Minimum wattage?

P

prmurat

Junior member
Location
San Francisco CA
As I am installing an Acewell on my SF2, I know that one of the LED currently available on the unit will not work for the charging light (not enough watts) I have 2 questions:
-I will ne installing a charging light somewhere but what is the minimum watts to have the reg working normally?
-If, parallele to the right Watts charhing light I plus one the LED, will it work OK for the reg? or should I install a diode on the LED plugging?

thanks!!
 
You don't need a diode on the LED however you will need a resistor in parallel with it to power up the field coil in the generator. I'm not sure what the original bulb wattage is but use the formula R=V x V/P ie Volts squared/Power ie if the original bulb is 3 watts the R in ohms = 12 x 12/3 ie around 48 ohms, nearest would be a 47 ohm resister which would be fine, it doesn't need to be accurate. Make sure you use a resistor of the same or more wattage as the bulb. In this case it would be a 47 ohm 5 watt resistor, or you could use 3 of 150 ohm 1 watt resistors in parallel making 50 ohms which would also be fine.

If the original bulb is a different wattage get back to me and I will work it out for whats available at the electronicsy shoppe.

Note the LED bulb should be a 12V one and get it to fit the bulb holder. I can't remember what the SF2 had as a bulb holder but suspect it's a bayonet style in which case it should be wired the right way around. It won't hurt anyting if it's the wrong way around, just won't light up.


Edit: or wot Stefan said.
 
Thank you guys... a 3 w it will be... there are no bulb holders or things like this as my Speedo/tach is this one:
http://www.acewell-meter.co.uk/Acewell%20Speedometers/ACE-2xxx/index.htm

2676389342_097044d703.jpg


(now it is wrinkle black) (hole on the left is for the contact). I'll be using the red triangle for charging light, don't know what to do with the neutral...
 
So I'm not sure how the red triangle light is wired in there, but I suspect it is one side grounded and the other side expecting 12V to light it up, in which case it won't work for the charging light. I could stand corrected though, would have to look up the acewell wiring instructions, it needs to be like this: +12v from ignition switch - top of LED/resistor in parallel - field of generator - ground (this is done internally in generator). So you may have issues. If you can link me to the acewell wiring instructions then I can check.

edit: the generator will work fine without any led indicator and just the resistor. You can tell it's charging by the voltmeter on the acewell.

 
I think you are referring to the light that has two bulbs back to back instead of a triangle in which case yep it is grounded on one side and wants 12V to light up which means you can't use it for a generator light without pulling the acewell apart and modifying it which I WOULDN"T recommend. Just run it without a light and check the voltmeter in the acewell occasionally.

Or mount another light on the nice faceplate you've made. Make it a 3W bulb and you won't need the resistor.
 
You lost me here... In this wiring I could use the oil pressure in parallele with a 3w bulb on the - side?
Knowing that the + is the general plugging of Acewell...

BTW what is the 61 (diode) for in the stock wiring (not that I have it my project!!)
 
Well that was quick and a waste of time like 99% of meetings, at least it was an online one so I was able to do more important things like this.....

So just add the resistor in between the switched 12V from the ignition switch and the oil light line then take the oil light line to the generator field and you should be good to go. I'm not sure what the terminal 61 is on the regulator is, will go find out, but this should work anyway.

See link below hopefully.

https://www.dropbox.com/s/c9k0hojywncfze0/acewell.pdf?m
 
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