How many batteries of 12 V x 125 Ah needed to continuously run a load of 1600 Watts using an inverter of 24 VDC x 120 VAC that uses a battery bank as primary source?

  1. For a load of 1600 W , at least an inverter size of 2400 W is needed. The existing inverter output is 3000 W, it will be able to run the load at near 55%. The question is to arrange a battery bank system so that the load can be run without depleting the batteries with a runtime of at least 10 hr.

  2. The batteries of 12 V x 125 Ah need to be in parallel and in series in order to provide the 24 DC source to the inverter. There are currently some 12 V x 125 Ah and more may be needed. We need a way to compute how many more batteries of 12 V x 125 Ah to arrange in order to satisfy the inverter input of 24 DC and the minimal runtime of 10 hr for the loads.

After some computations, I found that the minimal needed for the load for 10 hr will be around 1300 Ah that will give about 10 batteries of 12 V x 125 Ah

  1. What needs to be taking care is the remaining 14 hr runtime, an extra load of 1600 W that may be needed later on and how the batteries will be charged. Based on the specs , it’s an inverter/charger that can be auto switch. To avoid discharge of the batteries the inverter can be set to autocharge the batteries from a generator or solar panels for the remaining 14 hrs

If in need to add another load of 1600 W, since the first one works at 55% adding extra batteries in the bank may help in order to avoid overload of the inverter. This part of the problem is to find out how many batteries can be added to the system without risk.

The inverter Specs
Output Specifications:

•   Continuous Output Power: 3000 Watts
•   Surge Rating: 9000 Watts (20 Seconds)
•   Output Waveform: Pure Sine/Same as input (Bypass Mode)
•   Nominal Efficiency: >88% (Peak)

Input Specifications

•   Nominal Input Voltage: 24.0 VDC
•   Minimum Start Voltage: 20.0 VDC
•   Low Battery Alarm: 21.0 Vdc-22.0 VDC

Batteries Specs

•   Nominal Voltage : 12 V
•   20 Hr Capacity : 125 Ah
•   Charging Current : 8A-35 A
•   Charging Voltage : 14.4-14.9 V
•   Float Voltage : 13.5V-13.8 V

Bitcoin miner load specifications

•   Hash Rate: 11.5 TH/s. ...
•   Power Consumption: 1450 W +7% at the wall, with Bitmain's APW3 PSU, 93% efficiency)
•   Power Efficiency: 0.126 J/GH + 7% at the wall, with Bitmain's APW3 PSU, 93% efficiency)
•   Rated Voltage: 11.60 ~ 13.00 V.
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    \$\begingroup\$ I don't see how 3000W with 3 AGM batteries can be an option if the supply is 24 VDC. Correct your specs. \$\endgroup\$ – Jack Creasey Sep 16 '18 at 0:51
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    \$\begingroup\$ You miss my point, 3 *125AH cannot be arranged to provide 24V @ 125AH. You would need a minimum of 4 batteries ...2S2P configuration for a total of 24V @125AH. Nowhere do you say how long you want the convertor to run. Your batteries are specified at 20Hr discharge rate, so recommended current is about 6A. If you want to run at higher current, then use Peukerts Law to work it out: en.wikipedia.org/wiki/Peukert%27s_law \$\endgroup\$ – Jack Creasey Sep 16 '18 at 4:54
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    \$\begingroup\$ The issue of 3 batteries was explained to you here : electronics.stackexchange.com/q/396176/152903 \$\endgroup\$ – Solar Mike Sep 16 '18 at 7:26
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    \$\begingroup\$ And you keep repeating that question over and over without understanding the basics or listen to reason/Solar Mike. \$\endgroup\$ – winny Sep 16 '18 at 20:48
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    \$\begingroup\$ The load is a bitcoin miner in the previous question, so it likely runs at a very high portion of its maximum wattage. Clara, this is an improvement on your previous question, but you're missing some of the most important data. You can see that the inverter has >88% peak efficiency, which means that it's efficiency will be >88% when run at optimal load (most likely close to 3000W). We need an efficiency graph to figure out your actual power consumption at 1600W output. If you already own the miner, it would help if you grab a voltmeter and measure it's actual power use. \$\endgroup\$ – K H Sep 17 '18 at 1:36

Minimum of 4 Batteries. 2 in Series to give you 24V connected in parallel to another 2 in series. 12 x 125 A = 1500 W per battery for 1 hour of operation at inverter efficiency of 88% would give you 1320 Watts per hour final. So two connected in series would give you 24 V Nominal and 1320 Watts for 1 hour. Another set of two in parallel would give you a total of 24V nominal and 2640 Watts for 1 hour.

Your load is a continuous 1600 Watts. So in 1.65 hrs your batteries will be depleted. Most of us know batteries do not last till 0 %. If you want to maintain the batteries in good working condition 50% of battery depletion is recommended. Given the above scenario your 4 battery capacity to support the load is 1.65/2 = 0.8 hrs.

Please note the surge of your load is not mentioned. So you also need to account for it in addition.

My recommendation would be using 6 V batteries with higher AH rating. ( 235 HA or 435 AH ) ( 6V x 4 ) + (6V x4)

Later if you double the load with same inverter you need to resize the battery bank to support the load and making sure your inverter can handle the continuous/ surge load requirements of your load.

Ask further if you need clarification.

Sample system 48V. 6x 435 AH deep cycle lead acid batteries (20.88 kWh Total battery capacity) 48V Inverter charger 24 V 235 W solar panels ( 15 total, 3 in series x 5 sets)

The above system was used on a household with a typical draw of 10kWh per day and with typical charging of battery bank from Solar panels only. 4-6 hours of Sunlight is enough to charge the batteries from total depletion to full.

Most starters including myself have made the mistake of powering heavy load with 12V deep cycle batteries. If someone intend to operate heavy loads best deep cycle batteries would be 2V or 6 V. Reason is simple, these batteries has the volume and thick plates to sustain the load for longer time duration with less heat and chemical reaction.

Always remember anytime an inverter is used power is reduced by at least 10%. (published manufacturer data is always higher than actual data) Please pay attention to proper battery ventilation, if Lead Acid batteries.

Safety is paramount when building a system. Proper circuit protection both incoming and outgoing power is a must. So please have your system inspected by a qualified professional before operation.

Finally, do not be discouraged by technical information. We are all humans, and we learn well by making mistakes and learning form our mistakes.

  • \$\begingroup\$ I understand your point however as per the question I need to use in the design a battery bank constituted of 12v x 125AH since 3 batteries have already been requested. Need to purchase more in order to run the 1600W load continuously. \$\endgroup\$ – clarajune Sep 16 '18 at 3:04
  • \$\begingroup\$ Since you have mentioned your primary power source is battery, I strongly urge you to build a battery bank system that you can rely upon and economical in the long run. But if you are unable to change battery type, you need to buy another 12V 125 AH battery to complete your battery bank. Keep in mind your total usable time is only around 1 hour with 4 total in real life scenario. \$\endgroup\$ – Omegaone CA Sep 16 '18 at 4:36
  • \$\begingroup\$ In order to run the load for more than 1 hr I need to consider more batteries in my bank. I have already 3 x 12v x 125AH . How many more batteries to add on my bank so that batteries won’t get depleted ? \$\endgroup\$ – clarajune Sep 16 '18 at 13:54
  • \$\begingroup\$ First, you need to describe the load and average usage over 24 hr period. Second how you plan to recharge those battery bank and the time to charge. If you keep AH at 125 per battery, each time you add 4 more batteries you are adding 0.8 hrs to the total capacity at 50 % depletion rate. ( means you can charge the bank back to 100% from 50%) As you can see, adding 4 batteries at this rate is not ideal and costly and every 4 more batteries only get you 0.8 hrs of operating time. \$\endgroup\$ – Omegaone CA Sep 16 '18 at 14:07
  • \$\begingroup\$ By the way, my answers are based on reality and experience as a former IPP ( Independent Power Producer) for grid connected Net-Metering system. As I concentrate on building with efficiency in mind. Perhaps others can give you better suggestions, but if you find these information are useful please rate it appropriately as I am new here. \$\endgroup\$ – Omegaone CA Sep 16 '18 at 14:18
How many batteries of 12v x 125AH needed to continuously run a load of 1600W using an inverter of 24vdc x 120vac

First, you need to pay more attention to unit labels. AH stands for Ampere-Henries, which is not a relevant unit for batteries. On the other hand, you apparently mean "12 V", not "12v". Note the upper case V for volts, and the space between the number and its units. These things matter. NIST has a comprehensive document on all this.

In any case, you obviously need at least 2 batteries. That comes from requiring 24 V with 12 V batteries. You need two in series to get that.

The other criteria that can be derived from your specs is the minimum current capability of the batteries. (1.6 kW)/(24 V) = 67 A. Let's say the inverter is 80% efficient, so it really needs 83 A at 24 V. Therefore, your combined battery bank must be able to deliver at least 83 A at the total of 24 V.

With two 12 V batteries in series, both batteries need to be able to deliver 83 A.

The Ah spec you provided is not relevant since the spec above doesn't say how long the inverter must be able to run from the batteries while delivering 1.6 kW.

  • \$\begingroup\$ It’s always good to be meticulous ( meticuleux) specially when it comes to engineering. Thank you for pointing out the rules. \$\endgroup\$ – clarajune Sep 17 '18 at 18:00
  • \$\begingroup\$ @winny Thanks for your solution. In your proposition considering the inverter is 80% efficient how did you get 83 A at 24 V ? Also if I consider a runtime of 10 hr on batteries how that will change the design? \$\endgroup\$ – clarajune Sep 19 '18 at 14:24

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