Lotus Elan

Brise Starter Motor... Wiring detail

PostPost by: mark030358 » Mon May 23, 2011 7:19 pm

Hi,
I wired my new Brise starter in by retaining the origional bulkhead solenoid. Basically the main starter feed cable to the starter was connected to the solenoid contact at the starter.

Is this the right way to go as perhaps back Emf generated will briefly keep the coil energized after starting.

cheers

Mark
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PostPost by: john.p.clegg » Mon May 23, 2011 7:48 pm

Mark

Is it actually happening to you or are you worrying over nothing (yet),surely the bulkhead solenoid would do just the same job as the inbuilt solenoid......back emfs can't be generated in an open circuit...?

John :wink:
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PostPost by: mark030358 » Mon May 23, 2011 11:54 pm

Back emf is generated by a collapsing magnetic field and is approx the same as the supply voltage. So as my solenoid coil is linked to the starter (basically in parallel), as the magnetic field collapses an emf is induced can this hold in the solenoid???


ie

When the armature of a d.c. motor rotates under the influence of the driving torque, the armature conductors move through the magnetic field and hence e.m.f. is induced in them as in a generator The induced e.m.f. acts in opposite direction to the applied voltage V(Lenz?s law) and in known as back or counter e.m.f. Eb. The back e.m.f. Eb (= P ? ZN/60 A) is always less than the applied voltage V, although this difference is small when the motor is running under normal conditions.

Consider a shunt wound motor shown in Fig. (4.2). When d.c. voltage V is applied across the motor terminals, the field magnets are excited and armature conductors are supplied with current. Therefore, driving torque acts on the armature which begins to rotate. As the armature rotates, back e.m.f. Eb is induced which opposes the applied voltage V. The applied voltage V has to force current through the armature against the back e.m.f. Eb . The electric work done in overcoming and causing the current to flow against Eb is converted into mechanical energy developed in the armature. It follows, therefore, that energy conversion in a d.c. motor is only possible due to the production of back e.m.f. Eb.

Net voltage across armature circuit = V ? Eb

If Ra is the armature circuit resistance, then, Ia = (V ? Eb)/ Ra

Since V and Ra are usually fixed, the value of Eb will determine the current drawn by the motor. If the speed of the motor is high, then back e.m.f. Eb (= P ? ZN/60 A) is large and hence the motor will draw less armature current and viceversa.



Significance of Back E.M.F.

The presence of back e.m.f. makes the d.c. motor a self-regulating machine i.e., it makes the motor to draw as much armature current as is just sufficient to develop the torque required by the load.

Armature current,Ia = (V ? Eb)/ Ra

(i) When the motor is running on no load, small torque is required to overcome the friction and windage losses. Therefore, the armature current Ia is small and the back e.m.f. is nearly equal to the applied voltage.
(ii) If the motor is suddenly loaded, the first effect is to cause the armature to slow down. Therefore, the speed at which the armature conductors move through the field is reduced and hence the back e.m.f. Eb falls. The decreased back e.m.f. allows a larger current to flow through the armature and larger current means increased driving torque. Thus, the driving torque increases as the motor slows down. The motor will stop slowing down when the armature current is just sufficient to produce the increased torque required by the load.
(iii) If the load on the motor is decreased, the driving torque is momentarily in excess of the requirement so that armature is accelerated. As the armature speed increases, the back e.m.f. Eb also increases and causes the armature current Ia to decrease. The motor will stop accelerating when the armature current is just sufficient to produce the reduced torque required by the load.

It follows, therefore, that back e.m.f. in a d.c. motor regulates the flow of armature current i.e., it automatically changes the armature current to meet the load requirement

cheers

Mark
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PostPost by: john.p.clegg » Tue May 24, 2011 5:47 am

Mark


Hmmmmm...see what you mean,the solenoid contacts are both in series with the starter motor coil windings but the brise solenoid winding will be in parallel with the motor winding (via the brise solenoid contacts) so yes you could have a problem that way...

John :wink:
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PostPost by: mark030358 » Tue May 24, 2011 8:57 am

Think I'll fit a diode :)
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PostPost by: john.p.clegg » Tue May 24, 2011 12:33 pm

Mark

Sounds good to me....

John :wink:
P.S.
Is it actually happening to you or are you worrying over nothing (yet),
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PostPost by: mark030358 » Tue May 24, 2011 11:51 pm

It was happening ... I think :?

rewired the coil now do shold be ok...

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PostPost by: Panda » Wed May 25, 2011 2:47 am

Hi Mark,
I fitted a Brise starter about 12 months ago and simply fitted the wire going to the starter to the other
( live ) terminal on the original solenoid. i.e. both connections on the same solenoid terminal.
I then removed the solenoid pull in wire and extended it down to the Brise starter small terminal.
I had no problems with the wiring whatsoever. But the troubles I had with the pinion gear led me to scrapping the Brise and fitting another brand which has been terrific.
Has anyone else had any similar problems?

Alan
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PostPost by: alaric » Wed May 25, 2011 11:44 pm

Hi. I have a Brise. I wired the high current wire to the standard bulkhead solenoid as per the original starter motor, and connected the solenoid on the motor to the motor end of the high current wire. I figured there was no power in the backemf that's been discussed, and couldn't see how it could be an issue to wire as I have done. It's worked fine, and I've never noticed a problem with it. I also didn't reverse the ring gear on the flywheel as advised by Matty's, and as far as I am aware it has not damaged the starter or ring gear.

Regards.

Sean.
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