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  #1  
Old Fri 06 June 2008, 05:08
Gerald D
Just call me: Gerald (retired)
 
Cape Town
South Africa
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 6-wire or
8-wire motor. No choice is needed with a 4-wire motor.

6-wire motor wired full winding = 8-wire motor wired in series. For an
8-wire motor use the motor datasheet specified series current rating.
For a 6-wire motor use 1/2 the datasheet specified current.

6-wire motor wired half winding = 8-wire motor wired in parallel. For
an 8-wire motor use the motor datasheet specified parallel current
rating. For a 6-wire 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.

6-wire motors:
full winding inductance = 4 times the half winding inductance.

8-wire motors:
series winding inductance = 4 times the parallel winding inductance.

Explanation:

Motor power output doubles when you double the supply voltage. Motor
iron-loss 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 6-wire, 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 8-wire. 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
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  #2  
Old Wed 11 June 2008, 11:48
Gerald D
Just call me: Gerald (retired)
 
Cape Town
South Africa
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 (un-desireable).

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
ping-pong 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 ping-pong 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
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  #3  
Old Thu 19 June 2008, 07:53
DocTanner
Just call me: Don Ross
 
Blue Ridge, Texas
United States of America
Quote:
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.
Finally a very simple way to verify if you are getting the most out of the setup.

Donald W. Ross
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  #4  
Old Thu 19 June 2008, 12:20
Richards
Just call me: Mike
 
South Jordan, UT
United States of America
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.
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  #5  
Old Thu 19 June 2008, 12:50
Gerald D
Just call me: Gerald (retired)
 
Cape Town
South Africa
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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