Make Electronics

Switching Basics and More 69

Experiment 9: Time and Capacitors

Release the pushbutton, set aside your meter, and discharge the capacitor by
touching R2 across it for a second or two. Now substitute a 50K resistor for R1,
and repeat the measurement. The meter should count upward almost twice
as fast as before.

Voltage, Resistance, and Capacitance

Think of the resistor as a faucet, and the capacitor as a balloon that you are
trying to fill with water. When you screw down the faucet until only a trickle
comes through, the balloon will take longer to fill. But a slow flow of water will
still fill the balloon completely if you wait long enough, and (assuming the bal-
loon doesn’t burst) the process ends when the pressure inside the balloon is
equal to the water pressure in the pipe supplying the faucet. See Figure 2-79.

Similarly, in your circuit, if you wait long enough, eventually the voltage across
the capacitor should reach the same value as the voltage of the power supply.
In a 12-volt circuit, the capacitor should eventually acquire 12 volts (although
“eventually” may take longer than you think).

This may seem confusing, because earlier you learned that when you apply
voltage at one end of a resistor, you get less voltage coming out than you have
going in. Why should a resistor deliver the full voltage when it is paired with a
capacitor?

Forget the capacitor for a moment, and remember how you tested just two
1K resistors. In that situation, each resistor contained half the total resistance
of the circuit, so each resistor dropped half the voltage. If you held the nega-
tive probe of your meter against the negative side of your power supply and
touched the positive probe to the center point between the two resistors, you
would measure 6 volts. Figure 2-80 illustrates this.

Now, suppose you remove one of the 1K resistors and substitute a 9K resis-
tor. The total resistance in the circuit is now 10K, and therefore the 9K resistor
drops 90% of the 12 volts. That’s 10.8 volts. You should try this and check it
with your meter. (You are unlikely to find a 9K resistor, because this is not a
standard value. Substitute the nearest value you can find.)

Now suppose you remove the 9K resistor and substitute a 99K resistor. Its volt-
age drop will be 99% of the available voltage, or 11.88 volts. You can see where
this is heading: the larger the resistor, the larger its contribution to voltage
drop.

However, I noted previously that a capacitor blocks DC voltage completely. It
can accumulate an electrical charge, but no current passes through it. Therefore,
a capacitor behaves like a resistor that has infinite resistance to DC current.

(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 capaci-
tor steals almost the complete voltage drop in the circuit, and the voltage

Figure 2-79. When the faucet is closed

half-way, the balloon will take longer to fill,

but will still contain as much water and as

much pressure in the end.