Electrical – Capacitor size across power source selector relay

capacitorpowerpower supplyrelay

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:)

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.