How to power an embedded board using capacitors and solar panels?

Question

I am working on a battery-less system. The system has the following energy pipeline:

Solar panels -> DC-DC boost converter -> capacitor -> microcontroller.

Here the solar energy is harvested using solar panels. Then for a period of time, this energy is stored in the multilayer ceramic capacitor (MLCC). When the energy stored inside the capacitor reaches the maximum operating voltage of the microcontroller (5.5 V), then using an analog switch, the MLCC is connected to microcontroller. The microcontroller is powered ON and it will executes operations till there is sufficient energy inside the capacitor. And when the energy reaches the minimum operating voltage of the microcontroller (1.7 V), then it is powered off. This cycle repeats. It is fine if it the embedded board executes only for 1 ms and is powered off for 100 ms.

I want to know how to calculate the required capacitance and energy storage capacity of the capacitor? Is there a way to find the actual capacitance at different voltages due to DC bias?

Based on the datassheet of the nrf52840-dk MCU.

Answer 1

Capacitance is the amount of charge transfer per voltage loss. 1uF means you can take 1uC (that's 1uA x 1s, or 1mA x 1ms, or so on) to get a 1V drop.

You say the system uses 3.16 uA when powered. When not powered, I think you are saying it doesn't matter because the power will be cut off with an analog switch.

You want it to run for 1ms, that's 3.16nC. A 3.16nF capacitor (that's really tiny) would only drop 1 volt when that much charge was used. You can afford for the voltage to drop by 3.8 volts - not just 1 volt - so you only need 1/3.8th of that, which is about 0.83nF. I want to emphasize that's a really really tiny capacitor. If your PCB was rather big, you could even have that much parasitic capacitance already, without adding any capacitors at all.

(This calculation doesn't take into account whether the microcontroller uses more current at higher voltages, nor whether the capacitor loses capacitance at higher voltages.)

So, you may as well use a bigger capacitor and your microcontroller can run longer than 1ms at a time, or you can use more than 3.16uA.

The same math works for charging: assuming that you just connect the solar panel directly to the capacitor, and it acts as a constant current source (which solar panels do, approximately, when the voltage is below the MPP voltage), then to make it rise by 3.8V you need 3.8 times the capacitance, in current x time.