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#1




The latest way to look at stepper motors for power supply calcs . .
Copied from Gecko yahoo forum today:
I suggest you can pretty much ignore winding resistance and winding voltage. There are only two electrical specifications that matter: 1) Rated motor phase current. This is the more important spec. Use this rating to select the drive's current set resistor value. 1a) Series or parallel? You have this choice if you have a 6wire or 8wire motor. No choice is needed with a 4wire motor. 6wire motor wired full winding = 8wire motor wired in series. For an 8wire motor use the motor datasheet specified series current rating. For a 6wire motor use 1/2 the datasheet specified current. 6wire motor wired half winding = 8wire motor wired in parallel. For an 8wire motor use the motor datasheet specified parallel current rating. For a 6wire motor use the datasheet specified current. 2) Motor inductance. This is the other important spec. It determines the maximum practical power supply voltage for your motor. That voltage is 32 times the square root of the motor inductance in milliHenries. V = 32 * SQRT mH. 6wire motors: full winding inductance = 4 times the half winding inductance. 8wire motors: series winding inductance = 4 times the parallel winding inductance. Explanation: Motor power output doubles when you double the supply voltage. Motor ironloss heating quadruples when you double the power supply voltage. This means motor heating outraces motor power output, placing a maximum limit on power supply voltage. This limit can be calculated from the motor specifications. Back when most step motors were round, expensive and 6wire, I came up with a simple calculation that worked well: Maximum supply voltage equals 20 times the motor's rated voltage. The underlying principle always was the motor inductance but the equation hid it and the necessity of pulling a square root. Times have changed. Motors are square, inexpensive and mostly 8wire. They are much better than the best round motors. What makes them inexpensive also crashes the old, simple equation; winding fill. The round motors used nearly 100% wire fill (the windings filled nearly all the available space on the stator). That kept the resistance to inductance relationship constant (R = L^2). The newer motors don't have 100% fill because oftentimes a smaller gage wire is used. This disconnects the resistance to inductance relationship and makes the '20 times rated voltage' rule inaccurate. Mariss 
#2




A little discussion between Mariss and myself on his yahoo forum:
Marcus/Mariss: With regards to stepper motors, we know that increase in driver supply voltage leads to extra mechanical power output (desirable) and more motor heating (undesireable). Increasing the current set resistor value gives the same sort of result: extra power output (desirable) and more motor heating (un desireable) Changing the supply voltage is diffult to do / changing the resistor is easy. Shouldn't the standard practice then be to select a supply voltage which is too high, and then adjust the resistor value for a lower current if heating is a problem? Thanks Gerald What you suggest is not true. Torque is proportional to current. At low speeds, the drive sets the motor current and by extension the motor's torque. In a switching type drives (all Geckodrives), current is either increasing or decreasing; it can never be a steady DC value. Current is regulated by having it pingpong between an upper and a lower limit whose average value is the motor's rated current. The upper and lower limits determines the amplitude of what is called 'ripple current'. Ripple current is what causes 'iron losses' (eddy currents primarily). Motor heating is the sum of iron losses and copper losses. Copper loss is the heating caused anytime current passes through a resistance. Example: You have a 2 millihenry (2mH) motor rated a 4A, 0.5 Ohms and you have a 24VDC power supply. The rate of current change (current slope) is equal to V/L amperes per second. 24VDC / 2mH = 12,000 Amps per second. That means current through the coil would be 12,000A after 1 second, 24,000A after 2 seconds and so on. The drive switches voltage to the motor from +24V to 24V and back 20,000 times a second (20kHz). That means +24V for 25 millionths of a second (25uS) and 24V for another 25 millionths of a second. The process then repeats endlessly. If current can change by 12,000 Amperes in one second, it will change by 0.3A in 25uS. This means the peak to peak ripple current amplitude will be 0.3A (300mA). If the drive is set to 4A, the upper limit is 4.15A and the lower limit is 3.85A. The current will pingpong between these two limits 20,000 times a second. If you double the voltage to 48VDC, the current slope becomes 24,000 Amps/sec, the ripple amplitude becomes 0.6A and the upper/lower limits become 4.3A and 3.7A respectively. Iron losses (iron heating) increase with the square of the ripple current, making them 4 times more than what they were at 24VDC. What if you set the phase current to zero Amps? The upper/lower limits become +0.3A and 0.3A, the ripple current is unchanged as are the iron losses. All that changes is the I^2 * R heating (copper losses) go to zero (was 4A^2 * 0.5 Ohms or 8 Watts) at the expense of having zero torque. The iron losses for this motor at 48VDC are the majority cause of heating and they remain unchanged. The motor will be only slightly cooler. Mariss Thanks for that detailed reply Mariss. Here is an example of a typical quandry . . . . . Do the calc of voltage and get a result of 68V. Look at the options from local suppliers and see that he stocks 50V or 70V supplies. A cautious person would pick the 50V supply. I am tempted to pick the 70V and adjust the current if heat is a problem. What would you do? Thanks Gerald It is entirely dependent on the application. If yours requires 300 RPM from the motor then it makes no sense to use any voltage higher than 24VDC. If it requires 3,000 RPM then 80VDC is the only option. The G203V has a yellow LED indicator that helps with this. It lights when the motor is turning fast enough to develop full power at your supply voltage. If it doesn't light during a rapid move then: a) Your supply voltage is unnecessarily high. b) You could go faster. An alternate method of selecting the supply voltage for any of our drives is: a) Baseline your system. Use a 24VDC power supply and run only one axis with it. It should be the most highly stressed one. b) Keep increasing the speed on that axis until the motor doesn't run reliably any more (begins to stall). Mark the last reliable speed as your "wachyagot" speed. c) Divide that into your "wachyawant" speed. d) Multiply the result by 24VDC. That will be your optimal supply voltage. Example: You get 60 IPM from (b) and you really need 150 IPM. From (c) you divide 150 by 60 and get 2.5 as the result. From (d) 2.5 times 24VDC is 60VDC. That is the optimum voltage for that application. Mariss 
#3




Quote:
Donald W. Ross 
#4




I wish that Mariss had added one more paragraph to warn us that the optimal voltage (to get the desired speed) might be too high for the amount of allowable heat. As voltage goes up, so does the speed, but the temperature goes up even faster.
So, using his suggestion, we can see what voltage we would have to use to get the speed that we want, but, to me, that is a starting point to see whether the power supply, the stepper driver and the stepper motor can actually work together as a healthy unit at that voltage and at that temperature. 
#5




Mariss has really made it look too easy. We don't have the luxury of adjusting the voltage for each job that comes along.
Also, his approach assumes that the torque requirement does not increase with speed  that is not true for a cutter moving through wood. It might be true for jogging with no load against steady friction only. However, I get a clear message that a cool motor has not reached its full "potential" (pun intended) 
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