[DeTomaso] Starter Replacement Options

[Forest Goodhart]

Mike,
If you read carefully what he wrote he said that he would be jumpering from the cable connection at the starter to the small terminal on the unit mounted solenoid leaving the firewall solenoid as from the factory. When this is done the cable between the firewall solenoid and the starter is dead until the solenoid is energized. This then sends power to the starter and since the large cable is jumpered to it, it also energizes the starter mounted solenoid thus engaging the starter. This on the surface seems like a simple way to wire the starter. The problem is that because the starter is turning when you release the key the starter motor becomes a generator and feeds back to the solenoid keeping it energized for a few seconds until the start slows enough to drop out the solenoid. This was the method provided years ago by some of the vendors who could not explain why the starter would run on for a few seconds.
You have yours wired the correct way, as do I and many others, and that is why you have no problems.
Forest
 

________________________________
 From: "MikeLDrew at aol.com" <MikeLDrew at aol.com>
To: forestg at att.net; detomaso at poca.com 
Sent: Thursday, May 1, 2014 11:35 AM
Subject: Re: [DeTomaso] Starter Replacement Options
  



In a message dated 5/1/14 9 24 21, forestg at att.net writes:



If you do this the starter will run on after releasing the key from the start position leading to damage to the starter.
>

>>>Uh...what?

The main battery cables from the battery to the starter are connected on the non-switched side of the stock solenoid.   The only function here is to serve as a connector for the two cables (and provide power for the solenoid).

When the key is turned, the voltage crosses the stock solenoid and goes to the output lug.   In the past, this voltage then went down to the starter via the thick cable, but now a smaller wire is the only thing on this lug, and the power goes to the solenoid on the new starter, activating it.

As soon as you release the key from the spring-loaded 'start' position, power across the stock solenoid is interrupted, meaning that power from the output side lug to the solenoid on the new starter is interrupted.

So, there is no way for the starter to 'run on' as you suggest.

Mike (been wired this way for years, and no I'm not smart enough to figure this stuff out on my own, I read the diagram that came with the starter...) :>)
-------------- next part --------------
   Mike,
   If you read carefully what he wrote he said that he would be jumpering
   from the cable connection at the starter to the small terminal on the
   unit mounted solenoid leaving the firewall solenoid as from the
   factory. When this is done the cable between the firewall solenoid and
   the starter is dead until the solenoid is energized. This then sends
   power to the starter and since the large cable is jumpered to it, it
   also energizes the starter mounted solenoid thus engaging the starter.
   This on the surface seems like a simple way to wire the starter. The
   problem is that because the starter is turning when you release the key
   the starter motor becomes a generator and feeds back to the solenoid
   keeping it energized for a few seconds until the start slows enough to
   drop out the solenoid. This was the method provided years ago by some
   of the vendors who could not explain why the starter would run on for a
   few seconds.
   You have yours wired the correct way, as do I and many others, and that
   is why you have no problems.
   Forest
   From: "MikeLDrew at aol.com" <MikeLDrew at aol.com>
   To: forestg at att.net; detomaso at poca.com
   Sent: Thursday, May 1, 2014 11:35 AM
   Subject: Re: [DeTomaso] Starter Replacement Options
   In a message dated 5/1/14 9 24 21, forestg at att.net writes:

     If you do this the starter will run on after releasing the key from
     the start position leading to damage to the starter.

   >>>Uh...what?
   The main battery cables from the battery to the starter are connected
   on the non-switched side of the stock solenoid.  The only function here
   is to serve as a connector for the two cables (and provide power for
   the solenoid).
   When the key is turned, the voltage crosses the stock solenoid and goes
   to the output lug.  In the past, this voltage then went down to the
   starter via the thick cable, but now a smaller wire is the only thing
   on this lug, and the power goes to the solenoid on the new starter,
   activating it.
   As soon as you release the key from the spring-loaded 'start' position,
   power across the stock solenoid is interrupted, meaning that power from
   the output side lug to the solenoid on the new starter is interrupted.
   So, there is no way for the starter to 'run on' as you suggest.
   Mike (been wired this way for years, and no I'm not smart enough to
   figure this stuff out on my own, I read the diagram that came with the
   starter...) :>)