Charging from an outboard motor coil.

[graham]

I have just bought a fairly new Suzuki outboard with a charging coil and rectifier built in.

With the charge lead connected to a fully charged 85 ah leisure battery through a multimeter it was only reading 0.7 amps charging.

Would this be beecause the battery was fully charged and nothing was turned on or wouldnt that make a difference?

Suzuki claim the coil can charge 6 amps. Any advise appreciated as my electrical knowledge is very limited.

 

[aluijten]

Even with your limited knowledge you are exactly right!

The battery will only 'absorb' so much current, this is related to the charging state.
Try to switch on some additional loads, to see if the output current from the coil increases. Be careful though, many multimeters are not well equiped for high currents, and for sure not for a long time.

Arno

 

[bruce]

also, output is related to ob motor rpm...

 

[graham]

Thanks for that,I was hoping that would be the case.Tommorow I will "borrow" a flat battery from the pile next to the club skip and try that.

PS I did try it it at different RPM s.

 

[aluijten]

Actually the RPM's should not make that much of a difference if a regulator is used. The regulator changes the fieldstrength of the altenator, by this it will keep the output voltage at 13.8 volts. I'm not sure if it works this way with outboards though...

Arno

 

[KeithH]

Also remember that a multimeter, when used in ammeter mode, will affect the current. That is because an ammeter is (generally) a voltmeter measuring the voltage across a resistor. So incorporating the ammeter effectively adds a resistance in the circuit, and so will drop the current a little. How much drop will depend on the sensitivity of the instrument

I think it is safer to look at the charging voltage - and there are a number of threads on that issue.

 

[bonny]

These charging coils are very basic and have no regulation at all.
Simply put, the coil generates AC by the fact that the magnet in the flywheel is passing it. The AC is rectified to DC by, you guesed it the rectifier. The resulting voltage is proportional to the speed of rotation (care of A. Einstein E=MC^2).

Since there is no regulation, the open circuit voltage (ie not connected to battery) is related to the engine RPM and will probably reach 20-30V. This will reduce when connected to a battery due to the internal resistance of the coil and come down to 12-14V, even higher with a fully charged battery depending upon the battery size (AH rating). Since this resistance is constant then care of Ohms law the charging current will also be proportional to the rpm.

At tickover you would be doing well to get an amp, at full RPM you may get the 6A that Suzuki quote.

Because there is no regulation, the state of charge of the battery has a negligible effect on the current.

 

[GART]

Hi Bonny
I have a question you may be able to answer, my set up is the same as you describe that the outboard gives out 20+ volts at low amps and using Ohms law it works out at 6 amps @ 12v. I understand the battery works as a resistor and absorbs the current. But, if 20V + is hitting the same battery terminal that my GPS, log etc. are connected to, am I putting 20 Volts into my instruments?

 

[pampas]

The answer to your question when connected to a battery is NO

 

[bonny]

Hi Gart,
As long as your battery is in good shape and also the connections to it then the battery will limit the voltage to around 14V max. In effect the battery will act as a regulator and your instruments should be OK.

If however, your battery is a bit duff and goes high impedance (happens if acid level drops), or you have faulty connections to it then the voltage will go up and if this goes in excess of approx 16V (depends upon specific equipment) then I'm afraid the results will be terminal.

This happened to me some years ago when a wire broke on my battery terminal which was being charged from a 4HP Mercury. Bang went my mobile phone which was on charge and my nav lights.

 

[William_H]

Did some fiddling with my old Evinrude/Johnson 6HP with lighting coil. Into a charged battery via a bridge rectifier I got .7 amps. The output in this case is a rectified AC which means that while it is DC it rises and falls in voltage as per the AC that feeds it. This means that the part of the cycle where the voltage rises above the battery voltage is when current flows. As the current flows the voltage falls to that of the battery due to resistance of the coils but at least current goes in. The lower the battery voltage the more current in non linear manner. If you run lights directly from the O/B coil then the current flows all of the cycle of rising and falling so power output is much more hence the higher current rating of the charging system.
In the case of the Evinrude the coil is sized to give a voltage appropriate to running 12 volt lamps and my measurement indicates this is only around 13 or 14 volts no load but should support a few lamps up to near 12 volts. It is almost impossible to measure accurately because it is not smooth DC and digital voltmeters can give eroneous indications because they sample sometimes for short periods of time.
In theory at least if you run lamps off the battery and feed the charger into the battery depending on how the battery volts hold up you could get a net discharge of the battery where if you ran the lights directly from the O/B charger you could run them at reasonable brightness forever. This is because the charging capability is not good but power at lower voltage than a battery needs(to be charged) is good. regardless you will probably run the lights from the battery on charge and of course as the battery volts go down the charge current will rise a little. In other words it is all a bit of a variable feast it certainly would be good to fit an ampmeter to monitor what is going on and don't expect too much battery charging.
To Gart a battery is definitely not like a resistor. A resistor takes current in direct proportion to the voltage but a battery takes lots of current until the voltage rises the current being more in proportion to the difference between the charging voltage and the battery voltage. Even that is variable because the charging voltage must fall to that of the battery by internal resistance of the generator or the battery voltage must rise from internal rsistance of the battery. So the charge voltage must equal the battery voltage or near to it because they are connected together by a piece of wire. Now are you confused?
use an ampmeter and a volt meter and be grateful for what you get by way of charge but don't try to predict it. luv will