In one of my application, I am using Tadiran Lithium Thionyl Chloride Batteries MODEL TL-5934. In the data sheet , the max pulse current shows 50mA.

However, in my application, we have low duty cycle current spikes up to 70mA.

Our battery are draining twice the speed as I calculate. How will these current spikes affect the battery Life??

This is the datasheet of Lithion battery we used enter image description here

In this chart, it estimated that 0.2mA constant current draw will drain the Unit in about 4000 hours

enter image description here

This is our appication's current draw profile. 50000 points acquired by DMM in 50 sec. Constant current draw is very small probably 0.01mA, however, have lots of current spikes, which I suspected kill the battery much faster.

enter image description here

  • \$\begingroup\$ Now you know high pulse current degrades mAh capacity \$\endgroup\$ – Tony Stewart Sunnyskyguy EE75 May 3 '18 at 3:15
  • \$\begingroup\$ If you add a low ESR cap to reduce peak current to 3 mA may help . If battery ESR is ~150 Ohms then get a lower ESR battery because 5mA drops the voltage 750mV and will be below 3V or 2V or whatever your cutoff is much sooner. Use C = Ic dt/dV. with dV as low as feasible for your dt spike \$\endgroup\$ – Tony Stewart Sunnyskyguy EE75 May 3 '18 at 3:29
  • \$\begingroup\$ I suspect the problem is not in spikes and how they allegedly degrade the battery. How did you calculate the average charge drawn from the battery on this highly "spiky" signal? Are you sure your DMM captures the spikes accurately enough and has sufficient time-sampling resolution? \$\endgroup\$ – Ale..chenski May 3 '18 at 5:18
  • \$\begingroup\$ I.e. Ali means is the spike near 1ms and what is the tolerance on that pulse width \$\endgroup\$ – Tony Stewart Sunnyskyguy EE75 May 3 '18 at 13:30
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    \$\begingroup\$ Agreed and a good low ESR cap is needed when 100R*70mA= 7V The problem here is perhaps an unfiltered inductive load. Design fault. \$\endgroup\$ – Tony Stewart Sunnyskyguy EE75 May 3 '18 at 21:48

So according to the discharge characteristics you posted, the capacity actually goes up at higher currents, which is unusual. Granted the potential is a little lower so the total energy (Wh) might be roughly the same. In order to know how much discharge time you're losing because of the spikes, you'd need to know how wide they are. If you integrate the sample you collected and divide it by how long that sample is, you get a correction factor. Then take your 4000 hours and divide it by this factor to get the expected discharge time from the observed current profile. If the result is significantly different than what you actually get, then there's something else going on: either your profile is inaccurate or you're pulling more current than you think you are.

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  • \$\begingroup\$ Thanks for the input, Can you illustrate on how to get the correction factor. I had the data of 50000 points. the spikes width is about 30 data points. How do i go from there. THANKS \$\endgroup\$ – seanyu24 May 7 '18 at 21:06

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