Hi
can anyone suggest a simple working idea to make this work
I need. A circuit perhaps using a DPDT relay which will allow a mains powered 12v transform supply to a 12v air pump to switch to a backup 12v battery supply to the air pump in the event of mains power loss, and then back to the transformer when mains power is restored.
I would appreciate any suggestion.
Thanks
Les

 

Does the air pump need AC or DC supply?
How much current does the pump need?

 

Connect the two supplies to the load with appropriately sized Schottky diodes. If the supplies have output voltage adjustments you can compensate for the voltage drop in the Schottky diodes. The load will draw power from the supply with the highest potential.

 

Does the air pump need AC or DC supply?
How much current does the pump need?

Does the air pump need AC or DC supply?
How much current does the pump need?

 

Hi
Thank you for replying to my inquiry, below is the specifications of the pump, the idea is to run the pump by 240v input 12v dc output transformer, in the event of a power cut I want a 12v leisure battery to take over running the pump until the power is restored. I thought a NC DPDT relay held open by the power supply until a power cut which would then close connecting the battery might do the trick, or any idea you may have.
Thanks

Specifications:
Voltage: DC 12V
Power: 35W
Current: 3.5A
Pressure: 0.068Mpa
Flow: 68L/min
Diameter: 10mm / 0.4inch
Size: Approx. 220 * 110 * 140mm / 8.6 * 4.3 * 5.5inch
Weight: Approx. 1660g

 

Relay circuit will work fine doesn't matter what the battery voltage is.
Diodes will work as long as the 12 volt supply output voltage is higher then the battery voltage.

 

Check to see if the pump can handle a voltage of 13.8 volts. If so then you can use mains power to keep the battery charged. When power fails the battery will just take over the load. A diode to prevent back feeding the power supply would be all that is necessary. Obviously you'll want to be able to keep the battery charged and ready for use. A 12 volt lead acid battery likes a float charge of 13.8 volts.

In my garage I have a car radio hooked to a battery and a 13.8 volt charger. The charger keeps the battery topped off. The charger CAN run the radio but there's not enough power for good base tone. So the bigger supply is necessary. When I play the radio it mostly plays off the battery and the charger helps keep the charge topped off. When I shut the radio off the battery is charged back up to 13.8 volts and is ready for the next time I am in the garage working on a project or the car.

 

Relay circuit will work fine doesn't matter what the battery voltage is.
Diodes will work as long as the 12 volt supply output voltage is higher then the battery voltage.
View attachment 234731

D2 wouldn't be necessary as the supply can keep the battery charged if the voltage is set high enough to maintain a proper float charge. 12 volt supplies sometimes run a little higher than 12 volts. Nominal voltage for a lead acid battery is 12.6 volts. A charge voltage would have to be higher than that to maintain the battery. So if the charger can be adjusted then D2 would not be necessary.

 

Relay circuit will work fine doesn't matter what the battery voltage is.
Diodes will work as long as the 12 volt supply output voltage is higher then the battery voltage.
View attachment 234731

 

That’s great.
Thanks to all you guys for taking the time to advise on my question.
Les.

 

I am using a 18650 rechargable battery and a 5v supply to power a camera. Once the main power goes out, the battery takes over. When the power returns the battery charge 876 561 1580 on WhatsApp please need this project done. Will be using a breadboard

 

Welcome to AAC.

Recommend you start your own thread.

Caution: depending on how you're hooking things up - you could be setting your battery up for a firey explosion if you charge it the wrong way. Please post a diagram of your circuit.

 

I created a similar circuit to includ battery float charge for a friend's powered recliner chair. One series diode string to isolate the chair circuit and a battery connected directly to power the motor when the power fails. The diode string was needed to drop the supply voltage down to the float voltage that the battery needed. 24 volts from the battery works well to drive the motor, it takes a string of diodes to drop the 28 volts from the supply down to the specified float voltage. The speed difference in the motor is not enough to be noticed, anyway. The duty cycle of the motor is very low, perhaps 3 minutes per hour on a busy day. But my friend is stuck in the chair if the power fails, so this was important. And at ten amp-hours for the battery, drawing the four amps will cover most day long power outages that they get. Ten amp-hours, drawn at 4 minutes per hour will last a long time.