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ContinuousWave: Small Boat Electrical
Electrical Trouble Shooting: Accessory Wiring
|Author||Topic: Electrical Trouble Shooting: Accessory Wiring|
posted 04-07-2006 05:19 PM ET (US)
I own a 1991 Walk Around 21. I took it out recently and observed that whenever I turned on the navigation lights or wipers, my fish finder and chart plotter would go blank and turn off by themselves. I didn't have this problem last fall. The boat sat for two winter months and last week I went boating and noticed the occurrence. It happened again two days ago. What should I look into as far as fixing this minor glitch? TIA.
posted 04-07-2006 07:31 PM ET (US)
The Fish finder and GPS combined are very low current draw. MAYBE a half an amp,
quite possibly 0.2A. The lights and wipers are probably
The good news is the problem is reproduceable.
I'd start by eyeballing all the connections in the console, and loosening
Get a digital voltmeter.
Step 1.: Turn everything off. Measure the
Step 2: Find where power comes into the console (or dashboard).
Step 3: measure where the GPS and FF are wired in, everything
And BTW, you probably don't have a short (a connection where
Also, pull the fuses and put them back. I suspect the lights
posted 04-07-2006 10:56 PM ET (US)
You have a high resistance in the 12-volt distribution wiring.
posted 04-08-2006 12:25 AM ET (US)
pcchan - as Jimh mentions - you have a high resistance (often caused by corrosion) in your power distribution system. Correct the problem by cleaning and protecting connections - starting at the battery and working your way through the other circuits. Chances are, it is close to the battery. --- Jerry/Idaho
posted 04-08-2006 09:57 AM ET (US)
To explain further:
The voltage drop across a resistance is directly proportional to the current flow:
E = IR
E = electromotive force in volts
i.e., Ohm's law.
If a bad connection exists in your 12-volt distribution wiring, it adds resistance. Assume it adds 5-ohms of resistance.
The current drawn by the electronic devices is small. Let us say 0.2-ampere. The voltage drop across the bad connection will be
E = IR
Thus the 12-volt supply voltage to the electronic devices will drop to 11-volts. This is still enough for them to operate.
When the navigation lighting circuit is energized, the lamps draw about 2-amperes of current. The voltage drop across the high resistance will now be:
E = IR
Now the supply voltage drops to 1-volt. The electronics shut off.
This example is a bit simplistic, but it demonstrates the problem.
Also, if the above situation were to occur, consider the power being dissipated across the bad connection.
Power is proportional to current and voltage:
P = IE
By substitution and algebra, we can combine the two formulas to solve for power in terms of voltage and resistance:
P = IE and E=IR, or I=E/R
If 11-volts drop across a 5-ohm load, the power dissipated will be
P = 11^2/5
This is a great deal of heat and will soon make itself known.
In actual practice this won't be quite as described above. The current drawn by the lamps will be reduce by the drop in voltage supplied to them. For example, if each lamps is rated as a 10-watt lamp at 12-volts, this implies that three of them will have a resistance of
R = E^2/P
If there is a 5-ohm resistance in the distribution, the 12-volt supply will divide across them in proportion to their resistances, thus across the lamps will be
E = 12 x [ 4.8/(4.8 + 5) ]
and the voltage across the 5-ohm resistance will be
E = 12 x [ 5/(4.8+5)]
So again, the electronics will see the supply voltage drop from 11 volts to about 5.87 volts, and they shut off.
posted 04-09-2006 09:03 PM ET (US)
I’d look at the negative (mis-named “ground”) wiring as well. I’ve spent ages hunting the wrong side of the circuit for problems, especially with trailer lights. Often we only look at the red wires where the switches and fuses are connected, figuring the black wires somehow don’t count for as much. But if you load up a bad connection on the negative side with a high current device, then the resistance and resulting voltage drop will reduce the voltage seen by your entire system (including electronics). If you don’t feel like pursuing Jim’s math, just be sure your negative wire is as big and well connected to the battery as your positive wires.
posted 04-09-2006 10:51 PM ET (US)
I haven't had a chance to trouble shoot the problem yet, but I am grateful for all your suggestions and the technical aspect of it. I will work on it some time this week. Thank you again for always being so helpful.
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