fusible link questions

[Raymond C Drouillard]

>
>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.

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

___________________________________________________________________
You don't need to buy Internet access to use free Internet e-mail.
Get completely free e-mail from Juno at http://www.juno.com/getjuno.html
or call Juno at (800) 654-JUNO [654-5866]