5
\$\begingroup\$

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.

enter image description here

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 (3x4.7F, 2.3V caps in series). Calculated time = 8.2s, actual time = 8.5s
  • 2.35F (2x4.7F, 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.

\$\endgroup\$
  • \$\begingroup\$ Is there any difference in the voltage rating of the capacitors? I'm suprised you put the supercap before the regulator since this means more wastage and higher price due to the required higher voltage rating (unless you have to have a higher voltage for some reason). Also, the LDO could already have reverse circuit protection and if you place the supercap after the LDO, no need for LED. \$\endgroup\$ – Gustavo Litovsky Dec 10 '12 at 16:12
  • \$\begingroup\$ By the way, I noticed that there is a difference between the multiple capacitors and the single one, which is the voltage rating. This could be a big difference. Did you ensure that the capacitors had time to charge up? \$\endgroup\$ – Gustavo Litovsky Dec 10 '12 at 16:14
  • 1
    \$\begingroup\$ 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. electronics.stackexchange.com/questions/30762/… \$\endgroup\$ – Brian Drummond Dec 10 '12 at 16:15
  • 1
    \$\begingroup\$ @BrianDrummond: You're right - that's the issue. The 5.5V caps have an ESR of 30ohms, which at 200mA is a voltage drop of 6V, so it won't work, whereas the other caps have ESRs in the mOhms. If you rewrite this comment as an answer I can accept it. \$\endgroup\$ – Amr Bekhit Dec 10 '12 at 16:28
  • 1
    \$\begingroup\$ @Amr : your basic approach is fine! With the cap where it is you get good regulation from 4.6 downto 3.55V. I hope you can source a single 5V one with good ESR, but if not, 2 in series will work. If you end up using 2 in series, you probably want a couple of equal resistors across them to ensure they charge equally (to Vs/2 each) - a couple of K each should be fine, and only burn about 1ma. \$\endgroup\$ – Brian Drummond Dec 10 '12 at 17:07
10
\$\begingroup\$

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.

\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.