1 - Understanding what you’ve bought or are going to buy

First of all let me preface all of this by saying that I pretty much bought a motor at random (very cheap on eBay) so I’m not suggesting that this is a suitable motor for your project.

So here is what I purchased (at around \$11 for two)

http://docs-europe.electrocomponents.com/webdocs/0030/0900766b800305e5.pdf

http://docs-europe.electrocomponents.com/webdocs/001b/0900766b8001b258.pdf

which are less than informative. My motors arrived with no wiring/plugs etc so the colour charts were less than useful – at which I point I wondered if I would ever make them work!!

However:- this is what I could learn from the datasheets:-

• moves 7.5 degrees per step (ie  48 steps per revolution on the motor shaft)
• has 6 wires
• is a uni-polar motor
• has a 50:1 gearbox (so 50x48 or 2400 steps per revolution on the output shaft)
• holding torque = 100 Ncm
• working torque = 100 Ncm
• typical working torque = 65 Ncm
• Step rate at typical working torque=300Hz
• requires a 5V supply
• Resistance = 9.1 ohms per phase
• 550 mA per phase

So long as you have two of the last three parameters then you can calculate the other using the equation Volts =  Amps x Ohms

There is one potentially important statistic missing from these datasheets that can be calculated from the data given. If we can issue a maximum of 300 steps per second, (300Hz), and – due to the gearbox – we need to issue 2400 steps for a full revolution then the fastest we can turn the output shaft is (2400/300)  ie once every 8 seconds.  This may be rather slow for an output shaft if it is connected to wheels but could be fine if we are rotating the turret of a tank. So this is an important fact to ascertain before purchasing your motor.

So what have we learned?  The gearing, steps per revolution and the maximum frequency produce trade-offs between torque and speed. Whether the motor is uni-polar or bi-polar can also influence the required current draw, and torque, as discussed earlier.