Capacitor does not hold charge when switching power

Question

I'm trying to create a power source switch using a relay. This's supposedly will be used for my Raspberry Pi. So what I want to do here is :

• Create a backup power source with battery that will replace the main power if it's failed/off
• Add a capacitor to holds charge, so when the relay is switching, the power will not completely cut off

Here's my schematic:

I already tried to create this, but the problem is when the main power is cut off, the output still drops and then backs up to 5V. So I've assumed the capacitor is not doing its job.

But strangely, if I cut both of the power, the output still shows 5V (and slowly going down), so this means the capacitor actually holds a charge, right? Then why when the relay switching, the power still temporary went down? Do I did this incorrectly?

Answer 1

The cause of the problem could be the relay's switching characteristics.

The 'pull-in' voltage of a relay would be closer to its nominal voltage whereas its 'drop-out' voltage would be much lower.

After occurrence of a power failure, the relay's 'drop-out' characteristic would ensure that it stays on in spite of the decaying power supply voltage.

Hence the capacitor would also get discharged before switch-over to the battery could occur.

Here's an alternative scheme using a Schottky diode for switching.

The PSU would predominate, with it's voltage set marginally higher. In any case the probability of the PSU voltage exceeding that of the battery is quite high.

With availability of Schottky diodes having a forward voltage as low as 150mV, voltage drop during backup should not be an issue.

Answer 2

There are standard chips to do what you want, they're called ideal diodes: https://www.analog.com/en/products/ltc4413.html#product-overview

I suspect that one of these is way cheaper than that relay by itself.

As to why the relay doesn't work, it's simply too slow. At 2A, that capacitor should drop almost a volt per millisecond (I = C dv/dt). Most relays won't make contact before 10mS so you've got quite a bit of time where the relay has disengaged from one power source but hasn't engaged the other yet.

You might find something called a "make-before-break" relay that can mitigate this, but you will short the two power systems together briefly with one of those.

Answer 3

What voltage is your battery actually (when disconnected)? Or what chemistry & configuration of cells. A battery will not be 5.0V all the time. The voltage will vary by state of charge. Does your diagram represent "battery" as a combination of battery and 5V regulator?

If the battery is not actually 5.0V (for example, 4.8V), it will be an additional load on the capacitor at first. The capacitor charges the battery slightly, until the voltages equalize. Then the capacitor voltage should stay at the battery voltage ongoing, which would decrease slowly as the battery discharges. In this case the capacitor voltage would not reach 5.0V as it can only get as high as the battery voltage.

This setup is not so great because the capacitor will wear out faster with repeated inrush/outrush current way above spec. Capacitor current on switchover is limited only by power supply current capacity, battery internal resistance, copper trace resistance, etc. Try adding a small-value power resistor (say 0.1 ohm) between the relay and capacitor, to limit inrush current to the capacitor. You can monitor current through the resistor by the voltage across it (on oscilloscope). Should tell you something about how current is flowing in the circuit as the relay switches.