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#1
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Unipolar, Bipolar series, Bipolar parallel, choices for 2phase steppers
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. |
#2
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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 |
#3
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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." |
#4
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#5
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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!) |
#6
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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. -------------------------------------------------------------------- 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. -------------------------------------------------------------------- |
#7
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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. |
#8
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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 ------------------------ 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. ------------------------ 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. ------------------------ 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. |
#9
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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.
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#10
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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. |
#11
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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.
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#12
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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. |
#13
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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. |
#14
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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. |
#15
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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?
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#16
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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.
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#17
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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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#18
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uni-polar and bi-polar are not apples to apples.
Looking at the MotionKing motors; Series connected bi-polar is 4 times the mH, so the uni-polar number of 4.1mH would turn into 16.4mH The amps would be divided by 1.414 to get 2.82 amps. So, the MotionKing motors are really 2.82 amps and 16.4 mH, series connected bi-polar motors. sqrt(16.4) = 4.05 therefore 4.05 x 31.6 = 127 volts. that exceeds the 80 volt maximum, so the motors would have to be run on a voltage much lower than would be optimum. that means that either one would wire the motors a different way (single coil) or select a different motor. uni-polar motors can be connected in different ways. Depending on the way it is connected, the whole chart would have to be re-written. voltages- amps- mH... Using the MotionKing motors as single coil, (very common and very efficient use of the motor), the amps would be the same as the uni-polar chart, as would the mH and 64V selection you calculated. Since there is so much math behind this, I would highly recommend that anyone wanting to know more, read Mariss's Step_Motor_Basics.pdf http://www.geckodrive.com/photos/Step_motor_basics.pdf |
#19
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For the currently recommended motor (....A2A), it says the ratings are "per phase" but these are 6-wire motors, and the part number changes to B2A for a Unipolar connection. We can only use 4 of the six wires . . . .
Thus I am actually lost on how to connect the recommended motor and how many volts or amps it must get. As I said in that other thread "the risk that I don't know what I'm talking about" Where are Dirk Hazeleger and Mike Richards when you need them? Help! |
#20
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The Motionking 9801 motor is an 8-lead motor, so connecting it using Parallel wiring would probably be most efficient. The current rating for Parallel would be 4A * 1.4 = 5.6A. The Inductance would still be 4.1mH. The maximum voltage would be SQRT(4.1) * 31.6 = 63.98VDC. So, if four motors were installed, the power supply that I would use would have a minimum current rating of: 4 motors * 5.6 A * 0.66 efficiency factor = about 15A. The maximum voltage that I would use would be: SQRT(4.1) * 31.6 * 90% = about 57 V. So, just to be on the safe side, I would use a 40VAC toroidal 800VA transformer. After rectification, that would give me about 56VDC. That would allow the motors to pull maximum current and reach near maximum speed at a little less than maximum temperature.
EDITED: Gerald, you posted while I was typing. The PK296A2A-SGxx motor is the single shaft motor. The B2A is the dual shaft version. Both the A2A and the B2A are rated at 6mH Bipolar and 1.5mH Unipolar (half-coil). I use the PK296B2A-SG3.6 motors wired half-coil with power supplies ranging from 26VDC to 35VDC with excellent results. For half-coil connections with a Gecko G202 or G203, I use the Black wire and the Yellow wire for the A coil and the Red wire and the White wire for the B coil. i.e. Term #3 = Black wire, Term #4 = Yellow wire, Term #5 = Red wire, and Term #6 = White wire. |
#21
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Ouch, that is the second time the two model numbers made me think they were different electrically! Of course they are the same motor. And Dirk also said to use the unipolar configuration. So, the currently recommended "MechMate motor", the Oriental PK296_2A-7.2, should have a supply voltage of say 35 to 40 volts (37 from inductance formula, 28 from voltage formula) and 4 of those motors will want a power supply current of 8 Amps. That means around a 300 VA supply. Right?
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#22
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Mike, for that MotionKing motor, I would still go for the Unipolar, even though it has 8 wires. Still 56V but on a 600 Watt supply. Can't remember now why I favour unipolar over parallel . . . .
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#23
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Gerald, Bipolar series has good low end torque but poor higher speed torque, when compared to Bipolar parallel and Unipolar (half-coil). Bipolar parallel has excellent torque and excellent speed when compared to Bipolar series and Unipolar. Unipolar (half-coil) has the same speed characteristics as Bipolar parallel, but 1.4 X less torque. Unipolar also requires 1.4 X less current and produces significantly less heat.
My personal perference is to connect six-wire motors using Unipolar (half-coil) wiring and to connect eight-wire motors using Bipolar parallel wiring. By going that route, I get the best speed performance and higher torque at higher speeds from the six-wire motors and I get the best of everything from the eight-wire motors. The only eight-wire motor that I have is the PK299-F4.5. That is the best motor that I've ever used - bar none. It gets nice and toasty but it gives the same performance as the AS911 Alpha motors that I have on my Shopbot PRT-Alpha. The best part is that it costs about $360 with a Gecko G203v stepper controller (about $210 for the motor and $150 for the controller). The AS911 Alpha with controller costs $1,100. Not bad when you can get the same performance at 1/3 the cost. |
#24
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I’ve found the A1A motor performs much better using a half coil connection. The reason for this is the extremely high inductance of this particular motor. Lower inductance usually means higher speed. The half coil configuration reduces the inductance to one fourth of full coil connection. Although it looses some low-end torque it is really irrelevant. If you look at the torque curve published by Oriental you will notice it being very flat for the first 50 or so reducer rpm. You will also note the legend states permissible torque. This is the maximum rating for the gearbox. There is a lot more torque available than what it shows. Most of the torque lost going half coil will be in this speed range but there is a substantial gain in torque at the higher rpm range (compared to full coil bipolar series) .
It doesn’t make sense to me why all step motors don’t come standard with 8 wires. With 8 wires all wiring options are available. The A1A would sure be a candidate to run bipolar parallel if it had 8. Actually the A2A is pretty close to what the A1A would be if it could be wired bipolar parallel. Dirk |
#25
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Dirk, is there any other downside to bipolar parallel, aside from heat and big power supplies? "Roughness"?
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#26
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I have remembered why I am happy to tell everyone that unipolar (half-coil) is the way to go . . . . .
1. The nature of the load with CNC routing is generally that we need high torque when pushing the cutter at high speed through the wood. When we move the gantries and cars at low speed we need a lower torque than at the high speeds. The unipolar configuration gives us a good match to the motor load. 2. If we have lots of movement reversals, then the higher torque at low speed can be useful for accelerating. However, with my unipolar settings, I don't seem to "loose steps" by setting the accel. value too high. I can set the accelarations so high that the cut quality suffers from overshot corners, but the system still returns accurately to the "home position". The limit of my accel seems to be mechanical flex and not motor torque? (I know this can lead to another topic altogether). Remember that the 6-wire guys are not complaining about too little low-speed torque . . . . . they generally seem to be wanting more torque at the high speed end only (ie. to cut faster with bigger cutters) 3. Less heat at the motor and less heat in the control box. Smaller power supply and cooler geckos. 4. With unipolar half-coil, you could have a second chance if you burn half the motor ![]() ![]() |
#27
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Gerald,
I don’t know of a downside using bipolar parallel. The motors run as smooth as series. I’m not sure I would run the A2A at half coil, as its runs very well running full coil. I think there can be a point of diminishing returns when the motor’s inductance is already low. I would size my electronics to run at least bipolar series and if you have 8 wires I would consider sizing it for parallel. This would leave your options open and you can test the various wiring schemes easily by switching a few wires and changing your current resistor. Dirk |
#28
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Dirk and Gerald,
The torque charts on Oriental Motor's web site for the PK296A2A-SG7.2 motor with gearbox isn't very clear, but it looks like the Bipolar series connection gives the maximum of 40 in*lbs (640 oz*in) to about 50 RPM. With a 30-tooth gear, that would be about 4-ips. Using half-coil connections the motors gives that same torque at 100 RPM, or about 8-ips. As I see it, the problem with that motor and that gearbox and Bipolar series is that the gearbox limits the available torque to about 40 in*lbs. That means that gearbox can't handle the extra torque at low speeds that the Bipolar series connection produces. On the other hand, if the PK296-3AA motor were used (with an external gearbox or belt drive), then the Bipolar series connection would be superior to half-coil to about 100 RPM. But, if I needed a motor of that form factor, then the PK296-F4.5 motor, using Bipolar parallel connections, would produce more torque at 500 RPM than the PK296-3AA motor wired Bipolar series would produce at 50 RPM. There seems to be so many "ifs" involved when selecting motors and then trying to decide which wiring method to use. The good news is that the PK296A2A-SGxx motor, using either connection method will give very good results at any reasonable cutting speed. EDITED: When I first tried using a PK299-F4.5 motor wired parallel at 60-70 volts it was noisey and hot, as could be expected at that voltage. When I finally realized that I was over-driving the motor and that the maximum voltage should be about 50V, the motor quieted down. Now, I run the motor at 35V. At that voltage, it is smooth, powerful and quiet. (The reason that I tried it at 60-70 volts was that the Oriental Motor web site's torque chart for that motor lists a 60V power supply.) Last edited by Richards; Fri 12 October 2007 at 10:13.. |
#29
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I think Orientals rating on their gearboxes are very conservative. To my knowledge there hasn't been a failure since Shopbot started using them. You might also note that these are virtually the same box used on the Alpha's now. With the alpha's geared at 7.2 they can really do some damage if something hangs up. If it weren't for them going into Alpha mode to stop them this could be a real problem. Real world experience tells me the gearboxes will handle anything we can get the motors to give.
Dirk Last edited by Dirk; Fri 12 October 2007 at 10:39.. |
#30
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Quote:
If I conect like this: coil A -- black coil A- -- yellow coil B -- red coil B- --white what I need to doo with oher wire orange/green and brown/blue ??? Thanks for any help...... |
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