Volatage drop

[RobbieD]

Hi gurus,

I have a twin 130AH lithium set up in my car. So 260AH. The batteries are for start up and auxiliary components. The BMS automatically cuts power use to ensure the car is always able to be started. Anyway….I’m running 53mm2 cable 6m from front to rear. Same cable pos and neg, Victron shunt near batteries, with a victron 220a BP at the rear. My 3000w inverter is connected straight to battery side of the BP in the rear cargo area. Upright dometic fridge and various battery chargers are plugged in full time.

The batteries /system is cabable of draining at 200AH. All the online charts I used state a 1v drop at 200A across that cable over the 6m. The inverter cuts out if volts drop to 10.5 v.

The issue I’m having is, I’m getting a 2v drop from batteries to the inverter at anything above 150amp draw. Cable should be sufficient based on charts although I am contemplating going to a 70mm2. But I’m not 💯 convinced this is the issue. So the inverter shuts down, hitting its 10.5v threshold, but battery Bluetooth and victron shunt confirm batteries are down to 12.6v at the 150a draw. So in my head the batteries have plenty more to give but the voltage drop is killing it…..

This makes my induction cooking and onboard coffee machine die…. So frustrating. 1st world problems I know, but there has to be a solution.

Thanks in advance
Robert

 

[Mattb4]

I would suspect the issue is likely a poor connection somewhere between battery terminals and inverter terminals. Could be at your Victron shunt.

 

[RCinFLA]

53mm2 wire is about #1 gauge cable. Two 6 meter (19.7 ft on pos & neg) lines will have about a 1 volt drop at 200 amps.

You also have to add lugs, terminal compression connections, BMS series resistance, current shunt, and battery impedance.

1 volt drop for a 12v inverter is too much. For a 12v inverter you cannot afford much voltage drop. This is the problem with running a 12v inverter greater than about 1000-1500 watts in size. It is very difficult to deal with 275 amps of battery current draw for a 12vdc, 3 kVA inverter.

As the inverter DC input voltage drops, it draws more current to maintain AC output load. Inverter's power conversion efficiency also drops as its DC input voltage drops causing even more current to be drawn.

You also likely have poor terminal connections. Using cheap thin tubing wall gauge wire lugs adds more loss.

4/0 cable will give about half a volt drop in cable.

At a minimum, you should double up on existing cables. Also tape the positive and negative lines together to reduce run length inductance. Long line cable inductance causes voltage ringing on inverter DC input that can damage inverter due to current surge AC loads turning off and on.

 

[Mattb4]

53mm2 wire is 1-0 AWG equivalent. (53.5mm2) I do agree that many 12v inverters (mobile) that are rated for high wattage often will not deliver.

So the problem could be the quality of your inverter.

 

[RobbieD]

53mm2 wire is about #1 gauge cable. Two 6 meter (19.7 ft on pos & neg) lines will have about a 1 volt drop at 200 amps.

You also have to add lugs, terminal compression connections, BMS series resistance, current shunt, and battery impedance.

1 volt drop for a 12v inverter is too much. For a 12v inverter you cannot afford much voltage drop. This is the problem with running a 12v inverter greater than about 1000-1500 watts in size. It is very difficult to deal with 275 amps of battery current draw for a 12vdc, 3 kVA inverter.

As the inverter DC input voltage drops, it draws more current to maintain AC output load. Inverter's power conversion efficiency also drops as its DC input voltage drops causing even more current to be drawn.

You also likely have poor terminal connections. Using cheap thin tubing wall gauge wire lugs adds more loss.

Thanks to you and Matt,

are you saying that a 3kva inverter will draw 275A? If so I am definitely well below At 200A system supply. Getting nice big gauge lugs to suit the wire was also an issue at the time so I will look at up grading them also.

Would I be better swapping to the 70mm2 cable or running another of the 53mm2 to at least maximise the system. Or dropping back to a 2000w inverter, which will have to be at max to run the appliances when camping.

thanks for taking the time to help.
Robert

 

[RobbieD]

53mm2 wire is 1-0 AWG equivalent. (53.5mm2) I do agree that many 12v inverters (mobile) that are rated for high wattage often will not deliver.

So the problem could be the quality of your inverter.

Is there a way to test the inverter to see if it is the problem?

 

[RCinFLA]

3 kVA output / 10.5v / 85% efficiency = 336 amps.

3 kVA output / 12.0v / 92% efficiency = 272 amps.

3 kVA output / 12.5v / 94% efficiency = 255 amps.

 

[Mattb4]

Is there a way to test the inverter to see if it is the problem?

What inverter do you have?

 

[RobbieD]

3 kVA output / 10.5v / 85% efficiency = 336 amps.

3 kVA output / 12.0v / 92% efficiency = 272 amps.

3 kVA output / 12.5v / 94% efficiency = 255 amps.

Hmm I think it may not have much to do with the cable set up then

 

[RobbieD]

What inverter do you have?

Itech world 3000w company here in Australia, likely Chinese made though. They do get pretty good reviews here.

 

[Mattb4]

To adequately test your inverter you would need to connect it directly to your batteries with a short length of cable sized for full amperage requirements of 3000w including the stated efficiency of 85% Ex. (3000w/.85 = 3530w, 3530w/12v = 295a) and than have a 3000w resistive load to try and start. You would monitor voltage droop during loading.

 

[RobbieD]

To adequately test your inverter you would need to connect it directly to your batteries with a short length of cable sized for full amperage requirements of 3000w including the stated efficiency of 85% Ex. (3000w/.85 = 3530w, 3530w/12v = 295a) and than have a 3000w resistive load to try and start. You would monitor voltage droop during loading.

Ok, I can definitely try that. Does this change at all if the max I will use from the inverter is 2200w?

 

[HRTKD]

The Victron Battery Protect device is not supposed to be used on the circuit to an inverter.

 

[RobbieD]

Correct, inverter connected straight to battery side of the BP.

 

[HRTKD]

Victron has explicitly said that the Battery Protect is not for use with an inverter.

 

[RobbieD]

Hmmm, they, as in the retailer here said no issue as long as connected to the battery in side it wouldn’t affect it. But. Now that you say that I scan easily disconnect from the BP and test. Cheers.

 

[HRTKD]

See the last line in the screen shot. When I see "inverter", I think "inverter/charger". If you have just an inverter without a charger, that may work.

 

[RobbieD]

See the last line in the screen shot. When I see "inverter", I think "inverter/charger". If you have just an inverter without a charger, that may work.
View attachment 150616

Yeah definitely just a pure sign wave inverter. I’m going to bypass it to test the system now regardless. Thanks

 

[RCinFLA]

Just a DC solid state single direction switch. Not back-to-back MOSFET's. If you push charging current backwards through it greater than about 10 amps it will overheat MOSFET's body diodes and likely burn out MOSFET's. You also get a diode voltage drop that will drop battery absorb and float voltages relative to charger settings.

Bidirectional solid-state switch requires back-to-back MOSFET's. This doubles the Rds_ON resistance so requires 4x the number of MOSFET's for equivalent series resistance of switch.

RU6020R MOSFET is 60v Vds breakdown voltage (not enough for greater than 24v system) and Rds_ON resistance warm of about 3 milliohm. Eight in parallel yields 0.375 milliohms net.

200 amp spec for unit x 0.375 millohms produces 0.11 v drop not include long MOSFET leads, PCB, and connectors. About 15 watts of heating on MOSFET's. 15-20 watts is about right for heat sink size.

If eight MOSFET's were arranged as two back-to-back MOSFET's in series, by 4 pairs in parallel, resistance would be 1.5 milliohms. At 200 amps x 1.5 milliohms would be 0.3 v drop not including long MOSFET leads PCB and connectors. About 60 watts of heating on MOSFET's. It would cook.

 

[RobbieD]

Just a DC solid state single direction switch. Not back-to-back MOSFET's. If you push charging current backwards through it greater than about 10 amps it will overheat MOSFET's body diodes and likely burn out MOSFET's. You also get a diode voltage drop that will drop battery absorb and float voltages relative to charger settings.

Bidirectional solid-state switch requires back-to-back MOSFET's. This doubles the Rds_ON resistance so requires 4x the number of MOSFET's for equivalent series resistance of switch.

RU6020R MOSFET is 60v Vds breakdown voltage (not enough for greater than 24v system) and Rds_ON resistance warm of about 3 milliohm. Eight in parallel yields 0.375 milliohms net.

200 amp spec for unit x 0.375 millohms produces 0.11 v drop not include long MOSFET leads, PCB, and connectors. About 15 watts of heating on MOSFET's. 15-20 watts is about right for heat sink size.

If eight MOSFET's were arranged as two back-to-back MOSFET's in series, by 4 pairs in parallel, resistance would be 1.5 milliohms. At 200 amps x 1.5 milliohms would be 0.3 v drop not including long MOSFET leads PCB and connectors. About 60 watts of heating on MOSFET's. It would cook.

Thank you so much for the detail. Most of it is over my head. However based on me not running a reverse charge through the BP, do you seen an issue with the inverter connected on the in/battery side of the BP?

thanks again.
rob