So I was reading through TI's datasheet for their BQ25505 here, and came across the following:

8.1.2 Storage Element Selection

In order for the charge management circuitry to protect the storage element from over-charging or discharging, the storage element must be connected to VBAT pin and the system load tied to the VSTOR pin. Many types of elements can be used, such as capacitors, super capacitors or various battery chemistries. A storage element with 100uF equivalent capacitance is required to filter the pulse currents of the PFM switching charger. The equivalent capacitance of a battery can be computed as:

$$C_{EQ} = \frac{2 \ \cdot \ mAHr_{BAT(CHRGD)} \ \cdot \ 3600s/Hr}{V _{BAT(CHRGD)}}$$

I was looking for a bit of enlightenment for this formula.

Assuming a few things, (according to a battery datasheet here) I should have

\$mAHr_{BAT(CHRGD)} = 2,000\$


\$V _{BAT(CHRGD)} = 4.2\$

First off, am I correct in assuming \$V_{BAT(CHRGD)}\$ is 4.2V and not the nominal 3.7V?

Secondly, what is this \$C_{EQ}\$ formula giving me back? Why is this \$C_{EQ}\$?

If I get 3,428,571 it doesn't mean anything to me, hence my questions.


1 Answer 1


The formula gives you the capacitance that store the same amount of energy of a battery (ideally).

\$C_{EQ}\$ has the unit of mF.

So the number means a 3428571 mF (3428 Farad!!) capacitor is needed store the same amount of energy of the battery.


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