Take-home messages from this experiment include the following:
You can buy simple DC motors with reduction gearing built in, providing your choice of RPM. Literally hundreds of websites will sell you small motors for robotics projects.
When you reverse the voltage to a DC motor, the motor runs in reverse.
A DPDT relay can be wired so that when it closes its contacts, it reverses a power supply to a motor.
You can use two limit switches and a pair of diodes to stop a motor at two positions. In each of its stopping positions, the motor consumes no power and you won’t have the risk of it burning out.
What other projects can you imagine using this simple set of techniques?
Mechanical Power
In the United States, the turning force, or torque, of a motor is usually measured in pound-feet or ounce-inches. In Europe, the metric system is used to measure torque in dynes.
A pound-foot is easy to understand. Imagine a lever pivoted at one end, as shown in Figure 5-102. If the lever is one foot long, and you hang a one-pound weight at the end of it, the turning force is one pound-foot.
Figure 5-102. The rotational force created by a motor is known as “torque,” and in the United States it is measured in pound-feet (or ounce-inches, for small motors). In the metric system, torque is measured in dynes. Note that the torque created by a motor will vary according to the speed at which the motor is running.
Fundamentals
Wire gauges
If you’re going to power larger motors, or other components that take more current than LEDs or small relays, you really need to know about wire gauges. In particular, what’s the relationship between wire thickness and AWG (American Wire Gauge)? And what gauge of wire should you use for any given current?
You can find numerous charts and tables if you go online, but many of these sources contradict each other, especially on the topic of how much current is safe to run through each gauge of wire.
After making several comparisons (and testing some wire samples myself), I’ve compiled the table in Figure 5-103, which I recommend as a compromise. Note the following:
This table applies to solid-core copper wire.
For stranded wire, or copper that has been tinned (giving it a silver appearance), the number of ohms per foot will increase, the number of feet per ohm will decrease, and the maximum amperage will decrease, probably by around 20%.
Figure 5-103. American wire gauges (AWG) and their properties.
The maximum amperage assumes that the wire is insulated, preventing it from radiating heat as effectively as a bare conductor. I am also assuming that the wire is likely to be at least partially enclosed, inside a box or cabinet. At the amperages listed for each gauge of wire, you should expect the wire to become noticeably warm, and personally I would tend to use thicker wire instead of the maximums indicated in the table.
Most tables of this type only tell you the resistance of each gauge of wire in ohms per 1,000 feet. I have included that number but have also expressed the function the other way around, as the number of feet per ohm, as this doesn’t require you to do so much arithmetic with decimals.
Theory
Calculating voltage drop
Another fact that you often need to know is how much of a voltage drop a particular length of wire will introduce in a circuit. If you want to get maximum power from a motor, you don’t want to lose too much voltage in the wires that go to and from the motor.
Voltage drop is tricky, because it depends not only on the wire, but also on how heavily the circuit is loaded. Suppose that you are using 100 feet of 22-gauge wire, which has a resistance of about 1.5 ohms. If you attach it to a 12-volt battery and drive an LED and a series resistor offering a total effective resistance of about 1,200 ohms, the resistance of the wire is trivial by comparison. According to Ohm’s Law:
amps = volts / ohms
so the current through the circuit is only about 10mA.
Again, by Ohm’s Law:
volts = ohms × amps
so the wire with resistance of 1.5 ohms imposes a voltage drop of 1.5 × 0.01 = 0.015 volts.
Now suppose you’re running a motor. The coils in the motor create impedance, rather than resistance, but still if we measure how much current is going through the circuit, we can establish its effective resistance. Suppose the current is 1 amp. Repeating the second calculation:
volts = ohms × amps
So the voltage drop in the wire is now 1.5 × 1 = 1.5 volts! This is illustrated in Figure 5-104.
Bearing these factors in mind, I have compiled a table for you. I’ve rounded the numbers to just two digits, as variations in the wire that you use make any pretense of greater accuracy unrealistic.
To use this table, you need to know how much current is passing through your circuit. You can calculate it (by adding up all the resistances and dividing it into the voltage that you are applying) or you can simply measure the current with a meter. Just make sure that your units are consistent (all in ohms, amps, and volts, or milliohms, milliamps, and millivolts).
In the table, I have arbitrarily assumed a length of 10 feet of wire. Naturally you will have to make allowances for the actual length of wire in your circuit. The shorter the wire, the less the loss will be. A circuit with only 5 feet of wire, and the same amperage and voltage, will suffer half of the percentage loss shown in the table. A circuit with 15 feet of wire, and the same amperage and voltage, will suffer 1.5 times the percentage loss.
