# Do Inverters change the lithium-ion batterie's mA capacity and wattage? [closed]

let me say if I have a pack of lithium batteries connected in series and parallels, and i want to connect it to an Inverter, BUT...BEFORE I connect it to inverter, lets say it gives out 12 volts "hypothetically" some informations like 3.7 volt batteries, and 2500 mAh,
lets say then the pack gives 12 volts in DC and hypothetically again I calculate it ((still in DC)) is ..... W = V* Amp ... then it means wattage = 11.1 V * 5 Ah which is 55.5 W. then we got 55.5 watts, THEN....after we want to change the dc into ac it gives you 120 volts "please hypothetically" just answer me this, ARE the Wattage And Ampere are remains the same as DC? what would it be the result after AC?

## closed as unclear what you're asking by Leon Heller, Harry Svensson, Transistor, winny, laptop2dMay 18 '18 at 16:19

Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.

• Please clean up your question, it's hard to tell what you're even asking here. The power rating will not change (actually, it will decrease slightly, due to inefficiencies in the power electronics), but you seem to have amps and amp-hours mixed up so I'm not sure what you're asking there. – Hearth May 6 '18 at 17:30
• amp and amp-hour are definitely not the same. – Hearth May 6 '18 at 18:02
• 5 Ah * 11.1V = 55.5 Watt-hours. Not 55W. This means the pack can deliver 1W for 55 hours or 55W for 1 hour (conceptually... in reality it will be a bit less at 1 hour and a bit more at 55 hours). – mkeith May 6 '18 at 18:43
• Instead of generalizing, you better formulate your particular battery configuration and DC-AC converter properties, and then formulate your question/concern. – Ale..chenski May 6 '18 at 20:41
• ... and please use standard English sentence structure, proper capitalisation and punctuation so it is intelligible technical writing. – Transistor May 6 '18 at 20:57

I think this question needs some serious sorting out. It has several parts:

1. Battery cell. A "primary battery cell" stores ENERGY, which is a certain amount of Watts over certain time. A cell with 3.7V nominal and C=2500 mAh holds about 9.25 W-hours of energy. You can dump all 9 Watts in one hour, or you can sip 1 watt for 9 hours. All would depend on connected load. A 370 Ohm resistive load will work from this cell for 250 hours. However, every cell has certain limit on discharge rate. Normally the cells are made for 1C - 2C discharge rate (C is rated capacity of one cell), but some special battery packs used in Radio Control area can have discharge rates of 15C-25C-50C.

2. A Battery of cells. In general, the energy stored in any battery is a sum of individual cells, regardless of how they are connected, in series, in parallel, or properly mixed. So if you have a 6-cell battery, 2 in parallel, and three in series, you have a 55.5 W-Hour battery. Not "55W", but "55 Watt-Hours". Again, depending on load, you can get 100W in half hour, or 9 W in six hours, or 1 W for 55 hours. If your battery is 2P3S battery with typical 18650 cells, it likely can be discharged at 2C rate , or up to 10 A in your case. So, "hypothetically" you can get 120 W from 2P3S battery (with 2C discharge rate) for about 20 minutes. Or, more hypothetically, you can get 1200W for 2 minutes, if your cells can discharge at 20C rate, hypothetically.

3. Inverter. Inverters (AC-DC converters), or any other switching mode DC-DC devices of this sort, they CONVERT a input into output. They are limited by amount of POWER they can deal with, instant POWER. Ideal converters convert power at 100%, but real converters have some losses. A safe assumption is to have 20% losses. Say you have a 120-W capable converter that handles hypothetically any voltage level, and the 120 W is output rating. This means that if your output is 12 V, the converter will deliver up to 10 A of current to a 1.2-Ohms load. Or 1 A to 12-Ohms load, whichever is connected to its output. The input, however, will use 1.2 A at 120 V, or 20% more power than it delivers to output. But if you use, say mains AC and the inverter forever, you can transfer infinite amount of energy, watt-by-watt, hour by hour, not limited. So, P(in) = P(out)/0.8, and V(out)I(out) = V(in)I(in) *0.8. So no, "amperage" won't be the same on both ends of converter, it will be inversely proportional to voltage ratio.

4. If you plan to attach a 12 V DC to 120 V AC converter/inverter to your battery, it has very little to do with battery energy storage capacity. But same efficiency considerations and power limits will apply. So the 55.5 Whr battery with 12DC-120AC inverter will be able to deliver only 44.4 Whr of energy on 120 V side. 11 Whr will be dissipated as heat, lost in electronic components of the inverter. Again, if you attempt to get, say, 120 W out of this design (120 V at 1 A), the inverter must be rated as "120 W", and your battery will last only 22 minutes. So to get 1 A at 120 V output, your 12-V battery must be delivering about 13.5 A of current, sorry.

I hope this answers your question, "what would it be the result after AC?"

EDIT: considering the OP question literally, the answer is "no", nothing can change capacity of Li-Ion cells/battery. However, considering the [battery+inverter] as a "black box" with 120V output, the inverter will decrease the effective box capacity from 55 Wh to about 44 Wh first, due to internal losses (assumed at 20%). Second, the effective mAh rating of the box will be inversely proportional to output voltage. So the "box" capacity will be 44Wh/120V=366 mAh.

• could you calculate W= V*A for me 2 times? first when its DC and then for AC ? that's all I want to know – babylon May 6 '18 at 22:17
• @babylon, why don't you formulate your question in plain English, with complete sentences? Or show some block diagram, and point where you don't understand the relationships. Note, that "before inverter" would be called as "input to inverter", not "output". – Ale..chenski May 7 '18 at 0:58
• @babylon, and I am asking you to explain, what do you mean "after I change my voltage to AC". Add at least a block diagram to your question. Or a link to it. Please explain it, so we can understand your level of understanding and reply accordingly. – Ale..chenski May 7 '18 at 17:51
• @babylon, then I already answered this question with your particular numbers: since your battery is 55 Wh, the best output you can get is about 44Wh at output, total, until your battery is depleted. Since you "wanna" your output to be 120V (AC or DC, doesn't matter much, effectively), you can have 0.37 A at 120 V for 1 hour out of your setup. What else is unclear? – Ale..chenski May 7 '18 at 18:01
• @babylon, I am really getting tired of this. Your battery has capacity of 55 W-hours. However, you can draw 550 W for 6 minutes out of it. Or 1100W for 3 minutes. Do you understand this relationship? Do you understand the difference between instant power (Watts, which depends on load), and battery capacity, Watts*Hours, the amount of of stored energy? Which one do you call as "wattage"? I am starting to suspect that you have some high expectations about available "A" on 120-V side. Spell it please. – Ale..chenski May 8 '18 at 22:50

@Ali Chen already answered this as full as it needs to be, but based on the comments there still seems to be some confusion. I'll try to rephrase and expand a few things to see if it clicks.

First, you are calculating Watt-hours (Wh), not Watts (W). If your battery bank has 55.5Wh capacity, it can deliver 1W for 55.5h or 55.5W for 1h. The total potential Wh of your battery is a fixed absolute value rating that does not change based on what you run from it. This is partially why you're getting some push back on the question.

"If I attach an inverter to a 12V 55.5Wh battery, how many Watt-hours can I supply to a 120VAC load?" is a very different question.

If you want to know how long it will run and the available power to the load, you will need to know the total power consumed including inverter efficiency and load characteristics, as well as the current handling capabilities of everything involved. You didn't offer any relevant and necessary details here, so we can't answer that question without some hypotheticals and speculation.

100% efficiency equipment is only a theoretical ideal that doesn't exist in the real world. If you add anything other than a source to your circuit, you will lose power in the process.

To find the Watt-hours available to your load after the inverter, you will need to know the power lost in the process of the DC-AC conversion. Most low-cost low-frequency inverters are going to be in 75-85% efficiency range. 80% efficiency (20% loss) is a good starting baseline.

80% of 55.5Wh is 44.4Wh, so you will lose 11.1Wh capability between a 55.5Wh battery and a 80% efficiency inverter. You could think of it as an AC supply with 44.4Wh capacity. It's a little weird to rate things this way, but that seems to be what you're looking for.

You will have to determine the efficiency of your inverter, and that efficiency may change under different load characteristics. You will need to check the datasheet and probably do your own testing to confirm under your specific conditions. The inverter will also draw some small amount of power as long as it's on, even if no load is attached.

If you want to know the inverter efficiency, place a load across the inverter. Measure output load Wattage vs input source Wattage.

(Load W)/(Source W) *100 = Efficiency%.

Also, you will have to keep in mind that your inverter may have a cutout voltage that is surpassed long before your batteries are completely depleted. Your actual run time may be significantly less than the estimate you'll get by simply calculating how many total Watt-hours you theoretically have available.

You can add a boost converter and regulator if this is an issue, but those too will affect your efficiency and overall calculations (which I'm not going to get any further into now).