# Electronic – How Do I Limit My Motor Current While Maintaining Voltage Drop Across It

current-limitingdc motorvoltage divider

I am trying to build my knowledge of working with dc motors by using an IRF510 MOSFET to switch a motor on and off. Please refer to the schematic below to see how I hooked everything up.

\\$S_1\\$ = jumper I used to toggle the gate voltage to \\$0\\$ or \\$5V\\$
\\$R_P\\$ = pulldown resistor (\\$9.85k\Omega\\$, measured)
\\$R_M\\$ = electric motor resistance (\\$1\Omega\\$, measured)
\\$i_{DS}\\$ = drain current (\\$\approx750mA\\$, measured)
\\$R_{DS}\\$ = drain-source resistance when MOSFET is on (\\$0.6\Omega\\$, from \\$V_{DS}/i_{DS}\\$)
\\$V_{BATT, open}\\$ = open circuit voltage of 9V battery (\\$8.68V\\$)

The first thing that put me in a loop was when the battery voltage dropped after closing the switch. I took some more measurements to get a grip on what was going on:

\\$V_{BATT, closed}\\$ = battery voltage when \\$S_1\\$ is closed (\\$3.11V\\$, measured)
\\$V_{M}\\$ = motor voltage (\\$2.57V\\$, measured)
\\$V_{DS}\\$ = drain-source voltage (\\$0.54V\\$, measured)

After some research, I determined that the voltage drop was due to the internal resistance of the battery. This is what I was able to figure out after some more calculations:

\\$V_{R_i}\\$ = voltage across \\$R_i\\$ (\\$5.11V\\$, from \\$V_{BATT, open}-V_{BATT, closed}\\$)
\\$R_i\\$ = internal resistance of the battery (\\$2.87\Omega\\$, from voltage divider)

My gut reaction tells me to use a voltage divider to maintain the voltage across the motor. I want to do this because my motor is rated to operate between 5V and 9V. I also want to add a current-limiting resistor in series to prevent the current from getting so high that it burns out my circuit and drains my battery. Ideally, I can achieve the desired \\$V_M\\$ if I add a resistor in parallel to the motor (\\$R_P\\$) such that the equivalent resistance of the motor and its \\$R_P\\$ (\\$R_{EQ}\\$) is much larger than \\$R_i + R_{DS}\\$. However, since \\$R_M=1\Omega\\$, the best equivalent resistance I can achieve is \\$approx1\Omega\\$, which puts be back squarely where I started. Furthermore, the series current-limiting resistor will take the lion's share of the voltage drop, thus robbing the motor of the voltage it needs.

How may I achieve the voltage drop I want and limit the current? Any help would be greatly appreciated. This is the first time I posted to StackExchange so I apologize if I broke any protocol.