What Next? 247Experiment 28: Making a Coil React12V
DCFigure 5-34. The breadboarded version of the schematic in Figure 5-33 shows a simple
way to set it up for a quick demo. The green button is a tactile switch. The two red LEDs
should be placed so that the polarity of one is opposite to the polarity of the other.
When you press the button, one LED should flash briefly. When you release the
button, the other LED should flash.
What’s happening here? The coil possesses self-inductance, which means that
it reacts against any sudden change in the flow of electricity. First it fights it,
and during that brief moment, it blocks most of the current. Consequently, the
current looks for an alternative path and flows through D1, the lefthand LED
in the schematic. (D2 doesn’t respond, because it can pass current only in the
opposite direction.)
Meanwhile, the voltage pressure overcomes the coil’s self-inductance. When
the self-inductance disappears, the resistance of the coil is no more than 10
ohms—so now the electricity flows mostly through the coil, and because the
LED receives so little, it goes dark.
When you disconnect the power, the coil reacts again. It fights any sudden
changes. After the flow of electricity stops, the coil stubbornly sustains it for a
moment, because as the magnetic field collapses, it is turned back into elec-
tricity. This residual flow of current depletes itself through D2, the LED on the
right.
In other words, the coil stores some energy in its magnetic field. This is similar
to the way a capacitor stores energy between two metal plates, except that
the coil blocks the current initially and then lets it build up, whereas the ca-
pacitor sucks up current initially, and then blocks it.
The more turns of wire you have in your coil, the more self-inductance the coil
will have, causing your LEDs to flash more brightly.
Hot Resistors
You’ll be passing about 50mA
through the 220Ω resistor, while
the current is flowing. At 12 volts,
this works out at 0.6 watts. If you
use a 1/8-watt resistor, you will be
overloading it, and it will get quite
hot and may burn out. If you use a
1/4-watt resistor, it will still get hot,
but is unlikely to burn out, as long as
you don’t press the button for more
than a second or two.
Don’t run the circuit without the coil
of wire; you’ll be trying to pass more
than 50mA through the LEDs.