answers for log10.
Complete discharge in theory does not happen as the graph is asymptotic. In reality, capacitors are considered to be discharged after 5 RC seconds.
A 5000 mF capacitor is charged to 12.0 V and discharged through a 2000 W resistor.
What is the time constant?
What is the voltage after 13 s?
What is the half-life of the decay?
(d) How long would it take the capacitor to discharge to 2.0 V
Time constant =
= 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
V = 3.3 volts
0.693 × RC
= 0.693 × 10 = 6.93 s.
We need to rearrange the formula by taking natural
= V0 e –t/RC
V / V0
V - loge
V0 = -t/RC [When you divide two
numbers, you subtract their
0.693 – 2.485 = - t/10
Þ t = 1.792 × 10 = 17.9 s
capacitor, charged up to 12.0 V is connected to a 100 kW
What is the time constant?
What is the voltage after 10 s?
(c) How long does it take for the voltage to drop to 2.0 V?
The discharge of a capacitor is a required practical. You will use a circuit like this:
You will measure the voltage and/or the current at time intervals of, say, 10 s. You need to have time constant of something between 50 s and 100 s. You want to have a discharge that is slow enough for you to read the results with a reasonable degree of certainty. You will need to work out the value of the resistance and capacitance that will enable you to achieve this.
You will be required to plot graphs that will look like this:
The graph above was modelled on a spreadsheet using a 10 kW resistor.
(a) Show that the capacitance of the capacitor was 2200 mF.
(b) Discuss whether these would be suitable values to use if you were harvesting data using a timer.
You can, of course, use a data logger with a current sensor and a voltage sensor. Your tutor may well get you to do it that way.
You are required to draw a semi-log plot. Let's look at how we do this for the voltage. We know that:
We can rearrange this equation:
and take natural logs to give:
Remember that the logarithms of two multiplied numbers add up. If the one number is divided by a second, the logarithms subtract.
So we can plot lnV against t to give a graph like this:
It is a straight line of the form:
Notice the following:
Time constant is -1 × gradient;
The y-axis intercept is ln V0 (the starting voltage);
The x-axis intercept is the time taken for the voltage to fall to 1 volt (as ln 1 = 0).
The graph above was modelled on a spreadsheet using a 10 kW resistor and a 2200 mF capacitor.
(a) What does the y-axis intercept represent?
(b) Use the graph to show that the initial voltage was 6.0 V
(c) Use the graph to show that the time constant is 22 s.
(d) Calculate the original charge.
In the exam, you will answer questions on a graph like this.
Although the graph has been shown in the positive quadrant, the straight line continues below the x-axis as shown below:
What does the intercept at the x-axis represent? Justify your answer with a suitable calculation.
Charging a Capacitor (Extension)
We can measure the exponential rise in capacitor voltage using a very similar circuit to the one above.
The graph of voltage against time looks like this:
The equation for capacitor charge is:
After RC seconds the voltage is 63 % of the maximum voltage (100 % - 37 %). In theory the voltage never gets to exactly the intended voltage. In reality, it's so close that there is no difference. Electronic engineers say that the full charge comes after 5 RC seconds.
The charge in a capacitor being charged behaves in the same way as the voltage:
However the current falls exponentially:
As the charge increases, the flow of charge onto the plates is reduced.
Using Calculus to derive the equation (Extension)
A differential equation is where there is a function combined with a derivative, for example:
Differential equations describe the way that values change when a system changes by a constant fraction. The solution is not always easy, but a mathematical technique called the separation of variables is useful. To do this we:
1. Multiply both sides by dt;
2. Divide both sides by x.
This gives us:
Then we need to integrate both sides to get rid of the derivative.
The problem here is that the power rule doesn't work:
which is impossible. Instead there is another rule:
So we do the first integration:
And the second (using the power rule since k = kx0):
C and D are constants of two different values. We can, for the sake of this argument, roll the two constants into on by writing:
a = D - C
So we rewrite the equation as:
ln (x) = kt + a
Now we bring in the exponential number e (= 2.718...) on both sides to get rid of the natural logarithm (ln is a logarithm to the base e). The terms become exponents. (An exponent is a power to which a base number is raised, i.e. eexponent.) We rewrite the equation as:
e(ln(x)) = e(kt + a) = ekt × ea
The term ea is simply a constant. Since e(ln(x)) = x, so we now write the equation as:
Note that when two numbers add up, they multiply when they become exponents. This is the converse of when logarithms of two multiplied numbers add up.
If the constants C and D are zero, ea = e0 = 1. Therefore the expression becomes:
We can apply the method of separation of variables to the differential equation that we have here:
In this case the term 1/CR is the fractional amount by which the charge decays. So to solve this differential equation, we multiply by dt and divide by Q. So our equation becomes:
Now we integrate both sides:
The trick here is that 1 = t0 so the result of the integration is:
We can combine the constants by saying:
a = D - C
So we now write:
To get rid of the ln, we make both sides an exponent of the exponential number e (=2.718...). Now eln(Q) = Q, so our equation now becomes:
Now when t = 0, we can see that:
So far we have haven't attempted to define what ea is. But we can easily see that it is charge at time 0. This has the physics code Q0, so we can rewrite the equation as:
This is our final result. In some books or websites, you may see the equation written as:
where t is the physics code for the time constant.