Boxmag wiring diagram

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.


closed as unclear what you're asking by Olin Lathrop, PeterJ, Daniel Grillo, dim, Dmitry Grigoryev Jun 24 '16 at 13:02

Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.

  • \$\begingroup\$ Stack exchange doesn't use PHPbb type markup, there's a button with an image insert (or Ctrl-G) where you can upload an image from your PC if you want to edit your question. I was about to try editing it but the Photobucket links don't seem to be working. \$\endgroup\$ – PeterJ Jun 24 '16 at 10:18
  • \$\begingroup\$ Those links don't work so we can't tell what is being asked. Closing. \$\endgroup\$ – Olin Lathrop Jun 24 '16 at 10:55
  • \$\begingroup\$ Sorry, didn't know how that worked, most forum sites I'm familiar with do not have a direct upload ability. I have fixed the issue and added some additional information, please release my hold status. Thank you. \$\endgroup\$ – BoogerMc Jun 24 '16 at 15:03

I'm guessing at your schematic, but if I assume you simply place the resistor in series with the motor then a quick calculation shows that your resistor is severely underrated:

To drop the voltage from 7.4 to 3 volt there must be 4.4 volts across the resistor. Using the formula P = U² / R it follows that P = 3.9 Watt, way too much for your 5 ohm, 1/2 Watt resistor. It may be okay for short bursts (e.g. trigger pulls) but for repeated / long press it will overheat.

Also note that the DC resistance you measure across your motor with a multimeter is useless. When a motor runs it draws less current due to back EMK. The only proper way is to measure the current while the motor is running. You will also find that there is a non-linear relationship between voltage and current; so you have to experiment to find the proper value.

  • \$\begingroup\$ Your point about long bursts makes sense, I believe this may be part of the problem as the player whose equipment I am attempting to repair has a bad habit of laying on his trigger and not letting up. However, I have this same set up in other boxmags and have not experienced any similar issues, which is what confuses me. I have been experimenting with different resistors this morning and it seems the higher the resistance, the less likely the motor is to operate. I can get it to work with a 10 ohm resistor, but a 100 ohm or higher results in no motor operation. \$\endgroup\$ – BoogerMc Jun 24 '16 at 15:07
  • \$\begingroup\$ I, unfortunately, do not have any other resistors in between these values, but several over the 100 ohms and none of them worked either. Additionally, I placed the negative lead of my multimeter on the negative motor connection and the positive lead between the resistor and the motor, when I would activate the motor, the meter would begin climbing until it passed the 20 volt mark I had it set at, which I am guessing is the point you were making about reading voltage while the motor is operating, it is kind of useless, or I'm I connecting things incorrectly to get an accurate reading. \$\endgroup\$ – BoogerMc Jun 24 '16 at 15:18
  • \$\begingroup\$ I always get a little confused when it come to placing the leads for a proper reading. Hopefully the new images and information clear things up and make it easier to understand. Thank you for your input, it is helpful. \$\endgroup\$ – BoogerMc Jun 24 '16 at 15:19
  • \$\begingroup\$ 20 volt? That's odd... Anyway, I should start with a 10 ohm, 5 Watt resistor; if the motor runs too slow, go for a bit lower value (8.2, 6.8, 4.7 etc. you better order a bunch of them). And oh, you can get rid of the diode; it serves no purpose here. The motor doesn't run at 100 ohm because the current is then too low. \$\endgroup\$ – JvO Jun 25 '16 at 9:24
  • \$\begingroup\$ After reading your response, I did some research to better understand your comments; after all, I'm still a novice when it comes to electronics, circuits, etc. You stated I was dissipating about 3.9 watts, which is something I had not considered when calculating for my resistor. Based on that, I purchased several 10 watt and higher resistors and began experimenting. My results lead me to install two 10 ohm 10 watt resistors and tap the middle point to create a voltage divider to supply the motor with the necessary voltage. ... \$\endgroup\$ – BoogerMc Jun 26 '16 at 18:09

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