Part Number: BQ25570
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.
Product Marketing Engineer
Texas Instruments Inc
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In reply to Upal Patel:
In reply to Minh Lam95:
Thank you for your response. Please find below a part of my schematic related to BQ25570. The harvester module is simply a 1-stage multiplier circuit employing 2 schottky diodes.
In my opinion, after more tests, the extra discharging time of the application in case of higher input power might come from the 1st supercapacitor (C1). For higher input power, the harvested energy might be so high that more redundant energy is stored in C1. That causes more time for the application to operate.
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