Two sources of power:
12V switched power supply (primary)
12V deep cycle lead acid battery (secondary)
Note: In practice the switched power supply voltage will be adjusted to match the actual voltage across the battery when it is connected to the load.
12V Relay:
NO connected to secondary power supply
NC connected to primary power supply
Relay coil connected to primary power supply
Load (12V):
Surveillance cameras
Network attached storage
Router
Assume Total 10A
Simplifications:
safety precautions such as fuses ignored for now,
battery Ah, run time etc. ignored for now,
why am I doing this ignored for now.
Question:
When the relay switches from one power supply to the other there will be a temporary open circuit.
How do I calculate capacitor size to use?
The open circuit time is unknown but can I assume it is say 10ms?
Solution? Charge:
$$Q = 10\text{ms} \cdot 10\text{A} = 0.1\text{C}$$
$$Q = C \cdot V$$
Capacitor size:
$$C = \frac{Q}{V} = \frac{0.1}{12} = 8,333 \mu \text{F}$$
Now I realize that this means that at the end of 10ms the Voltage is zero. So I figured I'd ask for some advice:)