Electronic – arduino – 12v DC brushless PWM fan burnt out via Arduino Uno

The first choice to make is whether you want to switch the input to the 12 V power supply, or switch the 12 V to the pump. If the pump spends a lot of time off, then powering down everything is probably a good idea. The 12 V supply will take some idle current, which is a waste if the 12 V isn't used for long periods of time. On the other hand, if the pump is on a lot and is turned on and off a lot, it will be easier to switch the 12 V to the pump. You could then even run the microcontroller off of the 12V, perhaps even with a linear regulator if you can keep its current down.

To switch the line power into the power supply, a plain old relay would be the simplest choice. There are plenty of relays that can be controlled from 5V that are intended to switch line power. Your supply will take well under 1 A, so there will be a wide choice of relays.

To switch the 12 V power to the pump, I'd probably use a transistor as a low side switch. Since the 12 V is already isolated from the line, you can tie the - side of the 12 V supply to the processor ground and use a direct connection. Lots of stuff can switch 1 A at 12 V. Here is a simple circuit:

First off, you dont need a series resistor with the fan. If its rated for 1.68W at 12V, then as you calculated it, the current through it at 12V will be 140mA. The usual rule of thumb for saturating (turning it fully on/off) a transistor is that the base current should be 1/10 of the collector current.

That means that the base current would have to be around 14mA. That much of the current the arduino is probably capable to supply. But if its possible, I would rather use a MOSFET.

Ok, lets say that you use 2n3904 transistor, this is how you'd calculate the required base resistor. Because the BE junction of a transistor forms a diode, the voltage drop on BE will be around 0.7V. And if the emmiter of the transistor is at ground potential, then the voltage at the base would be around 0.7V. Following from this is a simple ohms law: \$R = \dfrac{U}{I}\$.

So: \$R = \dfrac{(U_{arduino-pin} - 0.7V)}{0.014A}\$

\$U_{arduino-pin}\$ is the maximum voltage arduino can provide on the output (usually 3.3V or 5V), depends on your supply voltage.

And this is how the schematic would look like:

^{simulate this circuit – Schematic created using CircuitLab}

Also note the diode D1, it is there to protect the transistor from the back EMF, genereted by the coil (motor in the fan) when the transistor turns off.