2
\$\begingroup\$

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?

\$\endgroup\$
  • \$\begingroup\$ The current should not vary with time. Perhaps your are interested in the charge? Current times Time. \$\endgroup\$ – russ_hensel Feb 28 '11 at 3:43
6
\$\begingroup\$

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.

\$\endgroup\$
  • 8
    \$\begingroup\$ wolframalpha.com/input/?i=.02+mA+8+years+to+mAh \$\endgroup\$ – endolith Feb 28 '11 at 5:45
  • 1
    \$\begingroup\$ @endolith, you are awesome! So is wolframalpha. \$\endgroup\$ – Kortuk Feb 28 '11 at 6:48
  • 2
    \$\begingroup\$ At some point in the 240mAh capacity discharge, the voltage will fall below the operating voltage for the flip flop, so it will go to an indeterminate or reset state. You need to examine the discharge curve to see what this is, so it's unlikely that you will get the full 240mAh because your circuit will not work all the way down to 0V. \$\endgroup\$ – Martin Feb 28 '11 at 8:25
  • 1
    \$\begingroup\$ @Martin, this is a good point, but I was mostly focused on showing that the current draw from his gate would be too much to last 8 years. \$\endgroup\$ – Kortuk Feb 28 '11 at 10:56
  • \$\begingroup\$ And you need still MORE margin if the product can ever get warmer than room temp (77F). In this class of device (CD4000), quiescent current increases by 2x if the temp rises ~20degF. \$\endgroup\$ – vandee Feb 28 '11 at 15:12
0
\$\begingroup\$

You've got an answer on the battery life, but I don't think that's the entire answer.

You should really take into account the self-discharge of the battery. That should be on the battery data sheet.

Also, your Iq seems very high for just a CD4013. Typical current draw at 25°C for a 4013 is 20nA, which is 1000 times lower. The guaranteed maximum is much higher at 2uA (10V/25°C) but that's still a lot lower than 20uA.

\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.