# Power a 12V water pump by a 12V battery through a relay

Hello everyone,
I am planning to control a water pump with my ARDUINO uno.
The water pump (link) has the following specs:

Working voltage: DC12V

I have a relay module (link) that I will use to connect the pump to a 12V battery (23A)

I have two doubts:

1. The Relay load voltage is AC250V 10A, DC30V 10A, are these value an upper limit? Do I need to use the mains or can I use my 12V 23A battery?

2. If I want to calculate the power consumption of the water pump, the load power 48W is intended per hour? I understand that 23A is the total current that the battery can provide but I do not seen any time variable anywhere. Could you please give me a hint for that?

Thank you so much

1. The relay figures are maximum limits (as the page you linked to says) so switching 12V 400mA should be fine.

2. Real confusion here. 12V @ 400mA is NOT 48W it is 4.8W so the specification is really suspect. Also a link to your battery is needed. Batteries are normally specified in Ah (where the h means hours) NOT A so we need to check the actual specification.

Steve

1. Since in each blue channel there are two series of numbers, such as 10A 30VDC - 10A 28VDC, I did not know if they were representing max and min. I am going to try first a simple arduino tutorial with a 9V battery to see if the relay works.

It is just defined as 23A.
You are right regarding the watts, I have just seen the label on the pump I have and it is different from the data in amazon, i.e 3.6W. I cannot find any Ah in the link I posted

Any clue?
cheers

23A is the battery type. It has nothing to do with the current (Amps) or capacity (Ah). A typical 23A-type battery has a capacity of around 35-40mA. It also has a maximum current of around 2mA constant. So it won't deliver 400mA but if it did it would only last for around 5 minutes before it was dead.

Now you see why we need links to the actual components.

Steve

slipstick:
23A is the battery type. It has nothing to do with the current (Amps) or capacity (Ah). A typical 23A-type battery has a capacity of around 35-40mA. It also has a maximum current of around 2mA constant. So it won't deliver 400mA but if it did it would only last for around 5 minutes before it was dead.

Now you see why we need links to the actual components.

Steve

I see, thanks
Any suggestion for a battery that can power the pump? A holder with more batteries? A car battery is too expensive, in that case I would go for a way to power it from the mains.
The pump is supposed to work 5 times for 5 minutes (every three days, but I guess this is not relevant).
thanks for any help

Yes, that is the battery for cigarette lighters and remote control fobs.

For this motor, you would want either a battery carrier for eight alkaline "AA" cells:

or for really serious use, two of these:

Now I am not sure whether the stall current is significantly greater than 400 mA but it says it is a "brushless" motor so it may be limited. Motors tend to have a much higher current draw when stalled or at the moment of starting up.

Question:
What is energy storage (milli amp hour or mAh) of these batteries?
55mAh
By Amazon Customer on 04 May 2018

Those are just a stack of 8 1.5V button batteries, I would go with a sealed lead-acid batt and charger.

There are many options. I'd probably use a 3S Lipo rechargeable battery but that's because I'm used to them and have some already. But they do need proper chargers to recharge them.

Steve

Thank you for all you replies.
Because of the temporary duration of my first project, I would try to go for cheaper options even though a rechargeable battery could be reused for more projects or long term duration systems.
I have found this one in the link which would be good for me since I do not need to order online.
It is a
Sealed Lead Acid Battery 12V 2.3Ah
If the pump is using 400mA, with this battery I should have a capacity for 5 hours. Am I wrong?

Do you reckon I can go ahead with this battery?

The pump is not a problem - 15 mins every 3 days. That's about 0.033 Ah per day.

The Uno is a major problem - it's minimum draw is about 45 mA, 1.08 Ah per day. So in two days your battery will be completely dead just from the draw of the Uno.

To lower your power use consider switching to a Pro Mini. After removing the regulator and the power LED you can go down to the µA range, so no problem there.

Please note that the capacity of this lead/acid battery is hardly more than that of a pack of 8 AA alkaline cells or 3 LiPo 18650 rechargable batteries.

wvmarle:
The pump is not a problem - 15 mins every 3 days. That's about 0.033 Ah per day.

Hi,
I have just purchased the battery I mentioned in the previous post. Hopefully, it would be fine. It is also rechargeable and it is just a little more expensive than purchasing 8 AA and a holder.
But now I am confused with your calculations, what have you done to end up with that number?

The Uno is a major problem - it's minimum draw is about 45 mA, 1.08 Ah per day. So in two days your battery will be completely dead just from the draw of the Uno.

To lower your power use consider switching to a Pro Mini. After removing the regulator and the power LED you can go down to the µA range, so no problem there.

Please note that the capacity of this lead/acid battery is hardly more than that of a pack of 8 AA alkaline cells or 3 LiPo 18650 rechargable batteries. Since 15 minutes is 1/4th hour: 400mA/4/3 days=0.033A=33mA. Is that what you did? so 400mA is actually 400mAh ?

Regarding the Arduino Uno, thank you for pointing it out. I was going to investigate this aspect later on. My initial idea was just to use a 9V battery and a sort of sleep mode if it does exist. Otherwise, I was not so worried because as a first attempt, I might plug it to the mains though the usb. I am tring to avoid it and any suggestion is much appreciated. In the future, the arduino mini is something I will take into consideration.

When I have all the components, I will create another post regarding this project on the proper section.
Cheers

Pump uses 400 mA, runs 1/4 hr in 3 days.
0.4 * 0.25 / 3 = 0.033 Ah/day.

The Uno won't sleep well. The processor does, but the rest not: the power LED, USB/TTL converter and regulator.

Don't use a 9V block battery if you have a much better 12V one on hand.

Hello,
I am working on my first project with Arduino and I would like to design a simple system to water my bonsai during my holidays.
I am looking for:

1. a temp solution
2. to avoid supply from the water and electrical mains (use of water can and battery)
3. use of the water pump every 3 days for 5 minutes for 3-4 weeks.

• Arduino Uno R3
• electric wires (2A)

I have started a preliminary discussion on the power consumption here
Sorry, I was not able to use fritzing and I draw a schematic by hand that you can find attached.

Is that right?

Thanks

Didn't look at things really closely, but you need to connect all the grounds. Do you have any specific problems or questions?

Let’s see the actual wiring.

That relay board has 5V relays, not 12V.

Make sure the relay jumper is removed.

Add a reversed biased kickback diode across the motor.

wvmarle:

Apologies,
I did not hide the link in my post, I though to continue in the Project section because now the topic is more general than the initial title
but is is ok

Power_Broker:
Didn't look at things really closely, but you need to connect all the grounds. Do you have any specific problems or questions?

I will try the wiring according that schematic soon (if it is correct).
Is it missing the connection ground arduino with ground relay?
What is the purpose of having common ground?

larryd:

Let’s see the actual wiring.

That relay board has 5V relays, not 12V.

Make sure the relay jumper is removed.

Add a reversed biased kickback diode across the motor.

Sorry I cannot visualize your imagine.
Regarding the diode, by using a relay I decoupled pump and arduino, Do I need a diode?

Cheers

Let's see the actual wiring.
That relay board has 5V relays, not 12V.
Make sure the relay jumper is removed.
Add a reversed biased kickback diode across the motor.

You need the diode to snub out inductive switching spikes.

Hi all,
I am catching up with the project. Before using a 12V battery and pump, I tried to devise a system for a 9V battery and 9V motor (modifying one Arduino starter project).

I attached two photos of the configuration I tried.

Clicking on the switch button, Arduino send a LOW signal to the Relay and the motor starts spinning.

I connect the positive of the 9V battery to the COM of the relay, I used a wire to connect the NO of the relay to the positive wire of the motor. Both negative of the motor and battery are connect to a common ground to Arduino GND.

I used a diode 1N407 between positive and negative of the motor.

Everything seems working; clicking the button, the IN1 led switches on, a tick is heard and the motor starts.

My questions are:

1. I did not remove the jumper VCC-JDVCC. Is the diode I added in the schematic enough to protect Arduino from the back EMF?
2. If I remove the jumper, do I need a third external power supply or can I use the 9V battery to power the relay and the motor?
3. Compare to other relay I did not find a GND from JD-VCC. Is that a problem?
4. Does JD-VCC supplies power to the coil or also to the motor? I mean, by leaving the jumper the motor is powered just from the 9V battery or the JD-VCC contributes to raise the voltage/intensity?

Thanks for any reply. When I am sure to do not damage anything, I will change the 9V battery with the 12V and the motor with the water pump.
Thanks