These ciruitries will not work.
You expect a staggering voltage output. Since there are three phases the voltage output of each phase will be lower than the sum of all of them. It's AC tech which varies a 'bit' from DC.

Secondly an adding of outputs will not work in this arrangement. Diodes or Triac must be capable of high currents up to 20A and more. They must be capable of handling voltage spikes- VCR's are very sensitive to voltage spikes.
They are-sadly- not the right ones for this job.

 

thanks for the input. i tested the last circuit i put up (post #24)with the regular diodes on a set of "normal" caps and a 12v ac transformer. with it i was only testing what one coil in the alternator would do.

the x-former was outputting about 13.8 volts no load. on the caps i was measuring right about 38 volts. this seems to line up nicely with my calculations.

13.8v x2 -0.7 x1.41 = 38

explained version:
13.8 volts x 2 (as each cap gets one half of the sine wave and puts them in series) - 0.7 (voltage drop of diode) x 1.41 (for converting ac to dc) = 37.93 volts

now i only need about 2.05 volts per alternator coil. math: 2.05v x2 -0.7 x1.41 = 4.8v

since i have six capacitors in series, and three alternator coils. i divide the coils into every two caps. this would be 4.8 volts x3 = 14.4 volts.

since the regulator is seeing the sum of the volts across the caps it will stop exciting the coils when the bank reaches ~14.4 volts

as to the diodes, i have a bunch of these (attached) that i was planing on using. any input on the reliability?

let me know if the math adds up

thanks
Jason

 

It's AC, dude- your calculations are nice, but do not work here.
You completely ignore phase shift. When you measure you measure RMS- but the reality is peak to peak. Guess what condensers do really not like?


the diode is nice, but- 180W in which time? Any ratings? Expect a huge- and I mean huge heat sink you need...

 

i dont think i need to worry about phase shift as i will be separating the coil into three separate windings. the capacitors wont see any AC as that is what the diodes are for. each capacitor will be charged by only one half if the AC cycle. for the positive half of the cycle one cap will get charged, for the negative half of the cycle the second cap gets charged. stick three of these in series and you have the required voltage needed to run the truck.

yes, i know that each of the windings will be 120 degrees apart but that is what the reserve capacitance is for, to carry over until the next charge cycle.

how large could the heatsink be? the ones in alternators are not that big and in theory will be dissipating the same amount of watts.

at 100 amps the voltage drop is 1.5 volts. that is 150 watts of heat to dissipate. this will only happen for a few seconds after startup. normal running current will be around 25 amps. 25 x 0.7 is only 17.5 watts. doesn't seem like it needs to be that large to me.

 

Either you don't get it or you don't want to hear it.

-^peak to vpeak is not the same as RMS.
- capacitors do have an extremly low internal resistance -as a result the current is extremly high.

As long as you "think" it is worthless to 'talk' about this.
This is not Kirchhoff's law, this is complex calculation:

û= 1.414 x 13.8V = 19.51V
So the real voltage is a bit different from your measurements and calculations...

 

Yes, i understand this. with my meter i was measuring RMS.

until fully charged, I have found that my capacitors will recover (recharged after cranking and starting the engine) in less then 10 seconds. so the duration of the high current is pretty short. once the caps are charged and the voltage is equal to the output of the alternator then no current will be flowing into the capacitors. the only current passing through the diodes is the amount needed to run the engine and accessories. this i have found to be roughly 25 or so amps

19.51 volts -0.7 volts through the diode = 18.81 volts into each test capacitor. since the circuit charged two caps at a time and in series the voltage doubled (think voltage Doubler) 18.81 volts x2 = 37.62 volts


i am not trying to argue with you or prove you are wrong, and i am sorry if it came across like that. i understand what you are trying to point out and i think we are both trying to say the same thing but using different words.

what i'm building is three voltage doublers in series. each powered by one of the windings in the alternator. the total string voltage will be limited by the regulator cutting the power to the field windings when it senses the voltage is at its preset value. each winding only needs to contribute about 2 volts AC P to P for the string of capacitors to total up tp 14.4 volts

this is based on existing circuits with the math already worked out, so i know that it can work.

if this make sense to you then great, if not then ask all of the questions you need

 

This circuitry is uneffective.
You simply waste efficiency and try to compensate losses during rectification.
Instead of using simply Diodes think about IUoUo charging. Sad thing is the capacity you have available is simply too small. To control the charge current isn't a simple thing to do-but a must. Capacitors of this high cvapacity do not like to be discharged and charged rapidly.

 

"This circuitry is uneffective.
You simply waste efficiency and try to compensate losses during rectification."

Can you elaborate on this, seems to me that it should be quite efficient with very little loses.

"Instead of using simply Diodes think about IUoUo charging."

for capacitors constant currant charging would be better. no need for any float voltages or top off charging. just set the voltage to a usable level that is safe for the capacitors. when the voltage of the capacitors are equal to the constant currant source then no current will flow.

"Capacitors of this high cvapacity do not like to be discharged and charged rapidly."

the ESR of these capacitors is still quite low. much better then any battery. i have deeply discharged and recharged them a dozen times within a few minutes and no detectable heating was present. i have also been running them in place of regular batteries in my truck since the beginning of this thread and have preformed well so far.

i am open to any other circuits that i could use in my truck that preferably wont cost a fortune and can use existing hardware. i just need something that will keep each capacitor equally charged so they last a long time

Thanks Jason

 

how will these last on a cold morning when the glow plugs cycle? I assume the Duramax uses glowplugs. I know the grid heater on my Cummins will pull some serious current during the "wait to start" portion of my starting sequence when cold out.

will they have enough available amperage to cycle the glow plugs, prime the fuel pump, run the electronics and crank the truck up?

 

Capacitors must be charged with a constant limited current. The difficulty is to achieve the ability to discharge quick and rapidly without damage.
Stereo applications are a different story.
Believe me you don't to have a self fragmenting high capacity capacitor next to you. I'm not concerned about the particles- it's the acid.

MAN DI trucks sometimes need up to 15 minutes of pre-glowing...