I am planing on controling water pump with my raspberry pi, and I didn't want to have 2 or more batteries so my thought was i could from 3.3v raspberry pi gpio output get 12v. with boos conventer to power the pump . I think Omh's law goes something like this : what you give you will get , so with P=U*I will "U" go up form 3,3 to 12v but "I" will drop and "P" will stay the same. So my question is does it work like that and if it does can my 12v pump start because of small currency. Sorry for bad english.

 

Trying to power a pump directly from something like an R-pi is probably not a good idea. The amount of current you can get from an R-pi gpio pin is likely limited to something in the 15 mA to 20 mA range. Even if you ganged a bunch of them together your total limit would probably still in the 50 mA to 100 mA range (you need to look at the specs -- I've never played with an R-pi, so I am just giving pretty typical numbers). Then that would get cut nearly in one-fourth even if your boost converter were 100% efficient, so your pump would have to be able to run on somewhere between 4 mA and 25 mA (probably closer to the lower end of that range). And to do that you would have to design and build (or find and buy) a suitable boost converter. Instead, how about starting with the 12 V battery that can run the pump and then use a regulator to get the 3.3 V for the R-pi? The R-pi's draw will probably be low enough that the loss in the regulator can be tolerated. If not, then there are LOTS of cheap 12 V to 3.3 V DC-DC converter chips out there.

 

Hello fan, welcome to the forums!

We will try to work around your English.

What you say is true, you need less current at 12 volts to get the same power as at 3.3 volts, but I don't think that is helpful to you.

A 12 volt pump needs 12 volts to run, so you need a 12 volt battery there. To get 3.3 to run the pi you could put a regulator there. A switching regulator may be best depending on how much current the pi needs, but the pi may need so little current you don't really gain anything over a simple linear regulator.

The pi output would drive a transistor or better a MOSFET to control the pump. You also need a diode across the pump as it will make a nasty spike when you turn it off. The diode catches that and keeps the MOSFET alive.

 

Use a 12V supply to power your pump. Use a different supply for the Pi.

The best way to switch the pump from the GPIO output on the Pi is to isolate the signal using an optical isolator. These have a LED on one side and a photo-transistor on the other.

The GPIO drives the LED. A series resistor might be needed. The photo transistor is used to switch a larger transistor that controls the pump.

It is possible to buy a device that contains all the parts in one package. Usually called an SSR. Example.

 

Use a 12V supply to power your pump. Use a different supply for the Pi.

The best way to switch the pump from the GPIO output on the Pi is to isolate the signal using an optical isolator. These have a LED on one side and a photo-transistor on the other.

The GPIO drives the LED. A series resistor might be needed. The photo transistor is used to switch a larger transistor that controls the pump.

It is possible to buy a device that contains all the parts in one package. Usually called an SSR. Example.

Note that this doesn't satisfy the TS's primary goal, which is to NOT use two supplies.

Other than that, though, good advice.

 

Note that this doesn't satisfy the TS's primary goal, which is to NOT use two supplies.

Very true!

I suggest getting the supply for the Pi from the 12V supply using an isolated DC-DC converter. 12V in, 5V out.
Keep the Pi isolated fron the noisy 12V pump motor.

 

Thanks you guys , you helped very much.
I was thinking a solution with pnp transistor but now I see there are a lot of better alternitives. Can someone draw me a scheme with 12v battery powering pi and a pump. Thanks

 

How much current does your Pi draw? That is an important design point. That is with any external loads too.

 

Idk really but it draws 5v and maymaybe 0.7A