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Protecting Alternator when Battery is suddenly removed

Sonia's Driver

New Member
TL; DR: I need to select a diode that will protect the alternator from a voltage spike when the battery is suddenly removed.

I have an endurance racecar with a DPST safety cutoff switch. The battery is connected to both input terminals. The alternator is connected to one output terminal, and all loads are connected to the second output terminal. For simplicity, the team regularly uses the cutoff switch to turn off the motor. The motor is a Subaru EJ25 from around 2011. The alternator doesn't have separate inputs for the field coil; it only has 1) a connection for the battery, 2) a connection to sense voltage elsewhere in the car, and 3) an output to a warning LED. (2 and 3 weren't connected to anything when I joined the team, and the regulator still works acceptably.)

I think I need a diode that will allow the field energy in the alternator to dissipate to ground when the cutoff switch is flipped and the alternator is simultaneously disconnected from the battery and all loads. I was thinking that I would just get a diode with breakdown voltage around 18 to 20 V, but when I searched I came up with $40 diodes on DigiKey, so I think I made some incorrect assumptions. Please help.

Sorry, I don't know how much energy the diode will need to dissipate. Also, it's possible that the mystery-box regulator has this function already built-in and I'm wasting my time, but I don't know - better to be safe than to burn through alternators...

Thanks!
 

Diver300

Well-Known Member
Most Helpful Member
Disconnecting the battery won't always stop the engine. If the alternator continues to generate enough power to keep the systems running, the engine will continue to run.

Is there are separate connection to the cutoff switch that stops the engine?

The problem is not the field energy. The problem is that the alternator is rotating, and there is a magnetic field from the field windings, so it continues to generate current with nowhere for that current to go. Therefore you may need to handle the full current of the alternator until the field winding current has reduced. The regulator will reduce the field current if the voltage is too large, but it will take time, maybe 200 - 500 ms.

Something like this:- https://www.digikey.co.uk/product-detail/en/littelfuse-inc/SLD16U-017/F5677CT-ND/3307067 seems to be designed to deal with the load dump. Some alternators now include something similar.

The other electronics in the car can be damaged as well as the alternator. The voltage surge with one of those TVS diodes fitted can still be quite big.
 

rjenkinsgb

Well-Known Member
Most Helpful Member
There are purpose made battery switches with ignition kill functions.
A quick google search finds one that also has an alternator protection function, switching in a load resistor as the battery is disconnected.

Data:
Item link:

Another example - even if they do not know the difference between a diode and a resistor!
 

Nigel Goodwin

Super Moderator
Most Helpful Member
The problem is not the field energy. The problem is that the alternator is rotating, and there is a magnetic field from the field windings, so it continues to generate current with nowhere for that current to go. Therefore you may need to handle the full current of the alternator until the field winding current has reduced. The regulator will reduce the field current if the voltage is too large, but it will take time, maybe 200 - 500 ms.
Back in the 70's, I was leaving a friends house about 10:00PM to go home, and as I kicked my motorbike up the battery lead became disconnected.

It was a far cruder alternator than a car, simply a permanent magnet rotating inside a coil, then a rectifier direct to the battery. So the only regulation was provided by the battery - so with the failure of that connection the voltage output rose excessively and blew every bulb on the bike. Obviously no where to get bulbs at the time of night, so I left the bike at my friends and walked home - about 7 miles over the hills.

While I'd like to think a car alternator wouldn't be 'that bad', there's a LOT more that could be damaged, so it's not a chance you really want to take.

Post #3 looks an excellent suggestion - and funnily enough I was watching 'Wheeler Dealers' yesterday where they converted a MK2 Ford Escort to a rally car - including fitting just such an FIA approved isolation switch.
 

Diver300

Well-Known Member
Most Helpful Member
I'm not convinced that a 3 Ohm resistor will do enough to stop a surge.

The standard test pulse (bottom of page 2 of https://www.littelfuse.com/~/media/...es/littelfuse_tvs_diode_sld_datasheet.pdf.pdf) has a peak of 87 V and an impedance of 0.5 Ohms worst case, so about 160 A of current. The 3 Ohm resistor will leave the peak voltage nearly as large if the series impedance is only 0.5 Ohms. It would still be nearly 75 V.

I think that the test pulse is a realistic worst case. Modern cars often have 150 - 220 A alternators, so significant power has to be absorbed to stop a voltage spike if that current is going into the battery.

If there are other loads, like fans, incandescent lights or heaters, then the surge is much smaller. Even if some current is flowing into the battery when it is disconnected, if there are other loads that will increase their current when the voltage goes up, the surge will be much reduced, but I don't think that the 4 - 6 Amps that will flow in a 3 Ohm resistor will really make much of a difference with 150 Amp alternators.
 

Sonia's Driver

New Member
Thanks for the replies everyone!

Diver300 - because the new cutoff switch is DPST, as soon as it's flipped the ignition coil and injectors will lose power and the engine will soon stop. The alternator will be isolated, so the only thing a voltage spike can damage is the voltage regulator. Because all loads are being disconnected, the voltage spike will be especially high because the only place the energy can go is reverse-bias through the rectifier diodes, if their breakdown voltage is lower than the voltage to create an arc. (Perhaps I should add a diode to the voltage sense wire as well, to prevent a damaged regulator from feeding power down that wire...)

Thanks, I think the SLD16U-017 is exactly what I need. At 75C it will be down-rated to 80%, and for a 500ms pulse it's rated to take around 1.7kW of peak power, so that's 1.36kW. I think the alternator is only rated for around 90A at ~5k RPM ... so (assuming 14v output) that's 1,260W, which is under the rated limit. Littelfuse mentions a duty cycle of 0.01%, so I'll just tell the team that the cutoff switch has a 100 second cool-down timer. :)
Questions:
  • I was planning to attach this TVS diode between the input to the alternator fuse and ground. Based on these specs, I don't think I need a resistor as well - is that correct?
  • The over-current failure mode is short, so should I add an LED in parallel to indicate that the protection diode is still good?
  • Should I care about the 1 to 20 nF of junction capacitance from figure 4? (I can't think why I would...)
  • Also, should I bother adding a diode on the sense wire as well?
Thank you!

rjenkinsgb - The linked FIA switch is inadequate because it doesn't simultaneously disconnect the alternator from the electrical loads, which is why I selected a DPST switch instead. (Our alternator doesn't have a separate input for the field windings, so it will keep generating voltage as long as the engine is turning.)

Nigel Goodwin - this car races on the 24 Hours of Lemons circuit, so our setup probably has a lot in common with that Wheeler Dealer episode.

narkeleptk - Diver300's suggestion was indeed a TVS diode.
 

rjenkinsgb

Well-Known Member
Most Helpful Member
The linked FIA switch is inadequate because it doesn't simultaneously disconnect the alternator from the electrical loads,
It disconnects the battery and kills the ignition system, so the engine stops.

No running engine = no alternator output.

What is the problem with that system?
 

Sonia's Driver

New Member
What is the problem with that system?
The only problem is that Lemons' rules state:
All electricity, including the battery, charging, and ignition circuits, must be interrupted by the kill switch.
The linked switch doesn't interrupt the circuit to the alternator, so it doesn't satisfy the rules.
More importantly, I already bought the DPST switch, so it's a moot question - adding a $2 diode is cheaper and better than dropping $30 on another switch.
 

Diver300

Well-Known Member
Most Helpful Member
I was planning to attach this TVS diode between the input to the alternator fuse and ground. Based on these specs, I don't think I need a resistor as well - is that correct?
You should connect the TVS diode to the alternator end of the fuse. The battery will be at the other end of the fuse, probably after the isolator.

The resistor wouldn't help. It would make the protection less effective.

The over-current failure mode is short, so should I add an LED in parallel to indicate that the protection diode is still good?
If the TVS diode fails short, the alternator fuse will blow when the isolator is turned on. An LED and series resistor in parallel with that fuse might be more useful.

Alternators are inherently short-circuit proof. If you short one out, the current coming from the alternator will not increase much. However, the output terminal of the alternator is usually connected to a battery, so shorting the output of the alternator to ground will result in lots of current from the battery. As you have a fuse protecting the wiring between the alternator and the battery, the battery current will be stopped quite quickly, and the alternator won't be harmed by a short.

I have seen a TVS fail after multiple load dumps. It was quite small as it was just protecting an ECU. It did short out, but the fuse feeding it was only 5 A so there wasn't any more damage.

Should I care about the 1 to 20 nF of junction capacitance from figure 4? (I can't think why I would...)
I don't know where you've seen figure 4, but the junction capacitance will make no difference. A huge junction capacitance would reduce the surge, but it would have to be billions of times bigger to make a significant difference.

Also, should I bother adding a diode on the sense wire as well?
There's no need. The sense wire is either connected directly to the main output, or it will be incapable of supplying current.

The voltage on the warning light connection will be limited by adding the TVS to the main output.

There would be no harm in having more than one TVS. It would have the advantage of reducing the stress on each TVS. The TVS diodes are fitted to cars but they are never intended to be used. The TVS diode's function is to protect the alternator and ECUs if the battery is disconnected once by accident. The car handbooks say to never disconnect the battery with the engine running, and it will rarely happen. If it does happen occasionally, the TVS will probably save a big repair bill, but it's not expected to work multiple times. The TVS that I had fail had suffered multiple load dumps, as a wire had come loose and was making intermittent connection, so it did eventually kill the TVS.

Where you are turning off the car regularly with the isolator, you may over-stress the TVS. Fitting a few in parallel will reduce the stress.
 

Sonia's Driver

New Member
As you have a fuse protecting the wiring between the alternator and the battery, the battery current will be stopped quite quickly, and the alternator won't be harmed by a short.
... oh yeah, that's right ... I wasn't thinking about the power from the battery ...

Where you are turning off the car regularly with the isolator, you may over-stress the TVS. Fitting a few in parallel will reduce the stress.
OK, they're relatively cheap so that sounds like a good plan. I'll also discourage people from unnecessarily killing the engine with the cutoff. It has to be demonstrated during tech inspection, but only once per race...

Thanks for your help!
 

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