I am currently working on a project on an automatic hand sanitiser dispenser. For this, I used LM358 and implemented an IR sensor. The output current of this IC (10mA) is passed to IRFZ44N and it amplified (70-80mA). But the problem is that the water pump requires current from 130 to 220mA. So, is there any way to amplify the current to the desired range? Or do breadboard implementation lead to current losses? If possible, should I make any changes to the load resistance to amplify the current? Here is the circuit diagram for the same.

 

Should that pump be between the mosfet drain and the 9V?

 

Is this a school project or just your personal habbit?

The D, S of MOSFET connected to 9V directly will damage the MOSFET.

 

How do you expect to drive the motor with the MOSFET connected between the 9V and ground?

Assuming you put the motor in series with the source of the MOSFET, then it should be able to readily drive the motor if it has sufficient gate voltage.
Measure the voltage at the MOSFET gate when ON.
It should be 7V or greater.

 

What is the 9V supply? If using a standard 9V battery, it may not be able to supply enough current due to the load (which causes voltage drop).
Also, the pump's internal resistance may be limiting the current with only a 9V supply. It may require a higher voltage supply to run properly.
Ohm's law prevails, V/R = I. You cannot change the current without changing at least one of the other two.

 

The notion of amplifying current is misguided. The only way a circuit can do that is to draw more current from the supply by making a "copy" of the low level signal. You cannot get something for nothing.

 

Your drawing shows the pump across the battery directly, and the fet shorting it out !!