fusible link questions

[Clarence L.Snyder]

Raymond C Drouillard wrote:
> 
> >
> >I was thinking more in terms of regulating the alternator to , say, a
> >steady 15 volts, and then cutting that down to whatever is ideal for the
> >battery in terms of voltage and current, based on temp, etc., for
> charging
> >it. Course, it would also be nice to have a device to step the battery
> >output back up, at least to critical devices, such as the ecu and
> >injectors, for those times when the alternator does not make a high
> enough
> >voltage for your chosen bus voltage.
> >
> >Again--dunno if I did not say it clearly, or if nobody was listening
> >carefully.
> >
> >Regards, Greg
> 
> Jacobs Electronics makes a switching regulater to step the voltage up for
> whatever you want to run - like solenoids, the ECM, headlamps, etc.
> 
> I had been considering a harebrain scheme similar to what you are
> discussing.  The main purpose was to increase the current output of a
> standard charger, however.
> 
> A switching regulater, when reducing voltage, will increase the current
> (much like a transformer).  My idea is to max out the field coil on a
> standard alternater, which will produce a high output voltage (> 100 V).
> The alternater can still supply its rated current at the higher voltage
> because the current is limited mostly by I2R losses.

Actually, the alternator is rated for maximum POWER output. 100 amps at
14 volts is 1400 watts, almost 2 HP. At an optimistic 80% efficiency, it
requires closer to 2.5 HP. Pushing the same current at higher voltage
quickly oveheats the alternator due to the increase in these I^2 R
losses, which increase at the rate of a power of two with voltage
increase - double the voltage, 4 times the heat with the same current.
> 
> If you use a 70A alternater and the output voltage goes up to 140 volts,
> you can use a switching regulater to step that down to 14 volts at 700 A.
>  Of course, some heavy duty components will have to be used.
> 
> This would be a tremendous help when running a winch and a bunch of
> off-road lights.  Under normal conditions, the alternater would be more
> efficient (they generally run around 60%) because the voltage dropped by
> the diodes would be a smaller fraction of the output voltage.
> 
> In a standard alternater, the current goes through two diodes (it uses a
> six diode three-phase bridge) with a drop of around two volts each.  That
> means that the stator coils are supplying 18 V and 4 V is being dropped
> by the diodes.  If the alternater is supplying thirty amps, it is
> supplying 30 * 14 = 420 watts.  The diodes are dissapating 30 * 4 = 120
> W.  Even if there were no other losses, we are talking about an
> efficiency of about 78%.
> 
> To get the same 30 amps out of the regulater, the alternater would only
> have to supply about three amps.  This would cut the I2R losses in the
> stator windings, and reduce the voltage drop on the diodes to about 0.7
> V.  The increased efficiency of the alternater would more than make up
> for the losses in the regulater.
> 
> There are some rubs with this scheme.  First of all, actually trying to
> draw 700 amps will be a real challenge.  Secondly, the alternater would
> require about 13 horespower plus efficiency losses to do this.  An
> improved belt and pulley system would have to be used.
> 
> I don't think that the current to the battery will have to be reduced
> under normal circumstances.  If the battery gets really drained, it might
> be an issue.  In that rare case, switching a heavy duty resister into the
> circuit (between the alternater and the battery, but not between the
> alternater and the rest of the load) would do the trick.
> 
> Ray Drouillard
> 
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