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I have a circuit powered by a 9V battery. The circuit will be subjected to significant vibrations. I want the circuit to not power off if the battery is disconnected for a second. Would a capacitor in parallel with the 9V battery do this? How do I know what capacitance I should use? Also, would the capacitor draw too large a current when the circuit is initially powered on?

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  • \$\begingroup\$ Yes it would, capacitor back up is common enough to power real time clocks etc. The capacitor you need depends on the current requirement of the load and the minimum voltage the load can operate at. \$\endgroup\$ – Puffafish Oct 3 '18 at 15:24
  • \$\begingroup\$ This tends to work when the battery voltage is quite a bit higher than the minimum required operating voltage. When the minimum required operating voltage is close to the battery voltage, then this approach doesn't work very well. I would suggest you simply use a different battery connector that is not subject to disconnection. \$\endgroup\$ – mkeith Oct 3 '18 at 16:06
  • \$\begingroup\$ This problem is exist in automotive too but 1000ms power cut is very long time. For calculating the Capacitor, The circuit's load specification and the minimum input voltage that is OK for your regulator to work with stability must be specified. Using big capacitors make high in rush currents and for controlling them you have cheap solution that is serial low Ohm resistors and more expensive solution that is using PTC and Thermistors. \$\endgroup\$ – BD_CE Oct 3 '18 at 16:12
  • \$\begingroup\$ I've seen thermistors being used to stop the in rush current into a cap during power up. The resistor is 2 ohms when cold, and like 0.3 ohms when warm (so that you aren't wasting energy during normal operation). Unfortunately, if it is warm and your connector disconnects for a long time (ie the cap is fully discharged), then you will still get the rush of current. \$\endgroup\$ – PJazz Oct 3 '18 at 16:45
  • \$\begingroup\$ Hannah, would you be able to edit your question, and add the extra details needed to provide a good answer? Mainly, how much current does your circuit use at 9V, and what is the lowest voltage that still works reliably? This will help to figure out a good solution for you. \$\endgroup\$ – mkeith Oct 4 '18 at 3:04
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Yes, it is possible to use a capacitor to keep the device powered during black outs and brown outs. I have done various projects (some withe real time clocks, others with active events on tamper etc).

The question is the current drawn by the load, and the voltage the load will work down to. If you look up a capacitor charge/discharge graph, you can see how the capacitor's voltage will drop as current is drawn from it.

If we know that the device will draw 1mA, and that the load will operate as intended until the voltage has drooped to 6V, we just need an equation to plug the numbers into... Luckily a simple google gives you a formula, or even a calculator.

Plugging in the numbers we've already talked about: V=9 down to 6, I=1mA, with an randomly selected capacitance of 2mF (a not uncommon, but large, capacitor value), gives us a time of just under 6 seconds... Call it 5 seconds to give some head room.

Now, one point I've glossed over, high value capacitors tend to be rated to 2.7V or so. This means that for your 9V supply, we'd need four in series. Capacitance then decreases as they are put in series, Ctotal = 1/((1/C1) + (1/C2)) etc. So you'll need to factor that into your total capacitance.

Four 2mF caps in series would give us 0.5mF, so now we're down to 1.5 seconds. So pretty close to your 1 second requirement mentioned in the question.

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  • \$\begingroup\$ I suggest you remove the "super" capacitor reference. 2 mF is nothing special and many of the super capacitors have limited maximum voltage ratings. It might send the OP off on the wrong track. \$\endgroup\$ – Transistor Oct 3 '18 at 15:48
  • \$\begingroup\$ Yes, will remove super. But paragraph 5 does mention about voltage limitations. \$\endgroup\$ – Puffafish Oct 3 '18 at 15:51

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