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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

Tony Stewart EE75
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seanyu24
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  • Now you know high pulse current degrades mAh capacity – Tony Stewart EE75 May 03 '18 at 03:15
  • 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 – Tony Stewart EE75 May 03 '18 at 03:29
  • 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? – Ale..chenski May 03 '18 at 05:18
  • I.e. Ali means is the spike near 1ms and what is the tolerance on that pulse width – Tony Stewart EE75 May 03 '18 at 13:30
  • @TonyStewartEEsince1975, this high spikes are produced one of transceiver on board. When it sending, it can draw supply current up to 70mA. So, it is kind what we expected. The question is how bad this spike negatively affect the battery life, especially it happened, every 20 sec and go over the battery max pulse current limit. – seanyu24 May 03 '18 at 18:56
  • @AliChen, Im using a agilent DMM, which can capture 50000 points in 50 sec. This is what exactly i did, so i think the data point are accurate. The spikes are because of the transceiver and diode on board. However, when i do the battery life calcultion. I calculate the average current consumption over this period of time. Those small and large spikes bring the average current draw from 0.08 mA to 0.16mA. But even based on the higher average current draw, the unit should drain much longer compare to our current time. So , thats why i asking, how badly this spikes affect the battery life. – seanyu24 May 03 '18 at 18:56
  • You need to report accurate battery current and pulse width (PW50) not 5mA on graph and 70mA comments otherwise how can we tell if you calculated properly? do you also have Vbat data? 1ms sample is not fast enough. This battery has very large ESR but also low leakage. – Tony Stewart EE75 May 03 '18 at 20:01
  • @seanyu24, Until you fully resolve each spike (have few samples over each one), you can't be sure what your DMM is measuring. Try to put some simple filter on the power rail, LC, 100 uF or something, to smooth out these crazy spikes, measure it again, and see if you have the same average draw. – Ale..chenski May 03 '18 at 20:12
  • @TonyStewartEEsince1975, from battery's chart, my estimate is that it has ESR of 100 Ohms. – Ale..chenski May 03 '18 at 20:14
  • I might have said 150 but its not-linear in any case a couple thousand times bigger than an 18650 cell. Yes a 50mV current shunt with LPF Amp that has a 60 second a time constant would be a good averge or simply LPF to 100Hz then sample at 1kHz and same for V and compute Wh or J or VI*t not just mA. So far I dont see any battery fault, just user err. – Tony Stewart EE75 May 03 '18 at 20:22
  • @TonyStewartEEsince1975, more, a normal 10-ms (100mH/200uF) filter will bring down current spikes into more comfortable "load" are for the battery, so the concern if "do spikes harm battery capacity" will go away automatically. – Ale..chenski May 03 '18 at 21:30
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    Agreed and a good low ESR cap is needed when 100R*70mA= 7V The problem here is perhaps an unfiltered inductive load. Design fault. – Tony Stewart EE75 May 03 '18 at 21:48
  • @Tony Stewart EE since 1975, thanks for the suggestion. If 1 ms sample rate is too big. which sample rate will you suggest? My DMM is only able to store 50000 points a time, so i might reduce the capture time to increase the sample rate. – seanyu24 May 04 '18 at 20:39
  • 10us just to capture spikes – Tony Stewart EE75 May 04 '18 at 20:41
  • @AliChen, thanks for the input, i have measured 50+ units, each one have this similar current pulse pattern.( 40~50mA pulse every 17 sec). However, i havent measured the Pulse width for each unit. A estimation would be 10 m sec. – seanyu24 May 04 '18 at 20:44
  • @TonyStewartEEsince1975, Thanks , will try. 10 us with 50000 points is only 0.5 seconds. I need to set the trigger to capture the current spikes. – seanyu24 May 04 '18 at 20:47
  • This battery can't provide 70 mA pulses. You NEED a serious LC filter with local capacitance between the battery and your device, with a cap big enough to manage your transmission demands. Take scope traces instead of using some goofy DMM, it is not the right tool for the job. – Ale..chenski May 04 '18 at 21:26
  • Obviously we cant tell without a schematic, but I suspect you have a cap with much lower ESR than the battery as it shud be. You can use a goofy DMM if that's all you have and you understand how to makde ideal test results, Are you capturing current with a 50mV shunt R and then capture Voltage at the same time Otherwise. you need to compromise your expectations for accuracy or combine many captures into 1 bigger dual trace file with sync.. – Tony Stewart EE75 May 04 '18 at 22:03

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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.

Mark Wolfman
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  • 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 – seanyu24 May 07 '18 at 21:06