Update on my starter problems.

[alvera]

Thanks for all the help received and it was the solenoid that was inoperative I took the starter and solenoid to Ribblesdale auto electrics in preston and he confirmed the starter motor was fine and it needed a new solenoid or he could rebuild it ( to keep it original he did it overnight and the cost was £25. I can't recommend them enough. They tell me they can overhaul any electrical motor including dynamos, alternators, starters etc. there phone number is 01772 796047

[Richard the old one]

Back in September I did ask if it was advantageous to fit a relay in my Dolomites starter circuit and I believe the consensus was that it would be a good move as it would extend the life of the various switches in the circuit and would ensure that a full battery voltage was applied to the solenoid.

I have been unable to find the resistance of the solenoid coils fitted in 1850s or Sprints to see what current they draw when they operate. In the smaller engine dolomites the current drawn is about 50amps for a very short period and I would like to be able to compare. Could someone provide the resistance readings please.

[MIG Wielder]

Just to reply to Richard's question....
Page 86-24 in the Sprint manual has the info you need.
" Pull in winding resistance = 0.25 to 0.27 ohms "
"Hold in resistance = 0.76 to 0.80 ohms "

Now the important thing is that the Pull-in coil is in series with the starter motor armature and the Hold in coil is in parallel with all this .
So ... Pull in current (worst case ) = 13.5 V / 0.25 ohms = 54A
Plus the hold-in current = 13.5V / 0.76 ohms = 17.7A
So worst case pull in current = 71.7A
The cold start current will be less of course.

So a standard 40A Halfords relay is not a lot of use.

HTH,
Tony.

[GrahamFountain]

Back in September I did ask if it was advantageous to fit a relay in my Dolomites starter circuit and I believe the consensus was that it would be a good move as it would extend the life of the various switches in the circuit and would ensure that a full battery voltage was applied to the solenoid.

I have been unable to find the resistance of the solenoid coils fitted in 1850s or Sprints to see what current they draw when they operate. In the smaller engine dolomites the current drawn is about 50amps for a very short period and I would like to be able to compare. Could someone provide the resistance readings please.

Oddly enough back in Aug , 2019 "Richard the old one" posted that the 1850 ROM says the Pull-in coil is 0.25 to 0.27 ohms and the hold-in coil 0.76 to 0.80 ohms. I checked the Sprint's ROM and that gives the same values.

So, for a battery with a reasonably low infernal resistance, about 45-50 Amps for pull and 10-15 Amps hold, i.e. 55 to 65 Amps initially, falling to just the 10-15 Amps of the hold coil once the solenoid has closed the switch twixt battery and starter motor.

But I think there may also be an issue in the back EMF the inductance of the hold coil puts across the ignition switch contacts as you turn the switch back to run. If that manages to cause arcing, if the voltage is over 327V when the contacts are just 7.5 um apart (6 to 8 us after they open assuming the current step and EMF pulse run up and down the lead at something like 2/3 C), it might cause an issue.

[GrahamFountain]

Pull in current (worst case ) = 13.5 V / 0.25 ohms = 54A
Plus the hold-in current = 13.5V / 0.76 ohms = 17.7A
So worst case pull in current = 71.7A
The cold start current will be less of course.

So a standard 40A Halfords relay is not a lot of use.

HTH,
Tony.

Isn't 13.5V across the coil a wee bit optimistic from a battery supplying even the 60 odd-ish Amps to the solenoid? I was working on 12 with just the two coils and 10 while the starter is cranking. But I admit it's guessing at the internal resistance of the battery.

Graham

[Carledo]

Pull in current (worst case ) = 13.5 V / 0.25 ohms = 54A
Plus the hold-in current = 13.5V / 0.76 ohms = 17.7A
So worst case pull in current = 71.7A
The cold start current will be less of course.

So a standard 40A Halfords relay is not a lot of use.

HTH,
Tony.

Isn't 13.5V across the coil a wee bit optimistic from a battery supplying even the 60 odd-ish Amps to the solenoid? I was working on 12 with just the two coils and 10 while the starter is cranking. But I admit it's guessing at the internal resistance of the battery.

Graham

Even in a perfect world, you won't get more than 12.5v cranking as the battery isn't being charged by the alternator at that point. I would suggest that around 10 is more likely in the real world! And the pullin is only a momentary surge.

I use standard 40a relays (not from Halfrauds though) for starter circuits and not wrecked one yet, they don't even get warm unless you're spinning it for around 10 mins on a fresh (huge) battery.

Steve

[cliftyhanger]

I am glad I stumbled across this post. It has helped me understand why my starter sometimes clicks. With such a low resistance any additional "issues" in the circuit will cause havoc. eg a tired ignition switch or whatever. I had assumed the satrted solenoid only pulled about 15A and then dropped to something insignificant.

I will have a head scratch about fitting a relay. I assume it is OK to use the main starter cable as the power supply for the relay, and the old "trigger" cable to do that job to the relay?

I will do the same for my spitfire, the solenoid trigger cable is quite thin, looks like 10A cable, though may be 17A. (I could compare with some samples, but a bit pointless if I fit a solenoid)

[GrahamFountain]

Doing it a bit more exactly, and taking maximum values for the open circuit voltage and minimum internal resistance of a new battery off the net as 12.9 Volts and 0.02 Ohms, with the pull-in coil at 0.25 Ohms and the hold coil at 0.76 Ohms, the maximum surge current, when there is no other load on the battery but the two starter coils is as follows:

12.9 / (1/(1/0.25 + 1/0.76)+0.02) = 12.9 / (0.188+0.02) = 62 Amps.

The sustained current is much harder to do exactly without a value for the minimum cranking current. But I think it's widely accepted that 10 V is about all you can expect as a supply voltage while cranking (implying a cranking current of 145 Amps with an internal resistance of 0.02 Ohms), and the maximum sustained current through the hold coil is just 10/0.76 = 13.2 Amps. It might be a bit more, but it's nowt like the 62 amps of the surge.

However, those are for a new, freshly charged battery at STP (possibly NTP) probably after a couple of attempts to start have warmed the solenoid coils - when the battery ain't fresh charged anymore. But first thing in a morning after the car's been stood all night, resistances will be higher and voltages lower and the currents smaller.

So, with those values, 62 Amps is top value for surge current.

Maybe someone else knows how fast the solenoid pulls in, so what the duration of that surge would be. Then, given the resistance of the switch contacts, we could calculate the approximate maximum power dissipated in the switch, and, given some details of its thermal properties, etc. what that means. But I'm guessing that's all a bit much to ask.

But I still wonder if the cause of damage to the ignition switch isn't, at least in part, arcing across the contacts when the current to the solenoid is shut off, when you let the key spring back to run position. Problem is, I have no idea of the inductance of the hold coil nor the resistance, capacitance, and inductance of the wiring from the switch to the coil (which will disperse the Heaviside step function of the switch current to a sigma function at the coil, and attenuate the return voltage as well). So, all I know for sure is di/dt at the switch, which is infinite and a guess that the resistance, capacitance, and inductance of the wiring are all bugger all. I'm also guessing at the speed of propagation in the wire at about 2/3 C, so the return pulse is about 5 to 10 us after the switch opens. But even so, that the back EMF at the switch is above the magic 327 Volts when the contacts are the 7.5 um of the Paschen minimum (at STP) apart seems reasonably credible to me.

So, as well as isolating the ignition switch from the surge current while the solenoid pulls in, a relay will protect it from the back EMF from the hold coil when the contacts open. I'm guessing that most good relays will be reasonably reliable even when the load's a bit inductive.

Graham

[Bumpa]

The relay I used is a "Panasonic, 12V dc Coil Automotive Relay SPST, 70A Switching Current Plug In" from uk.rs-online.com, search for 811-3097. It was £5.77 and works fine.