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Say I have a battery that can provide 150-200mA of current continuously, but I have a module which is active periodically, and requires a large pulse current as well as a continuous current that exceeds what the battery provides.

For argument's sake lets say the module is a GSM modem, with a pulse of 2A for 200us, and continuous current of 300mA for 5s.

  • What is the best way to utilise some sort of bulk capacitor or charge reservoir to power the module?

  • How might I do this without exceeding the battery's max output current to provide the charge?

Here's an example of a battery that illustrates the supply: http://www.farnell.com/datasheets/1445889.pdf

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  • \$\begingroup\$ You're missing at least one further spec, how much battery drop and at what rate is allowed over the current pulse? Answer either or both of those, and that will lead you straight to a value of capacitance. \$\endgroup\$ – Neil_UK May 10 '17 at 13:35
  • \$\begingroup\$ I have never seen a battery that can only provide 200 mA continuously but cannot provide 300 mA for 5 seconds. What kind of battery is it. Maybe you are misinterpreting the specifications. \$\endgroup\$ – mkeith May 10 '17 at 13:43
  • \$\begingroup\$ @mkeith, I'm sure the battery could provide more, but higher current draw reduces the overall capacity and causes the voltage to drop. \$\endgroup\$ – Sensors May 10 '17 at 13:50
  • \$\begingroup\$ I have never dealt with that type of battery before. I think you should ask the battery supplier about the 300 mA pulse. The 2 Amp pulse will have to be provided by a capacitor. You could also consider whether another battery type might be a better choice for your application. \$\endgroup\$ – mkeith May 10 '17 at 14:25
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Q=CV and differentiating I = C dv/dt so, if the rate of change of output voltage (due to discharge) across the capacitor is limited to (say) 0.1 volts in 5 seconds then: -

300 mA x 5 / 0.1 = C or C = 15 farads.

A 15 farad capacitor charged to (say) 5 volts, will drop to 4.9 volts after 5 seconds of supplying 300 mA to a load.

If the battery can supply 200 mA of that current then the capacitor only needs to supply 100 mA and this makes the capacitor smaller in value.

I'm sure you can now do the math for the 2A/200us example.

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  • \$\begingroup\$ wouldn't it be beneficial to use a boost converter, to drain the cap to zero? A 15F cap is a rather bulky device. \$\endgroup\$ – Christian May 10 '17 at 13:38
  • \$\begingroup\$ @Christian it probably would but, there isn't enough info in the question to provoke that sort of detail from me. Feel free to leave your own answer of course. \$\endgroup\$ – Andy aka May 10 '17 at 13:39
  • \$\begingroup\$ Thanks for the maths. How do I go about charging the capacitor and deal with the in-rush current which could exceed the battery's current capability though? Would a series resistor work well enough to limit this? Or can I rely on the Capacitor's ESR? \$\endgroup\$ – Sensors May 10 '17 at 13:41
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    \$\begingroup\$ To the OP, I think you are inventing problems for yourself. Please provide a link to the actual battery specifications. \$\endgroup\$ – mkeith May 10 '17 at 13:44
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    \$\begingroup\$ @Sensors if your battery can only supply 200mA it's quite like you can rely on the battery's ESR. \$\endgroup\$ – user253751 Jan 24 '18 at 6:17
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Actually, there are batteries (one example from Tadiran) that include both a primary Lithium cell (similar to your EVE battery type) and an internal "super cap"; they are specifically designed to solve the issue you are describing.

Of course, you could also separate the battery from the super cap, but as was pointed out, you'd still need a fairly large cap make it work. Tadiran doesn't specifically publish the values of their caps (only the pulse performance of the resulting battery), but you can see from the dimensions what they are dealing with.

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