Preventing Voltage Spike from Alternator Cut-Off

[wingssail]

Originally Posted by expozen

You mean wire it like shown below?

Turning off the alternator with either the key switch or the BMS relay is fine. It will shut off the field current and there will not be a diode killing spike, however it is good practice to stop the engine with the kill control before shutting off the key switch.

Of critical importance though is not to switch off the battery switch and the parallel switch with the engine/alternator running. This creates the diode killing spike because the alternator will still have full field current and full output voltage but nowhere to go.

A second issue is the location of the battery switch. The physical location. On a friend's yacht it was located near the floor near the companionway and it was possible to step on it, which they did while the engine was running, then they turned it back on, anyhow, somehow they blew ALL the electronics which were turned on at the time. Big disaster in the Solomon Islands. I gave them a spare depth sounder and a handheld GPS, it was all I had, and therefore all they had.

I recommend that you make you system absolutely simple so that it is totally transparent what is going on, and very obvious before you make a mistake.

[expozen]

Originally Posted by wingssail

however it is good practice to stop the engine with the kill control before shutting off the key switch.

Interesting! Why is that? I've done this before, but hadn't fully thought about it. Would there be a potential of damaging something?

[wingssail]

Originally Posted by expozen

Interesting! Why is that? I've done this before, but hadn't fully thought about it. Would there be a potential of damaging something?

Look, I'm not an expert like wotname or some of the others, but I just feel that the kill control which cuts fuel to the diesel, gives a gentle and gradual shut down, including slowing the alternator to zero. Switching off the key there is still the engine running and the alternator spinning and perhaps some residual magnetism in the field will cause some current to continue without any control.

In your case would the key switch shut off power to the BMS and therefore the relays?

I don't know, it's just me.

[john61ct]

No, even a "sudden" killswitch stopping alt field current is OK.

What is not, is suddenly isolating the high current source from its sink (the only bank).

[saillr]

Some years ago I rewired my alternator output to go through a DC shunt. I did not have a large wire crimper so decided to solder the new connectors on. What a mistake. The cable was a little on the short side and with the engine vibration the solder joint went cold and a spike ruined the diodes. Along with new diodes I was given a separate diode by the repair company and told to wire this across the output terminals and never have a spike ruin the diodes again. Very simple fix and true it has never happened again. I don't think the diode was anything special. Maybe 100V reverse voltage and capable of a few amps.

[Wotname]

Originally Posted by expozen

So, it seems like there is two different phenomenon at work here:

There's the potential voltage spike generated by the collapse of the magnetic field caused by the ceasing of current flowing through the alternator.

And there's a voltage spike generated by just ceasing of current in general, much like a water hammer effect............

It looks different but really it the same.

Clearly you are thinking about this but as yet, you don't know enough electrical theory to fully understand what is happening - please don't take this as condescending; none of us are born knowing about electricity, we all have to learn it if we are to understand it.

A quick tutorial if you are interested otherwise disregard.

First step - any time a conductor passes through an magnetic field, a voltage is induced (generated) in the conductor. It doesn't matter if it is the conductor moving or if it is the magnetic field that is moving. We say the magnetic field cuts the conductor. The size of the voltage is dependant on the strength of the field, the length of the conductor and the rate of cutting of the magnetic field. They are all proportional so long conductor = bigger voltage; stronger field = bigger voltage and faster cutting = bigger voltage

Second step - if both ends of the conductor are joined together, the voltage that was induced in the conductor allows current to flow though the conductor. We say current is flowing and the circuit is live. Again it is proportional; bigger voltage = bigger current

Third step (this one is important and not often considered) - every time current is flowing in a conductor, it creates a magnetic field around the conductor. No current = no magnetic field and some current = some magnetic field. They are proportional so small current = small field and big current = big field. If the current is steady (i.e. DC), the magnetic field remains stationary.

Fourth step - now consider a live DC circuit (i.e. one which is carrying a steady current) - say a simple light bulb which is turned on. A steady DC current is flowing in the conductors and therefore there is a stationary magnetic field around each conductor. Now turn the light bulb off with the switch. Current almost instantaneously drops to zero, therefore the magnetic field almost instantaneously drops to zero. As it collapses, it is cutting the very conductor that was creating it, therefore a voltage must be induced into that same conductor. But no current flows because the circuit was broken (opened) when the switch was turned off. If the voltage produced is big enough, it will ionise the air at the point of the open circuit (i.e. the switch) and we will see this as a spark.

What makes a difference is simply the size of this induced voltage.

Looking again at the above theory, the size of the induced voltage is dependant on the length of the conductor, the strength of the magnetic field (i.e. the size of the current flowing) and the rate of change of the relative movement between the magnetic field and the conductor. Again they are all proportional.

Inside the alternator are the stator coils and coils of wire = longer length.
The alternator maybe be generating say 30 amps of current i.e. a fair size magnetic field around the stator.
Disconnecting the output almost instantaneously stops the current flowing so the magnetic field collapses very fast cutting though all those coils. Thus the induced voltage spike will be hundreds or thousands of volts which is large enough kill the diode pack.

Going back to the light bulb circuit, the current flowing is small say 0.5 amps (for a 6W bulb), the length of the conductor is usually short and therefore the induced voltage is small (maybe tens of volts).

There is more to it, but the above is the simple explanation however it should be enough to understand the "why" of the induced voltage spike.

It shows up elsewhere and the following example maybe of interest.

Take a fully charged 12V battery that is in good condition. Clip a large gauge lead to one pole and take the other lead near the remaining pole. No current will flow while the lead is not connected to both poles. In fact, you have to get the lead really really close (sub-millimetres) before a spark occurs as there is only 12V available to ionise the air gap.

Now short out the battery with the same lead. Note this is bad idea but consider the theory for a moment. Very large amounts of current will flow in the lead. Depending on the battery and the jumper lead, this could be 1000+ amps and it creates a large magnetic field around the jumper lead. Now pull the lead off one terminal - the large magnetic field collapses and induces a voltage into the lead. Although the lead was short, the magnetic field was big so this induced voltage will be in the order of tens of thousands of volts. This large voltage is big enough to ionise a large air gap and we see this showing up by the large and long arc that can be created when pulling a short lead of the battery. Really we can only create a decent spark when we disconnect a battery, not when we connect one.

For anyone still reading (), it should be noted the polarity of the induced voltage is always opposite to that of the voltage that was creating the current flow in the first place. This why a suppression diode will work. In the case of the alternator, the output terminal +ve. when it is disconnected, the induced voltage at the terminal will be -ve. The suppression diode will normally be blocking when the voltage at the terminal is +ve but will become conducting when the induced voltage becomes -ve at the terminal.

Maybe someone can explain all of the above is a simpler and easier to understand way but this is the best I can do ATM.

[Lake-Effect]

Originally Posted by saillr

Some years ago I rewired my alternator output to go through a DC shunt. I did not have a large wire crimper so decided to solder the new connectors on. What a mistake. The cable was a little on the short side and with the engine vibration the solder joint went cold and a spike ruined the diodes.

Thanks for posting your experience. It seems people still need to be reminded that just soldering is not sufficient, especially for power connections.

Along with new diodes I was given a separate diode by the repair company and told to wire this across the output terminals and never have a spike ruin the diodes again. Very simple fix and true it has never happened again. I don't think the diode was anything special. Maybe 100V reverse voltage and capable of a few amps.

Do you recall the diode type or part number? Just wondering if it was just a simple rectifier to eat up reverse-polarity spikes, or maybe a zener diode that becomes a shunt if the battery is disconnected and the output shoots up beyond the zener voltage.

[smac999]

Originally Posted by expozen

Interesting! Why is that? I've done this before, but hadn't fully thought about it. Would there be a potential of damaging something?

Well most boats have a stop button. That is powered from the key switch. So if the key is off. You can not stop the engine. You probably have a mech pull cable. I would still kill before turning off key.

[expozen]

Originally Posted by smac999

Well most boats have a stop button. That is powered from the key switch. So if the key is off. You can not stop the engine. You probably have a mech pull cable. I would still kill before turning off key.

My boat has a pull cable. But strangely, I can turn the key off and remove it but the engine will still keep running until I pull the stop cable. I imaging that turning the key off while the engine is running would turn off the alternator, but I haven't verified that.

[expozen]

Originally Posted by Wotname

Clearly you are thinking about this but as yet, you don't know enough electrical theory to fully understand what is happening - please don't take this as condescending; none of us are born knowing about electricity, we all have to learn it if we are to understand it.

I know enough about the world to know that I rarely, if ever understand what's happening. I'm always trying to understand more. But it's sometimes hard to balance all the things to learn with all my other needs and wants.

I think that your explanation is fascinating. And I appreciate you writing such a response. It's given me a better understanding... but also given me more questions...

Originally Posted by Wotname

Now turn the light bulb off with the switch. Current almost instantaneously drops to zero, therefore the magnetic field almost instantaneously drops to zero. As it collapses, it is cutting the very conductor that was creating it, therefore a voltage must be induced into that same conductor. But no current flows because the circuit was broken (opened) when the switch was turned off.

Does this mean that a voltage would be induced on both sides of the point of disconnection? For instance, when an alternator is disconnected from a battery, the portion of the disconnected circuit on the alternator side would have a relatively large induced voltage because of the relatively long wire. And there would also be a voltage induced on the battery side of the circuit, from the point of disconnection to the battery. But that induced voltage would be relatively small because of the relatively short wire?

Also.... when a voltage is induced within this disconnected length of wire, I would take that to mean there is a difference in potential energy across either end (like there is now more electrons at one end of the wire than the other). So when the voltage is induced, would a current then flow across this length of wire from one end to the other? Because I imagine it would have to come to an equilibrium after the voltage has been induced. And while that current flows from one end of the wire to the other, would that itself then generate another weaker magnetic field around the wire, which would too collapse to generate a smaller voltage in the wire, and so on? Like, is this a cascading event?

[guyrj33]

Has anyone suggested a zener diode across the alternator outputs?
Xapstop is a commonly marketed one.

[Lake-Effect]

Originally Posted by guyrj33

Has anyone suggested a zener diode across the alternator outputs?
Xapstop is a commonly marketed one.

You mean Zap Stop. Here's a blast from the CF past. Thanks for reminding us of this.

[expozen]

Originally Posted by guyrj33

Has anyone suggested a zener diode across the alternator outputs?
Xapstop is a commonly marketed one.

The one on Balmar's site specifies that it's a 'sacrificial' one-time use device. Which does't seem like a real solution. As someone stated in another thread, they work the first time, no one knows it's fried, then the alternator diodes just fry the second time. I could see it being useful as protection in the event the fuse between the alternator output and the battery blew. You'd just have to know that if that fuse blew, you'd also need to replace the ZapStop at the same time.

[Wotname]

Originally Posted by expozen

I know enough about the world to know that I rarely, if ever understand what's happening. I'm always trying to understand more. But it's sometimes hard to balance all the things to learn with all my other needs and wants.

I think that your explanation is fascinating. And I appreciate you writing such a response. It's given me a better understanding... but also given me more questions...

Every answer raises ten questions!

Does this mean that a voltage would be induced on both sides of the point of disconnection? For instance, when an alternator is disconnected from a battery, the portion of the disconnected circuit on the alternator side would have a relatively large induced voltage because of the relatively long wire. And there would also be a voltage induced on the battery side of the circuit, from the point of disconnection to the battery. But that induced voltage would be relatively small because of the relatively short wire?

In essence, yes - remembering most of this wire length is in the stator coils of the alternator.

However there is more to the story but you will need a far deeper understanding of coils, inductors, inductance reactance and magnetic fields before it starts to make sense. I'm not smart enough to pass on all the ins and outs in a CF post.

Remember we intentionally disconnect live circuits all the time (every time we turn a working device off) and the induced voltages are harmless almost all the time. Really the alternator is a bit of a special case and it is only the diode pack that gets fried.

Originally Posted by expozen

Also.... when a voltage is induced within this disconnected length of wire, I would take that to mean there is a difference in potential energy across either end (like there is now more electrons at one end of the wire than the other). So when the voltage is induced, would a current then flow across this length of wire from one end to the other? Because I imagine it would have to come to an equilibrium after the voltage has been induced. And while that current flows from one end of the wire to the other, would that itself then generate another weaker magnetic field around the wire, which would too collapse to generate a smaller voltage in the wire, and so on? Like, is this a cascading event?

No, not really. Current isn't flowing in the conductor (in the conventional sense) thus no magnetic field is established around the conductor. The charge (voltage) is created as the charge carriers in the conductor are forced towards either end of the conductor while there is relative motion between the conductor and any magnetic field. Remove the field or the motion and the charge carriers simply return to being evenly distributed along the conductor.

ATM, time doesn't permit a longer explanation and also you are approaching the limit of my understanding of static EM fields. Maybe later

[Wotname]

Originally Posted by Lake-Effect

You mean Zap Stop. Here's a blast from the CF past. Thanks for reminding us of this.

Thanks for the reminder!

There is a few inaccuracies in that thread but they don't need addressing after all this time. Most of the "what happens" is correct but a lot of the "why does it happen" is not so correct.