Two sources of power:
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12V switched power supply (primary)
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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:
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NO connected to secondary power supply
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NC connected to primary power supply
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Relay coil connected to primary power supply
Load (12V):
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Surveillance cameras
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Network attached storage
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Router
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Assume Total 10A
Simplifications:
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safety precautions such as fuses ignored for now,
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battery Ah, run time etc. ignored for now,
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why am I doing this ignored for now.
Question:
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When the relay switches from one power supply to the other there will be a temporary open circuit.
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How do I calculate capacitor size to use?
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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:)
Best Answer
You are more or less close to the answer, but you should rather use the capacitor energy equation: \$E = \frac{1}{2} C * V^2\$
The Energy that you need to store can be calculated as
\$E_{needed} = Power * time = ( 12 V * 10 A ) * 10 ms\$
Then you have two instants of time, before and after the switching:
\$E_{before} = \frac{1}{2} C * V_{before}^2\$
\$E_{after} = \frac{1}{2} C * V_{after}^2\$
then \$E_{needed} = E_{before} - E_{after}\$
Also you want \$V_{after}\$ and \$V_{before}\$ close to each other, perhaps 1 V is ok (you must check the load specifications), so there is enough voltage to feed the load, \$V_{before}=12 V\$, \$V_{after}=11 V\$.
The voltage will always go down a "little", that is the rule when discharging capacitors, how "little" depends on how big the capacitor.