I am designing a battery powered circuit which performs RF communication. The goal is to use a 3V CR2032 battery. The RF circuit consumes 0.2mA as average current and lowest voltage it can work on is 2.8V. This average current is broken into various pulse current draws which are as follows:

  • 8mA for 2ms
  • 15mA for 1.5ms
  • [90mA for 0.2ms,15mA for 0.3ms,150mA for 0.2ms] X 4 times
  • Deep Sleep with current of 20uA

The part in bracket (step 4) happens 4 times as 4 Tx/Rx cycles. The circuit is doing this at every 2 secs. I am trying to put a capacitor so that these pulse currents can be taken care by a supercap. As per my calculations:

For an average current of 0.2mA, assuming we need capacitor to store enough power for 5 secs, Charge required in capacitor = 0.00028mAh = 1mC

Using C=Q/V at 3.1V, C = 1mC/3.1V = 0.32mF

I am planning to use a 330ohm resistor so the charging times comes to be: T=5*R*C = 5*330*0.32mF = 0.53 secs


  1. Are the above calculations and assumptions correct?
  2. I tried using a 1.5F capacitor but the performance was dismal, there was a continuous draw which started at 200uA and gradually went upto 1000uA bringing the voltage of battery<2.8V, I am assuming this is because the capacitor wasn't getting enough time to get charged.
  3. Please propose any better alternatives

Battery Datasheet : https://www.alldatasheet.com/datasheet-pdf/pdf/596824/PANASONICBATTERY/CR2032.html

Capacitor Datasheet : https://www.mouser.in/datasheet/2/40/AVX-SCM-1018838.pdf Version used by me works at Vmax of 5V and has quiescent current of 10uA, to be on safer side of calculations I am assuming 15uA

  • \$\begingroup\$ yes makes sense, the dropout voltage for circuit is 2.8V that makes it operate almost at the edge, but then I was thinking to put multiple CR2032 to get more capacity in the small voltage range I have. I am trying to prevent using a boost convertor to keep the circuit simpler and avoid any further current draws \$\endgroup\$
    – mosdkr
    Commented May 24, 2020 at 10:07
  • \$\begingroup\$ sorry missed the lower voltage earlier, please see that you cannot make your circuit work lower, I can flat out tell you it will not be able to use more than 50% of the battery capacity under those conditions, \$\endgroup\$
    – Reroute
    Commented May 24, 2020 at 10:09
  • \$\begingroup\$ yes I am considering that and have accepted that I won't be able to utilise the whole battery and would try to compensate by either putting a higher mAh battery like CR2450 or use multiple CR2032 cells. Do you think I should put a boost convertor \$\endgroup\$
    – mosdkr
    Commented May 24, 2020 at 10:11
  • \$\begingroup\$ It is not an issue with the capacity, it is an issue with the voltage and ESR inherent to these size batteries, you fall below 2.8V under small loads due to the batteries own discharge curve around 50% \$\endgroup\$
    – Reroute
    Commented May 24, 2020 at 10:12
  • \$\begingroup\$ assuming I utilise only 50% capacity of my battery pack, would the calculations for capacitor hold true? If they are I will probably think of putting a boost regulator \$\endgroup\$
    – mosdkr
    Commented May 24, 2020 at 10:15


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