Exactly how BoostFactor works?

DIY
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1

Can anyone explain exactly how BoostFactor works? The VE.Configure help says the following:

“This setting is a special setting for PowerAssist. When the Multi is charging and, due to a sudden load, the mains current exceeds the AC input current limit the Multi will switch to assist mode (when enabled, see PowerAssist).”

So far so good, no doubts.
Help continues:

“At that moment the current needed is unknown.”

This is where the doubts begin. The required current is not only known at the moment the current reaches the limit, it is also not known before that, right?
Help continues:

“The Multi makes an assumption of the magnitude of this current. This assumption is equal to the AC input current limit multiplied by this “Assist current boost factor” setting. The default factor is two.”

I understand the calculation, but what does it do with the result of this calculation? What impact does this calculation have on the PowerAssist function?
Help continues:

“This will prevent the circuit breaker from tripping because current provided by the inverter minus the current drawn by the load is always lower than the rating of the circuit breaker.”

And this is the most confusing part for me. First he says that this will prevent the circuit breaker from tripping, but what he just said was just that he assumes that the load can reach the value of the input current limit multiplied by the BoostFactor, without explaining what he does with this calculation. How does this calculation prevent the circuit breaker from tripping? Furthermore, he says that the inverter current minus the load current is always less than the circuit breaker current, but isn’t the load current the inverter current plus the grid current? Isn’t the inverter current less than the load current? So the inverter current minus the load current won’t result in a negative value? Is this text correct? Shouldn’t it be the other way around, that is, shouldn’t it be the load current minus the inverter current, which will be the grid current that is passing through the circuit breaker? And what exactly does he do to ensure that the current in the circuit breaker will not be exceeded? What does it do with the calculation of the input current limit value multiplied by the BoostFactor?

2

Hi @vcleto,

So, boostfactor is just the factor by which you multiplies the input current limit. In case of sudden load, like a motor starting, we know that it can draw a lot of current for a very short amount of time. This tool serves two purposes :

  1. prevent a circuit breaker from tripping as explained
  2. prevent stall for a generator

The circuit breaker mentioned it the one before the inverter, not after. You draw current from this breaker to feed your load, so if you add to your loads current from inverter, you draw less energy from the breaker.

If you have a 10 amps breaker before, if you start a 5 A motor, you are probably going to exceed your 10 A limit for a short period, which boostfactor will prevent but just sending “boostfactor x input current limit” directly to the output, which will lower load on AC in.

I hope it was clear enough.

3

Thanks for the answer, but unfortunately I had already understood what you said from the VE.Configure help text I referred to. What I want to know is what Multiplus does with this information that the load can reach the value of the input current limit multiplied by the Boost Factor. What exactly does it do to prevent the inverter input circuit breaker from tripping? How does this data impact the amount of current used from the grid. If you read my question again, you will notice that my questions about how exactly the Boost Factor works were not answered in your answer.

4

Actually this is independent of the boost factor, its only influenced by the max ac-in setting.

5

Hi,
Actually, I think I answered that by explaining (maybe not that well) how it limits drawn current from ACin. De facto, if you limit the drawn current, you cancel the risk of tripping your breaker.

To sum up, if you draw too much power from acin, power assist complete more than it should to unload ACin, resulting in back current to breaker if incorrectly set up. Boost factor is there to ajust the depth of multi’s response to sudden load and works as you described it.

I don’t know how to go further in this explanation, but I hope it helped anyway :slight_smile:

6

I think you can’t explain further because you also don’t know how the result of multiplying the Boost Factor by the input current limit is used so that the input circuit breaker doesn’t trip. You understood the same thing I understood and what is said in the VE.Configure help text, but you don’t know exactly how it works. If, for example, the input current limit is 10A, and the Boost Factor is 2, what exactly will Multiplus do when it knows that the maximum load current is 20A and that the input circuit breaker is 10A? What exactly will it do differently if the BoostFactor is changed to 1.5, or if it is changed to 3? How much current will it draw from the input and how much will it assist in each of these cases for a given load current X?

7

Are you wanting to change the multplier for any reason?

8

Actually, I made a few experiences with different sudden load with a generator on acin. It behave exactly as described. When multi can’t know exactly how much power is needed, it send send the fixed parameter, as a result : when boost factor is too high, the generator is fully unloaded, so its rotating speed increase a little bit, and when boost factor’s too low, the generator absorbs the load and stall because of the overload.

Really, it’s simple. Multi just sends power at a fixed value to compensate.

What I don’t know is if this system exists because ajusting time is too slow, or because it can’t read a pic of current accurately.

Let’s wait for one of victron expert to answer if you will.

9

The power assist comes from the core in it. Explained as simply as possible it is the reservoir for boosting a load using the DC from the battery side.

Even simpler leave it at default and see how it works. Usually it doesn’t need to be messed with.

You dont want less boost, as @wgomes mentioned now it won’t help enough.
You also don’t want too much as you can overload the system.

16

The boost factor is an interesting one as ess manipulates it as well dynamically.

The higher the boost factor the more the battery (or dc) assists. Lower means less battery use.

It’s basically a math equation to determine how much to assist.
Your boost factor requirement (in watts) = (Input limit)(Boostfactor)(AC voltage)

You would have to work out also based on other factors.

  1. Input current you are expecting to use at the time. (Taking into consideration input voltage)
  2. Load expected vs size of the inverter (you can’t assist ideally above the inverter power)
  3. How much higher than input current it is expected to be (considering if there is inrush as well or just a straight load)
  4. How much assistance from DC you may have. (Some batteries just can’t.)

This effect is also affected possibly by dynamic current limiter and weak AC.

I will say from experience the 2.0 has been great for almost all set ups i have done that were not ESS.

Things to watch for if you decide to experiment on it are how fast your battery drians, how often the inverter assists) and if you are close to or overloads. And then of course if you are using a genrator or weaker ac source if itnis overloading the source (will be visible with voltage drop - or more obvious breaker tripping or overheating)

33

Hello,
I would like to explain how it works ( I must admit not reading every single post so the answers could already be given).

The boost factor is there to set the correct level for power assist. beware, this is only to set the level for the initial amount of energy and will after this be not used by the system.

The point is that IF there is a AC input limit set ( let say 10A) there is a fuse inline which is 10A , and the load will be normally below this. When there is a load getting close to the setpoint, the charger will be throttled back to 0A to reduce the change the 10A will be reached. If the load goes up further the inverter needs to act actively. The point at the level the setting is crossed is, how much does the inverter need to inject?

There are two extremes possible. 1 the load is just 10,1A so the inverter hardly needs to do anything. 2 there is a motor with an extreme high current is needed, let say 30A.

So at the boost factor 2.0 the setpoint (10A) is multiplied by 2 so the initial action is that the inverter will inject 20A in this example. So, the fuse on the Ac input will not trip as this 20A will be used 10,1 for the load, and therefore a 9,9 A Burts travels to the input.

So all ok!

In situation 2 the load gets 20A from the inverter (if its able to do so of course) and 10A from the input , So all is fine to.

Therefore, in most cases factor 2.0 is OK. When not ? Well when there is a huge inverter power and the loads tend to be small compared to this.

So if there are 30kVa of inverters but the input setting is 16A or such and the gird isnt very firm voltage wise (long cabling or such) , then at a 20A load the injection will be 32A with so much power behind it that the system voltage will jump quite aggressive. then a lower factor setting might be better as its smoother.

At the case the (startup) load is much higher then average ( like an elevator system) it known that the factor 2.0 results in a current which together with the input current is still too low ( and therefore the AC in will be overloaded) , then a higher setting may be used , like factor 3.0. Then when power assist is started to available power goed up substantial.

in short, boost factor is only active for 0-50ms, after this the power assist is based on its own measurements on ACinput. the 2.0 setting is most of the time perfect, so dont touch it unless you have a situation like described.

Clear for all?

4 Likes
34

Thank you very much, Johannes! No, no one has answered anything yet about how exactly it works! I believe this is the first time that an explanation of how Boost Factor works has been published at this level of detail! I still have some important questions that I will address below, so that we can have a very complete and detailed explanation of how Boost Factor works!

Is this initial moment when the Boost Factor is triggered only when the load current reaches the input current limit? Isn’t it too late to wait until the limit is reached to apply the Boost Factor? Doesn’t it trigger the use of the Boost Factor a little before the limit is reached? How much sooner? Is the use of Boost Factor triggered based only on the current level, or does it also depend on the current growth rate?
If it is used for such a short time, it is unlikely that the effect of using the Boost Factor will be visible on the measurement panel, right? If the load continues to oscillate around the input current limit after using the Boost Factor, can it be used continuously?

Does this mean that 9.9A will be injected into the grid? But what if the system is configured off-grid, so that no current is injected into the grid? Does it still inject it?

Based on AC Input or AC Output measurement?

Does this mean that if the load remains above the input current limit, the Multiplus will continue to use the entire input current limit and only supplement what is above it?

Or does it use a little less than the input limit? How much less? If it uses less, does this amount of less depend on the Boost Factor?

Is it possible for Power Assist to remain on even if the load is slightly below the input current limit?

I have seen situations where Power Assist was in use even if the load was below the input current limit, and I have seen reports that the amount of less input current that was being used depended on the Boost Factor (with a lower Boost Factor, the Multiplus would continuously use more input current than with a higher Boost Factor).
What is the explanation for this, if the Boost Factor is only used for 50ms?

35

Is this initial moment when the Boost Factor is triggered only when the load current reaches the input current limit? Isn’t it too late to wait until the limit is reached to apply the Boost Factor?

JB its starting at the setpoint , before here the charge current is already throttles down of course. when the load remains below the setpoint, the power assist will not start.

Doesn’t it trigger the use of the Boost Factor a little before the limit is reached? How much sooner?

Jb No, not sooner.
Basicly every source can be overloaded a little without a big problem, thats happening at most start up currents or switching in a switch mode device. Every fuse ( even the fast version) will not trip that fast.

Is the use of Boost Factor triggered based only on the current level, or does it also depend on the current growth rate?

Jb On the level only. The slope is not taken into the account.
This would be nice of course , but as it is done now it work under the most critical circumstances. Only criteria which messed the power assist up is a high fluctuation in the incoming voltage as the inverter might find it hard to follow/keep synced

If it is used for such a short time, it is unlikely that the effect of using the Boost Factor will be visible on the measurement panel, right?

JB With a digital scop it can be seen, otherwise I doubt you will find it.

If the load continues to oscillate around the input current limit after using the Boost Factor, can it be used continuously?

Jb there is a minimal time that the power assist remains active , (2 sec) , so if the load drops and comes back the power assist will do the same again.
For stopping charge/doing nothing/powerassist no moving parts are involved, only the voltage setpoint which is adjusted contatsnt anyway. So there is no maximal usage for powerassist

So, the fuse on the Ac input will not trip as this 20A will be used 10,1 for the load, and therefore a 9,9 A Burts travels to the input.

Does this mean that 9.9A will be injected into the grid? But what if the system is configured off-grid, so that no current is injected into the grid? Does it still inject it?

JB Off grid , so a generator powered it you mean?, yes, then the initial pulse will travel there. This is the main reason that IF there is a generator and then specially ne with an inverter output on it, the ACinput voltage window should be ideally be Brough to limits that the pulse cannot raise the voltage to much. Standard thats 190-270Vac on a 230V model, so theoraticly the powerassist can raise the voltage to 270Vac to get the current to the desired level

This would not make a generator happy, therefore in these cases the ACin window should be lowered to lets say 245-250 to be sure the power assist will limit itself is boosting with a high voltage

Based on AC Input or AC Output measurement?

JB Based on the AC input ( you might be aware the output is not actively measured, what you see on a minitor is a calculation ACin power/DC power)

Does this mean that if the load remains above the input current limit, the Multiplus will continue to use the entire input current limit and only supplement what is above it?

Jb Yes ! If there would be a DC source feeding the battery, the Multi can be in power assist 24/7

Or does it use a little less than the input limit? How much less? If it uses less, does this amount of less depend on the Boost Factor?

Jb There is a small hysterisis built in to compensate for voltage fluctuation,
Normally a power assist will aim at 85-90% of the setting to stay well below the ACin setting and there is little/no change the fuse will trip

Is it possible for Power Assist to remain on even if the load is slightly below the input current limit? I have seen situations where Power Assist was in use even if the load was below the input current limit, and I have seen reports that the amount of less input current that was being used depended on the Boost Factor (with a lower Boost Factor, the Multiplus would continuously use more input current than with a higher Boost Factor). What is the explanation for this, if the Boost Factor is only used for 50ms?

Jb The boost factor is only the initial burst, after this is not looked at.
Yes the power assist can remain active if the load goes just below the setpoint. There are two major reasons, one is that there is a margin as far as setpoint, so at a 10A setting the power assist aims for 9A , so if the load remains between 9-10 it stays in power assist mode. but hardly uses battery power, Secondly there is the accuracy on the AC measurement. The power factor/losses on the dc and deviation in the measuring makes that the powerasist is not always exact at the level it needs to be (but close).

The larger the system , the wider the current range is on measuring ACinout, the less accurate the unit will be..

2 Likes
36

If the Multiplus can wait until the current reaches the limit to apply the Boost Factor, then why doesn’t it continue supplying only the current needed to serve the load? Why suddenly take over the entire load current?

This reinforces my previous question: why assume a current greater than the portion that exceeds the input current limit? Greater even than the entire load?

Not necessarily a generator, any network in which injection of current into the network is not permitted.

I think this explains why the inverter I used to power the Multiplus burned out… I tried using a charger and an inverter to power the Multiplus, to convert the voltage from 50 Hz to 60 Hz. The Multiplus was configured to not inject current into the grid, I didn’t imagine that it would inject it anyway… :frowning:

Are you saying that you want to lower the maximum voltage limit that Multiplus accepts at the input from 270V to 245V or 250V? Does it use this same parameter as the limit of the voltage that it can apply at the input? To prevent Multiplus from injecting current into the grid, wouldn’t it be necessary to set the Boost Factor to 1 or less?

37

#1. If the Multiplus can wait until the current reaches the limit to apply the Boost Factor, then why doesn’t it continue supplying only the current needed to serve the load? Why suddenly take over the entire load current?—> Because it’s not clear at the moment of crossing the setpoint line if the current is slightly over, so assist a little bit , Or if the crossing the setpoint is huge, so assist at full power. If you assume the needed power is only little but in fact its a motor needed full power, you are too late and the input fuse may trip, followed by an inverter which is overloaded.

38

#2 This reinforces my previous question: why assume a current greater than the portion that exceeds the input current limit? Greater even than the entire load? → as the inverter cannot look into the future, So it acts (injects power) → measures the result (were does the power go)–> adjust inverter power → measures ect…so you have to start from something … [quote=“Valdson, post:36, topic:26311, username:vcleto”]
think this explains why the inverter I used to power the Multiplus burned out… I tried using a charger and an inverter to power the Multiplus, to convert the voltage from 50 Hz to 60 Hz. The Multiplus was configured to not inject current into the grid, I didn’t imagine that it would inject it anyway… :frowning:
[/quote]

#4 think this explains why the inverter I used to power the Multiplus burned out… I tried using a charger and an inverter to power the Multiplus, to convert the voltage from 50 Hz to 60 Hz. The Multiplus was configured to not inject current into the grid, I didn’t imagine that it would inject it anyway… → depends on the voltage settings, as said before the power assist allows the ACinput voltage to go up to the max ACinput voltage setting before stopping further injection, default is 270Vac. So if the input inverter/generator or whatever cannot handle this it can cause an issue there indeed. Thats the reason the input voltage window needs to be set with care

39

#3 Not necessarily a generator, any network in which injection of current into the network is not permitted. → at most “online” electronics the energy is not always positive. Due to voltage fluctuations (at UPS systems) , energy generation (from a motor ) or any…the current can be momentarily be+/-. it’s not that energy in actual watts are supplied into the grid, in spike like this there is hardy energy. Its a millisecond behavior.

40

But it’s not just when the limit is reached that you don’t know how much extra power will be required! Before reaching the input current limit, you already don’t know! If you have to assume that high power may be required quickly, and if the inverter may not be able to supply this power in time and therefore needs to step up, by the time the current reaches the limit it’s already too late! It needs to do this a little earlier! And I can’t understand why it should assume the entire load current! In fact, more than the entire load current, if the Boost Factor is greater than 1! Why not assume only what is necessary to keep the input current below the input current limit, within a safety margin? Wouldn’t it make more sense for this safety margin to be configurable? For the configuration to establish an amount of current below the current limit, or a percentage of the current limit, from which Power Assist will be activated?

41

So the input voltage limit setting serves two different purposes? It serves not only to establish the voltage limit that will be accepted at the input, but also to establish the voltage limit that the inverter will apply to the input? Wouldn’t it be better to have two separate settings? I may want to accept up to 270V at the input, but I may want the inverter to not apply more than 230V at the input, for example! In order for me to continue accepting input voltage up to 270V, but limit the inverter so that it does not inject current into the grid, wouldn’t it be better to set the Boost Factor to 1, or less?

42

That would not be working
When setting a is not injecting above 240vac , but the voltage is 241 Vac from the grid , then powerassist doesn’t work anymore