[SOLVED] Lithium battery - power

[Audette]

I have a question about batteries and how much energy they store.

I am looking at using a Lithium battery for a portable instrument - the two choices I have are 14.8V 4.4Ah or 11.1V 4.4Ah.
The voltages I work with are 5V and 3.3V and generally the instrument will use around 400-500mA during normal usage. My question is, does the voltage have an effect on how many cycles my battery can go through? Someone once told me voltage makes very little difference to the energy within batteries and that it is the Ah specification that matters. However I never really got a proper explanation as to why this is and I am yet confused as to whether this is the case.

I was hoping my instrument should be able to cycle around 200 iterations using the 14.8V 4.4Ah battery. How would this be affected if I go to the 11.1V 4.4Ah battery - or even a lower voltage but same Ah.

Thanks in advance

 

[keith1200rs]

It is Watt Hours not Ampere Hours that matter. Watt hours = energy. So, provided you use switching regulators the 14.8V battery will last longer than the 11.1V.

The number of cycles doesn't depend on the voltage. The difference is simply that the 11.1V has 3 cells, the 14.8V has 4 cells.

You need to define how the current is split between the 5V and 3.3V to calculate how long the battery will last. Work out the power and assume maybe 80% efficiency for a switching regulator.

Keith.

 

[Audette]

When doing the power calculation for the battery, I have not used watt hours - but Amphours. Should this be changed?

I have used a formula that calculates the current x time - add 20% extra for 80% efficiency and divide the battery mAh value by the result. This I thought was the number of cycles. However, this does not take into account the voltage at all. Should I be therefore replace this with, current x voltage x time?

---------- Post added at 10:13 ---------- Previous post was at 10:13 ----------

When doing the power calculation for the battery, I have not used watt hours - but Amphours. Should this be changed?

I have used a formula that calculates the current x time - add 20% extra for 80% efficiency and divide the battery mAh value by the result. This I thought was the number of cycles. However, this does not take into account the voltage at all. Should I be therefore replace this with, current x voltage x time?

 

[keith1200rs]

You must take into account the voltage to get power (and energy). Power = volts x amps.

The amp hours are fine if you know your circuit takes a certain current and the voltage is fixed. If you start to vary the voltage and the circuit uses a switching regulator then the current will also change so you need to take account of that.

When you use current/time/voltage and add in 80% efficiency, what you should end up with is an amount of time that a circuit will last on a single charge.

As an example, if your circuit takes 5V at 1A you are taking 5V. An 80% efficiency regulator would take 6.25W to generate 5V (not 6W as you might think). If you are generating it from a 14.8V battery, then 6.25W works out as 422mA. So, your 4.4Ah battery will last 10.4 hours. If you re-did the calculation with the 11.1V battery you would take 563mA and so that battery would only last 7.8 hours.

The number of cycles a battery will last depends on the make and quality of the battery but is usually 500 to 1000 complete cycles. With LiIon partial discharges don't tend to reduce the life as much so 1000 half discharges is probably equivalent to 500 full charges in terms of the battery lifetime.

Keith

 

[Audette]

Thank you for the your detailed answer Keith.

Can you please confirm the following calculation based on what you said - but modified to include run time.

5V line, 0.3A together with 80% efficiency requires 1.88W.
3.3V lines, 0.3A together with 80% efficiency requires 1.24W.

the 5V line will be used for 30 seconds and 3.3V for 60 seconds for each sampling cycle.

so if we think of one cycle being 60seconds, the Watts used up during that time,

5V line = 1.88 * (30/60) = 0.94
3.3V line = 1.24 * (60/60) = 1.24

Therefore total wattage required per 60 seconds = 0.94 + 1.24 = 2.18

If the battery is a 14.8V, 4.4Ah battery,
the avg current from battery used = 2.18W/14.8V = 150mA

Therefore the number of hours battery will last = 4400mAh/150mA = 29 hours

Can I then convert this to number of cycles by saying = 29 * 60* 60(convert to seconds) /60 (sampling cycle time) = 1740

Therefore can I assume that at best the battery will give me 1740 of 60second cycles?

 

[keith1200rs]

You calculations are fine up to the 29 hours. Your next bit is actually correct, but a complicated way of going from 29 hours to 1740 minutes. You calculate down to seconds then convert it back to minutes (* 60 / 60) but the answer is correct.

The cycles here are not to be confused with the battery lifetime i.e. the number of charge/discharge cycles before the battery is holding charge well and needs replacing with a new one. That would typically be 500 to 1000 charge/discharge cycles.

Keith.

 

[Audette]

Thanks. I shall now get a battery and test and see how accurate the calculation is