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Basic DC Electricity and It's Application in Old Trucks
A common modification in early trucks is the conversion
of the original 6 VDC system over to 12 VDC. Often this causes some confusion
to the oletrucker who is not familiar with the principles of DC electricity
and current flow. The easiest analogy is to compare the flow of
current to that of water flowing through a garden hose. The flow of current
is similar to the flow of the water through the hose, the voltage is similar
to water pressure, and the resistance is like the diameter of the hose
that the water is flowing through. For a given diameter garden hose if
you increase the water pressure (i.e. raise the voltage), you increase
the water flow (increase the current). If, on the other hand you kept
the water pressure the same but decreased the diameter of the garden hose
(increasing the resistance), you would reduce the water flow (reduce the
current).
In electricity the principles stated above are represented mathematically by a formula known as Ohm's Law. Ohm's Law states that the voltage is equal to the current flowing through a circuit times the resistance of the circuit, or
| E | = Voltage (Volts DC) |
| I | = Current (Amps DC) |
| R | = Resistance (Ohms) |
Solving Ohms Law for the other variables yields:
| I = E / R | and | R = E / I |
When ever energy is converted into another form there is some inefficiency. No circuit is 100% efficient. The wasted energy is dissipated as heat (power). As long as the dissipated power is within the design considerations of the circuit components there is no problem. When circuit designs considerations are exceeded conditions can arise which result in the destruction of circuit components due to excess power dissipation. Such excess conditions include increasing the operating voltage or current requirements of the circuit. Power dissipation can be determined by Watt's Law, which states
P = I x E
Where
| P | = Power (Watts DC) |
| I | = Current (Amps DC) |
| E | = Voltage (Volts DC) |
Solving Watts Law for the other variables yields:
| I = P / E | and | E = P / I |
another useful derivation of this formua is obtained
by substituting Ohm's Law (E=IxR) into Watt's Law which results in
P = I X I XR or
P = I2 x R
III. What does this mean in regard to 6 VDC to 12 VDC conversions?
Actually it's pretty simple, suppose you have a stock
1950 Chevrolet 3100 which came from the factory with a 6 VDC system.
Let's also suppose that our stock Chevy uses 60 watts of power when its
running down the road with headlights on, heater motor blowing, and radio
playing. Note that all my numbers are theoretical, I have no idea
what the actual amount of power used by such a setup would be. At
any rate, by applying Watt's Law from above
| I = P / E |
| I = 60 Watts / 6 VDC |
| I = 10 Amps DC |
we can determine that our Chevy is drawing 10 amps of
current to keep itself functioning. Now suppose we decide we are
tired of that old 6 V system and we want to upgrade to 12 VDC.
After the installation of the 12 V system we cruise down the road still
using 60 Watts of power but now we can calculate that
| I = P / E |
| I = 60 Watts / 12 VDC |
| I = 5 Amps DC |
So we cut our current requirements in half while still
maintaining the 60 Watts in power that we needed to operate the vehicle.
This is one reason why automobile manufacturers changed from 6 VDC to
12 VDC. The reduced current demand of the 12 V system allows them
to use a smaller gage of wire which is both cheaper and easier to work
with.
Important Note
The above statement regarding half the current would only be true
assuming that all 6V devices (heater motors and light bulbs) were replaced
with 12V equivalents. If you are using the old 6V device in conjunction
with a voltage reducer then the affected circuit would actually use twice
the power that you did before because the device still uses the same amount
of power plus you have a voltage reducer dropping the other 6V from the
12V system.
IV. How does this affect heater blower motors?
The circuit below (Fig. 1) is a simplified schematic of the heater blower circuit in a truck equipped with a 6V system.
 |
Amp Under Normal Operation |
When the switch is closed, the motor begins to operate
and draws 1 amp of current (hypothetical value). Since there are
no other devices in series with the motor in this circuit, the entire
6 V supply will be applied across the motor terminals. With this
information we can calculated the effective load resistance of the motor
using Ohm's Law from above.
| R = E / I |
| R = 6 VDC / 1 Amp |
| R = 6 Ohms |
Now we know that our motor presents a 6 ohm load to the
supply voltage. We can also figure out how much power our motor
uses from this information. For this calculation we use Watt's Law.
| P = I2 x R |
| P = 1 Amp2 x 6 Ohms |
| P = 6 Watts |
So our motor uses 6 Watts of power under normal operation.
Suppose we convert to a 12 VDC system without making any modifications
to the blower motor circuit. This would result in the the circuit
shown below (Fig. 2).
 |
Has An Effective Resistance of 6 Ohms |
Calculating the amp draw of the motor under this
condition yields
| I = E / R |
| I = 12 VDC / 6 Ohms |
| I = 2 Amps |
and taking that and calculating power we find
| P = I2 x R |
| P = 2 Amps2 x 6 |
| P = 24 Watts |
So we find under these circumstances
that we are drawing twice the current we would under normal circumstances
and dissipating 4 times the power!! The
motor would quickly fail unless something is added to the circuit to bring
it back into normal operating parameters.
That something is what is commonly called the voltage reducer.
In actuality it is nothing more than a resistor which is designed to have
a voltage drop across it of 6 VDC. The schematic below illustrates
the same blower motor circuit with a voltage reducer installed.
|
Has An Effective Resistance of 6 Ohms and Voltage Reducer has a voltage drop of 6 VDC. |
Now that we have installed the voltage reducer with it's 6 V drop the other 6 Volts is dropped across the motor. Note that the voltage reducer must have the same resistance value as the effective resistance of the motor in order to drop the voltage to the required 6 VDC. The motor is now operating at the voltage and current it was designed for while the reducer dissipates the remaining power in the circuit.
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Me
Bill Bailey
Last Revised : 8/15/99
Thanks to Brad Rusnak for his comments concerning this write-up