TheRecklessEngineer Posted March 9, 2013 Share Posted March 9, 2013 Having looked at some schematics of various alternator regulators, it would seem that the output current is limited by the current drawn by the field coil when the full voltage is applied across it. You could therefore de-rate it by simply sticking a small (but power rated) resistor in series with the field winding as was suggested by someone earlier in this thread. Quote Link to comment Share on other sites More sharing options...
zoltan Posted March 9, 2013 Author Share Posted March 9, 2013 What current assumption do you think I need to make when calculating the resistor value? I suppose my problem is the current spike so should I set a current ceiling and use that value of resistor? Or do I use a variable resistor and use that to tinker? Is there a pitfalls with variable resistors that I'm not aware of. ( I'm out of my comfort zone here ) Quote Link to comment Share on other sites More sharing options...
TheRecklessEngineer Posted March 9, 2013 Share Posted March 9, 2013 Yes, finding a variable resistor with enough power dissipation will probably be expensive and quite large! Measure the resistance of the field winding. I=V/R will give you the current through it. It's a proportional relationship between the current in the field and the current in the stator, so to reduce the charging current by 20%, you need to reduce the current through it by 20% - or indeed the voltage across it by 20%. That 20% will be dropped across your resistor. So it'll be something around 14V across the field winding, you'll reduce that to 11.2V (14*0.8), dropping 2.8V (14*0.2) across the resistor. The new field current can be calculated from Inew=11.2/Rfield. Then you can calculate a value for your resistor using R=2.8/Inew. P=R*Inew^2 will give you the power that the resistor is dissipating so you can buy one rated accordingly. Or just post up your field winding resistance and I'll work out a suitable value for you. Quote Link to comment Share on other sites More sharing options...
minimaquinas Posted March 12, 2013 Share Posted March 12, 2013 i have a 1956 conyers with a jap engine which starts electricaly with power from the battery to be charged if there is any life in it but also has a pull start i could be for sale also i have a ex army 12 volt charger not sure what engine it has but a great peice of kit Quote Link to comment Share on other sites More sharing options...
SORNagain Posted December 23, 2013 Share Posted December 23, 2013 Hey, Zoltan. Any progress on this thing? The engineering is too good to waste... Quote Link to comment Share on other sites More sharing options...
zoltan Posted December 28, 2013 Author Share Posted December 28, 2013 I've given up on this for the time being. I needed to charge our batteries for our trip away over Christmas so I bought a 1kw Yamaha petrol generator which works straight away. I'm thinking of selling the project if anyone fancies taking it on. Basically, it needs sorting out on the electrical side, the hardware is all in place. It stands me in at about £140 so far. Quote Link to comment Share on other sites More sharing options...
zoltan Posted December 30, 2013 Author Share Posted December 30, 2013 What a difference a day makes, I'm back in love with it since it bailed me out of a flat battery this morning. It does have a use afterall! Quote Link to comment Share on other sites More sharing options...
simonr Posted December 30, 2013 Share Posted December 30, 2013 I think there is a potentially simpler way of controlling the Alternator. Alternators either sense the battery voltage through the charge light or a wire connecting directly to the battery. If you have the former type, you can control the output voltage and hence current by changing the voltage the alternator sees. Initially replace the charge light with a potentiometer whose ends are connected across the battery and the wiper to the alternator. By turning the pot, you will see the load increase & decrease. You know how the speed regulators work on lawn mowers? Using a flap which is moved by the airflow from a fan. I propose you use something similar to change the potentiometer setting. It would take a little playing with to get the range right - but I think it will work. Si Quote Link to comment Share on other sites More sharing options...
TheRecklessEngineer Posted December 30, 2013 Share Posted December 30, 2013 You won't be able to decrease the current below a maximum though Si. You can't put more voltage into your pot (thus apparently increasing the battery voltage and decreasing the charging current) than you have. My vote is as above. An appropriately rated resistor in series with the field winding. Post up your field winding resistance and I'll do the maths if you like. Quote Link to comment Share on other sites More sharing options...
zoltan Posted December 30, 2013 Author Share Posted December 30, 2013 Measuring off of the slip rings I get a reading of 3.3 ohms, does that sound right? Quote Link to comment Share on other sites More sharing options...
simonr Posted December 31, 2013 Share Posted December 31, 2013 You won't be able to decrease the current below a maximum though Si. You can't put more voltage into your pot (thus apparently increasing the battery voltage and decreasing the charging current) than you have. The charge light forms half of a potential divider - so the regulator does not 'see' the full battery voltage. If you reduce the resistance of the bulb, the alternator sees a higher voltage and reduces the output current to reduce the battery voltage to (what it sees as) 14.6v. I've found that you can persuade an alternator to spit out up to about 30v this way. Si Quote Link to comment Share on other sites More sharing options...
TheRecklessEngineer Posted December 31, 2013 Share Posted December 31, 2013 Aha, I see. That makes sense. Does this make the wattage of the bulb critical in maintaining the right voltage? From Zoltan's measurement, a 0.8 ohm resistor rated at 10W in series with the field winding should drop the peak charging current by 20% or so. Assumptions in the maths not misgiving. Something like this: http://uk.rs-online.com/web/p/panel-mount-fixed-resistors/7547219/ It's going to get warm so put it somewhere where air can get at it. Quote Link to comment Share on other sites More sharing options...
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