Electronic – Finding practical solutions for discharging and charging a capacitor

capacitorchargingdischargepower electronicssupercapacitor

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.

Best Answer

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.