18650 cell will use ______ amps every _____mins?

I am looking at how to utilise the BQ77915 Ultra-Low Power Primary Protector with Autonomous Cell Balancing by TI, using eight 18650 cells, each with a BQ77915 IC allowing 3 series - 5 series cells.

Looking through the design details the example says:

The battery allows 4 A continuous current.

If I decide to use five 18650 cells in series, 3600 mAh @ 0.2C discharge for one BQ77915 device, is this 3.6 A * 5 = 18 A every 12 mins?

I don't quite understand what "@ 0.2C" means. From reading online sources, I understand 3600 mAh is 3600 mA per hour but still don't understand @ 0.2C.

• @0.2C means ”at a discharge current of 0.2 times the cell capacity”. With a 3.6 Ah cell, this implies ”at a current of 720 mA”.
– jms
Commented Feb 6 at 1:47
• mA per hour is an entirely different (and much less useful) unit than mAh, which is mA times hours. Commented Feb 6 at 4:27

The ampere is a measure of flow of charge per unit time. Talking in terms of "amps every x minutes" does not make sense.

From reading online sources, understand 3600mAh is 3600mA per hour but still don't understand @0.2C.

No, it's 3600 mA for one hour. 3600 mA × 1 h = 3600 mAh.
Or 1800 mA × 2 h = 3600 mAh.
Or 7200 mA × 0.5 h = 3600 mAh1.

My question is, If I decide to use 5 * 18650 cells in series, 3600mAh @0.2C discharge for one BQ77915 device, is this 3.6A * 5 = 18A every 12 mins?

Again, "18 A every 12 mins" doesn't make sense. Instead, since all five cells are in series they will all pass the same current and have the same Ah capacity. 3600 mAh @ 0.2 C = 3600 × 0.2 = 720 mA discharge rate. At 18 A the discharge rate would be $$\ \frac {18}{3.6} = 5\ \text C \$$ (which would be 25 times the 0.2 C discharge rate).

The fact that there are five cells in series will raise the voltage to five times the cell voltage but does not increase the current capability.

Looking through the design details the example says "The battery allows 4A continuous current."

Then that's the most current you can draw from one, two ... or five series connected cells. If the cells are 3600 mAh then we can work out the discharge time as $$\ t = \frac {capacity}{current} = \frac {3.6}4 = 0.9 \ \text h = 54 \ \text{min} \$$ (subject to note 1 below). You might want to work on a time of 30 minutes for such a high discharge rate but the datasheet should help you work that out. Don't forget to allow for aging of the cells too.

Note that the above result is read as "4 A for 54 minutes".

1 I'm simplifying here. At higher discharge rates the battery becomes less efficient so effective capacity is reduced. Hence the 'C' rates.

If I decide to use five 18650 cells in series, 3600 mAh @ 0.2C discharge for one BQ77915 device, is this 3.6 A * 5 = 18 A every 12 mins?

3600 mAh is not a current, but a capacity. It means a battery can deliver a current of 3600 mA for one hour (or 1800 mA for two hours, etc.) before it is empty. It is not mA / hour, but mA × hour.

The actual real-life capacity is a little more complicated than stated above. The 0.2C discharge rate (meaning a discharge rate of 0.2 × 3600 = 720 mA) is mentioned because a battery's capacity depends on how much current you pull from it; the higher the current, the lower the capacity.

So what it says is: at a discharge current of 720 mA, the battery's capacity is 3600 mAh, i.e. you can use the battery for 5 hours when you pull 720 mAh from it; the capacity will be larger for less current, and smaller for a higher current. At, say, 1000 mA, you will see a smaller capacity than 3600 mAh, and the battery will not last 3.6 h at that current.

Five batteries in series give five times the voltage, not five times more current, so capacity in Ah stays the same, but it is at a higher voltage, so capacity in Wh is five times higher.