# Floating electrolytic capacitor with cathode connected

I have a board where I have some electrolytic capacitors on the input side of a voltage regulator. Under normal operation everything is ok. But the board can also during programming of the MCU be fed from the outputside of the regulator. This would leave the cathode (-) of the electrolytes (that might hold charge from previous operation) connected to ground and the anode (+) would be floating.

Please see the following conceptual circuit (not actual circuit!):

simulate this circuit – Schematic created using CircuitLab

During normal operation V1 is connected and supplying 12V and charging C1. During programming, V1 is disconnected and V2 is connected. The ground level may have shifted and C1 could possibly contain charge.

Do I risk damaging C1 when connecting V2 as the other terminal is floating and one is grounded to a new ground level?

Should I add a diode from OUT to IN to make the positive terminal of C1 not float when the device is being programmed?

• I'd be far more worried about damaging U1 by feeding voltage into its output. C1 will be fine and probably discharged by U1 long before you connect anything else. Jul 18 '17 at 12:37
• BE breakdown voltage is much higher than V2 so that's not a concern. Jul 18 '17 at 14:05
• Does C1 count as floating? It is connected to the regulator. Jul 18 '17 at 14:50

C1 will be fine. Floating is OK, it doesn't count as negative bias for electrolytic capacitors.

U1 might be damaged. Check the datasheet for the specific 7805 from the manufacturer you plan to use for recommendations. You may need to add a diode across it.

You may also want a capacitor on the output to improve stability and reduce noise. Again, check the datasheet for recommendations.

simulate this circuit – Schematic created using CircuitLab

• As I said it's not the actual circuit and I have the recommended output cap already. The datasheet of my particular IC has no mention of a shunt diode. Likewise datasheets for similar IC from TI says shunt diode is not necessary for 7805 as 5v is less than the 7v be breakdown voltage of the series pass transistor. Jul 18 '17 at 14:10

You can leave a capacitor in floating condition but capacitor discharged as soon as you disconnect V1 Supply via 7805 regulator. This is applicable if you are not using reverse supply from output side.

If you leave a capacitor in floating sage(connected to input of 7805 only) and giving supply from the output side of the regulator to the output pin, base and emitter of series pass transistor which is present inside of 7805 also who is responsible for output will get damaged and you may damage that IC.

To prevent this problem you need connect a shunt diode from output to input of 7805 so that you are able to supply from the opposite side which won't affect output pin.

In both of way you only need to worry about IC rather then capacitor.

Also, in your diagram, V2 value is 5V. This is not enough for input voltage range of the IC 7805. Minimum input voltage to the IC is 7V.