Optional Extra 0 Posted August 2, 2012 Share Posted August 2, 2012 Hi I have a 10hp outboard on my farr 727. I want to know how to make up a charging lead to connect to the charge coil on my outboard (which has the two leads in the cowling) and to then connect it to my house battery. I am aware that you need a diode in the coil so that current only flows from the outboard to the battery. Has anyone made one of these up for there outboards? Thanks Aidan (Optional Extra) Link to post Share on other sites
wheels 543 Posted August 2, 2012 Share Posted August 2, 2012 There should be an 8A output socket on the side of the motor case. Get a plug to fit that and wire to plug. Use a meter to determine Polarity. I should already be rectified. Link to post Share on other sites
Guest Posted August 2, 2012 Share Posted August 2, 2012 try not to let the bateries get too low with this type of system, they are realy only speced to top up a start, or basic nav light battery. what typically happens is that the motor system will chuck out whats needed, but the wiring can't handle the draw Link to post Share on other sites
Guest Posted August 2, 2012 Share Posted August 2, 2012 I should already be rectified. Damn straight. Link to post Share on other sites
Optional Extra 0 Posted August 2, 2012 Author Share Posted August 2, 2012 There are two wires coming out from the charge coil. So i thought that you would have a postive and nevagtive wire running to the battery? Link to post Share on other sites
wheels 543 Posted August 2, 2012 Share Posted August 2, 2012 Yes, positive goes to Positive on the battery and neg to neg of battery. If it is not rectified, then a simple diode is all that is needed. You need a Diode capable of handling the full current of the coil. A multi meter will tell you if it has been rectified or not. Link to post Share on other sites
Farrari 4 Posted August 2, 2012 Share Posted August 2, 2012 One quick way to tell if you already have a diode or not is to measure the resistance of the charging unit in both polarities. Put your multi-meter on the kilo-ohm scale and measure the resitance accross both leads. Reverse the meters leads and see if the value changes significantly. If it is the same both ways then you will need a diode that is rated for the current draw. Link to post Share on other sites
Guest Posted August 2, 2012 Share Posted August 2, 2012 I should already be rectified. Damn straight. which naturally leads to..........is this AC or DC output? Link to post Share on other sites
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