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New to electronics, have a little question.

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ghouck

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I have a little project I'm trying to get figured out, and I need a little help. I have a snowmobile, which has no battery (pull start). It does have a stator / regulator-rectifier that powers all the lights, gauges at ~14 volts. I installed another set of gauges that are electronic (exhaust gas temperature gauges). The problem is that when the engine is shut off, or the RPMs get too low for the system to put out full voltage, the EGT gauges shut off. My plan was to install a small (0.8 or1.2 amp/hour) 12v battery, with a diode so the system will charge it, but the diode will prevent the battery from draining back through the electrical system (Charging system -> Diode -> Battery -> EGT Gauges in the + side). That all seems to be OK, ,, but, , the machine's charging system at ~40watts will charge that small battery too quickly, as that sized battery is reccomended to be charged at around 4-6 watts. It is my understanding that a resistor could be used to limit the current to the 0.3 - 0.5 amps it should be limited to, but I don't know a) if a resistor will really work in this capacity, or b) how to calculate the ohms of said resistor. Any help will be appreciated. Thanks, Greg.
 
The alternator on the snowmobile will have to have some sort of regulator to stop the voltage rising too high when the revs are high and there isn't much load.

As a result, it is likely that you won't need to limit the current.

However, if you add a resistor of about 3.3 Ω in series, the current will be limited to about 1/2 amp.

The current through a resistor is the voltage across it divided by the resistance.
In this case, the voltage across the resistor is the alternator voltage - the battery voltage - the diode drop. The current though the resistor must be significantly more than the current taken by the gauges.

You might need to do some measurements as the alternator voltage might vary.
 
3.3Ω seems kind of low to me. Maybe something like this, with a resistor, diode and switch (so you could turn off the gauges before the battery runs completely flat) -
 

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Thanks Diver, I appreciate it. I'm not worried about the voltage being too high, I am concerned that the rate of charge will be higher than the battery can handle. It is my undersrtanding that too small of a battery in a charging system will draw more current than it can handle and damage itself. Is there any truth to that?

Also I was thinking of another diode parelell to the resistor allowing current to bypass the resistor in the event that the engine is shut off and the gauges are running off the battery.

I haven't quite gotten a grasp of how the 1/2 amp was calculated, but I'm gonna get it figured out.

Thanks, Greg


[Edit] Duffy, why wouldn't I just replace the switch in your diagram with a diode? or is there something I'm missing?

Also, on this model sled, an electric starter can be installed, and the only thing added is the battery, the starter, and the starter switch, so the electrical system is already set up to charge a 12v battery, but it's a much bigger battery than I am planning on using. The larger battery is too much weight to add for what I'm trying to accomplish. Thanks.
 
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