Make Electronics

What Next? 283

Experiment 32: A Little Robot Cart

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 mo-
tor 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 varia-
tions 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. So, to use the table:

1. Divide your length of wire by 10. (Make sure that you
   measure the length in feet.)


2. Use the result to multiply the number in the table.
   The table also arbitrarily assumes that you have a 12-volt
   supply. Again, you will have to make allowances if you are
   using a different voltage. So, to use the table:


3. Divide 12 by the actual voltage of your power supply.


4. Use the result to multiply the number in the table.
   I can summarize those two steps like this:
   Percent voltage lost = P × (12 / V) × (L / 10)
   where P is the number from the table, V is your power-
   supply voltage, and L is the length of your wire.

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100 feet

22 gauge wire

1.5 ohms 100 feet

22 gauge wire

1.5 ohms

Effective

resistance

1200 ohms

Approx. 10 mA current

15mV voltage drop

Approx. 1A current

1.5 volts voltage drop

Effective

resistance

10.5 ohms

Figure 5-104. The voltage drop imposed by wiring will depend

on the current and the resistance in the circuit. The drop will

be greatest when the resistance of the circuit is low and the

amperage is high.