Transmit: 12mA for about 1ms per beacon. Probably transmit 3 beacons per second for one second per event.Wake: 6mA for 2ms. One wake per event.Events are very far spaced apart.
C = I x T / VI = excess current to be provided.T = time to provide this extra current.V = acceptable drop in voltage during this period.C = capacitance in Farad to meet this requirement.C = .012A * .005S/0.5V = .00012F = 120uF
Your "batteries" will have fairly high internal resistance so I wouldn't expect any problems putting them in parallel. I have no idea how much energy you can get out of a potato or lemon battery...
But the series combination of potato cells won't have the same voltage, so it's a little bit like putting batteries with different voltages in parallel. I'm not sure what the practical impact is of having different voltage sources in parallel.
Potatoes in series, feeding a Buck Converter -- most efficient way to do it!With Series Potatoes, voltages will add up, and when Buck converted, current will multiply according to the following formula: IOUT = IIN*(eff/100)*VIN/VOUTWhere: VIN is the total series voltage of the potatoes eff is the efficiency of the Buck Converter in % VOUT is the Output Voltage of the Buck Converter IIN of course, is the current supplied by the series arrangement of potatoes -- which will be lower than the output current (if the output voltage is lower than the total potato stack voltage, less PS efficiency).So, the more potatoes in series, the more current you can get out of this thing!
I really like that! Is the potato battery a standard schematic symbol?