In the above diagram, I am trying to reduce the input voltage of 7.4v to no more than 3v, which is the maximum rating for the M2 motor. This is for a boxmag used with an Airsoft gun, specifically an M16 style 5,000 round boxmag. The idea is to operate the gun and the boxmag off the trigger switch of the gun. I have had this circuit in use in other boxmags, but recent events has shown that a potential problem with overheating of the resistor can occur and thus result in frying the resistor.
My current components are listed below, Motor (M2) Diode (1N4001) (Had to replace the original Schottky diode) Resistor 5 ohm, 1/2 Watt, 5% tolerance Switch DPDT
The switch must be able to achieve two modes of operation, 1] constant power for priming the boxmag, and 2] selective power for operation off the gun's trigger switch. The diode acts as a control gate to prevent damage to the motor, or that is how I understand it anyways.
I ran some calculations using available data to determine the resistor's rating, these are the calculations below,
The motor has these statistics, Voltage range: 1.5 to 3 volts Current: 0.18 to 0.25 amps w/o load, 0.70 amps +/- 15% at max efficiency Output: 0.31 watts
If I understood everything correctly, the device (M2) has a resistance of 10.57 ohms, which when entered into the formulas and calculated out, should have brought me to 2.37 volts at the motor, but since the circuit keeps frying the resistor, I'm not so sure anymore.
Therefore, my question is, how can I address this issue such that I will achieve the correct results without risking frying the resistor?
Any assistance to achieve my goals is appreciated. Thank you.