(Actually the insulation inside the capacitor allows a little bit of “leakage,” but a perfect capacitor would have infinite resistance.)
The value of any resistor that you put in series with the capacitor is trivial by comparison. No matter how high the value of the resistor is, the capacitor still provides much more resistance in the circuit. This means that the capacitor steals almost the complete voltage drop in the circuit, and the voltage difference between one end of the resistor and the other will be zero (assuming that we ignore little imperfections in the components). Figure 2-80 may help to clarify this concept.
Figure 2-80. When two resistances are in series, the larger one drops the voltage more than the smaller one. If the larger resistance becomes infinite (as in the case of a capacitor), the smaller one no longer has any measurable contribution to the voltage drop, and the voltage is almost exactly the same at both ends.
You should try this using real resistors and capacitors—although if you do, you will run into a little problem. When you use your meter in its “DC volts” mode, it diverts a little of the current in the circuit—just a tiny taste—in the process of measuring it. The meter steals such a small amount, it doesn’t affect the reading significantly when you are checking voltage across a resistor. The internal resistance of the meter is higher than the values of most resistors. However, remember that the internal resistance of a capacitor is almost infinite. Now the internal resistance of the meter becomes significant. Because you can never have an ideal meter, any more than you can have an ideal capacitor or resistor, your meter will always interfere with the circuit slightly, and you will get only an approximate indication.
If you try to measure the voltage on a capacitor that has been charged but is now not connected to anything else, you’ll see the number slowly falling, as the capacitor discharges itself through the meter.
Theory
The time constant
You may be wondering if there’s a way to predict exactly how much time it takes for various capacitors to charge, when they are paired with various resistors. Is there a formula to calculate this?
Of course, the answer is yes, but the way we measure it is a bit tricky, because a capacitor doesn’t charge at a constant rate. It accumulates the first volt very quickly, the second volt not quite as quickly, the third volt even less quickly—and so on. You can imagine the electrons accumulating on the plate of a capacitor like people walking into an auditorium and looking for a place to sit. The fewer seats that are left, the longer people take to find them.
The way we describe this is with something called a “time constant.” The definition is very simple:
TC = R × C
where TC is the time constant, in seconds, and a capacitor of C farads is being charged through a resistor of R ohms.
Going back to the circuit you just tested, try using it again, this time with a 1K resistor and the 1,000 µF capacitor. We have to change those numbers to farads and ohms before we can put them in the formula. Well, 1,000 µF is 0.001 farads, and 1K is 1,000 ohms, so the formula looks like this:
TC = 1,000 × 0.001
In other words, TC = 1—a lesson that could not be much easier to remember:
A 1K resistor in series with a 1,000 µF capacitor has a time constant of 1.
Does this mean that the capacitor will be fully charged in 1 second? No, it’s not that simple. TC, the time constant, is the time it takes for a capacitor to acquire 63% of the voltage being supplied to it, if it starts with zero volts.
(Why 63%? The answer to that question is too complicated for this book, and you’ll have to read about time constants elsewhere if you want to know more. Be prepared for differential equations.) Here’s a formal definition for future reference:
TC, the time constant, is the time it takes for a capacitor to acquire 63% of the difference between its current charge and the voltage being applied to it. When TC=1, the capacitor acquires 63% of its full charge in 1 second. When TC=2, the capacitor acquires 63% of its full charge in 2 seconds. And so on.
What happens if you continue to apply the voltage? History repeats itself. The capacitor accumulates another 63% of the remaining difference between its current charge, and the voltage being applied to it.
Imagine someone eating a cake. In his first bite he’s ravenously hungry, and eats 63% of the cake in one second. In his second bite, not wanting to seem too greedy, he takes just another 63% of the cake that is left—and because he’s not feeling so hungry anymore, he requires the same time to eat it as he took to eat the first bite. In his third bite, he takes 63% of what still remains, and still takes the same amount of time. And so on. He is behaving like a capacitor eating electricity (Figure 2-81).
Figure 2-81. If our gourmet always eats just 63% of the cake still on the plate, he “charges up” his stomach in the same way that a capacitor charges itself. No matter how long he keeps at it, his stomach is never completely filled.
Theory
The time constant (continued)
The cake eater will always have a few crumbs to eat, because he never takes 100% of the remainder. Likewise, the capacitor will never acquire a full charge. In a perfect world of perfect components, this process would continue for an infinite time.
In the real world, we say rather arbitrarily:
After 5 × TC the capacitor will be so nearly fully charged, we won’t care about the difference.
In the table is a calculation (rounded to two decimal places) showing the charge accumulating on a capacitor in a 12-volt circuit where the time constant is 1 second.
Here’s how to understand the
