Briggs AC light circuit

[SmilinSam 3,932]

I'm swapping a Briggs 16hp in place of the KT17 twin on a 917 Allis. Can I just tape off the white Ac lighting circuit wire and not use it seeing as how the 917 is wired to run the lights off the 12v system?

[D-17_Dave 12]

Yes. The light circut is just an A/C 12 volt max output line and capping it won't hurt anything. If you have one lying around you could swap the stator w/ a 15 amp stator negateing the low amp charging circut.

[Al 6]

Hi, It is correct that you can just tape it off. Only problem is that the charging system won't keep up with the lights if you use the lights a lot. I have added a diode to the lighting wire and used it to also charge the battery and it will them keep up with the lights. Daves suggestion is better, put in the other stator. Al Eden

[D-17_Dave 12]

Al, wouldn't adding a diode cut the 12 volt A/C line to 6 volts D/C or will it just cut the sign wave in half and only charge "half" the time?

[SmilinSam 3,932]

It would be easier for me just to run a separate wire to the light switch fromthe ac circuit and run it the way its supposed to be run and just unplug the original wire to the lights from the junction board.

[Al 6]

Dave, The diode will only conduct the positive peaks. If a full wave bridge rectifier is used it would be much more efficient, and would then need to have a regulator after it was rectified. As I said you are correct the stator is better. The 1 amp system is just a diode in series with the stator lead, as I have done the same with the light coil to utilize this unused power. Hope I can explain a little about AC. If you have 120 volts AC in your house the 120 volts is RMS. (root mean square) voltage. With sinosoidal AC if you graph the voltage, it goes from 0 up to 166 volts peak then it goes back to 0 and goes to 166 negative then back to 0. The peak to peak voltage would 332 volts. The RMS voltage is the equivalent DC power 120V. It is like you sawed off the peaks at the 120 volt level and tipped them upside down in the valleys. If the sine wave is not distorted then the peak voltage will 1.414 times the RMS value. and the peak to peak will be 2.828 times the RMS value. If you take the peak voltage and it is a true undistorted sine wave the RMS voltage will be .707 of the peak voltage. RMS is really a measure of the true heating effect power measurement, or the heating effect an equivalent DC voltage would supply. Wish I had a pad and pencil or a blackboard. If you cut my arms off I couldn't call for help. There is some theory on my website under the tech notes. I want to finish a bunch of that up by adding some practical trouble shooting training, but have such a full plate on repower paperwork that I don't have time to finish it right now. Feel free to check it out, an acquaintance of mine at the Briggs factory education facility called me about a year ago and asked if I cared if he sent people to it. Fine with me. Go to edensltd.com and open the tech notes. If this ac thing is confusing to you hollar and I will try to expand on it. Sorry I didn't do better on this. Al Eden

[BLT 717]

Isn't the 15Amp circuit an AC circuit that needs to rectified and regulated?

[D-17_Dave 12]

Sam, you could do this but engine rpm affects the voltage output. It also maxes out at 12 volts so even wide open the lights aren't very bright. As the rpm's drop the brightness does also. I ussually find these to be less than a crappy solution to have lights. If your not going to run the lights much like for mowing or some other long useage chore I'd leave it as it is and run it.

[ambler 0]

I thought the flywheel wouldn't support a full stator. I have an engine like sams and would like to upgrade to 16 amp dc.

[Al 6]

DAve, You are essentially correct and the 1 amp or 3 amp circuit does the same thing. It only charges at top rpms as the lights circuit with a diode will. As Bob asked, the 15 amp system must be regulated. This because the full wave rectified output from the stator is significantly higher than 14 volts and would try to charge at a higher rate than the 15 amps. What happens is the regulator controls the output from the stator and limits the output from the regulator rectifier to about 14.5 volts since this is about 2+ volts higher than the battery voltage, the current flows to the battery and charges it. If there were no regulator, the voltage as the battery became charged would continue to go up and would try to go to maybe 20 volts, and this would overheat and boil the electrolyte out of the battery and destroy it. With these low amp systems the reason they are able to use them unregulated is they have limited output voltage from the stator and are only half wave rectified as you mentioned above. In reality they only charge near full throttle and since they only replace the power the starter used they eventually replace it. If lights are used even at full rpm the charging system capacity will not replace the energy the lights are using and the battery will discharge even though the charging system is still trying to charge the battrery when the unit is being used. Same thing happens with units with electric clutches, these charging systems can't be used with electric clutches. In the Vangards, there 3 sizes of magnets that get into the process also. Al Eden

[HubbardRA 19]

I agree with Al's explanation. I once took the output from the 15 amp alternator on a K341 Kohler, hooked it to a full wave bridge rectifier, then hooked the DC output from the rectifier thru a voltage regulator from a starter-generator. This setup worked great the whole time I had it.