# How to estimate Li Ion Battery SOC?

In a simple model of a renewable energy system, I have a residential inverer and Lithium ion battery system with a round-trip efficiency (for inverter and battery, i.e. ac to ac) of 89% (at 25 degrees and a given charge / discharge power). The model needs to calculate the SOC of the battery (to determine when it's fully charged, etc.) which requires me to distinguish between charging and discharging efficiencies. Some studies assume the inefficiency is in charging, so SOC = 89% of charging energy, but this seems unlikely. Is it reasonable to assume that the losses for charging and discharging are approximately equal, so the charge and discharge efficiencies would both be approximately 94.3% (square root of 89%) or is there some other rule of thumb?

To clarify: this is a techno-economic model and is only concerned with energy flows in the system. If n(charge) is the charging efficiency and n(discharge) the discharge efficiency, all I know (from the manufacturer) is n(cycle) = n(charge) * n(discharge) = 0.89

For example, in time period t1, energy delta(E) is sent to the battery-inverter system and - if the battery has unused capacity - increases the energy stored in the battery by delta(E) * n(charge). The change in the state of charge is then:

delta(SOC) = delta(E) * n(charge) / usable capacity

I need to estimate SOC to know whether the battery has capacity to take delta(E).

• I think if you know Peukert's Law you can begin to compute but then the age or condition of the batteries, affects probability of failure to supply as new. What tolerance and confidence value can you define on your need in order to accurately define. – Sunnyskyguy EE75 Jun 14 '18 at 11:57

## 3 Answers

It is not entirely clear what you are getting at. What you SHOULD do if your model is detailed enough is periodically measure battery current and sum during charge and discharge to keep track of Amp-hours. Charge amp-hours and discharge amp-hours are very close to equal (very high Coulombic efficiency). The energy loss due to charge and discharge is mostly due to voltage differences. The voltage during discharge is lower than the voltage during charge, so even though the cumulative charge is the same, the energy is not.

The figures you are quoting may involve DC-DC converter losses or inverter losses. It is hard to tell from your question. But if the round-trip efficiency for the battery itself is 89%, that is probably due entirely to voltage difference. I don't really see how it can be apportioned between charge and discharge, nor do I see why it would be necessary to apportion it that way. Maybe it would be more clear if you provided more detail about your model. Maybe your battery model could just use an ideal battery with a small resistor in series.

• Thanks for your reply @mkeith. I have edited the question to (hopefully) clarify. – doctorer Jun 14 '18 at 4:35
• Just pretend all the loss happens on the charge cycle. It is not going to matter in the end, but doing it this way will make the accounting simple. Splitting it up will just complicate matters. Alternatively, if this seems non-intuitive to you, then do it the other way and assume all the loss happens during discharge. Either way you will get the same result. – mkeith Jun 14 '18 at 4:50
• Thanks @mkeith. Either way works fine for the energy flows, but not for the SOC. I need a value for the SOC (or the energy stored in the battery) to know how much can be added to or drawn from the. e.g. If I assume the loss is all in charging, the SOC will appear lower than it would be in reality, so my model will effectively increase the battery capacity, and visa versa if all the loss is on discharge – doctorer Jun 14 '18 at 5:04
• Well, it all depends on how you model everything. Normally, when we talk about capacity of batteries, we are talking about the useful discharge capacity. So let's do it that way. You have a 100 Wh battery. It is at 50% SOC. This means it can presently deliver 50 Wh during discharge. But to charge it fully, you need to deliver 50/0.89 Wh. Get it? Let's say it is at 50% SOC. You deliver 10 Wh to the battery. Multiply the 10Wh by 0.89 before adding it to the battery. So 50 Wh + 8.9Wh = 58.9Wh. So the new SOC is 58.9%. Get it? – mkeith Jun 14 '18 at 5:17
• Ah, OK, so it's all about the meaning of capacity = "useful discharge capacity" not "energy stored". In your example, a battery that has a _rated capacity of 100 Wh will take 112 Wh to fully charge it, will "actually" store some unknown amount between 100 and 112 Wh, and then will discharge to give 100Wh? In my case the inverter-battery system has "AC Energy" quoted which presumably will be the useful discharge capacity. thank you – doctorer Jun 14 '18 at 5:28

Lithium batteries have a charge efficiency around 99% (insane right?). So the rest of the inefficiencies are going to be from the charge controller (which is essentially a DC to DC converter with different control hardware).

The rule of thumb I like to use for DC to is 85% for a reasonable low end ( there are some that are worse than this) and 95% for the high end. In low current conditions the efficiency is much worse, but we are charging a battery, so the current is high.

So if we take 90% by 99% we get 0.891 or 89% to get the charge into the battery, to take it out, I'd knock off another 80%. So 89% by 80% is 0.71 or 71% for a non-detailed number.

This is only for DC to DC, if you need to convert to AC then the numbers might be lower. Once you actually start designing the system you can plug in real efficiency numbers.

The battery is highly efficient. Li-ion has 99 percent charge efficiency, and the discharge loss is small. In comparison, the energy efficiency of the fuel cell is 20 to 60 percent, and the ICE is 25 to 30 percent. At optimal air intake speed and temperature, the GE90-115 on the Boeing 777 jetliner achieves an efficiency of 37 percent. The charge efficiency of a battery is connected with the ability to accept charge. (See BU-808b: What causes Li-ion to die? under Coulombinc Efficiency.)
Source: http://batteryuniversity.com/learn/article/comparing_the_battery_with_other_power_sources

On to battery capacity in watt hours:

If you aren't sure if you need the capacity in WH or just AH well ah is easily found by just finding the capacity that relates to the cutoff voltage you want in the discharge curve.

If you do want WH though: If we say the voltage range is 4.1 to 3.1, and the capacity is 3.2ah well then you get a rectangle with a triangle on top of it that you can find the area of and know your Watt Hours. WH = (3.2*3.1) + 1/2(3.2* (4.1-3.1))

So after you calculate your battery size in watts, multiply it by the efficiency to find the total capacity in watts.

• Coulombic efficiency and energy efficiency are not the same. 100% Coulombic efficiency means that charge amp-hours = discharge amp-hours. Energy efficiency will always be lower than Coloumbic efficiency because charge voltage is always higher than discharge voltage. The OP's 89% is energy efficiency. It is not clear whether it includes battery losses only or also includes inverter and charger losses. OP does not seem to know. But it does seem that the 89% was presented to the OP as a given. – mkeith Jun 14 '18 at 5:04
• The 89% is for inverter and battery losses for the complete charge-discharge cycle, i.e. ac to ac. I have amended the question accordingly – doctorer Jun 14 '18 at 5:05
• Then your good to go, you losses are going to be 89% – Voltage Spike Jun 14 '18 at 5:28
• Clearly, but my original question! – doctorer Jun 14 '18 at 5:29
• Gotcha, there you go – Voltage Spike Jun 14 '18 at 5:34

The question arose from a misunderstanding, but maybe useful to leave it here. The meaning of "battery capacity" is not "energy stored in the battery" but instead "useful discharge capacity", i.e. the electrical energy that can be delivered after the charge and discharge losses. e.g. if the battery capacity is 13.2kWh, the battery actually stores more chemical energy than this (though we don't know exactly how much), but can deliver 13.2kWh.

Therefore, the model can use the simplifying assumption that all losses are on the charging cycle (n(charge) = n(cycle) = 89% and n(discharge) = 100%). Then we don't need to know the actual energy stored, just the useful deliverable energy.

e.g.if the SOC is 80%, the useful deliverable energy stored is 0.8 * 13.2kWh = 10.56kWh and the amount of additional energy I can send to the battery is 20% * rated capacity / n(cycle) = 0.2 * 13.2kWh / 0.89 = 2.97kWh.

With thanks to @mkeith for helping me work through this.