DC BATTERY STRING AMPERAGE

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ETEL64

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Senior Electrical Designer
Hello all, first time on the forum. Please be gentle.:oops:

Here is what I have:

(84) 13V, 80AH batteries in series. So I calculate that as 1092V and to 80AH stays the same.

Here is the question:

What is the amperage out of this string in order for me to size the cables to the load?

I am not looking for the answer but how do I get the answer.

Thank you
 
You will have several factors:

1 the current drawn by the load. Say you have a 100kW load, and a minimum string voltage of 950V, then you will need to be able to carry a load current of 105A

2 voltage drop in the connections of the battery system

3 short circuit protection and cable sizing to match the OCPD used for short circuit protection

Keep in mind that the battery itself has a 'nominal' voltage, but the actual operating voltage will change depending upon the state of charge and load conditions.

-Jon
 
ETEL64 said:
(84) 13V, 80AH batteries in series. So I calculate that as 1092V and to 80AH stays the same.

Here is the question:

What is the amperage out of this string in order for me to size the cables to the load?
Click to expand...
As @winnie said, the load determines the current. Be SURE you have suitable short circuit protection. Fuses for 1092V DC are not your ordinary fuse. And be sure your wire is rated for the voltage; standard 1000V cable normally used for AC would seem to be ok since the peak voltage of sinusoidal AC is 1.414 times the RMS, but verify suitability with an ENGINEER, not salesman, of the manufacturer.

What he didn't mention is that you need some balancing system to maintain that string. Just passing a current through to charge won't EXACTLY maintain equal batteries, and it will worsen over time. The system designer must consider and plan for this.
 
I DID NOT WRITE THIS. Found while looking on another forum while also wondering about your question and how that works. Below is what I’ve found I believe to be some help.



If you have a 12V battery and you're asking how much amperage can it kick out, the answer is however much or little it has to to satisfy Ohm's law, V = IR. The less resistance you have in a circuit, the more current will flow and vice versa. The absolute extreme of this would be if you had zero resistance (an ideal short circuit), then the poor battery would try to crank out infinite current to maintain the relationship. That means kaboom.

Of course, there will always be some resistance in the real world so your battery will probably only have to try to crank out thousands of amps - still kaboom.

To answer your question: How many amps a battery supplies depends entirely on the voltage of the battery and the resistance in the circuit. It is not a fixed value for any one battery or class of batteries. Even the resistance of the circuit is not necessarily a fixed value, it would depend on factors like the level of corrosion in the terminals and the temperature of the conducting wires. If you want a ballpark of how much current your battery sometimes supplies, check the cold crank amperage rating.
 
Dsg319 said:
If you have a 12V battery and you're asking how much amperage can it kick out, the answer is however much or little it has to to satisfy Ohm's law, V = IR. The less resistance you have in a circuit, the more current will flow and vice versa. The absolute extreme of this would be if you had zero resistance (an ideal short circuit), then the poor battery would try to crank out infinite current to maintain the relationship. That means kaboom.

Of course, there will always be some resistance in the real world so your battery will probably only have to try to crank out thousands of amps - still kaboom.
Click to expand...

In addition to the resistance of the external circuit, the battery itself has its 'internal resistance'. For lead-acid batteries this can be quite low.

As 72 Ah 12V 'Optima' brand AGM Lead-Acid battery has an internal resistance of 0.003 ohm https://www.optimabatteries.com/products/yellowtop-h6

Short circuit current of 4000A and power dissipation into the short of 48kW (not into the external circuit because the approximation here is no resistance in the external circuit, so that 48kW is going into heating up the battery itself).

-Jon
 
So your fuse needs a high enough interrupt current rating.

My Honda Civic manual had a locked rotor starter motor spec of 400A at 4V.
12v - 4v = 8v
8v/400A = 0.02 ohms but this depends on the battery, the current draw and the temperature.
 
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in stand alone system usually I select wire sizes based on charge controller max output and total system voltage that charge controller can handle, your system seems have comparatively less current but way to much higher voltage. the charge controller usually limits the amount of charge and discharge, So you have to size the wires based on maximum discharge current set point in charge controller.
 
How many batteries in series are you allowed to connect per the battery manufacturer? The differential voltage across each terminal on each battery is only 12v, but the voltage between any point in the system can be up to 1/2 or full system voltage to the ground if there is a one side fault anywhere in the system depending on if the system is grounded or not. The maximum of 1/2 the system voltage would be if it is center grounded and setup as a +/- system.

Suppose the minus side faults to the ground. Now the battery on the other has a potential of 1kV to ground your typical battery case is not designed to have 1kV applied between its guts and the outside world. Insulation rating is usually not specified, but that doesn't mean unlimited.
 
ETEL64 said:
Hello all, first time on the forum. Please be gentle.:oops:

Here is what I have:

(84) 13V, 80AH batteries in series. So I calculate that as 1092V and to 80AH stays the same.

Here is the question:

What is the amperage out of this string in order for me to size the cables to the load?

I am not looking for the answer but how do I get the answer.

Thank you
Click to expand...
Batteries push as hard as possible a circuit (use all their Volts). How quickly a circuit goes, however, depends on its strength. A high strength circuit doesn't work as quickly (i.e., it uses less Amps).
With a battery of the same power (Voltage), but a more or less total energy (mAh), the duration of a device changes. There are no risks or other effects on its behavior.
With a different battery (voltage) you can calculate solar panel battery, how a circuit behaves can be altered and how the device could be damaged.
We probably assume that it has a fairly simple DC motor circuit in it, like your daughter's automobile. How fast a battery tries to turn on a dc engine depends on the strength (Voltage). You will twice the voltage (as you have noticed) - you also twice the battery's amps. Amps are drawn. This way the 12V would run out two times faster (due to faster power output (greater amps)), if your 6V and 12V batteries both had the same mAh.
Thus, the car's more fun, not so long...
However, there is also a risk that faster running with more batteries means more power is consumed and heated up,
 
Two cautions:

The DC current a fuse or breaker can safely interrupt is radically different -- and less -- than AC.
When a fuse starts to blow, an arc forms inside. With AC, the arc current goes to zero 120 times per second and the arc needs to be re-ignited. With DC, the arc is steady and current continues to flow until something else happens. If the fuse is being used within its normal voltage range, the arc will go out after the fuse element is fully melted, (or the circuit-breaker contacts are fully opened) but if the voltage exceeds the rating, the arc will persist, perhaps indefinitely. Yes, it is possible to blow a fuse and not shut off the current to the load.
That's why you see fuses rated for, for example, "125 volts AC/32 volts DC". Often they won't have a DC rating at all -- that does NOT mean that they can safely be used on a DC circuit at the same voltage as their AC rating.

The amount of current a battery can deliver is temperature dependent, and the most-prominent specification is often "Cold-Cranking Amps", which is the amount of current a battery can sustain for 30 s at -18℃ without the terminal voltage dropping below 7.2 volts. Into a bolted short circuit, expect double that much current, and at room temperature, double again.
 
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