Keep blowing my starter solenoid up.

[cyclops222]

After 3 years the copper in the battery cable actually corroded away to causing intermittent slow starts.

 

[Lou C]

Ok had a similar issue but not so extreme (charging voltage 15.5 at batteries) sone years back. And I learned the cause & solution from a tech guy at ARCO & it also came in instructions for a new alternator I bought from them. Apparently on older OMCs resistance develops in the voltage sense circuit. So the alt sees lower voltage that what’s actually in the system and increases it to compensate. Their cure was to disconnect the sense wire from the alt and tape it up safely. Then run a jumper wire from the large B+ terminal to the S terminal. When I did this my charging voltage dropped to normal 14.2V at the batteries. Or go on a fishing expedition and find the high resistance connection! Actually thinking about it the best set up might be to run the sense wire from the alt to common terminal on the battery switch since it’s closer to the batteries…

 

[Lou C]

PS
Looking at a few Volvo wiring diagrams that’s how many of them have the sense wire connected….short wire from B+ to S terminal….

 

[nola mike]

You have that backwards. Higher voltage requires less current for a given power load.

Yes, but in something with a fixed resistance (light bulbs, poor connections) the current drawn is proportional to the voltage.

 

[JustJason]

Explain the me mechanism of this
My understanding is that V = I x R

It's true that V = I X R If you are trying to solve for voltage.
If you are trying to solve for current, it is I = V / R
If you are trying to solve for resistance, it's R=V / I

So, let's do a couple of equations.

Let's say we have a light bulb that is 1 ohm of resistance, and we are using 12 Volts, what are the amps drawn by the bulb? The proper equation is I = 12 volts / 1 ohm of resistance. 12 divided by 1 is 12. The light bulb draws 12 amps. (It must be a spicy bulb).

Now let's double our voltage to 24, do the amps go up or down?

Ohms law says I = 24 / 1. 24 divided by 1 is 24. So, the light bulb now draws 24 amps.

Did the current go up or down with an increase of voltage?

Keep in mind Ohms law only a applies to linear loads, like a resistor, IE the little filament inside of a light bulb. If you want to figure out the math on a non-linear load, like a starter motor, Watts law is the proper law to apply.

Watts law is:
P (Watts) = Voltage X Current.
Voltage = Power (watts) divided by current.
Current = Power (watts) divided by voltage.

Notice there is no resistance here, because there is no linear resistor. Electric motors are not linear resistors. Electric motors are rated in watts drawn. While the wattage rating of starter motors vary, your customary GM starter draws about 1.5 kilowatts, which to keep the math easier, is the same as 1500 watts.

So, let's say we have a bad ground, or corroded connection, causing a drop in voltage to a starter motor. Does the current consumed by the starter motor go up or down?

Let's do the equations.

We are solving for current, so the correct equation is Current = Power (watts) divided by voltage.

Current (Uknown) = Power 1500 watts / 12 Volts
1500/12 = 125. This means our 1500 watt rated starter motor, at a fixed 12 volts, draw 125 amps.

We all know battery voltage drops when a heavy load is applied, and then the voltage inside the battery recovers when the load is removed. So, what does our 1500 watt rated starter motor draw at 10 volts?

Same equation. Current (unknown) = Power 1500 watts / 10 volts.
1500/10 = 150.

So, at 12V, our 1500-watt starter draws 125 amps of current. At 10V our 1500-watt starter draws 150 amps of current.

The takeaway here is this:

Ohms law equations prove that when the voltage goes up, the current goes up. When the voltage goes down, the current goes down.

Watts law is just the opposite. When the voltage goes down, the current goes up, when the voltage goes up, the current goes down.

 

[JustJason]

saved

 

[JustJason]

update on my issue. after doing some checking with a voltmeter and test light I found that my alternator is putting out to the ignition switch 16.8 volts just off of idle on the hose. Way to high. Thinking this is what's burning up the contacts in my solenoid. Thoughts?

No, it's not, you have 2 separate problems. The contacts on the solenoid are open when the engine is running, not closed, and 16.8 V is not going to jump those contacts and burn them.

First, you are having an issue with your slave solenoid, not the starter solenoid; they are 2 different things.

The way it works is when you turn the key to start, voltage leaves the key switch and goes into and then out of the neutral safety switch inside of the shifter. After the voltage leaves the neutral safety switch, it travels down the harness and goes to the slave solenoid. Once that voltage hits the slave solenoid, the contacts inside f the slave solenoids close, and send voltage (from a different source, and not the key switch) to the starter solenoid, telling the starter solenoid contacts to close, and turn on the starter.

What you have to do here is first duplicate the problem. When it will not start, you need to have a helper repeatedly turn the key to start and see if that slave solenoid is clicking or not. If it's not clicking, you need to check for power on one of the small terminals on the slave solenoid and check for ground on the other solenoid. It's missing 1 or both.

As jumping the terminals on the slave solenoid turns the starter on, we know the voltage at the big terminals on the slave solenoid is fine, if there was an issue there, the starter would not turn on.

You probably have a bad key switch or a bad neutral switch, or loose wiring between them; that's usually what it is.

The alternator overcharge is a separate issue. On that, you need to disconnect the exciter wire and see if the voltage drops down to battery voltage. If it does not, the alternator is bad. If it does, you likely have low voltage/corrosion at the exciter wire.

 

[kd4pbs]

Explain the me mechanism of this
My understanding is that V = I x R

That is correct. And I=V/R, and R=V/I.
Resistance doesn't change. The load resistance is the load resistance. There are exceptions, like with devices that have certain types of DC-DC converters, but for some of the loads on our vessels, they can be characterized as a simple resistor. If that device exhibits 12 ohms across its power input terminals, it will have that whether it has 0 volts across them or 20 volts across them. When you increase voltage to that widget, the current also increases since that 12 ohms doesn't change.
At 12 volts, that 12 ohm load load will draw 1 amp, dissipating 12 watts. At 15 volts, that same load (the resistance doesn't change) is now drawing 1.25 amps, an increase of 25%. It's also being forced to dissipate 18.75 watts now, an increase of 56.25% over what it dissipated at 12 volts. Things can get toasty and smoky quick with this kind of stuff.
What really throws the monkey wrench into the ointment is that most semiconductor devices (the radio, tachometer, ignition system, etc.) have a non-linear load resistance, where up to a point things increase fairly linearly. Past that point, the resistance increases rapidly, along with the current and the power. And oddly enough, so does the cost of replacing that factory installed smoke we just let out running a stereo and a chart plotter designed for a maximum of 14.4 volts at the level of 16.8 volts
It is often said in my industry that the $800 power supply blew up so that it could protect the 25¢ fuse that is still working fine. rpatton got fortunate!

 

[kd4pbs]

Actually thinking about it the best set up might be to run the sense wire from the alt to common terminal on the battery switch since it’s closer to the batteries…

Lou - this is basically what properly installed multi-battery systems with battery isolators do. In my Regal (5.7 Gi V-P DP-SM) I've got two batteries which are charged through a battery isolator. There's a sense line coming from the isolator that runs back to the sense terminal on the alternator. I have run the charge circuit from the alternator directly to the isolator, bypassing the round engine connector. This way, the batteries are charged regardless of the position of the battery selector switch. Running that sense wire and dedicated charge wiring even to a single battery setup works quite well. It's also smart to run a separate dedicated charging circuit from the alternator, because most of the time it's ran through the round engine connector which was never designed to handle the charging current modern alternators can produce. I've seen many of the alternator circuits on those connectors in very poor shape.

 

[kd4pbs]

So, at 12V, our 1500-watt starter draws 125 amps of current. At 10V our 1500-watt starter draws 150 amps of current.

DC and universal motors draw less current with less input voltage.
AC induction and synchro motors do indeed draw more current the lower the applied voltage.
If this were not the case, if your battery is very worn out to just 3 volts when cranking, and you engage the starter, sparks, melted wires, and mayhem would ensue.

 

[kd4pbs]

You have that backwards. Higher voltage requires less current for a given power load.

As long as the power load is designed for that higher voltage, you are 100% correct sir. I don't think rpatton swapped out all of his 12V accessories for 16.8V accessories though

 

[dingbat]

So if we put 24 volts on 12 vdc bow lights ?

You would instantly blow the filament making R = 0.0
To Solve for current (I):
I = V/R
I =24/0
I = 0.00

Let's assume you have a five (5) watt bulb rated up to 24 VDC .......
5 watts @ 12V= 1.3A
5 watts @ 24V = 0.3A

We are talking working voltages in a boat. No electrical design.

Ohm's law deals with the relationships between current, voltage and resistance.
How do you decided what wire size and or fuse to use w/o applying electrical design principals?

 

[kenny nunez]

Another place to look is behind the instrument panel. Touch every wire terminal to check for a loose terminal.
Of all the marine relay/solenoids I have experience with the small ones are the most troublesome. I have had defective ones right out the box on some occasions. The older style large ones last the longest.

 

[JustJason]

You would instantly blow the filament making R = 0.0
To Solve for current (I):
I = V/R
I =24/0
I = 0.00

Let's assume you have a five (5) watt bulb rated up to 24 VDC .......
5 watts @ 12V= 1.3A
5 watts @ 24V = 0.3A


Ohm's law deals with the relationships between current, voltage and resistance.
How do you decided what wire size and or fuse to use w/o applying electrical design principals?

I think you mean R = Infinity......

R = 0.0 is a dead short

 

[kd4pbs]

You would instantly blow the filament making R = 0.0

That would actually be R=∞

Update... dang - I need to remember to read ALL the updated posts!

Jason... that would have to be a superconductor for sure!

 

[cyclops222]

A little knowledge can be misleading.

 

[kd4pbs]

Let's assume you have a five (5) watt bulb rated up to 24 VDC .......
5 watts @ 12V= 1.3A
5 watts @ 24V = 0.3A

For some reason, my numbers differ greatly.
I=P/V

 

[nola mike]

For some reason, my numbers differ greatly.
I=P/V

Same general concept. But I don't remember the point of this thread anymore.

 

[dingbat]

For some reason, my numbers differ greatly.
I=P/V

No idea how my calculator came up with the numbers above.....lol but the answer is still the same...... Higher voltage = less current