Electronic – How to calculate the total quiescent device current draw for eight years if the the device had a 20 µA quiescent current

batteries

I have a D-type flip flop (CD4013B) connected to a lithium battery with 240 mAh capacity, and it is stored for eight years at room temperature.

May I know how much quiescent current will be drawn from the battery during eight years storage time?

Let me expand

Maybe let me put to you all a better view of my current scenario:

  1. The quiescent current of this device is ~20 µA

  2. I have four batteries connected in 2S and 2P (that is, two in series and then parallel with another two in series) => 480 mAh @ 8 VDC

  3. Each battery has a capacity of 240 mAh @ 4 VDC and from the graph of capacity vs OCV, it will have a capacity loss of 20% after eight years storage at 55 °C.

  4. My circuit can work at minimum voltage of 5.7 VDC

Hence, I would like to know how much current is consumed from the IC and the self-discharge of the batteries during the eight years storage.

Will the batteries be able to operate my circuit after eight years?

Best Answer

The total discharge caused by your device will be 20uA * (8 years). The issue is you need to convert to mAh to match units.

.02 mA * 8 years * 365.25 days/year * 24 hours/day = 1402.56 mAh are required for this

Your battery is only 240mAh capacity, so it will only last a fraction of this time.

You can approach this with algebra also.

240mAh / .02 mA = 12000 hours = 500 days of lifetime from your battery

This is actually still not perfect. You need to factor in the self discharge rate of your battery. This will be specific to the battery you have chosen and can be very high, possibly dwarfing your 20uA limit and placing you in the mA range.

Let me know if there is something you need more information on.

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