Long Cable Run to Inverter

j-bear

New member
Hey Everyone,

Haven't been able to find exactly what I'm looking for on here so creating a new post. I have a 2000W PSW Renogy Inverter to power my induction cooktop. When I built out the van, I put my house batteries under the hood, which causes for an approximate 9 ft wire run (18 ft roundtrip) to my inverter which is located under the driver seat. I am using 1/0 gauge wires for this run which I incorrectly thought was sufficient. Now, I'm thinking I need 4/0 gauge wire for this run, but before spending a decent amount on that I wanted to double check. I wish I could move the inverter closer but haven't been able to find a solution for that.

Here is the troubleshooting I did: Check battery voltage (12.9-13V) then turn on inverter. Battery voltage stays constant.
Then turn on induction stove and set to 700W. Voltage drops to 11.6V.
Then turn induction stove up to 1000W. Voltage drops to 11.3 and inverter starts beeping (assumed to be the low voltage alarm).
If induction stove is turned up to 1200W, the inverter shuts down (assuming low voltage protection).
Once stove is off, the battery goes back up to 12.9V.
So I think my issue is the wire size causing too much resistance and lowering the voltage.

So my question is would setting this up with 4/0 gauge wire solve my problem of low voltage? Thank you.
 
D

Deleted member 86082

Guest
Hey Everyone,

Haven't been able to find exactly what I'm looking for on here so creating a new post. I have a 2000W PSW Renogy Inverter to power my induction cooktop. When I built out the van, I put my house batteries under the hood, which causes for an approximate 9 ft wire run (18 ft roundtrip) to my inverter which is located under the driver seat. I am using 1/0 gauge wires for this run which I incorrectly thought was sufficient. Now, I'm thinking I need 4/0 gauge wire for this run, but before spending a decent amount on that I wanted to double check. I wish I could move the inverter closer but haven't been able to find a solution for that.

Here is the troubleshooting I did: Check battery voltage (12.9-13V) then turn on inverter. Battery voltage stays constant.
Then turn on induction stove and set to 700W. Voltage drops to 11.6V.
Then turn induction stove up to 1000W. Voltage drops to 11.3 and inverter starts beeping (assumed to be the low voltage alarm).
If induction stove is turned up to 1200W, the inverter shuts down (assuming low voltage protection).
Once stove is off, the battery goes back up to 12.9V.
So I think my issue is the wire size causing too much resistance and lowering the voltage.

So my question is would setting this up with 4/0 gauge wire solve my problem of low voltage? Thank you.
The wire size you have is sufficient for the current. What type of battery do you have? The problem may be with internal resistance in the battery.
 

OrioN

2008 2500 170" EXT
The wire size you have is sufficient for the current. What type of battery do you have? The problem may be with internal resistance in the battery.
Or, the size of the battery/bank is insufficient and/or the battery/bank has lost capacity.
 
Last edited:

marklg

Well-known member
Check the connections. Bad crimps can lose a lot of voltage. Check if the cable ends are getting hot.

That said, I would go with the biggest practical cable. You can't have too much voltage. I have 4/0 going from my inverter to a junction and four 1/0 cables to each battery. My batteries are not together.

What kind of batteries do you have and have you checked the voltage on each one? Have you checked if the grounds are coming up in voltage? That can happen too.

Regards,

Mark
 

j-bear

New member
I have (2) 6V Duracell Batteries wired in series with 220Ah capacity. I will check the cables for bad crimps. Seems like that could be an issue. I will also check each battery individually for correct voltage. Together they are sitting at 12.9-13V.
 

j-bear

New member
Alright so each battery is sitting at 6.6V. Combined since I checked this morning they are up at 13.2V in series. When I measure the voltage at the connections to the inverter they are at 12.7V. Is that a sign of bad crimps?
 

j-bear

New member
Check the connections. Bad crimps can lose a lot of voltage. Check if the cable ends are getting hot.

That said, I would go with the biggest practical cable. You can't have too much voltage. I have 4/0 going from my inverter to a junction and four 1/0 cables to each battery. My batteries are not together.

What kind of batteries do you have and have you checked the voltage on each one? Have you checked if the grounds are coming up in voltage? That can happen too.

Regards,

Mark
How do you go about checking if the grounds are coming up in voltage?
 

autostaretx

Erratic Member
So that 13.2 -to- 12.7 volt reading was with very little load, or with the cooktop running?
(if so, what current?). That's a half-volt drop, and if you know the current you can calculate the round-trip resistance.
V= A * R, so R = V / A ... "V" being the half-volt, "A" being the current.

For diagnosing high-current systems, a clamp-on DC amp meter (ammeter) really helps.

Clamp-clamping.jpg
--dick
 

marklg

Well-known member
How do you go about checking if the grounds are coming up in voltage?
Pick a ground point, maybe the ground at one battery, if it is connected to the chassis there. Call that - at Battery 1. Connect the minus lead of your DC voltmeter there and leave it there.

Take the plus lead of the voltmeter and touch it to every other point in the circuit. From what I gather from your description these would be:

+ at Battery 1
- at Battery 2
+ at Battery 2
+ at the inverter
- at the inverter.

Ideally when the plus lead of the meter is on - at the inverter the voltmeter should read zero, but if the ground is not good, you will read some voltage.

Ideally, the reading at + at Battery 2 should be the same as + at the inverter.

Ideally, the reading at + at Battery 1 should be the same as - at Battery 2.

I will bet none of the above will meet the ideal case, but which ones are farthest from ideal may point out the worst cable.

I hope this is explained well enough. A picture of your system would be helpful.

Regards,

Mark
 

j-bear

New member
Pick a ground point, maybe the ground at one battery, if it is connected to the chassis there. Call that - at Battery 1. Connect the minus lead of your DC voltmeter there and leave it there.

Take the plus lead of the voltmeter and touch it to every other point in the circuit. From what I gather from your description these would be:

+ at Battery 1
- at Battery 2
+ at Battery 2
+ at the inverter
- at the inverter.

Ideally when the plus lead of the meter is on - at the inverter the voltmeter should read zero, but if the ground is not good, you will read some voltage.

Ideally, the reading at + at Battery 2 should be the same as + at the inverter.

Ideally, the reading at + at Battery 1 should be the same as - at Battery 2.

I will bet none of the above will meet the ideal case, but which ones are farthest from ideal may point out the worst cable.

I hope this is explained well enough. A picture of your system would be helpful.

Regards,

Mark
Hey Mark, thanks for the detailed info. Tested those locations today and got the below readings:
(this is with no load)

Minus lead on - Battery 2

Plus lead on:
+ Battery 1 = 6.32V
- Battery 2 = -6.32V
+ Battery 2 = 0
+ Inverter = 6.32V
- Inverter = -6.31V

Not sure how there would be any reading at + of Battery 2 would get a reading when it is connected in series to the - of Battery 1 so the reading between those two points should be zero.

Not sure if this tells me my crimps are good or if this shouldve been tried under a load?
 

j-bear

New member
Also, measured 12.63V at the batteries and 12.63V at the inverter this morning so that .5V drop is gone. Now I'm really confused.
 

marklg

Well-known member
Hey Mark, thanks for the detailed info. Tested those locations today and got the below readings:
(this is with no load)

Minus lead on - Battery 2

Plus lead on:
+ Battery 1 = 6.32V
- Battery 2 = -6.32V
+ Battery 2 = 0
+ Inverter = 6.32V
- Inverter = -6.31V

Not sure how there would be any reading at + of Battery 2 would get a reading when it is connected in series to the - of Battery 1 so the reading between those two points should be zero.

Not sure if this tells me my crimps are good or if this shouldve been tried under a load?
You must have put the minus lead of the DVM on the + at Battery 2 to get those readings and they are as expected with no load.

Here is a picture that might help. This is what I am thinking you have.

Bats_Inverter.png

Could you repeat with the minus lead of the DVM on

- Battery 2

and touch the plus lead of the DVM to all these points:
+ of Battery 1
- of Battery 1
+ of Battery 2
+ of the inverter
- of the inverter.

in both the no load and load conditions? Under loaded conditions this should show which connections or the batteries are bad.

Regards,

Mark
 

j-bear

New member
You must have put the minus lead of the DVM on the + at Battery 2 to get those readings and they are as expected with no load.

Here is a picture that might help. This is what I am thinking you have.

View attachment 146432

Could you repeat with the minus lead of the DVM on

- Battery 2

and touch the plus lead of the DVM to all these points:
+ of Battery 1
- of Battery 1
+ of Battery 2
+ of the inverter
- of the inverter.

in both the no load and load conditions? Under loaded conditions this should show which connections or the batteries are bad.

Regards,

Mark
Hey thanks Mark.

Based on your diagram, when i took those first measurements I had the minus lead of the DVM on the - of battery 1.

New measurements are as follows with the minus lead of the DVM on the - of battery 2.
+ Battery 1 = 12.65V
- Battery 1 = 6.32V
+ Battery 2 = 6.32V
+ Inverter = 12.65V
- Inverter = 0

Then I got the following measurements with a load of 700W on the inverter
+ Battery 1 = 11.42
- Battery 1 = 5.83
+ Battery 2 = 5.88
+ Inverter = 11.11
- Inverter = 0.13
 

marklg

Well-known member
Hey thanks Mark.

Based on your diagram, when i took those first measurements I had the minus lead of the DVM on the - of battery 1.

New measurements are as follows with the minus lead of the DVM on the - of battery 2.
+ Battery 1 = 12.65V
- Battery 1 = 6.32V
+ Battery 2 = 6.32V
+ Inverter = 12.65V
- Inverter = 0

Then I got the following measurements with a load of 700W on the inverter
+ Battery 1 = 11.42
- Battery 1 = 5.83
+ Battery 2 = 5.88
+ Inverter = 11.11
- Inverter = 0.13
Your batteries look suspect. Across the batteries(+ Bat 1 to - Bat 2) it goes down to 11.42 V from 12.65. a loss of 1.23V from no load to load. Across the Inverter it is losing 11.42-11.11=0.31 V on the + side wiring and .13 V on the - side wiring. So, your ground is better than the + wiring.

So, loss from no load to load:

Batteries 1.23V of which only .05 V is the wire between the batteries.
Ground wiring 0.13V
+ side wiring 0.31V

Sure looks like your batteries are the biggest culprit. The + side wiring could be better too if you want to use all of your 2KW inverter, but those batteries won't cut it to begin with. Exactly what model number are the batteries and how old are they?

If you are limited, replace the batteries first, then the plus side wiring to the inverter next. I think others will have to pipe up on what the best lead acid batteries are. I've got LiFePO4s. They are more expensive but they last way longer and keep up a higher voltage.

I aim for a couple tenths of a volt loss max in any single item. If you wanted to use all 2 kW, the losses above would triple.

My system is way over designed, but I can pull 2.5 KW and all is fine. Sometimes it is worth it for the no hassle factor. I can toast my toast and the wife can run the microwave or Keurig at the same time.

Regards,

Mark
 

autostaretx

Erratic Member
That really is an excessive drop on your batteries ... i only have one 67 AH 12v battery, and it doesn't drop that far with a 700 watt load.
Your batteries may have sulphated plates, reducing their current capacity.
+ Battery 1 = 11.42
- Battery 1 = 5.83
+ Battery 2 = 5.88
+ Inverter = 11.11
- Inverter = 0.13
OK ... you've got 0.13 voltage drop on the negative feed, and 0.31 volt drop on the positive, a total of 0.44 volts.
Plus you're losing 0.05 v between the two batts.

Are you using the Sprinter's frame as your negative feed to the inverter?
If not, the positive feed is losing three times more energy than your negative.

IS that 700 watt load a "700 watt" microwave? If so, it's drawing more.
But let's use "700 w" in this calculation: P = V*A, 700 = 12 *A, A= 700/12 = 58.3 amps being drawn.
Your 0.44+0.05 voltage drop is stealing 0.49 * 58 = 28 watts (not tooo bad).

The real problem is that your battery bank is not providing the current you need.
(but they do appear to be "balanced", so it's not a case of one spectacularly bad battery)

--dick
 

j-bear

New member
Your batteries look suspect. Across the batteries(+ Bat 1 to - Bat 2) it goes down to 11.42 V from 12.65. a loss of 1.23V from no load to load. Across the Inverter it is losing 11.42-11.11=0.31 V on the + side wiring and .13 V on the - side wiring. So, your ground is better than the + wiring.

So, loss from no load to load:

Batteries 1.23V of which only .05 V is the wire between the batteries.
Ground wiring 0.13V
+ side wiring 0.31V

Sure looks like your batteries are the biggest culprit. The + side wiring could be better too if you want to use all of your 2KW inverter, but those batteries won't cut it to begin with. Exactly what model number are the batteries and how old are they?

If you are limited, replace the batteries first, then the plus side wiring to the inverter next. I think others will have to pipe up on what the best lead acid batteries are. I've got LiFePO4s. They are more expensive but they last way longer and keep up a higher voltage.

I aim for a couple tenths of a volt loss max in any single item. If you wanted to use all 2 kW, the losses above would triple.

My system is way over designed, but I can pull 2.5 KW and all is fine. Sometimes it is worth it for the no hassle factor. I can toast my toast and the wife can run the microwave or Keurig at the same time.

Regards,

Mark
Hi Mark,

Thanks for the input!

The batteries are Duracell GC2 golf cart batteries. They are about 2 years old. Below is the link to them and I attached a picture as well before I mounted them under the hood.
IMG_4006.JPG
Supposedly, they have a 215 AH rating which would have been plenty for what I was planning on using them for but I recently decided to add in the inverter and induction stove due to concerns with propane.

I also ran another test on my inverter by using the cables that came with the inverter (double 4 gauge cables 3ft. long) and had a similar issue. I was able to get the induction stove up to 1000W with the batteries dropping to 10.6V and then when I go all the way up to 1400W is when the inverter shuts down for under voltage protection. So a bit of an increase from the 700W limit but I think that gives me my answer, that the batteries are the main issue here.

Looks like I will be using a portable propane stove for now until my batteries actually die (which may be sooner than later since I probably did some harm to them with all the inverter testing) and then upgrade my battery system, possibly to LiFePO4s but need to look into what I can store under the hood of the van.

Cheers,

Jacob
 

j-bear

New member
That really is an excessive drop on your batteries ... i only have one 67 AH 12v battery, and it doesn't drop that far with a 700 watt load.
Your batteries may have sulphated plates, reducing their current capacity.
+ Battery 1 = 11.42
- Battery 1 = 5.83
+ Battery 2 = 5.88
+ Inverter = 11.11
- Inverter = 0.13
OK ... you've got 0.13 voltage drop on the negative feed, and 0.31 volt drop on the positive, a total of 0.44 volts.
Plus you're losing 0.05 v between the two batts.

Are you using the Sprinter's frame as your negative feed to the inverter?
If not, the positive feed is losing three times more energy than your negative.

IS that 700 watt load a "700 watt" microwave? If so, it's drawing more.
But let's use "700 w" in this calculation: P = V*A, 700 = 12 *A, A= 700/12 = 58.3 amps being drawn.
Your 0.44+0.05 voltage drop is stealing 0.49 * 58 = 28 watts (not tooo bad).

The real problem is that your battery bank is not providing the current you need.
(but they do appear to be "balanced", so it's not a case of one spectacularly bad battery)

--dick
Hi Dick,

yeah seems to be an excessive drop. I am running a 1/0 gauge wire for the negative feed straight to the battery for the inverter. The inverter is also separately grounded to the frame of the sprinter with an 8 gauge wire (10 gauge recommended) and that didn't seem to be an issue although I can explore testing it further.

The 700 Watt Load is from an induction stove top. It controls the temperature settings by adjusting the wattage.

Thanks,

Jacob
 

Top Bottom