I'm looking for an alternative for a backup battery i'm planning to use.

PCB is powered by 24V DC power supply. I'm using a 3.3V LDO in combination with 3,6V battery to power mcu and other components (~20 mA):

current schematic

It would be more convenient to use a supercap. The application only needs to run for a couple of seconds (@20mA). I'm worried about inrush current so my proposed solutions would be adding a resistor (1). However this will result in a bigger capacitor. Besides my Vsense line is no longer available to detect lost +24V supply:

LDO with supercap and series resistor

So what do you think about using a supercap with series resistor (for charging) and a schottky diode for uncharching? +24V supply failure will be detected on the +24v side:

enter image description here

  • \$\begingroup\$ why don't you put Vsense on the +24V line if that is what you want to detect failure on?? The LDO output will stay high for a long time due to the battery/supercap. \$\endgroup\$ – KyranF Mar 26 '15 at 16:33
  • \$\begingroup\$ @KyranF, that's true. I'll edit my post and remove D3 in the last image. Main question is still regarding charge (through R18) and discharge (through D1) of the supercap. Is that ok with R18 and D1? \$\endgroup\$ – user2252031 Mar 26 '15 at 17:02
  • \$\begingroup\$ What other power inputs/supplies connect to VDD? Note that the capacitor will still discharge through R18 into VDD, just slower than through D1 - but D1 will drop voltage, so it may be ignored/useless anyway. if you put D1 where you had D3 before, it would work (and stop other supplies connected to VDD from trying to back-charge the capacitor. \$\endgroup\$ – KyranF Mar 26 '15 at 17:10
  • \$\begingroup\$ You can also get away with R18 being smaller, like 33 or 22 Ohms if you wanted. Currently 47R at 3.3V will charge the cap at 70mA to begin with \$\endgroup\$ – KyranF Mar 26 '15 at 17:11
  • \$\begingroup\$ There are no other power supplies connected to VDD. Just a mcu, some extra ic's (of which a RS485 transceiver ic). So I think there would be no need to put D1 where D3 was. I could simulate how the discharge current through R18 and D1 would go. \$\endgroup\$ – user2252031 Mar 26 '15 at 18:43

Consider putting the capacitor (along with its inrush limiting resistor and discharging diode) before the LDO. You probably want to run the whole lot from another diode to prevent it trying to drive the 24V supply.

Then you can use a lower value capacitor, because it can sag from 24V to somewhere around 5V before the LDO regulator starts to notice. Its voltage rating has to be higher, though, so I can't say whether there is an overall cost saving.

  • 1
    \$\begingroup\$ Would the trade off in package size from capacity versus voltage rating be worth it? I wonder what the size difference would be like.. also the huge cap would be wasting the difference in voltage while running the LDO, right at the time you do not want to be wasting energy. It's more efficient energy-wise for it to be on the output I suppose due to that \$\endgroup\$ – KyranF Mar 26 '15 at 17:32

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