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  #1  
Old Sun 10 August 2008, 02:17
baseball43v3r
Just call me: John #96
 
Glendora (California)
United States of America
making sure a power supply is correct

ok after reading through the post on choosing a power supply i think i have done the math correctly but i have a question or two. first the setup. i'm am adding 4 Keling motors to the build list but i'm slightly confused on the math.

the inductance says 3.3 so currently the voltage needed would be 58.13. now the watts is where i get slightly confused. The amp rating is (i believe) 4.3 times that by 4 and its 17.2. now this is where i get tricked up, the post continues to say that it only needs roughly 2/3 of that through the geckos. so after multiplication by .67 i get 11.52 which i multiply by 58.13 for a grand total of 669 watts. is this correct? i cant decide if it should have been 17.2 * 58.13 or not.

if my numbers are correct, i was looking on the antec site and they have no power supplies in the 58 volt range, only in the 63 volt range, would be going much higher hurt in terms of heat? or is it actually neccesary if i need to control other peripherals (ie boards)? also if my numbers are correct how much farther above my total wattage should i go that allows room for error without killing someone?
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  #2  
Old Sun 10 August 2008, 04:46
Richards
Just call me: Mike
 
South Jordan, UT
United States of America
John,

From the data sheet, each motor, if wired half-coil, would draw 4.3A. If you wired the motors parallel (using all eight wires), each motor would draw 6.1A.

So, depending on how you wire the four motors, multiply 4.3 X 4 or 6.1A X 4. The Geckodrive rule of thumb is that you can derate the current by 33% (multiply by 0.66) if the stepper motors are the old round type. Many people routinely derate the square motors, too, depending on how hard the motors are used.

The MAXIMUM voltage would be 32 X SQRT(3.3) = 58.13. Personally, I would use a power supply that gives about 75% of that voltage, or around 45VDC. A 30VAC toroidal transformer will give about 42VDC after rectification (bridge diode) and filtering (capacitor). Either a 500VA (motors wired half-coil) or a 750VA (motors wired parallel) toroidal transformer would work.

AnTek has the PS-8N42 model that should work with those motors wired half-coil or bipolar parallel.
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  #3  
Old Sun 10 August 2008, 16:41
baseball43v3r
Just call me: John #96
 
Glendora (California)
United States of America
i'm going to run these half coil (unipolar) so the 58 sounds right to me.

So you are finding out the maxiums that it would consume in each category and assume it will consume slightly less? i think that makes sense in a weird way.

how much farther above the voltage could i go? because the model you listed doesn't seem to be available (ie not on the prices list) and if i needed to step up one size i could. I'll shoot them an email in a minute though to see if they actually have the one you listed in stock.

I would build my own power supply but alas i have a girlfriend and if I were to die from assembling it she might not be too happy.
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  #4  
Old Sun 10 August 2008, 19:54
Robert M
Just call me: Robert
 
Lac-Brome, Qc
Canada
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Mike, May I ask you to explain a bit on why you seem to use 75% a the maximum voltage from the new mH equation ?
By math, itís back to +/- 25 x vs 32 x mH like the old way (25x by rated voltage) !
Robert
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  #5  
Old Mon 11 August 2008, 05:00
javeria
Just call me: Irfan #33
 
Bangalore
India
HI Robert - now I am getting confused - look at my post on my thread where I did the PS calculations - I would think now that I am wrong - right -wrong-

http://www.mechmate.com/forums/showt...5&postcount=25

well- ???
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  #6  
Old Mon 11 August 2008, 05:24
Robert M
Just call me: Robert
 
Lac-Brome, Qc
Canada
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Irfan, Iím not a good candidate to tell you right or wrong. I have NO test experience and knowledge on this.
The 32 x sqrt mH is the new way to calculate. Iím just curious to find out why this 75% lower fartor and if you like to play with numbers, this 75% come close to a 25 x sqrt mH !
Nothing more !!
Sorry to confuse anyone
Robert
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  #7  
Old Mon 11 August 2008, 06:58
Richards
Just call me: Mike
 
South Jordan, UT
United States of America
I reduce the voltage to 75% of the calculated maximum voltage because of heat. Stepper motors can run very hot without damage, but I don't like to run electronic equipment hot. If given the choice between getting the last oz*in of torque out of a motor, that already has more than enough torque, and running that motor 25-degrees cooler, I would run the motor cooler. It's just a personal preference.

Some people use the 32 X SQRT(inductance) figure as the lowest voltage to use. I use it as the highest voltage that can be used, kind of like the 120 mph figure on my truck's speedometer. Although my truck could probably nearly reach 120 mph, I never drive it faster than 80 mph. No speeding tickets and long engine life.
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  #8  
Old Tue 12 August 2008, 04:09
javeria
Just call me: Irfan #33
 
Bangalore
India
Ah Mike got it now. I might as well get my calculations down by 20% on voltage and in the way save some more cash on the power supply transformer.
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  #9  
Old Sat 16 August 2008, 13:00
baseball43v3r
Just call me: John #96
 
Glendora (California)
United States of America
ok i think i understand all of that. now i have another question for you technical minds. i have been looking through other threads like crazy trying to find information about this but no luck. but how exactly do you connect the power supply to an external (ie my garage, or more importantly a wall outlet for the kitchen table project) source for those here in the states? i'm at a loss right now, and i assume there is a fairly standard way of doing it and i think its probably staring me in the face but i cannot find it!
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  #10  
Old Sat 16 August 2008, 14:21
Richards
Just call me: Mike
 
South Jordan, UT
United States of America
John,

Review the schematics that Gerald has posted. The wiring from your main panel determines the type of transformer that you use in the United States. Most people use a 120VAC connection; however, some might choose to use a 240VAC connection. You need to use an On/Off switch and a fuse between the panel and the power supply. In the finished product, you would need to add a "disconnect" between the power supply and the on/off switch. The disconnect would disconnect all power from the box and the on/off switch would disconnect the power between the disconnect and the power supply.

In other words, the disconnect cuts all power to the box. An on/off switch and a fuse would turn off power to EACH powered circuit inside the box. A power supply would also normally have fuses on the OUTPUTS of each voltage level to protect the power supply from a short circuit on any of the output voltage lines.
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  #11  
Old Sat 16 August 2008, 22:18
baseball43v3r
Just call me: John #96
 
Glendora (California)
United States of America
i've read through a couple of gerald's threads on this and i still feel a little lost. on his kitchen table project there are wires leading off the table presumably to a wall outlet but i dont know where they are going to. do you have any reading you can direct me to or an example picture of what i should be looking for exactly? sorry for the bother on this one but i seem to be a little clueless about this

is all it is is the wires connected from the power supply to a fuse an on/off switch which is plugged into the wall?
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  #12  
Old Sun 17 August 2008, 00:10
Gerald D
Just call me: Gerald (retired)
 
Cape Town
South Africa
John, for your personal safety, and our peace of mind, you need to get an electrician near you to help you out. We are dealing with issues here that can be lethal.
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  #13  
Old Sun 17 August 2008, 00:24
baseball43v3r
Just call me: John #96
 
Glendora (California)
United States of America
when it comes time for the electronics part of this project i will definitely be including a electrician to do the electrical portion, as i dont feel comfortable with it anyways. i was mainly asking for my own general knowledge to try and understand the in's and outs of the machine.
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  #14  
Old Mon 25 August 2008, 13:27
Richards
Just call me: Mike
 
South Jordan, UT
United States of America
There still seems to be a lot of confusion about selecting power supplies to drive stepper motors. Unfortunately, there is no "one best answer" that works for everyone. I use my machine to cut cabinet parts, so I want the highest cutting speed. Others use their machines to cut delicate 3-D parts, so they may never need speed.

Basically, the motor that you use determines the power supply that you're going to need. A motor has two ratings that are important when selecting a power supply. The inductance rating helps you select the proper voltage and the current rating tells you how much current you'll require.

Unfortunately, how you wire the motor also determines the motor's inductance and the current that the motor will draw. Wiring a motor with the coils in Series give the highest torque, but limits the speed. Wiring a motor half-coil, give the best speed, but only about 70% of the torque. Wiring an 8-lead motor with the coils parallel give you the same speed as half-coil and excellent torque, but it requires 2X the current as half-coil.

We'll assume that most people have six-wire motors and that they're going to wire the motors half-coil.

To find the MAXIMUM voltage, use this formula: 32 X SQRT(inductance) = voltage.

Remember, that is the MAXIMUM voltage for most motors. That means that the motor is going to run HOT, too hot to touch. I don't run motors at the maximum voltage. Instead, I multiply the voltage by 75% and get a voltage that still gives good speed, but a lot less heat.

Let's assume that you need a voltage that is about 35VDC. You can either buy a 35VDC power supply or you can build your own.

Selecting the proper transformer to get 35VDC requires that you multiply the DC voltage by 0.7 (which is 1 / SQRT(2.0)). In this case, you'll need a transformer that gives you about 25VAC.

Add together all the current ratings of the stepper motors to find out how much current you're going to need. We'll assume that each motor requires 3A and that you're going to use four motors.

4 X 3A = 12A

To find the VA that the transformer must provide, multiply the AC voltage by the current.

25VAC X 12A = 300VA

Finally, you'll need to compute the size of capacitor that you'll need.

(80,000 X 12A) / 35VDC = 27,428uF

Those are the basic calculations. To get 35VDC at 12A you'll need a 25VAC transformer, a full-wave bridge rectifier, and a capacitor rated at 25,000 to 30,000 uF (connect several small capacitors in parallel to get the desired total capacitance).

It's good practice to over-rate the voltage of the capacitor by about 15 or 20 Volts, so a 50V, 30,000uF cap would be good.

I've used those basic math formulas when building a "bunch" of power supplies for stepper motors. I never worry about trying to get "exactly" the right components because stepper motors are very forgiving. I've used transformers ranging from 18VAC to 30VAC with PK296B2A-SG3.6 motors. All the transformers worked perfectly. The VA rating of the transformers ranged from 250VA to 500VA, again with perfect performance. I've used capacitors ranging from 11,000uF to 30,000uF, and the motors ran just fine.

So, I do the computations to find what the "ideal" power supply would be then I go to my electronics parts bin to see what I have on hand. I build up a power supply with the parts on hand and then modify things as necessary after I've had a chance to test the motors.

Having adequate current available from the power supply assures that the motor will work at maximum possible torque. Having adequate voltage means that the motor will run at the highest practical speed. The motor will only draw as much current as it needs from the power supply, so having a power supply that gives 1.5X to 2X the current that you need will do no damage (but it will cost more and it will take longer to drain when the switch is turned off). Some people use an old recommendation from Mariss (at Gecko) to multiply the current by 0.66. Mariss now recommends that for SQUARE motors that you do not de-rate the current. I've found that it all depends on how hard you drive the motors. Rarely are all four motors running at full current all the time, but I still use 100% of the computed current as the "ideal" when building a power supply.

Feeding the motors too much voltage WILL cause problems. You'll get excessive heat. If the voltage is high enough, you'll fry the motor(s).
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  #15  
Old Mon 25 August 2008, 13:43
Robert M
Just call me: Robert
 
Lac-Brome, Qc
Canada
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Mike,
Your generosity to this community and to others Iím sure is a testimony of your goodness and willingness to help should not come to us with out gratitude. I speak for my self but for other Iím certain, WE ARE FOR EVER VERY GREATFULL and owe you tremendous respect for what you give & share.
Letís face it with modesty, with out people like you, many, ( me for example) would still battle in some very trouble waters.
I can only wish to become & give back some help of sort to you and other as much as you came to rescue when it was ask upon or better yet, even when not ask !
Merci infiniment
Amicalement, Robert
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  #16  
Old Mon 25 August 2008, 16:07
Marc Shlaes
Just call me: Marc
 
Cleveland, OH
United States of America
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I second that! The way you just presented that Mike is the best you've done. (At least for my poor brain.) I am going to build a simple spreadsheet in excel that does what you just stated. When I post it, probably late tomorrow, I hope you will take a minute and see if I got it right before I (or anyone else) use it.

Thanks so much.
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  #17  
Old Mon 25 August 2008, 19:25
Doug_Ford
Just call me: Doug #3
 
Conway (Arkansas)
United States of America
Yeah. Thanks Mike. You're truly an amazing guy.
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  #18  
Old Tue 26 August 2008, 07:37
Richards
Just call me: Mike
 
South Jordan, UT
United States of America
As usual, I've proven that I need to always wear my glasses when I'm writing. I wrote, "multiply the DC voltage by 07 (which is 1 / SQRT(2.0))". It should be 0.7. (since fixed by GD)

I've received a few emails asking about how to figure Unipolar inductance or current when all you have is Bipolar Series data or Bipolar Series inductance or current when all you have is Unipolar data. I think that Gerald answered that, but I haven't found the post.

When you know the Bipolar Series data and need to compute the Unipolar data:

Unipolar current = Bipolar series current X 1.4
Unipolar inductance = Bipolar series inductance X 0.25

Bipolar series current = Unipolar current X 0.7
Bipolar series inductance = Unipolar inductance X 4.0

Bipolar parallel inductance = Unipolar inductance
Bipolar parallel current = Unipolar current X 1.4

Bipolar parallel current = Bipolar series current X 2.0
Bipolar parallel inductance = Bipolar series inductance X 0.25

Sharp readers might be scratching their heads and asking what the difference is between Unipolar and Half-Coil. Actually, with Gecko stepper drivers, you can't wire the motor Unipolar. Mariss has said to use the Unipolar data for Half-Coil wiring. You'll also note that Unipolar has 1.4X the current that you would assume that a single coil could handle. Mariss isn't bothered by that, so I'm not bothered by that. I have rated my PK296B2A-SG3.6 motors at the Unipolar current ratings throughout most of my testing and the motors run just fine. (Maybe that's why de-rating the voltage by 25% keeps the motors cooler; however, voltage relates to speed and stepper motors are not normally run at high speeds when cutting. For instance, I can jog my Alpha easily at speeds up to 30-ips, but I cut at 8-ips maximum except when I'm surfacing a new sheet of MDF for the spoil board.)
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  #19  
Old Tue 26 August 2008, 09:50
Gerald D
Just call me: Gerald (retired)
 
Cape Town
South Africa
I personally don't subscribe to de-rating the power supply voltage by 25% below the result of the 32XSQRT(inductance) calculation because that takes away 25% of the motor torque. Okay, if one knows that you won't ever need the upper 25% of the motor's torque capability, then you can discard that torque with a lower supply voltage.

For the last couple of weeks we have had 48V driving our half-coil PK296A2A-SG7.2 motors, which is 22% higher than the recommendation of Mariss and 63% higher than the recommendation by Mike. (The G203V's were fitted with 36k resistors for 3amp). For the type of cutting/duty cycle that was being executed, the motor temps havn't been caught to be too hot.....yet.

It is winter here now, and all types of cutting jobs havn't been tackled yet, so I don't have the confidence to say the test is complete and that 48V is a good voltage. But, I need to order 5 transformers pronto, and they will be made to spec. Therefore, I am entirely comfortable to order 5 transformers at 40 Volts(DC). It is based on testing and measuring for our conditions - I cannot justify taking the voltage (and torque) down by another 25%.
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  #20  
Old Tue 26 August 2008, 11:53
Richards
Just call me: Mike
 
South Jordan, UT
United States of America
Gerald, I think that we're still in basic agreement. I reduce the voltage because of excessive heat. If you're not getting excessive heat, then you have no problem. My favorite voltage for the PK296B2A-SG3.6 motors is about 35VDC. That voltage keeps the temperature of my motors at 146-degrees F. or lower.

Current and voltage are interrelated, but, generally speaking, the current drawn by the motor is the greatest factor that determines torque and voltage is the greatest factor that determine top speed. So, at moderate speeds (200 to 400 rpm), your torque should be about the same even if the voltage ranged from 35V to 50V.
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  #21  
Old Tue 26 August 2008, 12:54
Gerald D
Just call me: Gerald (retired)
 
Cape Town
South Africa
Mike, I think it all revolves around our personal perceptions of "excessive heat". You want a peak of 145F [60C] and I am comfortable to see 185F [80C] for short periods. Also, at your altitude, you have less air for convection cooling and your motors will run a little hotter there.

A safe choice for a first time builder is to keep to a lower voltage. There might be something in your way of cutting/moving, or your atmospheric air, or your thickness of dust layers, or your choice of drive, that is going to make your motors cook. I have the luxury of two machines that I can measure and tweak for higher voltages. When all 7 machines are running one day, a cooked motor won't bring the whole show to a standstill.

But, having raised the possibility of cooked motors, I must observe that such things are very rarely (if at all?) reported at multiple forums. This observation is one of the factors pushing me nearer the limits
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  #22  
Old Tue 26 August 2008, 14:07
domino11
Just call me: Heath
 
Cornwall, Ontario
Canada
Gerald,
Are you keeping all 7 Machines? I thought I remember you saying you were selling the original 2 when the 4 new ones were built. Sean must be really busy these days.
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  #23  
Old Tue 26 August 2008, 22:48
Gerald D
Just call me: Gerald (retired)
 
Cape Town
South Africa
The original 1st machine has a bunch of buyers lined up for it, and that is the only one that is under discussion for being sold. Its table, at 8' x 4' is just too small (actually too narrow) - which means the gantry is too short. The pressure to sell it also comes from a lack of space and to generate some finance for the additional machines. One of the new machines is going to my house, where it shouldn't be working too hard, and could serve as emergency spares in a pinch.
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  #24  
Old Wed 27 August 2008, 07:09
Marc Shlaes
Just call me: Marc
 
Cleveland, OH
United States of America
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OK, I built the spreadsheet and it works well... Then Mike had to go and say a bunch more . I will put the bipolar / unipolar conversion stuff in it but one thing he said confused me again.

Sharp readers might be scratching their heads and asking what the difference is between Unipolar and Half-Coil. Actually, with Gecko stepper drivers, you can't wire the motor Unipolar. Mariss has said to use the Unipolar data for Half-Coil wiring.

I have searched the Internet and this forum for one place that explains:
  • Unipolar
  • Bipolar
  • Series
  • Parallel
  • Half-Coil
in a way that a simple guy like me could understand so that the spreadsheet makes that clear to guys like me. Haven't found it yet. There are places on both the Internet and this forum that comes close but a fair amount still went by me. I'll work on it a bit more today and then post the spreadsheet.
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  #25  
Old Wed 27 August 2008, 14:16
Richards
Just call me: Mike
 
South Jordan, UT
United States of America
Stepper motors that work with Gecko stepper drivers must have two coils, an A coil and a B coil; however, stepper motors are normally built with different configurations to allow the best performance for the required job.

A four-wire motor is automatically a Bipolar series motor. It can only be wired one way to a Gecko stepper driver.

A six-wire motor can be wired either Bipolar series or half-coil when used with a Gecko stepper driver. It cannot be wired Unipolar. The Gecko stepper driver does not have that option. Other drivers may have terminals to connect all six leads on a six-wire motor. Those stepper drivers are true Unipolar stepper drivers. The Oriental Motor CSD2120-T stepper driver is an example of a Unipolar stepper driver.

An eight-lead motor actually has four coils inside the motor, 2-each A coils and 2-each B coils. It can be wired half-coil, Bipolar series or Bipolar parallel, but it can't be wired Unipolar with a Gecko stepper driver.

An Oriental Motor six-lead motor has the A coil connected to three leads, The end leads are Black and Green with a Yellow center-tapped lead. The B coil has three leads. The end leads are Red and Blue with a White center-tapped lead. If you connect the Black lead to the A terminal, the Green wire to the /A terminal, the Red lead to the B terminal and the Blue wire to the /B terminal, you are using the Bipolar series wiring method. You will get more torque and less top speed if you use the Bipolar series wiring method.

If you connect the Black lead to the A terminal, the Yellow lead to the /A terminal, the Red lead to the B terminal and the White wire to the /B terminal, then you are wiring the motor using the half-coil method. You will get about 4X the top speed but only about 70% of the torque when compared to the Bipolar series wiring method. (I normally wire all my motors half-coil). Also, note that ANY end lead, Black or Green, Red or Blue can be used with the center-tap lead, but direction may be reversed. Just look at the Oriental Motor wiring chart that is published at www.orientalmotor.com and you'll be able to see how the coils are connected.

An eight lead motor gives the most options for wiring. Either A coil can be used with either B coil and you'll have a half-coil connection. Both A coils can be connected in series and both B coils can be connected in series and you'll have a Bipolar series connection. Both A coils can be connected in parallel and both B coils can be connected in parallel and you have a Bipolar parallel connection.

That might help with understanding how the motors can be connected, but it doesn't answer the question of which method to use. Because the PK296A2A-SGxx motor only has six-leads, you have to either wire them Bipolar series or half-coil. I wire mine half-coil. They run smoother at higher speeds and they have much greater torque at higher speeds. If I had a need to run slow with all possible torque, I would wire the motors Bipolar series. However, given the mass that has to be moved, I doubt that you would actually see a lot of difference in performance with either wiring method. I have not read of any complaints from those who have connected their motors Bipolar series nor have I read of any complaints from those who have connected their motors half-coil. If you have PK296A2A-SGxx motors and your power supply puts out 30-40VDC, try half-coil. If it puts out 60-75 volts, try Bipolar series.

Mach 3 has tremendous capability to tune the motor settings to get the most out of your motors without stalling.

Experiment. Experiment. Experiment. (But, please tell the rest of us what works for you.)
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  #26  
Old Wed 15 October 2008, 18:52
dragonfinder1
Just call me: Dave #49
 
Astoria, Oregon
United States of America
Am I doing this right?

Mike

Letís see if Iím starting to understand this power supply thing.

My supply voltage would be 120.
Motors would be wired half coil.
Reduction 3:1 belt drive
The motors I would use are the MotionKing 34HS9801
Their specs say
Amps 4.0
mH 4.1

So,
4.1 sqrt * 32 = 64.8 * .75 = 48.6 VDC
4.0 * 4 * .67= 10.72 Amps
10.72 * 48.6 = 520 VA

Iím not sure if the MotionKing specs are given for half coil or not

Kobus is using these motors and he said his power supply is 300VA and 42 VAC ( Thatís probably before the .75 factor )

If this is correct, would these motors run from the Gecko G-540?
The G-540 is rated to 50V and 3.5 A. I would probably have to add a current limiting resistor .
Am I thinking along the correct lines, or have I gone astray?

Dave.
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  #27  
Old Wed 15 October 2008, 19:12
bradm
Just call me: Brad #10
 
Somerville(MA)
United States of America
Mostly Right!

Dave, you're doing great.

I think you have two challenges with those motors and a G540: The G540 won't put out the 4A / phase (3.5A max), and the inductance is outside the Gecko's sweet spot.

As Geckodrive puts it in my owner's manual: "All NEMA-17, most NEMA-23 and a few NEMA-34 motors are acceptable. The motors preferably should be square in cross-section, not round. The motors can be 4, 6 or 8-wire
motors. 5-wire motors cannot be used with the G540. Choose a motor that has a rated current of 3.5A or less. Choose a motor that has a rated winding inductance of 2.5mH to 3mH if maximum power output (>100W mechanical) is a requirement. Never use a power supply voltage greater than 32 times the square-root of the motor inductance expressed in milli-Henries (mH)"

If my understanding of Geckos is correct, you won't need a current limiting resistor, you simply won't get more than 3.5A out of the Gecko, which may lead to compromised motor performance.
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  #28  
Old Wed 15 October 2008, 19:45
dragonfinder1
Just call me: Dave #49
 
Astoria, Oregon
United States of America
My thinking

Brad

Here my reason for thinking this way. First I have to say I don't know newton-meters well at all.

The motors are rated for 490 n.cm. Like I said I don't know nm's, but if my calks are right, that would about 694 oz/in, and the motors will have a 3:1 rear ratio. ( I will add them )

Even if the motors only achive 50% of the rated torque, that still gives me around a 1000 oz/in. That should be enough to break a 10mm bit.

Even if its a push for the power, I think I would like to try the G-540 unless someone says that would be a really bad idea. I am prepared to buy the G203v's to power the motors. The G-540 costs $300 usd, so if it doesn't work out and I don't ruin anything else, I think I could live with that. I've made a lot bigger mistakes than that before and if it works, we have another way to power the MechMate.

If this is really a bad idea, I'm sure more experts will come to my rescue

Dave
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  #29  
Old Thu 16 October 2008, 06:51
bradm
Just call me: Brad #10
 
Somerville(MA)
United States of America
I agree with that

Dave, I'm running a G540 with OM motors and a 7.2 geardown, and I'm happy. So it looks to me like your reasoning is sound.
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  #30  
Old Thu 16 October 2008, 08:15
Marc Shlaes
Just call me: Marc
 
Cleveland, OH
United States of America
Send a message via Skype™ to Marc Shlaes
Here is a pretty nice Motor Parameter Conversion Program.

It does a number of things including torque.

http://www.linengineering.com/flash/Calculator.swf
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