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Gerald_D
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Username: Gerald_d
Post Number: 819
Registered: 11-2005
| | Posted on Tuesday, February 06, 2007 - 09:15 am: | 
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Data sheets sometimes do not give all the ratings for various wiring configurations....
To find the rated Voltage of a stepper motors, multiply the Resistance by the rated Current. You need the Voltage rating to select the voltage of your power supply.
Ratings are typically only given for the Unipolar situation (using a single coil in each phase). With a 4-wire motor there can be no confusion as there are no alternatives:
- BIpolar (similar to "bipolar series")
With a 6-wire motor you can wire them either:
- UNIpolar (use half the motor), or
- BIpolar series (UNIvoltage X 1.4 and UNIcurrent ÷ 1.4)
(only 4 wires will be connected - 2 wires will not be connected)
With a 8-wire motor you can wire them either:
- UNIpolar (use half the motor), or
- BIpolar series (UNIvoltage X 1.4 and UNIcurrent ÷ 1.4), or
- BIpolar parallel(UNIvoltage ÷ 1.4 and UNIcurrent X 1.4)
Makes you go crazy! Nobody agrees on the best way to connect them for a CNC router - it depends on how fast the motor has to work. But for a router, the motors have to work fast and slow, so I end up with the middle choice, UNIpolar. That means you only connect 4 wires if there are more than 4 available. The wiring is easier, and you know you have a second chance if you burn half the motor! 
The wires which are not connected, must be separately(individually) insulated. |
Gerald_D
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Username: Gerald_d
Post Number: 822
Registered: 11-2005
| | Posted on Tuesday, February 06, 2007 - 08:07 pm: |
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The following diagram is from the Oriental Motor site. I have added the letters A to F for easier identification. Wire colours differ between the various manufacturers, but the principles stay the same:
Let's examine the first post in terms of wiring:
.... With a 4-wire motor there can be no confusion as there are no alternatives:
- BIpolar (similar to "bipolar series") Diagram A
With a 6-wire motor you can wire them either:
- UNIpolar (use half the motor), Diagram B, using yellow and white wires (center-taps), and then using black or green, and red or blue. ie. use half-windings only.
- BIpolar series (UNIvoltage X 1.4 and UNIcurrent ÷ 1.4) Diagram C. Yellow and white not connected
(only 4 wires will be connected - 2 wires will not be connected)
With a 8-wire motor you can wire them either:
- UNIpolar (use half the motor), Diagram D, using black/yellow or orange/green for the one coil, and red/white or brown/blue for the other coil
- BIpolar series (UNIvoltage X 1.4 and UNIcurrent ÷ 1.4), Diagram E: Yellow/orange are connected to each other, and they are not connected to the driver. Same with white/brown. The 4 wires going to the driver are black/green and red/blue
- BIpolar parallel(UNIvoltage ÷ 1.4 and UNIcurrent X 1.4) Diagram F: the four connections to the driver are clear. All 8 wires go to the driver, in pairs |
Gerald_D
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Username: Gerald_d
Post Number: 824
Registered: 11-2005
| | Posted on Tuesday, February 06, 2007 - 08:57 pm: |
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Difference in performance between series or parallel for an 8 wire motor, quoting Mariss:
"You have "two motors in one" when you have an 8-wire motor.
1) "Both" motors have the same low-speed torque.
2) The series connection phase current is 1/2 the parallel connected
phase current and runs out of low-speed torque at 1/2 the speed of the
parallel connected motor for the same supply voltage.
3) The series connected motor runs cooler than a parallel connected
motor for the same supply voltage.
4) Use a series connection and/or a low power supply voltage for
low-speed applications. Use a parallel connection and/or a high power
supply voltage for high-speed applications." |
Gerald_D
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Username: Gerald_d
Post Number: 829
Registered: 11-2005
| | Posted on Wednesday, February 07, 2007 - 08:25 pm: |
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From Oriental Motor reference:
 |
eloid
Registered
Username: Eloid
Post Number: 6
Registered: 02-2007
| | Posted on Wednesday, March 21, 2007 - 03:03 am: |
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can any motor be wire Unipolar, Bipolar
what the advantage or in general the better choice for large 5x9 foot table |
Mike Richards
Registered
Username: Richards
Post Number: 88
Registered: 05-2006
| | Posted on Friday, March 23, 2007 - 07:45 am: |
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If you're wondering why motor manufacturers make things so complicated - such as a Unipolar connection can only use 0.7X the voltage of a Bipolar connection but lets 1.4X the current flow through the motor, it all has to do with WATTAGE (heat). For example, a mythical motor that is wired with Bipolar connections and is rated for 10V at 10 ohms would let 1 amp of current flow through the windings. That also means that 1A X 10V would create 10 watts of heat.
If we hooked that same motor up using only 1/2 the coil, and still used a 10V power supply, we would have 10V / 5 ohms which would let 2A flow through the coil. The problem is that 10V X 2A = 20W of heat. Twice the heat means shorter life for the motor. However, if we reduced the voltage to 7V and still had 5 ohms of resistance, we would have 1.4A flowing through the coil. BUT, 1.4A X 7V is approximately 10 watts of heat.
That's where the magic numbers (0.7 and 1.41) come from. They're the numbers that are used to keep the WATTAGE within the motor's specifications.
(Maybe everyone else already knew the theory behind the practice, but I must have been sleeping when that particular concept was presented at the Oriental Motor training class that I attended many years ago. It all became clear today when I tested some new G203 stepper drivers with various Oriental Motors stepper motors. When I connected the PK299-F4.5B motor up and started playing with voltages the theory suddenly crystalized. I found that when I increased the voltage and then read the motor's temperature with an infared thermometer, that wattage - heat - builds in a hurry especially when voltages are raised to the limit!) |
Gerald_D
Registered
Username: Gerald_d
Post Number: 1231
Registered: 11-2005
| | Posted on Saturday, April 07, 2007 - 03:25 pm: |
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Condensed notes lifted from the Gecko-drive forum today:
(half-coil, half-winding are interchangeable with each other and with unipolar above)
MR
You said: Eight-wire motors are about 3% more efficient when parallel connected than an equivalent half-winding connected six-wire motor, but are considerably more complicated to hook up. There is no significant difference between a parallel connection and a half-winding connection.."
Would you please explain?
MF
Let's say you have a 4A parallel rated 8-wire motor and a 40VDC power supply. Each of the four coils have a 1 Ohm resistance.
If the motor is connected half-winding (1 coil), the resistive voltage drop is 4V, leaving an effective 36V across the coil inductor. If the motor were parallel connected, the resistance would be 0.5 Ohms, the resistive voltage drop would be 2V and the effective coil inductor voltage becomes 38V. The half-coil motor 'sees' 94.7% the voltage of a parallel-connected motor.
This difference is of no consequence because it occurs while the drive is operating as a current source. Inductive reactance increases with motor speed until it limits motor current and not the drive. This transition is coincident with the end of constant torque region of the motor's speed-torque curve. The winding's RMS current is 2A at this point.
This results in a resistive voltage drop of 1V for a parallel connection and 2V for a half-winding connection leaving 39V and 38V across the inductor respectively. The half-winding connection 'sees' 38/39 or 97.4% the voltage of the parallel connection. The difference, 2.6%, is insignificant. Raising the half-winding supply voltage from 40V to 41V would cancel the difference.
MR
The point of concern that is still the center of the debate is the relationship between a motor using a parallel connection compared to one using a serial connection.
If we use the Oriental Motor's PK299-F4.5 motor as an example, the specs for a parallel connection are: 880 oz*in holding torque, 6.3A/phase, 1.9V, 0.33 ohms/phase and 2.5mH/phase. The specs for the same motor connected serially are: 880 oz*in holding torque, 3.18A/phase, 3.9V, 1.32 ohms/phase and 10mH/phase.
The graphs published by Oriental Motors that show that that particular motor has good low speed torque when used either serially or parallel, but that torque with a serial connection drops off very quickly to 400 oz*in at about 100 RPM, while that same motor with a parallel connection is still producing about 400 oz*in of torque at 1,000 RPM.
Can you explain how to determine when to use the serial, half-coil and parallel connections.
MF
The underlying principles are very simple. For purposes of discussion take as a given from my previous post:
Series 8-wire = full-winding 6-wire. Parallel 8-wire = half-winding 6-wire
In 1,2,3 fashion:
1) Series inductance = 4 times parallel inductance. Inductance goes up as the square of the number of turns of wire in a winding. Current passes thru twice as many turns of wire in series as it does in parallel.
2) Torque is proportional to ampere-turns. Current passes thru twice as many turns in series as it does in parallel so the required is 1/2 what is required in parallel.
3) Inductance has a property called 'inductive reactance'. Reactance is a resistance to current and it is measured in Ohms. Unlike resistance, inductive reactance is proportional to frequency. Double the frequency (step pulse rate) across an inductor and its reactance doubles.
4) Inductive reactance obeys Ohm's Law. At a given frequency, current is proportional to voltage (I = V / R). Doubling the voltage doubles the current. From (2), this doubles torque as well. Power is torque times RPM. Doubling voltage doubles torque which doubles the power output. Motor power output is proportional to power supply voltage.
5) From (1), series has 4 times the inductance as parallel. From (4), series inductive current is 1/4th the parallel current. From (2), torque is proportional to ampere-turns. In series, 1/4th the parallel current passes thru 2 times the parallel turns of wire, making series torque 1/2 the parallel torque at a given speed and supply voltage.
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Series motor power output is 1/2 the parallel motor power output for the same speed and supply voltage. In fact, power output is:
Power = V / SQRT L
where V is the supply voltage and L is the motor inductance.
A series motor outputs exactly the same power as a parallel motor run at 1/2 the series motor supply voltage.
-------------------------------------------------------------------- |
Mike Richards
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Username: Richards
Post Number: 102
Registered: 05-2006
| | Posted on Saturday, April 07, 2007 - 06:20 pm: |
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One little oddity that I didn't expect pops up when we use Mariss's Power - V / SQRT L formula.
Using the data sheet from Oriental Motor's PK299-F4.5 motor, so that we all have the same reference data, if we use a 50V power supply for serial connection, a 35V for unipolor (half-coil) connection and a 25V for parallel connection and inductance figures of 10mH for serial, 2.5mH for half-coil and 2.5mH for parallel, we get: 50V / 3.16 = 15.81. 35V / 1.58 = 22.13. 25V / 1.58 = 15.81. Those results show that the most 'powerful' connection would be half-coil.
I don't know if that would actually be the case and I'm right in the middle of rebuilding my test setup, so I can't even test anything until early next week. In the meantime, if anyone else has run torque tests on a motor connected using all three wiring modes and has also adjusted the voltage as per Mariss's formula, I would like to see the results. |
Mike Richards
Registered
Username: Richards
Post Number: 103
Registered: 05-2006
| | Posted on Thursday, April 12, 2007 - 05:36 am: |
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Mariss posted some new formulas on the Yahoo GeckDrive forum that really helps in determining how well a motor should work. Here are the latest two formulas:
1) W = (Vs * Th) / (10^3 * L * Ir)
Where:
W = motor power output in Watts mechanical
Vs = supply voltage
Th = rated holding torque in in-oz
L = winding inductance in Henries
Ir = rated phase current in Amperes
2) RPM = (0.191 * Vs) / (L * Is)
Where:
RPM = the motor's corner speed
Vs = supply voltage
L = winding inductance in Henries
Is = drive's set phase current in Amperes
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If we use the Oriental Motor PK296A1A-SG3.6 motor as an example and a 70V power supply, the highest speed at which the motor should be run before losing too much torque, also known as the corner speed, is:
Series winding: (0.191 * 70) / (0.0308 * 1) = 434 RPM. Then 434 / 3.6 = 120 revolutions of the spur gear. 120 RPM would equal between 375 and 565 inches per minute depending on the number of teeth in the spur gear.
Half-coil winding: (0.191 * 70) / (0.0077 * 1.5) = 1157 RPM. Then 1157 / 3.6 = 321 revolutions of the spur gear. 321 would equal between 1000 and 1500 inches per minute depending on the number of teeth in the spur gear.
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PK299-01AA Motor
Series winding: (0.191 * 70) / (0.056 * 1.4) = 170 RPM. Then 170 revolutions of the spur gear would equal between 530 and 800 inches per minute depending on the number of teeth in the spur gear.
Half-coil winding: (0.191 * 70) / (0.014 * 2.0) = 477 RPM. Then 477 revolutions of the spur gear would equal between 1500 and 2200 inches per minute depending on the number of teeth in the spur gear.
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Mariss's new formulas seem to match the results that I've seen in numerous tests over the past 18 months. Although I don't have that particular motor, the forumla shows that for the PK296B2A-SG3.6 that I do have, the corner speed is about 1400 RPM at 33V. In my tests, I've noticed that at speeds of 1500 and below, that there is very little heating, but as the speed increases above 1500 RPM, the motor quickly gets too hot to touch. In Mariss's white paper on stepper motor theory, he predicted that the motor would get hot if run faster than the corner speed.
Of course, a lot of different things affect the feed speed on any machine, so you may get significantly different results on your machines, but once you have some working numbers from one motor, you would be able to substitute another motor from a reliable source with confidence in the expected results. |
Håvard
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Username: Soulvoid
Post Number: 15
Registered: 02-2007
| | Posted on Tuesday, April 17, 2007 - 09:20 pm: |
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How about bipolar parallel? It would be interesting to compare the results for both power and corner speed, however the inductance is not shown in the motor specs table for parallel. Looking at the torque/speed curve, I'd guess that bipolar parallel would give the most power. |
Mike Richards
Registered
Username: Richards
Post Number: 105
Registered: 05-2006
| | Posted on Tuesday, April 17, 2007 - 11:11 pm: |
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Bipolar parallel requires motors with eight leads. The standand PK296 and PK299 motors come with six leads. However, using Oriental Motor's web site for basic information, you would use a power supply that has 1/2 the voltage as you would use for a Series Winding and you would use 2X the current that you would use for a Series winding. Finally, you would use the same inductance figure as you would use for a unipolar (half-coil) winding.
PK299-F4.5 Motor with 20X power supply
Parallel winding: (0.191 * 38) / (0.0025 * 6.3) = 460 RPM. Then 460 revolutions of the spur gear would equal between 1400 and 2100 inches per minute depending on the number of teeth in the spur gear. |
Håvard
Registered
Username: Soulvoid
Post Number: 16
Registered: 02-2007
| | Posted on Wednesday, April 18, 2007 - 11:49 am: |
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I keep wondering if using less expensive motors (lower holding torque) and bipolar parallel is an inexpensive solution. It sure looks like it from these calculations as you get roughly the same corner speed and almost twice the holding torque as the unipolar/half winding rating. |
Mike Richards
Registered
Username: Richards
Post Number: 106
Registered: 05-2006
| | Posted on Wednesday, April 18, 2007 - 03:32 pm: |
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A motor connected as half-coil and with the power supply adjusted to give 20X the motor's rated voltage will produce about 70% of the torque of the same motor connected as parallel. However, the corner speed will be about 2X higher for the half-coil connection.
In the example below, the half-coil connection could easily spin fast enough to allow you to use a belt-drive gearbox. At 2:1 or 3:1, you would still have lots of speed and 2X or 3X the torque.
PK299-F4.5 Motor with 20X power supply
Parallel winding @ 38V: (0.191 * 38) / (0.0025 * 6.3) = 460 RPM. Then 460 revolutions of the spur gear would equal between 1400 and 2100 inches per minute depending on the number of teeth in the spur gear.
Half-coil winding @ 56V: (0.191 * 56) / (0.0025 * 4.5) = 950 RPM. Then 950 revolutions of the spur gear would equal between 2900 and 4400 inches per minute depending on the number of teeth in the spur gear. |
Håvard
Registered
Username: Soulvoid
Post Number: 17
Registered: 02-2007
| | Posted on Thursday, April 19, 2007 - 12:15 am: |
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Given
RPM = (0.191 * Vs) / (L * Is)
RPM = the motor's corner speed
V = supply voltage
L = winding inductance in Henries
Is = drive's set phase current in Amperes
And
- BIpolar series (UNIvoltage X 1.4 and UNIcurrent ÷ 1.4),
- BIpolar parallel(UNIvoltage ÷ 1.4 and UNIcurrent X 1.4)
And
Running the motors at 20x voltage
And
Inductance for bipolar series being 4 times that of halwinding, bipolar parallel being the same as halfwinding
Then we get
RPM (for half winding)= (0.191 * Vs) / (L * Is)
RPMs(for bipolar series) = (0.191 * Vs*1.4) / (L*4 * Is/1.4)
RPMp(for bipolar parallel) = (0.191 * Vs/1.4) / (L * Is*1.4)
With some elimination of the common values then
RPMs = 1.4/(4/1.4) * RPM = 0.49*RPM
RPMp = (1/1.4)/1.4 * RPM = 0.51*RPM
If this is correct (my math is a bit rusty and it's late) the corner speed of both kinds of bipolar connections are roughly half of the half winding connection. If we use the values 1.41 as suggested by Mike R, they are very much exactly half. I wonder if the graphs show are wrong or if it's just the falloff beyond the corner speed that is much less with bipolar series because of the lesser inductance.
For
W = (Vs * Th) / (10^3 * L * Ir)
W = motor power output in Watts mechanical
Vs = supply voltage
Th = rated holding torque in in-oz
L = winding inductance in Henries
Ir = rated phase current in Amperes
Bipolar holding torque is 1.41 times that of the unipolar.
W = (Vs * Th) / (10^3 * L * Ir)
Ws = (Vs*1.4*Th*1.4)/(10^3 * L*4 * Ir/1.4)
Wp = (Vs/1.4*Th*1.4)/(10^3 * L * Ir*1.4)
Ws = (1.4*1.4)/(4/1.4) * W = 0.68 * W
Wp = 1/1.4 * W = 0.71 * W
I'm guessing we have the same rounding error here as well, so they are both roughly 0.7 of the half winding connection...
Also interesting to note that the corner speed seems to fall in the midband resonance field(5-15RPS) making midband resonance compensation essential to get the most out of the motors. However without a gearbox and with pi inches to a revolution you still get 12 inches pr. sec with 4RPS or 240RPM and 720 inches pr. minute with a theoretical accuracy of 0.04mm even with a driver without midband resonance handling. It requires an efficient cutting tool to get the job done at that speed.
A belt driven ratio of 3 with small motor will be an inexpensive solution with high accuracy, however it adds a bit of complexity.
I think those formulas posted by mariss is the best tools for comparing motors that I've seen. |
Mike Richards
Registered
Username: Richards
Post Number: 107
Registered: 05-2006
| | Posted on Saturday, April 21, 2007 - 12:04 am: |
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Havard,
You've done an excellent job of explaining the relationship between the three types of wiring. I also agree about the low speed resonance problem. The biggest factor (to me) in running the numbers is that holding torque is greatly reduced a very short time after a motor stops - to keep the motor from over-heating. Using a gear box multiplies the torque. Adding 3:1 belt-driven transmissions to my Alpha was the single greatest improvement to that machine. Chatter was virtually eliminated. My best guess is that the additional torque produced by the transmission fixed the most critical problem. |
Håvard
Registered
Username: Soulvoid
Post Number: 19
Registered: 02-2007
| | Posted on Thursday, April 26, 2007 - 11:45 pm: |
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So basically, because you are using two windings instead of one you create twice the amount of heat (for the same current) and since the motor can only handle a given temperature you must reduce the current/power when running in bipolar? |
Mike Richards
Registered
Username: Richards
Post Number: 112
Registered: 05-2006
| | Posted on Friday, April 27, 2007 - 05:00 am: |
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That's the way it works. Current is Voltage / Resistance. Wattage (heat) is Current X Voltage. The various ratings for Series, half-coil (Unipolar) and Parallel give an almost consistent Wattage reading for a particular motor. The goal is to get the most power (Torque) out of a motor without over-heating things. |
Håvard
Registered
Username: Soulvoid
Post Number: 21
Registered: 02-2007
| | Posted on Monday, April 30, 2007 - 02:03 am: |
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Just thinking out loud here, how large a power supply do you really need? If the voltage is sufficient to get a corner speed that is beyond the practical cutting and jogging speed, are there any benefits to having a higher voltage? |
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