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|Resistors and Capacitors Together|
Now that we have looked at each of the three basic types of electronic components, we need to explore how they behave in various combinations. As we do so, remember that while each component still retains its basic properties, the combination can have its own characteristics, which may not seem intuitive at first.
On this page, we'll begin by considering a resistor and capacitor working together in a circuit, and see how the resistor affects the charging and discharging of the capacitor.
Consider the circuit shown to the right. Initially, we will have switch S in position 2. Capacitor C is fully discharged and no current flows through R and C. The circuit is quiescent at this point.
Now we move the switch to position 1. This connects the series combination of R and C to the battery. Current can flow through the circuit, and the capacitor will begin to charge. The question is, how fast?
Keep in mind that the voltage across a capacitor cannot change instantaneously. Therefore, at that first instant the entire battery voltage, E, appears across resistor R, and the charging current for C is determined by Ohm's Law: I = E/R.
However, now the capacitor voltage, VC begins to increase. This reduces the voltage, VR, that remains across the resistor. Therefore the charging current will be reduced slightly, and the capacitor will charge more slowly than before. This will continue until the capacitor has charged to the voltage E, and there is no further current flow in this circuit.
But this isn't quite enough. We can see that the values of R and C will affect the amount of time it takes for C to become fully charged. But we'd like to be able to state the appropriate equation so we can determine not only the charging time but also the way in which VC and the circuit current will change while C is charging.
To accomplish this, we go back to some basic definitions. First, we note that by moving one coulomb of electric charge from one plate of a 1 farad capacitor to the other, we will change the voltage between plates by 1 volt. We can change the size of the capacitance, adjust the voltage, and thereby adjust the amount of charge required to make the change. However, the basic equation still holds: E = q/C, where q is the electric charge in coulombs.
The other definition is for the current flowing in the circuit: one Ampere of current consists of one coulomb of charge passing a given point in a circuit in one second.
In this circuit, however, the charging current is constantly changing as the voltage across C increases. Therefore, we must look at a steadily decreasing rate of charge. This brings us to a bit of differential calculus. If you aren't familiar with this, don't worry; you'll be able to make use of the results. But for completeness, we include the appropriate expression here:
|iC = C||dvC|
Qualitatively, the current flowing through the capacitor is directly proportional to the value of the capacitor itself (high value capacitors charge more slowly), and is directly proportional to the change in capacitor voltage over time. The use of differential calculus allows us to track the changing current on an instant-by-instant basis.
But the current charging the capacitor is also the current flowing through R. And the voltage across R is whatever part of E that hasn't already been built up as the charge on C. Therefore, we can apply Ohm's Law here:
|iC = iR =||E - vC|
Solving differential equations is beyond the scope of this page. However, we can present the final equation that describes the capacitor voltage at any time t, for any values of R and C, and any battery voltage E:
Here, ε is the base for natural logarithms, with a value of approximately 2.7182818. Because we are using it in this fashion, the equation above is known as an exponential equation.
At the moment switch S is closed, time t = 0. Since ε0 = 1, we see that:
This is exactly what we would expect, since the capacitor is completely discharged at the start of the sequence. But what does the rest of the charging curve look like? Let's plot this expression over time, as the capacitor charges, and see how it behaves.
The figure to the right shows the capacitor voltage, vC, as a percentage of E as the capacitor charges over time. Thus, it is the plot of the equation:
|vC||= 1 - ε(-t/RC)|
Note the product RC in the exponent of ε. This is very important, because it shows that both R and C control the charging time equally. Also, because that exponent is (-t/RC), if we set RC = t, the exponent will be -1. Therefore, we show time in this graph as multiples of RC. In addition, the RC product is identified as the time constant for this circuit.
At first glance, this would seem to be very strange. How can the product of a resistance in ohms and a capacitance in farads possibly give us a time in seconds? To understand how this is possible, we go back to the basic definitions and some dimensional analysis.
Resistance opposes the flow of current through a circuit. By Ohm's Law, R = E/I. Thus, 1 ohm may also be expressed as 1 volt/ampere.
Current is a measure of the amount of charge flowing through a circuit in a given amount of time. By primary definition, 1 ampere is equal to 1 coulomb/second.
Capacitance is the capacity to hold an electrical charge. A capacitance of 1 farad will exhibit a change of 1 volt if 1 coulomb of charge is moved from one plate to the other. Hence, 1 farad can be expressed as 1 coulomb/volt.
Putting these three basic definitions together we get the following progression:
|=||volts × seconds||×||coulombs|
Thus, we see that the RC product is indeed a measure of time, and can properly be described as the time constant of this circuit. This in turn means that this curve can be used to determine the voltage to which any capacitor will charge through any resistance, over any period of time, towards any source voltage. It is the general curve describing the voltage across a charging capacitor, over time.
Theoretically, C will never fully charge to the source voltage, E. In the first time constant, C charges to 63.2% of the source voltage. During the second time constant, C charges to 86.5% of the source voltage, which is also 63.2% of the remaining voltage difference between E and vC. This continues indefinitely, with vC continually approaching, but never quite reaching, the full value of E. However, at the end of 5 time constants (5RC), vC has reached 99.3% of E. This is considered close enough for practical purposes, and the capacitor is deemed fully charged at the end of this period of time.
Now that we have charged our capacitor, what happens if we move switch S in our original circuit to position 2 (We have repeated the circuit to the right for easier reference)? We have disconnected resistor R from the source, E, and connected it in parallel with capacitor C instead.
At this point, the capacitor has a discharge path, so it will begin discharging through R. However, as vC continues to drop, the discharge current likewise decreases, in accordance with Ohm's Law. Therefore, it is logical to assume that the capacitor discharge curve will probably follow some of the same rules as the capacitor charge curve above. However, the resistor voltage is now the same as the capacitor voltage, since R and C are now in parallel. So what is the resulting equation?
The figure to the right shows the appropriate RC discharge curve. This graph shows the function:
Here, E refers to the starting voltage on the capacitor, which need not be the same as the battery voltage E in the schematic diagram above. In fact, if you look at the capacitor voltage after it has partly discharged, the same curve applies.
As with the charge curve, the discharge curve is exponential in shape. The RC time constant still applies; the capacitor is deemed to be fully discharged at the completion of five time constants.
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