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I am wondering can I charge and discharge a supercapacitor at the same time like a battery.

I have 1.3V going into a 10F 2.7V capacitor in series with a 0.5 ohm resistor. I calculated it will take about 30 seconds to charge to "full". And I will have a 2.6A peak current.

The minimum voltage I need is 0.9 when I discharge the capacitor. I have a 500mA current load (1700mW) that I need for startup for 1 second.

I calculated that it takes 8 seconds to discharge to 0.9V (which is more than enough time). I then need a constant 0.085mAs after the 1 second power surge.

  1. Can I charge and discharge at the same time?
  2. Will I be able to charge the capacitor fast enough to keep the voltage above 0.7V after the power surge?
  3. Will I be able to keep my load running after the power surge when my load only requires 0.085mAs?
  4. I see that every 1 second my voltage from the capacitor drops 0.05V, so when I get to 1 second I have 2500mW of power left. If my load needs 1700mW, do I subtract that and I'm left with 800mW of power to deliver to my system while I charge my cap?

I am trying to figure out if a capcitor can supply a consistent wattage even when not completely full.

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  • \$\begingroup\$ By "mAs" do you really mean milli-ampere-seconds or is that a typo? It's just "mA" for milli-amperes. \$\endgroup\$
    – Transistor
    Jul 7, 2020 at 1:56
  • \$\begingroup\$ @Transistor I mean milli-amp per second \$\endgroup\$
    – Sherman
    Jul 7, 2020 at 17:06
  • \$\begingroup\$ Amperes and mA are defined as charge per unit time so they come with seconds built-in. The units mA/s (milliamperes per second) suggests a rate of change of current. E.g., "The current is decreasing from its initial value of 100 mA at 2 mA/s." But that's not what you mean. Your sentence should read "... when my load only requires 0.085 mA?" \$\endgroup\$
    – Transistor
    Jul 7, 2020 at 17:15
  • \$\begingroup\$ Okay that makes sense. I read somewhere else someone was using mAs. But my load needs 0.085mA because seconds are built-in. Thank you! @Transistor \$\endgroup\$
    – Sherman
    Jul 7, 2020 at 17:27
  • \$\begingroup\$ A schematic is better than words. You can add one in using the CircuitLab button on the editor toolbar. Double-click a component to edit its properties. 'R' = rotate, 'H' = horizontal flip. 'V' = vertical flip. Note that when you use the CircuitLab button on the editor toolbar and "Save and Insert" on the editor an editable schematic is saved in your post. That makes it easy for us to copy and edit in our answers. You don't need a CircuitLab account, no screengrabs, no image uploads, no background grid. \$\endgroup\$
    – Transistor
    Jul 7, 2020 at 17:30

1 Answer 1

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Can I charge and discharge at the same time?

No, current only flows one way. The remaining current would need to come from the charger. So if 2A was going to the caps and the load needed 1A, the source to charge the caps would need to source 3A.

I see that every 1 second my voltage from the capacitor drops 0.05V, so when I get to 1 second I have 2500mW of power left. If my load needs 1700mW, do I subtract that and I'm left with 800mW of power to deliver to my system while I charge my cap?

No, because you need to deal with energy in joules. Watts is joules per second, if you used 2500mW of power for 1 second it would be 2500mJ, that is what needs to be subtracted from the capacitor.

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  • \$\begingroup\$ Okay. So if my source can only provide 1.3V with 50mA so 65mW (0.065 mJ/sec), but my load needs 1.7mJ/sec. I used my 10 F capacitor to provide 3.38mJ/sec. I can use the remaining 1.68mJ/sec to have the capacitor to be charged? And my load after that 1 second only needs 30mAh, so 0.85uA for 1 second at 3.3V so I calculated that my load needs at least 0.0275mJ/sec to stay on. \$\endgroup\$
    – Sherman
    Jul 1, 2020 at 19:54

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