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

[FilterGuy]

FYI, info here: https://en.wikipedia.org/wiki/Inrush_current_limiter

Yup, I have seen that. Unfortunately, with the kind of current we are talking about, the two solutions offered don't really work. At 100 amps, even 0.1 ohm would burn 100^2 x .1 = 1kW. Whatever is used to precharge the caps must not be in the circuit for 'normal' use.

 

[PHoganDive]

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

 

[Pierre]

Deleted

 

[Pierre]

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 ?

 

[Pierre]

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.

 

[RCinFLA]

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.

 

[RCinFLA]

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.

 

[Pierre]

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.

 

[FilterGuy]

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)

 

[RCinFLA]

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.

 

[John Frum]

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

 

[FilterGuy]

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.

 

[PHoganDive]

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.

 

[FilterGuy]

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.

 

[Airtime]

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/

 

[RCinFLA]

All the parallel MOSFET's in the BMS series switch have a very large gate capacitance and drain to gate capacitance. When turned back on after a BMS shutdown event the gate driver circuit will take some time to get the MOSFET's gate voltage up to produce the series switch lowest resistance state. It will not be a well controlled ramp up but it will result in the MOSFET series switch having a softstart like effect.

In most cases this should be enough of a ramp down in series resistance that a BMS reactivation will not cause enough current surge to trip the DC breaker while charging the inverter capacitors.

 

[RCinFLA]

The most likely reason for the BMS to shut off discharge is if the battery or cell is too low.

Not so sure about that. You want to avoid BMS shutting down because it removes all power to inverter and perhaps your battery monitor. Inverter low battery shutdown should be setup to shut down inverter output before BMS low battery does.

To me, especially with a low cost BMS marginal series pass MOSFET resistance, the most likely reason for BMS shut off is the temp sensor on the BMS shutting it down because BMS got too hot from high current drain from inverter.

 

[FilterGuy]

Not so sure about that. You want to avoid BMS shutting down because it removes all power to inverter and perhaps your battery monitor. Inverter low battery shutdown should be setup to shut down inverter output before BMS low battery does.

All true. And that is what I try to set up.

To me, especially with a low cost BMS marginal series pass MOSFET resistance, the most likely reason for BMS shut off is the temp sensor on the BMS shutting it down because BMS got too hot from a few minutes of high current drain from inverter.

OK.... but I have not had that experience. Part of the reason fro that is that when I use a FET based BMS, I try to get one that is rated for double the expected max continuous current).

 

[BiduleOhm]

Yep, or

All the parallel MOSFET's in the BMS series switch have a very large gate capacitance and drain to gate capacitance. When turned back on after a BMS shutdown event the gate driver circuit will take some time to get the MOSFET's gate voltage up to produce the series switch lowest resistance state. It will not be a well controlled ramp up but it will result in the MOSFET series switch having a softstart like effect.

In most cases this should be enough of a ramp down in series resistance that a BMS reactivation will not cause enough current surge to trip the DC breaker while charging the inverter capacitors.

Unless they use a very very weak gate driver the switch time would be in a few dozens of µs at most, with the mosfets in the linear region for even a shorter time, so the soft start effect would be negligible IMHO.

 

[JoeHam]

For inverters larger then about a 300 watts, any ON/OFF switch within inverter will be just shutting down the electronics controlling the inverter's MOSFET's drivers. They do not use a large and expensive switch (ie. your DC breaker/switch), so capacitors are always on the battery lines coming into the inverter.

For a 3kW to 8kW inverter they typically have 7,000 to 10,000 uF's consisting of several paralleled capacitors. Throwing 48v on these caps is not that tough on caps. They are more stressed by the constant ripple current of normal inverter operation, but the large charging surge current is hard on the DC breaker if it is used to turn on inverter. It causes pitting on its contact surfaces, like little spot welding pock marks that can increase breaker switch resistance and eventually cause it to start arcing on large inverter current draw. High surge current could cause the breaker's contacts to stick together by the spot welding action and not allow the breaker to pop open for its primary safety function. And there is aways the possiblity the high surge current immediately pops the breaker back open compounding the pitting process.

Discharging time on these caps is variable depending on models of inverter. Some inverters put a couple mA current bleed resistor across these caps to accelerate their discharge time. A day or more to as little as 5 minutes might not be unexpected.

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 possiblity 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.

This info made me wonder about my 3kw inverters. Both 48v 3kw inverters.

My WZRELB HF inverter definitely sparks and makes a pop that is impressive after being disconnected from the battery for a week or so. Now I always use a precharge resistor if it’s been disconnected for more than a couple days.

My Sigineer LF inverter seems different. I left it disconnected for a couple weeks then reconnected yesterday without a precharge. No spark or pop. Does the huge transformer help somehow?

Anyway thanks for sparking the idea (see what I did there) to try the Sigineer without precharge. I may learn something ?

 

[FilterGuy]

My Sigineer LF inverter seems different. I left it disconnected for a couple weeks then reconnected yesterday without a precharge.

That is very interesting. Even LF inverters have the capacitors. If you are correct, it would be the first one I have ever heard of that did not have a capacitor bank directly across the DC.

 

[JoeHam]

That is very interesting. Even LF inverters have the capacitors. If you are correct, it would be the first one I have ever heard of that did not have a capacitor bank directly across the DC.

I did not mean to imply there was no capacitor.

Just an observation that I don’t understand.

 

[BiduleOhm]

If there's absolutely no load on the caps then they probably stayed charged (especially with some dielectric absorption in addition to that) enough to not have a huge inrush current when you connected the inverter.

 

[RCinFLA]

Without capacitance across battery line at inverter DC input, the battery cable inductance would cause high frequency transient voltage spikes that would eventually destroy the inverter MOSFET's. It would also drive the feedback control system crazy trying to figure out what inverter PWM drive is correct for matching incoming DC voltage level to the output AC voltage and power requirement.

It is the job of the caps to remove high frequency voltage transients caused by DC cable inductance. Also the reason to tape battery lines together in parallel to minimize the battery line inductance as the greater the line inductance the greater the ripple current on the filter caps which can destroy them over time.

 

[JoeHam]

If there's absolutely no load on the caps then they probably stayed charged (especially with some dielectric absorption in addition to that) enough to not have a huge inrush current when you connected the inverter.

Good to know that I can go over 2 weeks without the caps discharging then if that’s true.