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 Author Topic:   Electric Starting Motors jimh posted 07-02-2009 08:57 AM ET (US)         The typical electric starter motor on an outboard engine is a series-wound DC motor. The mechanical power output of the motor is directly proportional to the current flow in its windings, which in turn is proportional to the applied voltage. (See http://en.wikipedia.org/wiki/Brushed_DC_electric_motor .) The more voltage applied to the motor, the more mechanical power it can create to start an outboard motor.The voltage applied to a starter motor in an outboard engine is supplied by the engine starting battery or cranking battery. The terminal voltage at the battery is proportional to its state of charge and inversely proportional to the current flow. The higher the state of charge, the higher the battery voltage. The higher the current flow, the lower the battery voltage.The starter motor is connected to the battery by electrical conductors, a solenoid relay, and perhaps some distribution switches and buses. The total resistance of these circuit elements will be in series with the motor resistance. Whenever any current flows through this series resistance, a voltage drop occurs. The voltage drop is proportional to both current flow and resistance. The higher the current flow, the greater the voltage drop; the higher the series resistance, the greater the voltage drop.The current in the starter motor is inversely proportional to its rotational speed. This means that when the motor is initially started and is turning slowly, the current demand will be the greatest.We all have likely experienced the situation of an electric starting motor being unable to crank over the engine it is supposed to start. The ignition key is turned to start, the solenoid engages, but the engine does not crank. This is due to the voltage at the starter motor becoming too low to generate enough mechanical power to crank the engine. The electric motor is stalled.To overcome this stalled situation. more voltage must be supplied to the starting motor. There are only two circuit elements which can affect this: the battery itself, and the voltage drop across the series resistance of the other circuit elements. Increasing the battery voltage requires increasing the state of charge. Since in a small boat the only source of charging current for the battery is the outboard engine, there is no possibility for charging the battery until the engine is started. We cannot effectively change the battery voltage in this situation, unless we change to a new battery or connect another battery in parallel.The voltage drop across the series resistance of other conductors in the circuit can be controlled. By using large conductors with very low resistance, the voltage drop can be minimized. Because the current flow at initial starting of the electric motor is very high, perhaps as much as 500-amperes, the voltage drop in the conductors can become significant. For example, if the conductors are made from 4-AWG wire and are 30-feet in length (total), the resistance in the conductors will be4-AWG wire = 0.25-ohm/1,000-feet x 30-feet = 0.0075 ohmVoltage drop = 0.0075 x 500Voltage drop = 3.75This means that 3.75-volt of the battery terminal voltage is lost across the conductors. If the battery were fully charged and had a terminal voltage of 12.9-volts, only 9.15 volts would be applied to the starter motor.If the wire size in our example is increased to 2-AWG, the resistance becomes2-AWG = 0.15-ohm/1,000-feet x 30-feet = 0.0045and the voltage drop reduces toVoltage drop = 0.0045 x 500Voltage drop = 2.25The starter motor will be supplied with 1.5-volts more than previously. This additional voltage may mean the difference between getting the engine to crank over or not. Also note that the 1.5-volt difference is as great as the change in battery terminal voltage from full charge (12.9-volts) to almost no charge (11.4-volts).

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