alt.energy.renewable

Todd,

Your voltage regulator circuit could use a 3 amp 100v ( fast recovery would
be preferrable but not absolutely necessary because you are not "hard
switching" ) to clamp the energy that is stored in the field coil. When your
2N3773 transistor turns off the voltage jumps up to the avalanche the
transistor. Very few transistors can stand this for very long. By adding the
diode you will also notice the 2N3773 runs cooler when the voltage is at the
regulator set point. The cathode of the diode goes to the plus field
connection and the anode to the 2N3773 collector. Because your regulator is
mostly running in the linear mode the voltage across the 2N3773 does not see
a lots of voltage. However if a heavy load is removed ( similar to load dump
in autos ) the above condition occurs and your 2N3773 shorts out.

Ray



"Todd" wrote in message
news:3a0350be-5ab2-4606-bcf6-97e05d8b4681@b64g2000hsa.googlegroups.com...
> On Mar 7, 4:38 pm, "Ray King" wrote:
>> The loss in the alternator causes heat to be developed in the diesel. The
>> diesel engine is only 30% efficient. If you are driving 30 watts that is
>> un
>> necessary in the load that produces about 100 extra watts in the diesel
>> engine minimum.
>
> Combustion causes heat to be developed in the diesel far exceeding any
> heat caused by the alternator. The current in the armature causes
> negligible (e.g. 2W) load on the diesel. It's the batteries that are
> the load. The smaller the load, the less heat the diesel generates ...
> i.e. the less work it has to do. With no battery (i.e. charging ) load
> (e.g. when the batteries are fully charged), the load due to the
> alternator is just 2W. The efficiency of the engine has nothing to do
> with anything.
>
> /T