DW3000 uses CR2032 power supply, how much capacitance is required?

Hi,Qorvo team
I want to used CR2032 to power the DW3000 to keep RX for 1 second.

F=C/U, C=I*T

F=I*T / U

80mA*1s/3.0V=26mF

Do I need to use a 26mF capacitor?Is that right?

T=47uF*3.0V/80mA=1.7625ms

The 47uF capacitor only RX 1.7625ms

Is that true?

Thanks,KK

Hi @KOv01
1 second for CR2032 is s long time. Typically the battery can provide a current less than 1mA… And 47mF is a huge capacitor bank (probably bigger than the CR2032 battery). You should not use ceramic capacitors because of voltage derating.

The RC constant is just a part of it - normally it is considered that the capacitor bank is charged after 3 * T when it reach cca 95% of nominal voltage. Take a look at this plot https://www.allaboutcircuits.com/tools/resistor-capacitor-time-constant-calculator/ You need to calculate how long it will take to discharge from fully charged state (3V) to minimum DW3000 voltage and then you need to tune the capacitor bank size to be able to handle this load for a given period.

However it is a question if it makes a sense to use DCDC boost to increase the voltage to DW3000 maximum voltage to have more energy stored in the capacitors → smaller capacitor bank.

Cheers
JK