DC voltage drop

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Jon1980

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I'm looking to install a power inverter for my truck. I've been researching them and looking at their reviews. I know it needs to connect directly to the battery or battery's, needs to be fused. But what gets me is the size of wire required! If I wanted 2400 watt at 120 that would be 200 amps DC, which is way over kill. Just a reference, what I'm reading and what I know on AC doesn't compute. I know DC doesn't carry as far but I don't see why I would need to run 500 to my back seat?

what is the calculation on DC voltage drop?


Thanks!
 
Jon1980 said:
I'm looking to install a power inverter for my truck. I've been researching them and looking at their reviews. I know it needs to connect directly to the battery or battery's, needs to be fused. But what gets me is the size of wire required! If I wanted 2400 watt at 120 that would be 200 amps DC, which is way over kill. Just a reference, what I'm reading and what I know on AC doesn't compute. I know DC doesn't carry as far but I don't see why I would need to run 500 to my back seat?

what is the calculation on DC voltage drop?


Thanks!
Click to expand...

The voltage drop calculation for DC is exactly the same as for AC in situations where the inductance of the wire and skin effect changes in current distribution are not issues.
Two things you need to realize when doing the calculation for 12V DC in your car are:

1. The voltage drop in the wire is for only one direction if you are using the frame of the car as the return (ground) conductor. Most calculators for residential/commercial automatically figure in the the voltage drop for two wires of the length you specify IF you select single phase or one wire of the length you specify if you select three phase.
If there is an option for DC you need to find out whether the calculator you use does the one-way or two-way voltage drop.
2. A tolerably low voltage drop for 120AC (say 6 volts) is an absolutely unacceptable voltage drop when you are starting out with only a 12V supply.
This generally means that the wire needed for a given current for a 12V DC application will be much larger than for 120V AC.

One other result from item 2 is that it is much better to put the inverter as close as possible to the battery even if it means running a longer AC line to whatever the inverter is powering. Unfortunately between the heat and the lack of available space it is not always possible to put the inverter under the hood. But putting it in the trunk to keep it out of the way has big drawbacks.
 
Jon1980 said:
I'm looking to install a power inverter for my truck. .... 2400 watt at 120 ....
Do you really need 2400 Watts ?
If so, look at other options:
a. small genset in truck bed
b. engine mounted generator - for a 200A 12 V draw you will need a bigger alternator anyway plus need to leave the truck running when using it or you wont be able to start the truck if you run the battery too low.
c. mount an extra 12V golf cart battery back with the inverter and just run 4 AWG back to that assuming you have a 100 A alternator in the truck already.

d. other? .
Click to expand...
Click to expand...
 
[If I wanted 2400 watt at 120 that would be 200 amps DC,]

If you really wanted 2400W at 120v,
then yes you would need 200A at 12vdc.

no lies its maths !

How ever in reality many of the bigger invertors like 2400w
will use 24vdc input rather than 12vdc,
this then cuts that 200A in half.
 
Power loss in the wire goes up with the square of the current, and the current is inversely proportional to the voltage. So at 12V the power loss in the wire is 100 times what it would be at 120V.
 
Thanks for the replies.

I work in the oil and gas industry and often don't have access to power. We haul around box trailers with a gen and threading tools. For two guys to load it up in a 3/4 ton pickup just to fab up strut racks is quite a job. We have large 120/240v generators. Just need to run a drill motor, band saw or charge bateries. So 6 amps 120v, still looking around 83 amps for 1000 watt.



Sent from my SAMSUNG-SM-G530A using Tapatalk
 
Jon1980 said:
I know DC doesn't carry as far
Thanks!
Click to expand...

Only because the standard DC utilization voltages that are common in practice, are significantly lower than standard AC voltages.

Single phase two-wire AC at the same RMS voltage and RMS current as its DC counterpart, will be identical in distance performance.
 
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GoldDigger said:
The voltage drop calculation for DC is exactly the same as for AC in situations where the inductance of the wire and skin effect changes in current distribution are not issues.
Two things you need to realize when doing the calculation for 12V DC in your car are:

1. The voltage drop in the wire is for only one direction if you are using the frame of the car as the return (ground) conductor. Most calculators for residential/commercial automatically figure in the the voltage drop for two wires of the length you specify IF you select single phase or one wire of the length you specify if you select three phase.
If there is an option for DC you need to find out whether the calculator you use does the one-way or two-way voltage drop.
2. A tolerably low voltage drop for 120AC (say 6 volts) is an absolutely unacceptable voltage drop when you are starting out with only a 12V supply.
This generally means that the wire needed for a given current for a 12V DC application will be much larger than for 120V AC.

One other result from item 2 is that it is much better to put the inverter as close as possible to the battery even if it means running a longer AC line to whatever the inverter is powering. Unfortunately between the heat and the lack of available space it is not always possible to put the inverter under the hood. But putting it in the trunk to keep it out of the way has big drawbacks.
Click to expand...


Unless you own a Mazda Miata like me! :lol:
 
If the space is available, mounting it under the hood next to the battery is a good solution. The cables will be short and the manufacturer has already assured that the battery is in a relatively cool zone. (not applicable if the battery is inside a shroud or duct) Another option is not to mount it at all, but hook it up with Andersen Powerpole® connectors. ( http://www.andersonpower.com/us/en/products/powerpole/pp120.aspx ) When you're ready to use it, park the truck, open the hood, put the inverter on the fender and plug it in.

If you're going to use it more than intermittently, you'll want a fast-idle device. Automotive alternators don't deliver their full rated power until they're turning about 4000-5000 rev/min; only half or two-thirds is available at idle.

Don't use solid wire in a vehicle. It's not flexible enough to tolerate engine vibration.
 
Carultch said:
Only because the standard DC utilization voltages that are common in practice, are significantly lower than standard AC voltages.

Single phase two-wire AC at the same RMS voltage and RMS current as its DC counterpart, will be identical in distance performance.
Click to expand...

A better evaluation would be Vdrop and P delivered:
Vrms = Vdc
same load R
same line parameters L, R and C

DC has a slight but tangible advantage
 
Ingenieur said:
A better evaluation would be Vdrop and P delivered:
Vrms = Vdc
same load R
same line parameters L, R and C

DC has a slight but tangible advantage
Click to expand...

That's a good point too, as L and C affect AC but do not affect DC. Impedance due to L = 0 in DC, and impedance due to C = infinity. Whereas in AC, both impedances are functions of frequency.

If L and C are neglibible, as usually is the case for wire sizes small enough to be in the positive AWG scale, 2-wire AC is equivalent to 2-wire DC in its ohmic distance performance and copper utilization, given same voltage, same load, and same total wire resistance.
 
Besoeker said:
More nonsense.
RMS has no relevance to DC.
Click to expand...
Formally, the same definition which is used to determine RMS value of V or I applies perfectly well (although trivially) to DC.
As long as you do not define RMS in terms of an average over one cycle of a periodic waveform, which is not defined for DC.
The concept of RMS can be applied equally well to a non-periodic but known waveform.
 
GoldDigger said:
Formally, the same definition which is used to determine RMS value of V or I applies perfectly well (although trivially) to DC.
As long as you do not define RMS in terms of an average over one cycle of a periodic waveform, which is not defined for DC.
The concept of RMS can be applied equally well to a non-periodic but known waveform.
Click to expand...
With DC there is nothing to be squared and nothing to take the mean of.
 
Besoeker said:
More nonsense.
RMS has no relevance to DC.
Click to expand...

the evaluation should use equal values for ac rms and dc
eg use 100 vac rms and 100 vdc

you need to lose the boner you have for me
as usually you disect one small thing and ignore the gist of the entire post
all for confontation, ego and snark

will the rms of a rectified voltage give a dc value?
is that 'relevance' ?
 
Ingenieur said:
the evaluation should use equal values for ac rms and dc
eg use 100 vac rms and 100 vdc

you need to lose the boner you have for me
as usually you disect one small thing and ignore the gist of the entire post
all for confontation, ego and snark

will the rms of a rectified voltage give a dc value?
is that 'relevance' ?
Click to expand...
The DC value is mean, not RMS. I thought you might have known that.
 
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