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Do I need a larger battery monitor

MikeV

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Oct 16, 2021
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MY SETUP: I am upgrading from 2 lead acid batteries to 2, 100ah LiPO4 drop in batteries, each of the new batteries have a Max continuous discharge current of 100A. I plan to add a 2000w inverter to the system that will only run off a dedicated plug. Currently I have a cheap 100A battery monitoring system that is shunt based.

Question: Do I need to replace the 100A battery monitoring system? I will never go over 100A discharge unless I am using the inverter. What risk am I taking on off I leave it and how will it affect the accuracy of the device.

Thank you for any information.
 
Post a link to the shunt that you have. If the shunt itself can only handle 100A then that's real risky. Think of the shunt just like another piece of wire. Would you use a wire that could fail at 100A to wire your batteries to your inverter?

If you wire your 2 100Ah batteries in parallel for a 12V system then your combined batteries can handle 200A max continuous discharge which is needed to handle a full 2000W. You would also want a 250A fuse at the battery. So clearly a 100A shunt is very bad.

If you wire the batteries in series for 24V then your combined batteries are still 100A max continuous discharge but that is all that is need for 2000W at 24V. You would want a 125A fuse at the battery. So a 100A shunt is still bad since the fuse MUST always be the weakest link in the system.
 
Thank you both. The batteries will be in parallel, thus a 12v 200 Ah system.

I will upgrade the monitor to at least a 200A fuse but might as well go bigger. The problem I have read is that there is larger error on a system that is mismatched, ie if I install a 500A shunt but only pull 10A on average.

Thank you as well for the fuse information. I will ensure the fuses are installed where needed
 
I will upgrade the monitor to at least a 200A fuse
Is this a typo? Did you mean 200A shunt?

You need at least a 250A shunt. The shunt needs to handle more than the fuse.

BTW - Also make sure you are using at least 2/0AWG wire for the battery and inverter wiring.
 
Ideally you need a quality 350 or 500 amp shunt in the system. The low cost unit shunts tend to overheat at high currents, so if you use the existing shunt with the expected current in excess of its rating it will possibly melt.

The unit you have is matched to a 100 amp/75mV shunt so it is/may not be possible to operate with a larger rated shunt, as the calibration for current and power will be incorrect.
The options you have is to replace the DORK with the complete higher rated unit with the 350 amp shunt, or to connect the inverter power negative direct to the battery terminal, bypassing the 100 amp shunt. Of course this will result in power/current taken by the inverter to be disregarded by the monitor.

My advice would be to replace the monitor with a system rated for 350 or 500 amps. If cost is not a problem the Victron smart shunt , ( Bluetooth display on phone), or the Victron BMV 7xx series ( display and/or Bluetooth), have a 500 amp rated shunt . I think the units from DORK and similar, offer a 350 amp shunt option.

I have found running a 2000 watt inverter in a 12v system is not a problem with suitable batteries and adequate cables/fuses. A 24 volt system may be slightly more efficient but where there are 12v appliances/loads in addition to inverter powered items, a 12v system is easier and less costly to implement and the overall efficiency will be similar.

Mike
 
With a 2000W inverter you will draw 150 plus amperes from 12 volts system. My experience; I have a Victron Battery Monitor with a 500 ampere shunt. My largest draw is the 2000w inverter driving a 1000W microwave. All has been well these many years.
 
Just wondering where you got the 0.85? Ohm's law would be 2000/12 = 166.66666666...
either way the plug on an inverter is only good for 15amps AC that's why there are two plugs on a 2000w and three on a 3000w so unless you are using more than one plug you can't pull more than 1800w
 
Just wondering where you got the 0.85?
The 0.85 is to account for the inverter inefficiencies. For example, if you try to draw 100W from the inverter, the battery needs to provide 100W / 0.85 = 118W. The 85% might be a bit low for some inverters in real life but using that number will give a safe number resulting in better wire and fuses.

either way the plug on an inverter is only good for 15amps AC
The calculations I was showing was for the DC battery connections on an inverter, not the AC side. This whole thread is about the DC side.

And, FYI, not all inverters have plugs. Many inverters support much more than 15A. 3000W at 110VAC would be 27A, for example.
 
Happy New Years all I wanted to give you all an update.

I ended up purchasing a victron smart shunt 500A. So the issue with the charge controller is no more. Also purchased a victron distributor, fuses, and 2/0 wire. This should make the process clean but costly.

Along the way I ran into a new issue I need some help with. In my rig there is a battery solenoid that connects the house batteries with the engine battery. While I was tracing wires in an effort to relocate my house batteries I found the positive wire runs to the solenoid first, and then to my converter. Also there is a fuse box that has item that I believe run off the engine battery. I am concerned about connecting lithium to this solenoid.

QUESTION: should I bypass the solenoid all together? I do not want my alternator charging my New lithium batteries, and I carry a jumper pack if needed for the engine battery.

If I bypass this, what am I losing? Do these fuses get power from the house batteries?
 

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2000W / 12V / 0.85 = 196A
My exact measurements. 2000W psw inverter, full load of 1000W microwave, 152 ampere on Victron Monitor. Any surge is very small and very short.
My exact statement. "With a 2000W inverter you will draw 150 plus amperes from 12 volts system" Please note the key word "Plus".
The exact meaning of my post. Use a larger shunt. Victron 500 amp works very well for me.

My apologies to the OP for this rant not directed at you. I would remove the solenoid. Or at least the cables connecting it to the starter battery. Best to disconnect both cables completely. Disconnect both battery negative cables first before working on the solenoid cables. Reconnect the negatives last. One thing don't do, don't bypass the solenoid, disconnect it. The converter end probably connects to the battery connection in the converter. You have to trace wires to find what powers those fuses.
 
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