I have a buck converter.

  • Vin = 12V

  • Vout = 5V

  • Iload = 0.5A

I have one dropout requirement. The input voltage will drop from 12V to 4V for 10s. Including ramp up/down time of 10ms. During this time, my 5V output should be stable.

Q: How to calculate the value for my input/output reservoir capacitors value so that I have constant 5V during dropout time. Is there any formula?

  • 1
    \$\begingroup\$ Did you know C=Ic*dt/dV But was the 10s ramp with a 0.5A load? \$\endgroup\$ – Tony Stewart Sunnyskyguy EE75 Jul 25 '19 at 6:16
  • \$\begingroup\$ yes. Can you please provide me the calculation. I am not getting \$\endgroup\$ – Newbie Jul 25 '19 at 7:00
  • \$\begingroup\$ if the Vin must remain between 12v and 8volt, with input load of 0.25 amps, for 10 seconds, you need about 1 farad. \$\endgroup\$ – analogsystemsrf Jul 25 '19 at 7:41
  • \$\begingroup\$ what is the formula you used? and what capacitor? input or output? can someone explain it a little more briefly and simple language \$\endgroup\$ – Newbie Jul 25 '19 at 8:14
  • 1
    \$\begingroup\$ You won’t get a constant 5 volts so, be reasonable and think how much droop you can tolerate. \$\endgroup\$ – Andy aka Jul 25 '19 at 8:55

Let's try to keep it simple by making some assumptions:

The formula, as quoted, is C = I*dt/dV. Assuming the average input current is 0.25A, the voltage drop is 4V (i.e. from 12V to 8V) and the time is 10s, then C = 0.25*10/4 or 0.63F. Note that this is the MINIMUM value which will meet this requirement, and that it ignores the probable extra current demand from the converter as the input voltage drops. The efficiency of the converter is also a variable. Furthermore, you need to isolate the input from the capacitor during the dropout time - a diode will do the job, with a slight loss in overall efficiency.

In short, the suggested value of 1F appears reasonable. Have you considered a backup battery instead?

  • \$\begingroup\$ Thank you for the answer. But how can we place 1F big capacitor in the place. we have space constraint. No, we have not considered back up battery \$\endgroup\$ – Newbie Jul 25 '19 at 13:03
  • \$\begingroup\$ Now you are moving the goalposts. The original question has been answered. \$\endgroup\$ – henros Jul 25 '19 at 14:47
  • \$\begingroup\$ ok sorry. thank you \$\endgroup\$ – Newbie Jul 25 '19 at 16:41

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