I'm designing a circuit where I need a supercapacitor to continue powering the circuit for a few seconds after the main power has been removed. A simplified schematic of the circuit is shown below.
The circuit consumes 200mA and needs to remain on for 3 seconds. I've calculated the required capacitance as follows:
- Regulator dropout voltage is 250mV, therefore min input voltage = 3.3V + 0.25 = 3.55V
- Diode Vf is 0.4V, therefore max input voltage is 4.6V
- Capacitor V = 4.6V-3.55V = 1.05V
- Combining the Q=CV and Q=It equations results in C=It/V. Plugging in the numbers results in a required capacitance of 0.2*3/1.05 = 0.57F
I've tested the circuit out with a few supercaps and here are my results (time measured using a stopwatch and stopped when multimeter shows that voltage has reached 3.55V):
- 1.6F (3×4.7F, 2.3V caps in series). Calculated time = 8.2s, actual time = 8.5s
- 2.35F (2×4.7F, 2.3V caps in series). Calculated time = 12.3s, actual time = 13.2s
- 0.5F (2x1F 2.7V caps in series). Calculated time = 2.6s, actual time = 3s
- A single 1.5F, 5.5V cap. Calculated time = 7.9s, actual time = "instantaneous".
- A single 1F, 5.5V cap. Calculated time = 5.3s, actual time = "instantaneous".
- A single 0.33F, 5.5V cap. Calculated time = 1.73s actual time = "instantaneous".
What I'm confused about is why the single 1.5F and 1F capacitors don't seem to behave at all according to my calculations, whereas the capacitors formed for more than one in series work just as calculated.
Best Answer
I was intrigued by this problem; but I came across this answer which might explain it : look up the ESR (equivalent series resistance) - for your supercaps - some are as high as 80 ohms which would explain what you are seeing.