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I have an alternator that charges at ~58 V at 100 A. If its load (a lithium bank) cuts out suddenly (bms cutoff), I need a way to shunt power for 200 ms to a dummy load to prevent the current from falling to 0 A and voltage spiking to 230~280 V (after about 100 V it begins burning out the alternator diodes and destroying everything else connected in parallel, so it doesn't rise to infinity).

Ideally the voltage would be kept from going beyond 75 VDC. 200 ms is how long it takes the alternator field to dissipate after the regulator quickly cuts power to field, so that's how long I need to shunt energy away from the main circuit.

For 12 V and 24 V alternators, this is an $89 device made by a company called Sterling Power, but nothing exists for 48 V. I've been told the physics of 48 V make it much more difficult.

A manufacturer of a 48 V alternator regulator gave me the following data for a 200 V alternator "load dump" scenario. My alternator is 100 A max, so I only need that much protection:

"You will want to see about suppressing the voltage to below 100 V or so max, else will be out of the range of viable to most devices to survive. The detail here is a multi-joule 200 ms or so event, not just a HV/LC event – most devices already have protection for that in them. Suppressing a 200 A 48 V alternator for 200 ms at 100 V comes out to 3,000 joules, if I have done my math correctly. That is the goal here. 100 V suppressing, with a 3 kJ rating."

He cites 100 V but, just to be clear, that's not where the spike begins. It begins at around 58 V x 200 A in his example, and voltage then rises as current falls, so by the time it reaches 100 V, current will only be 116 A. He likely calculated 60 x 200 x 0.2 = 2400 and then rounded it up to 3000 joules for headroom.

For my situation, it begins at 58 V x 100 A and lasts 0.2 s, so 58 x 100 x 0.2 = 1160 joules. And maybe we round up to 1500 joules to give some headroom. That's how much energy I need to shunt to a dummy load once the voltage crosses 75 VDC.

I found many voltage suppression devices for 48 V (Transtector was great about looking into their diverse 48 V suppression offerings) but their devices are for microseconds events, not 200 ms. So it's really getting into the area of sustained, albeit brief overvoltage, and not so much a transient.

Any ideas? I thought of maybe a 72 VDC converter with a 65 VDC minimum input (90 Vmax) but wasn't sure if it would be fast enough to begin shunting current to a dummy load attached to its output.

There must be a way to build a large inductor or an RC network that could absorb that kind of energy after the voltage crosses a set threshold.

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    \$\begingroup\$ Why do you have to shunt this energy? Why not just disconnect it? \$\endgroup\$
    – Finbarr
    Oct 19 at 20:58
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    \$\begingroup\$ @BeachInMexico Just to be clear on the calculation, I get \$200 \:\text{A} \cdot 100 \:\text{V}\cdot 200\:\text{ms}=4\:\text{kJ}\$. And this means a pulse average of \$20\:\text{kW}\$ for \$200\:\text{ms}\$. Makes me wonder what your source impedance is, in practice. Because that's what's needed to figure out what to do. If the source impedance is exactly zero (which it is not), then of course there is nothing you can do. \$\endgroup\$
    – jonk
    Oct 19 at 21:32
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    \$\begingroup\$ That's the amount of time it takes the alternator's field to dissipate after the regulator cuts power to the field. \$\endgroup\$ Oct 19 at 22:25
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    \$\begingroup\$ Off the top of my head, one solution might be to have a zener that triggers an SCR that dumps the energy into suitably sized resistor(s). \$\endgroup\$
    – Kartman
    Oct 20 at 0:23
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    \$\begingroup\$ @BeachInMexico I'm sure you've already been looking into various surge arrester technologies. So I won't duplicate your work. But while thinking through those, you might also consider a way to include (hybrid design) a series RC pair, as well, where the C is fully rated for the voltages involved and also matched with the alternator inductance (if I'm following you, at all.) I'm sure there are pages discussing practical relationships for calculating R and C values. I don't know of any off-hand, but if I knew more details about the circumstances I might be able to think of a suitable approach. \$\endgroup\$
    – jonk
    Oct 20 at 0:49
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It seems perfectly feasible to dump the load into a 1 kW wirewound resistor using a mosfet.

Example resistor: Vishay p/n FSE100022ER500KE

This is a 1 kW wirewound resistor rated for 10x overload for 5 seconds. This will have no problem handling 100 Amps for 200 ms as long as you don't do it over and over. Cost is around 80 US dollars in single unit quantity.

Example MOSFET: Infineon p/n IPTC019N10NM5ATMA1

This is a 100 V MOSFET with 2 mOhm Rds(on). Ignoring switching losses, the dissipation will only be 20 Watts (assuming Id = 100 amps) for 200 ms. That should be no problem. Cost is under 10 US dollars in single unit quantity.

Those are the key components. You will also need a comparator, maybe a gate driver for the MOSFET, and a voltage reference. And, I guess, probably a 10 or 12 V regulator of some type for the gate driver and reference and comparator. You will want to make sure the MOSFET latches on and stays on long enough to dump all the energy. So the 12 V rail for the comparator might need a diode and capacitor to hold it up at least until after the field collapses.

With this power level, you will need to do some failure mode analysis (regardless of what solution you choose). What if the load dump mechanism fails in the "on" position? Do you want to detect that fault and suppress operation of the alternator? Blow a fuse and just have no load dump? Etc. That is your responsibility because you know your system. I don't.

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  • \$\begingroup\$ Excellent! Thank you so much mkeith. Good news is I have a stable source of 12V from the vehicle's starter battery that isn't part of this 2nd alternator circuit, so it's very easy to deliver a fused 1A 12V rail for the comparator and FET. I've yet to wire up a comparator, but I'll start learning. With other devices from Transtector, I've looked at failing short and open, the former being most likely to damage the alternator, the latter being back to square one with a voltage spike. I'll do some more research on this on. Any recommendations on a good comparator to pair with the Infineon FET? \$\endgroup\$ Oct 20 at 13:41
  • \$\begingroup\$ Power transistors fail short. Never seen one fail open. Not to say that it can NEVER happen. But I don't think you need to consider both cases. Does this circuit need to minimize power consumption? For instance is it OK if it uses a few mA at 12 V all the time? \$\endgroup\$
    – mkeith
    Oct 20 at 17:56
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    \$\begingroup\$ Hmmm. There is a lot to this circuit. But the idea is that, if necessary, you add capacitance to the output so that when the BMS cuts off, the voltage ramp rate is not too fast. For example 1 V per microsecond would probably be acceptable. Let's say the comparator switches at 60 or 65 V or so, and 10 microseconds or less later, the MOSFET is on. Once the mosfet is on, the voltage will no longer increase. In fact, it will probably decrease. \$\endgroup\$
    – mkeith
    Oct 21 at 4:14
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    \$\begingroup\$ This answer is just a sketch. I think maybe it would be best to study a bit and then ask another question once you know more. However, I would not directly connect the compartor output to the MOSFET gate. I would use a gate driver to make sure the safe operating area is not violated. The gate driver will switch the MOSFET with authority. \$\endgroup\$
    – mkeith
    Oct 21 at 4:16
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    \$\begingroup\$ @mkeith - saw a review of a consumer PC power supply unit where a power transistor exploded during one of their tests. I would have to assume an exploded transistor is an open circuit :) \$\endgroup\$
    – user253751
    Oct 21 at 8:59
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While it's certainly possible to design a shunt load to absorb that much energy, it would require a somewhat large and expensive pulse-rated resistor and a nontrivial semiconductor to switch it in.

Instead I would take the approach of accelerating the regulator's woefully slow load-dump response: Use an op-amp or comparator to detect an overvoltage condition (above 70 V, say), and respond by short-circuiting the alternator's field winding. This only needs a modest FET, BJT or SCR since the field current will only be single-digit amps. You'll want to add a circuit breaker or PTC resetable fuse on the upstream field supply unless it's already current-limited. Depending on the dynamics you may need to make the field short-circuit latch for a few hundred ms even after the overvoltage trigger condition has gone away, or use hysteresis.

This is known as a "crowbar" overvoltage protection scheme and is common in aircraft power systems to protect expensive avionics from alternator regulator failures. Slightly different application here but it should work just as well.

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    \$\begingroup\$ If you short circuit the field it will increase the time it takes for the magnetic field to die away. I don't know what the time constant of the field winding is (L/R). It may be appropriate to crowbar the output of the alternator - often there is so much leakage inductance that the current into a short circuit will not be much more than its full output current. The energy would then be dissipated within the stator windings and laminations. \$\endgroup\$ Oct 20 at 1:45
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    \$\begingroup\$ Thank you! So the idea here is to destroy the alternator but protect other parallel gear, right? My understanding is the regulator is extremely fast in responding but the stored magnetic energy has to go somewhere, and it takes about 200ms for the field to collapse no matter what you do. But the alternator costs ~$2,000, the parallel gear costs about $4-5k so it's better than nothing. I'd much rather find a way to shunt the energy into some giant caps and/or nichrome windings, or maybe even a giant inductor. Thoughts? \$\endgroup\$ Oct 20 at 2:17
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    \$\begingroup\$ @BeachInMexico Wire-wound resistors are known for their pulse capacity. That's the first place I'd look. \$\endgroup\$
    – jonk
    Oct 20 at 5:39
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    \$\begingroup\$ The goal is to destroy neither the alternator nor the load on the bus. The field supply circuit breaker will trip, that's all, and can be reset. Kevin: Yes, the field current won't disappear instantly, but the field winding inductance is typically only a mH or two, and the resistance is several ohms, so it will die away in a couple ms. \$\endgroup\$ Oct 20 at 16:52
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    \$\begingroup\$ Agreed - it's not a lot for a coil of wire, but it would be a lot for most semiconductors or film resistors. \$\endgroup\$ Oct 20 at 21:01
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Three alternative solutions:

  1. It is perfectly safe to short an alternator output instead of dumping its excessive energy somewhere. Most permanent-magnet alternators are even regulated by PWM shorting. Here, you only need to short it for these 200ms until the field dissipates.

Option: you may short all the alternator windings upstream to the rectifier to the minus rail, it may be simpler in regard to transistor selection.

  1. If you don't have other sources of power (like, solar/wind/etc) you may get better results by making BMS disconnect the alternator field instead.

  2. You may do an alternator field flyback circuit that is capable of quickly dissipating the field energy. It is 1/20 or even 1/100 of the alternator output and the task is proportionately easier.

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  • \$\begingroup\$ Thank you Fraxinus! Regarding 1) won't that damage the alternator? Regarding 2) I do have another stable 12V power source (vehicle starter battery, this 2nd alterantor is totally separate), but the BMS is not capable of communicating. Believe me, I'm working on getting the manufacturer to add a 2 second warning, and next time I'll pay out the nose and just get a CAN enabled BMS that can send a signal to the regulator to shut down the alternator. But for the time being... And 3) Can you rough up a schematic of that so I have a sense of what that would look like? Many thanks! \$\endgroup\$ Oct 20 at 13:47
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    \$\begingroup\$ In regard to 1: No, I already wrote that it is safe. The current will be not much higher than the nominal working current. In regard to 2: it boils down to disconnecting all charging sources. The 12V circuits are not interesting in this regard. 3: I'll try to add a picture later. \$\endgroup\$
    – fraxinus
    Oct 20 at 13:53
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I would think a thyristor/SCR crowbar circuit on a heatsink could survive for 200ms.

Except instead of crowbarring a short, crowbar into a resistance. If you're really worried you can use water submerged resistors (like hot water tank heating elements). We use something like this at my work for a continuous 4kW loads. Our setup is 4kW and is two 10 gallon buckets and four 1500W hot water heating elements. Since yours is a pulse you could probably just stuff all the heating elements into a single bucket. I don't even think you need the water to be honest but it's a safety net during testing. Also works as a dummy test load. You could make one bucket that draws the same current as your battery when charging and then crowbar into the other bucket.

This is a traditional crowbar circuit except a dump resistance has been added. They often don't have this and are just a short. It clamps the circuit to near zero which causes an overcurrent which blows the fuse.

The dump resistance is in case the thyristor can't survive the surge by dissipating the power within itself. Unlike MOVs and TVSs which must do so and are far more stressed.

You may not need a dump resistance at all if the thyristor can handle the surge until the fuse blows and crowbars often do not use one. Clamp voltage will be far lower that way.

schematic

simulate this circuit – Schematic created using CircuitLab

Even if the thyristor can't handle it, as long as it is big enough so it does not fail by exploding and leaving an open-circuit, it will fail short which is what you want anyways.

So you basically want the voltage to stay at 58V or lower when the crowbar latches. A couple of volts will be dropped across the thyristor which leaves 56V at 100A. Which means you need 0.56 Ohms or lower.

For example, the large thyristors such as the VS-T90RIA10 come in packages that accept screw terminal wiring and bolt directly to the heatsink. They cost between $50-$200 depending on the part. The part mentioned here costs about $50.

https://www.mouser.ca/ProductDetail/Vishay-Semiconductors/VS-T90RIA10?qs=u%252Bh7cTrMUz%2FI0KVPJO4JIg%3D%3D

And in the datasheet it states it can handle a non-repetitive half-sine wave surge of more than 600A for 200ms. enter image description here

I don't think it should cost more than $200USD: thyristor, some hot water elements, some buckets, and an aluminum block to act as a heatsink for the thyristor. No point in an actual heatsink with fins since the pulse doesn't last long enough for air cooling. You just need thermal capacity so the heat from the thyristor has somewhere to go other than the thyristor.

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    \$\begingroup\$ Excellent DKNguyen! Thank you! Between you and mkeith I suddenly have to viable solutions to begin studying. I really appreciate it. \$\endgroup\$ Oct 20 at 13:44
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    \$\begingroup\$ @BeachInMexico Thyristors/SCRs are far more rugged than MOSFETs by the way. But they latch on. To turn off something else must interrupt the current like an aC czero-cross or some other device. \$\endgroup\$
    – DKNguyen
    Oct 20 at 13:47
  • \$\begingroup\$ By latching on you mean the circuit would remain closed once activated, right? That might still work since I believe the regulator would shut down charging, and if I can build an alert mechanism into the system, I could then shut off the battery bank so it didn't wake up and signal the regulator to resume field current. I'd like to know if an event like this happened anyways, and would plan to immediately discontinue using the alternator until I figured out the cause and how to prevent it. Would I need to replace the thyristor if it latched on, dissipated the energy, then couldn't be reset? \$\endgroup\$ Oct 20 at 13:50
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    \$\begingroup\$ @BeachInMexico Yes. A thyristor can't turn off current under its own power once it is already flowing through itself. You need something else to do it. No replace. Just how it works. It's not a reverse fuse. Crowbar circuits often have a output series fuse so the short overcurrents through the fuse while clamping the load voltage, then the fuse blows leaves the source open. \$\endgroup\$
    – DKNguyen
    Oct 20 at 13:56
  • \$\begingroup\$ Okay, good to know. Any chance you could rough up a schematic. I've got some of the pieces but can't quite see it all together. I really appreciate your help! \$\endgroup\$ Oct 20 at 13:58

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