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prime the caps in my inverter by connecting it directly to my converter

BMS with MOSFET as series cutout switch doesn't have the problems of mechanical contacts, but another good reason a solenoid contactor cutout is a bad idea. There might be a small possibility that when the BMS switch re-engages there is enough surge current created to pop the DC circuit breaker. I would expect the series resistance of BMS MOSFET's, battery line resistance, DC breaker resistance, and any current shunt resistance is normally enough to prevent popping circuit breaker. If you have this problem your DC breaker amperage size is probably not high enough or breaker is defective.
A circuit breaker is slow enough that, like you say, unless it is already under-sized it should not pop.
The current detection inside a BMS however can be *very* fast and can see a huge surge as a short circuit and cut out. (@Will Prowse had that happen in the video he released yesterday. )
 
Shoot I missed the point again.
In the above scenario the charger will likely prime the caps before the battery is reconnected.
As @FilterGuy had already explained.
Actually, with a seperate port BMS, it seems like the charger may not prime the caps.
1593707995867.png

Even with a common port BMS, there are BMS cut-out scenarios that make me wonder about potential problems:
In particular, if the BMS cuts off discharge due to low temp, it is likely the charger is also turned off so when the temp rises enough for discharge to turn back on there is a potential for a surge. (The same scenario could apply to recovery from an Over-temp shutdown) .

Having said all of that.... I have not experienced any problems like this..... but I have never let my cells get too cold or too hot. I don't know how large of a worry this really is.

I would expect the series resistance of BMS MOSFET's, battery line resistance, DC breaker resistance, and any current shunt resistance is normally enough to prevent popping circuit breaker.
As I mentioned above, I agree that a typical DC circuit breaker will not pop (assuming it is properly sized in the first place)
The two concerns I have are 1) the BMS Over-current sense poping and 2) stress from the surge on the FETs and other components.

It is a good point that the parasitic resistance of the circuit will cut the peak surge. It does not take very much resistance to knock down the peak. The first-approximation for the surge is that there is zero resistance and the surge is initially infinite. However, that is not real life. There is always resistance. I just don't know what a good 'total system resistance' estimate would be.

In @Will Prowse demonstration videos he typically has an absolute bare bones set-up. When he hooks up the inverter without pre-charging, it pops the over-current in the BMS. However, in my real-life set ups I have at least a fuse a shunt and a buss bar (or two) in the circuit. I work hard to minimize the resistance in all the connections.... but each one of these things will add resistance.....and reduce the peak surge current.

Note: I can't bring myself to purposfully create surge scenarios to test all of this.... I am too cheap to endanger my equipment.

So, is there a 'real' problem or am I just worrying too much? I don't know. However, if the problem is real, I would expect to hear stories from folks that experienced it.... but so far all of the discussion has been about a theoretical problem. Consequently, I suspect it is not a huge problem.

Note: In Will's videos and on this forum there is a recent trend to design systems where the BMS turns off the charger or inverter rather than directly cut the current. With this type of design the caps stay charged all the time.... even when the BMS shuts down the inverter or charger.
 
Note: In Will's videos and on this forum there is a recent trend to design systems where the BMS turns off the charger or inverter rather than directly cut the current. With this type of design the caps stay charged all the time.... even when the BMS shuts down the inverter or charger.

That is what I have been thinking about
Code:
usage model:
    isolated ac ups
design:
    24 volt system
    1000-1500 watt hardwire capable inverter
    Discreet common off the shelf components
    commodity 100 amp atc fuse block
    commodity separate port bms
    battery_wire = 6 awg stranded thhn
    inverter_wire = 6 awg stranded thhn
    charger wire = 10 awg stranded thhn
diagram:
primitives {
    mrbf = marine rated battery fuse 100 amps
    lvd = low voltage disconnect(victron battery protect)
    |F| = fused screw terminals 30 amps
    |U| = un-fused screw terminals
    |S| = spare
    | = unfused 1/4 inch post 100 amps
    ... = repeat last object
}
battery {
    pos = mrbf<->cell_in_series<->...
    neg = shunt_sampler<->cell_in_series<->...
}
system {
                    |F|->lvd_in->lvd_remote_pos->inverter_remote_switch
                    |F|->shunt_power
                    |F|<-charger
                    |S|
                    |S|
                    |S|
    battery<->pos<->|->inverter
    battery<->neg<->|<-inverter
                    |S|
                    |U|<->premises_ground
                    |U|<-inverter_ground
                    |U|<-lvd_ground
                    |U|<-charger
                    |U|<->bms_b<->|->bms_c->charger
                                  |<-bms_p<-lvd_remote_neg<-inverter_remote_switch
}
 
What about the slow start cct’s used on aircon units ? Never looked at one but there must be some inductor in there that resists the initial current flow and thereafter it is only the resistance of the coil that will have a small volt drop.
your talking about s soft start, but those are for motors.

I get a spark when I connect batteries. there's even a spark in breakers, just FYI. Your just controlling where the spark happens.
 
Simple circuit to stop surge current. DC Solid state relay typically turns on with about 3 volts of input minimum (up to about 30 volts max on most). Zener diode keeps solid state relay from turning on and applying full current until inverter caps are charged up to approx 2 volts less than minimum supply voltage. You can series 2 zeners to get closer to proper voltage if needed (use one a few volts less than needed, and choose the second one so the sum of the 2 voltages add up to the needed voltage, you can get zeners in the 3.3 to 10 volt range in a few tenths of a volt steps, check the 1N47xx series of zeners). https://www.vishay.com/docs/85816/1n4728a.pdf

Inveter Spark Suppress Circuit.jpg
 
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Simple circuit to stop surge current. DC Solid state relay typically turns on with about 3 volts of input minimum (up to about 30 volts max on most). Zener diode keeps solid state relay from turning on and applying full current until inverter caps are charged up to approx 2 volts less than minimum supply voltage. You can series 2 zeners to get closer to proper voltage if needed (use one a few volts less than needed, and choose the second one so the sum of the 2 voltages add up to the needed voltage, you can get zeners in the 3.3 to 10 volt range in a few tenths of a volt steps, check the 1N47xx series of zeners). https://www.vishay.com/docs/85816/1n4728a.pdf

View attachment 16721
This will only work with SSR's and not contactor type relays as the voltage across R1 and L1 combination will be too low to energize the relay ?
 
Cap pre-charge.JPG

How about this : We use the RC time constant of R1 and the inverter caps for the voltage to build up across RL1 and then to bypass R1.
RL1 will disconnect once the battery is disconnected and the inverter caps have discharged.
Few refinements to follow e.g. snubber , flyback diode etc. This design will also help to discharge the inverter caps.
Please fault this idea.
 
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A circuit breaker is slow enough that, like you say, unless it is already under-sized it should not pop.
The current detection inside a BMS however can be *very* fast and can see a huge surge as a short circuit and cut out. (@Will Prowse had that happen in the video he released yesterday. )
There is two mechanisms for tripping within a breaker. One is heating bi-metal strip to trip latch. It is the slow average current trip. The second is magnetic field tripping the latch. The magnetic field trip is a fast trip. It is set higher in current than average heat based trip and intended for high surge current, like a momentary short on line and it is very quick trip. Its absolute trip current tolerance is not real good so there will be some variance even on same model and amperage breaker.

Search youtube for How a Circuit Breaker Works in Slow Motion

Difference between a DC and AC rated breaker is primarily the DC rated breakers have larger and/or harder metal surface contacts and stronger pull back spring due to DC arcing is more likely to weld contacts together.

You generally do not want to engage or switch off breakers under high current flow as there will always be some damage to contacts. (Obviously this rule does not apply if you see smoke coming out of your inverter) At their rated current tripping, a few trips will kill a breaker, whether AC or DC breaker. They are usually designed so if contact is damaged to point to cause it to heat during normal current flow the heat will make it to the bi-metal strip causing it to trip open the breaker.
 
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View attachment 16731

How about this : We use the RC time constant of R1 and the inverter caps for the voltage to build up across RL1 and then to bypass R1.
RL1 will disconnect once the battery is disconnected and the inverter caps have discharged.
Few refinements to follow e.g. snubber , flyback diode etc. This design will also help to discharge the inverter caps.
Please fault this idea.

The resistor permanently across breaker is somewhat okay although I would not recommend it as 'some level of voltage' on inverter when breaker is off might cause some issues, from blowing out something to causing processor to repeatedly attempt a starup sequence.

Coil across DC, definitely not a good idea.

If you want to get fancy, put a 5 ohm resistor in series with a 10 amp capable push button in parallel with breaker (without compromising the high current connections on breaker). That will charge capacitors in inverter in about half a second.
 
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The resistor permanently across breaker is somewhat okay although I would not recommend it as 'some level of voltage' on inverter when breaker is off might cause some issues, from blowing out something to causing processor to repeatedly attempt a starup sequence.

Coil across DC, definitely not a good idea.

If you want to get fancy, put a 5 ohm resistor in series with a 10 amp capable push button in parallel with breaker (without compromising the high current connections on breaker). That will charge capacitors in inverter in about half a second.
Kindly clarify 'Coil across DC' statement. Hundreds of applications use this all the time.
I do not want to be fancy , I want to be smart - your suggestion still requires manual intervention.
Why don't you come up with an idea to solve the problem without manual intervention.
 
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Simple circuit to stop surge current. DC Solid state relay typically turns on with about 3 volts of input minimum (up to about 30 volts max on most). Zener diode keeps solid state relay from turning on and applying full current until inverter caps are charged up to approx 2 volts less than minimum supply voltage. You can series 2 zeners to get closer to proper voltage if needed (use one a few volts less than needed, and choose the second one so the sum of the 2 voltages add up to the needed voltage, you can get zeners in the 3.3 to 10 volt range in a few tenths of a volt steps, check the 1N47xx series of zeners). https://www.vishay.com/docs/85816/1n4728a.pdf

View attachment 16721
I like the circuit. Ideally, the Zener would allow the SSR to turn on at just below the voltage that the inverter decides it can turn on. This way, if the inverter switch is 'on' you don't get a problem with the inverter trying to drain the capacitors faster than the resistor can charge them. Or worse, you could get an oscillation where the inverter turns on, drains the caps and turns back off (and then repeats) before the SSR ever turns on. Even if the capacitors are only 2/3 charged when the SSR turns on, the peak surge will be dramatically less.

Having said that, I am still not a fan of using an SSR. If you are going to use one. Make sure to use a high quality one that is not going to generate a lot of heat (losses)
 
Rough numbers: Assuming best quality, minimum series resistance values.


Battery resistance = 3 milliohms (likely higher unless a large AH capacity battery array)

BMS series MOSFET = 1.0 milliohm (better then cheap BMS)

Power cables = 1.7 milliohm (10 foot length pair of #2/0 wire)

Breaker = 0.5 milliohms

500 amp shunt = 0.1 milliohm

Misc inverter chokes, series lines, & terminals = 1 milliohm

Inverter capacitor ESR = 10 milliohm (highest quality, four 3,300 ufd, 40 milliohm ESR, in parallel)


Total series resistance = 17.3 milliohms (dominated by capacitors' ESR)


58.4 vdc applied voltage (16 x 3.65v cells)

Initial (t=0) instaneous surge current = 58.4 v / 17.3 milliohms = 3,400 amps

Time until surge current drops below 250 amps = 0.55 milliseconds
Time until surge current drops below 100 amps = 0.75 milliseconds
Time until surge current drops below 10 amps = 1.2 milliseconds
Time until surge current drops below 1 amp = 1.7 milliseconds.
Time until surge current drops below 0.1 amp = 2.1 milliseconds


The peak current may amaze you, this is probably worse case, but your system will still be quite high. Keep in mind the time is very short, only fraction of a millisec.

I did not include battery cable inductance which will be about 2 micro-Henries for this example (assuming they are taped together). That will cut peak current down a bit.

An average breaker contact can take 20 to 50 millisec to totally settle out open or close contacts so they have a lot of contact 'grinding' time.

Bottomline, do not use your breaker to turn on power.
 
Rough numbers: Assuming best quality, minimum series resistance values.


Battery resistance = 3 milliohms (likely higher unless a large AH capacity battery array)

BMS series MOSFET = 1.0 milliohm (better then cheap BMS)

Power cables = 1.7 milliohm (10 foot length pair of #2/0 wire)

Breaker = 0.5 milliohms

500 amp shunt = 0.1 milliohm

Misc inverter chokes, series lines, & terminals = 1 milliohm

Inverter capacitor ESR = 10 milliohm (highest quality, four 3,300 ufd, 40 milliohm ESR, in parallel)


Total series resistance = 17.3 milliohms (dominated by capacitors' ESR)


58.4 vdc applied voltage (16 x 3.65v cells)

Initial (t=0) instaneous surge current = 58.4 v / 17.3 milliohms = 3,400 amps

Time until surge current drops below 250 amps = 0.55 milliseconds
Time until surge current drops below 100 amps = 0.75 milliseconds
Time until surge current drops below 10 amps = 1.2 milliseconds
Time until surge current drops below 1 amp = 1.7 milliseconds.
Time until surge current drops below 0.1 amp = 2.1 milliseconds


The peak current may amaze you, this is probably worse case, but your system will still be quite high. Keep in mind the time is very short, only fraction of a millisec.

I did not include battery cable inductance which will be about 2 micro-Henries for this example (assuming they are taped together). That will cut peak current down a bit.

An average breaker contact can take 20 to 50 millisec to totally settle out open or close contacts so they have a lot of contact 'grinding' time.

Bottomline, do not use your breaker to turn on power.

I was under the impression that each connection point was ~.5 ohms. @Airtime
 
I was under the impression that each connection point was ~.5 ohms. @Airtime
Proper connections are *way* lower resistance than that.

If it was .5 ohms, it would heat up dramatically. P=I^2 x R At just 50 amps, we would see a 50^2 x .5 = 1250W

At .5 milliOhms it would be 50^2 x .0005 = 1.25W

Resistance is the devil in these circuits.
 
This will only work with SSR's and not contactor type relays as the voltage across R1 and L1 combination will be too low to energize the relay ?
SSR input (L1) isn't really an inductor, just happened to be the first reasonable symbol I came across (still on the learning curve of this schematic program...and it was 2AM). SSR's typically use an LED on the input to drive the output switching circuit, which also gives them electrical isolation between the input and output. You probably could use a contactor type relay, but they will have much higher 'on' power usage for the relay coil than a SSR does. DC SSRs typically use MOSFETs to do the heavy current switching, and they require almost nothing for gate drive to keep them in the 'on' state (the LED on the input likely uses more power). You'd probably also have to have a more complicated driver circuit for it than the simple zener and resistor setup as well.
 
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This discusion is convincing me that the right way to design systems is to use BMSs like chargery or electrodaucus where you can use controll signals to turn loads/chargers on/off. This gets difficult with mobile systems that have a lot of different load and charge devices, but it would still be my goal moving forward.
 
I was under the impression that each connection point was ~.5 ohms. @Airtime
Milliohms, not ohms. And my understanding is that is high for new connections, more of a number to use to estimate what it might degrade to over time. For fresh new connections using tinned lugs, I'm seeing less than 0.1milliohms per connection.

This source says to plan around 0.25 milliohms per connection in your design:
https://marinehowto.com/fusing-termination-voltage-drop/
 
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