Inverter with Inbuilt LVD

[paulrogers6]

Dear All

I'm considering buying an inverter (Victron Energie Phoenix 12/1200) that contains an in-built Low Voltage Disconnect.

The supplier states that the cut of voltage is set at 9.6 volts, the reason being that during use, if the a large load is placed on the inverter the voltage from the batteries will drop and anthing higher than this will cause the inverter to keep switching on and off.

To me tho' this seems very low and would likely leave the batteries in a damaged state by the time it cut out.  I thought that LVD's were generally set to cut out at a much higher voltage than this and batteries should not be discharged to below 80% without damaging them.

Any thoughts?  Any one had problems with inverters cutting out as loads are applied to them?  What would happen to a 12v battery if it were discharged to 9.6v?

Thanks

Paul

[boB]

That 9.6V LVD level may be time dependent...  i.e., it may only let the battery Voltage drop that low for only a moment.  That would be too of voltage to let the battery stay that low for very long.   I would imagine that Victron knows this, but, maybe not ?

boB

[Flux]

I tend to agree, after all Victron Energy have been at the high end of this market for long enough to know what they are doing.

This battrey voltage/state of charge issue is a real nightmare. On full load with a big inverter you probably do get transients down to 9v with a battery near its discharge limit. On the other hand if running the inverter on near no load then 11.5v may be a more reasonable low discharge figure.

If you buy a big inverter then the manufacturer can only assume that you will use it at full load. Victron's full systems include an energy meter ( Ah meter) and the user would normally monitor these things. I don't consider a LVD to be a normal working tool but it nice to have something to protect the batteries from total discharge if something goes wrong. If you want it as a normal tool then you may have to use something with adjustable cut off voltage and set it to suit your load conditions. Set at something like 10.5v you will almost certainly get inverter tripping if well loaded with a low battery.

Flux

[paulrogers6]

boB/Flux

Many thanks for the comments.

Would the voltage drop only be percieved by the Inverter's LVD, eg does the load on the inverter cause it to "see" the voltage drop to 9v?

What I'm getting at is, if I were to use an external LVD (say Glen's unit connected to the on off switch of the inverter) would this still see the same 9v at the battery?

Example switching voltages for an LVD based on Glen's circuit are 12.2/12.8V.  This suggests they are not affected by the voltage drops (tho of course no mention is made of an inverter - which might be the driver behind the large drops).  Is this case for an external LVD/inverter? or even for an LVD connected to 12V loads?

[ghurd]

I am not familiar with Victron inverters.

Inverters tend to carry large loads through relatively undersized wires.

The wires cause a voltage drop, so the battery voltage is higher than the inverter sees.

The large loads can pull the battery voltage lower than it will be after the large load is removed (the battery voltage bounces back up a ways).

If the draw is low, both of those factors are nearly negated, and the inverter will run until the battery is 9V.

Certainly could use my circuit connected to the inverters switch.

My idea with the circuit as an LVD was for items not worth an expensive adjustable factory made LVD,

or as a pre-dump useful use of some soon to be dumped power (attic fan, small pond aeration, etc).

My circuit has separate wires to the battery terminals for sensing the actual battery voltage.

Power carrying wires must be separate from the voltage sensing wires or the inverter type inaccuracies are introduced.

G-

[boB]

Ghurd pretty much completed this one.

Sampling from the battery terminals should get rid of the voltage drop problem and get a better indication of battery health.  Still should have a time constant in there somewhere though.   Let it drop for semi-short periods, like 20 seconds or something like that before tripping the LVD.

Basically a large R-C-D-R time Constant circuit  (Diode & Resistor in series so that the C charges to the higher voltage faster when recovering) and will also respond to the average voltage rather than instantaneous voltage, etc.

i.e. Fast attack, slow decay, sort of.

Sorry, my brain is somewhat still in bed this morning.

boB

[Madscientist267]

Tending to agree with all of this with one addition -

I generally don't like automatic low end cutoff (particularly if I can't easily adjust the threshold) because of all the discrepancies that pop up under different circumstances. Discharging a battery with 60% SOC with heavy load and a light drain on a battery that is actually discharged can appear very similar... there's really no [practical] way to see the difference electronically, unless you want to develop some kind of crazy sensing array that measures all kinds of things and makes an 'intelligent' decision. It would be more expensive than it was worth IMHO.

As Flux pointed out, its good for a last resort if you're not available to intervene manually, but isn't really good for everyday limiting.

I prefer an alarm (with ability to mute of course) to let me know that terminal voltage is out of bounds, and that lets me decide if this is a normal condition for what is going on or not.

If I'm sucking 100A from a 200AH battery with a decent charge on it, I would still expect it to be pulling down a bit, and so it's not really an issue at that time, and I would mute the alarm.

If I'm pulling 1A from the same battery, and the alarm is going off, it's dead and I need to do something else for juice before I damage it.

If I continue to ignore the alarm status, and voltage dips below 9, then I should be considered an idiot (just as if I had not payed the electric bill), and power should be automatically disconnected until I resolve the problem.

Yes?

[paulrogers6]

Hey Guys

Thanks for the comments.

Manually checking the voltage is not an option, unfortunately as most of the time I'm away from the house.  It seems a shame to spend a lot of money on an Inverter and then add an LVD but there you go!!

On the subject of voltage drop, presumably if the LVD is a distance away from the Batteries (say 10') then the wires used to sense the voltage presumably need to be of a large enough gauge to ensure there is no voltage drop "seen" by the LVD?

I currently plan to have a busbar in the "control room".  The batteries will be connected to the bus bar, along with the Inverter. If the LVD sensor were connected to this (rather than the battery terminals) would it still "see" the battery terminal voltage, or would the inverter load cause the voltage at the busbar to drop in the same way as the inbuilt LVD problem?

Thanks

Paul

[ghurd]

I can not speak for other LVD circuits, but mine has a very minor change in voltage drop in the sensor wires between connected and disconnected.  AWG #22 and 10' is no problem.

Also, with mine, the power circuits can be very far from the sensor circuit, and the 2 halves only need connected with a small single wire.  But mine does not have any delay to speak of (a bit of delay and it can be increased considerably, but nothing like a whole second, let alone 20 seconds).

The inverter will cause drop in the buss bar.  How much depends on how heavy the buss bar to battery wires are.  Make the battery cables short and heavy and I doubt it will cause a problem with any LVD using separate sensor wires.

It shouldn't be difficult to connect the sensor wires directly to the battery terminals.

G-

[paulrogers6]

Glen

Many thanks for the reply.  I'll run a couple of wires from the LVD directly to the battery terminals.

Thanks for all your help ..... again.

Regards

Paul