Dear all,
My question looks silly. But I got stuck with this issue. My circuit's block diagram is shown below. VSTOR (and VBAT altogether) of BQ25570 is connected to a supercapacitor for later use. VSTOR is only connected to the load when it reaches 3.3V (by using a switch, not shown in the diagram). With different input powers to the harvester module, although with THE SAME VSTOR voltage (3.3V) and the same load, I observed different discharging time in both cases! Please see fig.1 (0.630W) and fig. 2 (3W) attached.
The thing is, with the very basic equation of a capacitor (E = 1/2*C*V^2), the stored energy in a cap must be the same if the voltage is at the same level. But with these different charging time (figure 1 and figure 2, input power is 0.63W and 3W respectively), the stored energy at the cap C2 in these two cases are totally different.
One more issue I got, related to this issue. I was trying to estimate the current flow which represents the power consumption of my circuit by this equation: I = C2 *dV/dt (C2 = 6.8mF, V = VSTOR over time) ==> V drops, I rises up, and vice versa. The problem is, the amplitude I got, with that calculation, is up to 500mA with a drop of about 20mV in VSTOR. It does not make sense at all for my circuit, especially all the components are the ultra-low power ones.
From my observation as above mentioned, my conclusion is that: can not extract the current amplitude from VSTOR by this equation, because the energy is not only stored in C2, but also in somewhere in BQ25570 which I don't know. Am I right?
Thank you very much for your time. Any suggestion would be highly appreciated.
