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  #1  
Old Wed 02 February 2011, 18:32
Richards
Just call me: Mike
 
South Jordan, UT
United States of America
How much voltage do I really need?

We've all been using the formula:

32 X SQRT( Inductance ) = MAXIMUM voltage as the basis for selecting a power supply. In reality, that formula tells how high of a voltage we can use before the insulation melts in the motor; it doesn't really tell us what the best power supply is.

I just completed some interesting tests where I played with various motors and various power supplies. After running those test, I decided to find out what the minimum voltage would be to still give me 750 RPM (shaft speed, before a belt-drive or gearbox) with a PK299-F4.5 stepper (2.5 mH inductance) and a PK296B2A-SG3.6 stepper (1.5 mH inductance).

The PK299-F4.5 ran at over 2,000 RPM with a 47VDC power supply computed as near maximum using Mariss's formula. It sounded great. It made a great impression. But, in reality, everything over 750 RPM was just for show. At speeds higher than 750 RPM, the torque chart shows too little torque to be of any practical value.

I set Mach 3 to 45,000 pps, with 2,000 steps per inch and 750 inches per minute as the top speed, then I used my Variac to turn down the voltage until the motor could not hit 750 RPM. I went all the way down to 24VDC, and the motor was still running perfectly. 24VDC was the lower limit because my test bench has G202 stepper drivers that require a minimum of 24VDC to work properly.

The PK296B2A-SG3.6 motor did just as well. I limited it's top speed to 1,500 RPM so that the gearbox wouldn't be damaged. At 47VDC, it hit that speed with no problem. Then, I changed the parameters in Mach 3 and limited it to 750 RPM. At 24VDC, it was still running perfectly at 750 RPM.

My conclusion is that I don't have to worry about getting a perfect match between the motor and the power supply on a CNC router. With the motors that have been suggested for years (which is basically any motor that has an inductance of 2.5mH or lower), a 24v to 35v power supply will probably work just fine.

-----

I selected 2,000 steps per inch in Mach 3 as a simple way to figure RPM. A Gecko stepper driver requires 2,000 steps per revolution, so with 2,000 set as the number of steps per unit (inch), I could just dial in the number of inches per minute that I wanted and know that I had also told the program what RPM to use.
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  #2  
Old Wed 02 February 2011, 20:46
Gerald D
Just call me: Gerald (retired)
 
Cape Town
South Africa
Mike, I don't agree with you. Under real life loaded conditions, with various motors on various CNC tables running 40 hours a week, we have found that the higher the voltage, the better. (By better, I mean getting more production and less scrap). We are actually using transformers that give more voltage than the formula, but I don't want to get into an involved list of which voltage we use on which motor.

The only argument for reducing voltage is to limit heat. While we often see mention on forums of steppers feeling hot, I can't recall a single mention of an actual burnt out motor. I have to conclude that burnt motors are rare and that most users are running voltages well below the limits.

Your argument is based purely on observations of speed. But voltage also has a direct relationship to torque . . . ie. the ability to push a cutter hard enough without losing steps. It doesn't matter what speed the motor is working at, the simple fact is that a drop in voltage has a proportional drop in torque.
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  #3  
Old Wed 02 February 2011, 21:28
Richards
Just call me: Mike
 
South Jordan, UT
United States of America
Gerald,

The Oriental Motor torque charts support both of us. A PK299-F4.5A motor has the same holding torque whether it's driven with a 24 volt, 48 volt or 75 volt power supply. Taking 200 oz*in as the minimum usable torque (600 oz*in when used with a 3:1 belt-drive, or the same as the holding torque of a typical PK299 sized motor without a belt-drive), the 24V supply can go 400 RPM, the 48V supply can go 1,100 RPM and the 75V supply can go 1,500 RPM.

With a 1.5-inch pitch diameter pinion, a 3:1 belt-drive will be running 628 inches per minute at 400 RPM. That's 2X faster than the speeds that I used to cut MDF and plywood. With a 3:1 belt-drive the cutting torque is the same as the holding torque that I had on my original PRT-Alpha.

I doubt if many people would be disappointed with 600 oz*in at 10-ips. A $25,000 Shopbot PRS-Alpha can't deliver that much torque at that speed, but a $208 motor connected to 3:1 belt-drive through a $150 stepper driver driven by a 24VDC power supply can.

That same PK299-F4.5A motor without a belt-drive still has over 500 oz*in of torque at 100 RPM with a 24V power supply. Using that same 1.5-inch pinion gear gives you 7.8-ips at 100 RPM. That's still not bad at all and it's still much better performance than I had with my non-geared, non-belted Shopbot PRT-Alpha.

The PK299-01A motors that Shopbot sold you cannot be compared to the PK299-F4.5A motors that I'm writing about. Their inductance was much too high to be used effectively with a Gecko stepper driver.
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  #4  
Old Wed 02 February 2011, 23:08
Gerald D
Just call me: Gerald (retired)
 
Cape Town
South Africa
The numbers make my head hurt - all that I know for sure is that increasing the voltage makes the machine better and over 10 years with 20 motors we havn't burnt a motor yet.
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  #5  
Old Thu 03 February 2011, 00:40
Alan_c
Just call me: Alan (#11)
 
Cape Town (Western Cape)
South Africa
Send a message via Skype™ to Alan_c
Gerald I am with you on the head hurting bit, I have been mulling over this subject over the past few weeks. - Is there not a relationship between input volts and the limiting resistor on the gecko with regards to the torque, volts gives speed and amps gives torque?

Let me give an example, I have PK299-F4.5A motors on my machine, a 50V AC transformer and G203V geckos - now the transformer gives 70V DC after the rectifier which is too much for the specified motors (ideally should be about 50V DC) I have resistors limiting my current to about 2A because I dont want to cook the motors. If I change the transformer (by paralleling it instead of series connection) to give me 35V DC and up the resistors to give me about 4A I will still get enough speed for my requirements as well as almost doubling my torque and having cooler motors. This scenario will not work with all motors of course, the lower inductance motors can run at higher voltages. - or am I completely off the track here?
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  #6  
Old Thu 03 February 2011, 07:59
Richards
Just call me: Mike
 
South Jordan, UT
United States of America
Alan, I prefer to run my PK299-F4.5 motors at 35VDC. At that voltage, I can wire them either half-coil @ 4.5A or parallel @ 6A. Their heat buildup is moderate and they have lots of torque at the speeds that I need.

Watts (heat) is Voltage X Amps. 70V X 2A would produce the same heat as 35V X 4A. (I'm ignoring heat build-up because of iron saturation when running the motors at high speeds. I'm assuming that you would use the same speed with both voltages.)

In theory, by limiting the current flowing through the motor, any motor could be run at any voltage (up to Gecko's limit of 80VDC). I prefer to keep the voltage at or below the maximum voltage computed using the formula: 32 X SQRT( Inductance )

Gerald's case is different, assuming that he is using PK296A2A-SG7.2 motors. Those motors are rated at 3A, when in reality they are 4.5A motors. Oriental Motor uses 3A maximum to protect the gear boxes. If they were run at 4.5A at 39VDC, they would get as hot as if they were run at 58VDC @ 3A. They can be run at higher voltage because their current is being limited to less current than the motor was designed to handle.
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  #7  
Old Thu 03 February 2011, 08:26
smreish
Just call me: Sean - #5, 28, 58 and others
 
Orlando, Florida
United States of America
Note:
I run my PK296A2A-SG7.2 motors @ 56V @ 3A without getting to warm and have excellent response (both top end speed and holding) Going strong after 3 years now.
Machines #5, #12 and #28 (and the UNCSA MechTorch) all have similar set ups to my knowledge
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  #8  
Old Thu 03 February 2011, 10:19
Richards
Just call me: Mike
 
South Jordan, UT
United States of America
Sean,

At 56V and 3A, your motor could develop 168 watts, if they were driven hard. The motor (which is actually the PK296-03AA) can handle 4.5A at 39V, Oriental Motor derated the PK296A2A-SG7.2 because of the gearbox, not because of limitations of the motor. 4.5A X 39VDC is 175 watts. So, at 168 watts, you're within the permissible operating envelope.

Heat depends on the current actually being pulled by the motor. If the motor is not being driven hard, it will pull less current. If the motor is being driven hard and is nearly at its stall speed, it will pull more current. Just before a motor faults, it will pull maximum current, so, most of the time, a motor on a CNC machine is only pulling a fraction of it's rated current.
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  #9  
Old Thu 03 February 2011, 12:46
smreish
Just call me: Sean - #5, 28, 58 and others
 
Orlando, Florida
United States of America
Mike,
Thanks as always. My feedback was more "tactile" than "scientific" for the new readers of the forum. I guess the point I was making silently in my head while typing earlier was "the system as stated is engineered well beyond reasonable daily use and really isn't working as hard as it could". Your input is awesome as always!

Have a great day.
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  #10  
Old Sun 07 July 2013, 01:19
IPenev
Just call me: Ivan
 
Varna
Bulgaria
Hi
I have been reading and I need some help understanding the changes of the voltage and the current from the PS till the motors. As far as I read, the power supply is calculated for proper current and voltage.
The current is calculated by: Amps (all motors)*0.67
The voltage is calculated by: 32*sq(inductance of the motor) Do we sum the inducance of all motors ?
V*A=W
Where I am confused is that we are delivering the calculated W to the controllers, I read that the controllers are adjusting the current they receive, convert it and give it out as a completely different function of High and Low (this is clear). My question is do we have to consider some changes in the voltage as well? Or, the voltage does not change. Because I did not read any consideration for the controller when we calculate the PS.
Thank you in advance
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  #11  
Old Sun 07 July 2013, 05:19
KenC
Just call me: Ken
 
Klang
Malaysia
32*sq(PHASE inductance of the motor IN mH)
Quote:
Originally Posted by IPenev View Post
Where I am confused is that we are delivering the calculated W to the controllers, I read that the controllers are adjusting the current they receive, convert it and give it out as a completely different function of High and Low (this is clear). My question is do we have to consider some changes in the voltage as well? Or, the voltage does not change. Because I did not read any consideration for the controller when we calculate the PS.
Thank you in advance
You are almost there, keep it up

Before I get to the 2nd part, I wish to correct something, you don't deliver power to the controller, you deliver power to the Motor Driver.
The controller in this case is Mach3 which lives happily in the PC.

The simple answer is No.

The not so simple answer is as follows.

The calculated W is the theoretical Wmax the motor will require.
You don't intentionally deliver the full W to the motor driver (if it does, you are running thing dangerously close to frying something...), the motor driver only draw as much W the stepper motor required from the PSU. & the motor only draws what it needs to move the loads you put it up against (speed, acceleration, load mass, opposing forces...etc).
There will be some voltage drop along the electronics chain, but that isn't important, as a matter of fact, the voltage drop is a good thing for those who run everything at the max ceiling. Nothing which you should concern when designing a PSU.
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  #12  
Old Mon 08 July 2013, 10:31
IPenev
Just call me: Ivan
 
Varna
Bulgaria
Thank you much Ken, you read my mind, I meant the Motor Driver above
Bottom line is we do not consider the Motor Driver, it adjust the Voltage and the Current depending on the controller. We just make sure that we give it enough of Ams and Volts to "work with".
Is there a case that a Motor Driver has been burned, and I am not talking about the ones that need some resistance calculated before connecting the wires to it or overloading it with more the 80V (max criteria). I did not meet any limits for current to the Motor Drive.
I also read that some are trying to power more units with PS that gives out more then one connection. For example a PS that give a 9V, 25V and 35V (or something like that) could it be used to power the Controller, a fan and the Motor Drives. Is this recommended, because the two systems should be grounded, one the Motor Drives with the PS and the other to the PC ?
Thanks in advance

Last edited by IPenev; Mon 08 July 2013 at 10:35..
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  #13  
Old Mon 08 July 2013, 12:31
KenC
Just call me: Ken
 
Klang
Malaysia
No, you really don't want to choke the supply from the PSU to the driver, that is why er use unregulated power supply here... u want lowest possible PSU output impedence.
The safety net we put here is make sure your PSU don't pump more V than the driver can take. The design philosophy is to make sure it will never happen...

Yes, you can use a single transformer with multiple secondary winding to do everything including your laundry... So long as you know what you are doing with your grounding/earthing scheme of the various type of signal & power circuit in your set up... which is not that difficult after having your share of hair pulling moments.
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  #14  
Old Mon 08 July 2013, 13:11
IPenev
Just call me: Ivan
 
Varna
Bulgaria
hahah (for the laundry part), OK, Thanks, I guest it is time to get some parts
Thanks again Ken
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  #15  
Old Tue 09 July 2013, 09:06
IPenev
Just call me: Ivan
 
Varna
Bulgaria
Hi
I have been asking about the PS and particularly how is the increase of the voltage in DC happening, I learned the diode arrangement. I rad that when the voltage is converted from AC to DC, the voltage increases by 1.4 factor. Does anybody know how this factor came along ? Is it discovered experimentally or there is some formula behind it ?
The other thing is that the voltage increases when converted to DC but when we put a load to the circuit the voltage decreases. Do we consider that in our calculations.
I assume that we are calculating the VA (Watts) at DC current and then divide the voltage by 1.4 to get the write transformer ?
I appreciate any feedback, I am not interested in just building the MM, I would like to understand each step of the way
Thank you in advance
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  #16  
Old Tue 09 July 2013, 09:48
bradm
Just call me: Brad #10
 
Somerville(MA)
United States of America
Actually, it's the square root of 2 which is approximately 1.4. Also see https://en.wikipedia.org/wiki/Root_mean_square



Note that 1 / 1.41442715 = 0.707.

You do not include the voltage drop in your calculations.

You do divide the voltage by 1.4 (multiply by 0.707) to get your desired transformer
voltage.
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  #17  
Old Tue 09 July 2013, 19:27
David Bryant
Just call me: David #99
 
Western Australia
Australia
Hi All

with regard to the increase in voltage operating better:
http://www.anaheimautomation.com/man...otor-guide.php
shows:

"Q: While increasing speed, why do stepper motors lose torque?
A: Inductance is the leading cause for motors losing torque at high speeds. The electrical time constant, τ, is the amount of time it takes a motor winding to charge up to 63% of its rated value given a resistance, R, and inductance, L. With τ = R/L, at low speeds, high inductance is not an issue since current can easily flow through the motor windings quickly. However, at high speeds, sufficient current cannot pass through the windings quick enough before the current is switched to the next phase, thereby reducing the torque provided by the motor. Therefore, it is the current and number of turns in the windings which determines the maximum output torque in a motor, while the applied voltage to the motor and the inductance value of the winding will affect on the speed at which a given amount of torque can be produced.

Q: Why does increasing the voltage increase the torque if stepper motors are not voltage driven?
A: Voltage can be viewed as forcing current through the coil windings. By increasing voltage, pressure to force current through the coil also increases. This in turn causes the current to build faster in the winding and is able to produce a larger magnetic field. This larger magnetic field is what produces more torque.

Q: What temperatures are stepper motors able to run at?
A: Most stepper motors include Class B insulation. This allows the motor to sustain temperatures of up to 130 C. Therefore, with an ambient temperature of 40 C, the stepper motor has a temperature rise allowance of 90 C allowing for stepper motors to run at high temperatures.

Q: Is it possible to get more torque by running the stepper motor at double its rated current?
A: It is possible to increase torque by increasing the current but by doing so, it weakens the motors ability to run smoother."



So get the stepper driver voltage as close to the maximum voltage that it can handle.

Run the current up to or over the rated value, within temp rise limits and smoothness requirements.

Some stepper drivers allow a holding current of half the running value when stationary that would reduce heat build up in some situations.

Remember that (Slightly confusingly) the simple DC situation of P=VI is not directly applicable for the stepper motors as the voltage is high while current is building up and will be low while the stepper coils is held at the maximum current.
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  #18  
Old Thu 11 July 2013, 11:17
Richards
Just call me: Mike
 
South Jordan, UT
United States of America
Posts have been edited or removed from this thread past this point - see VOLTAGE: the Mike Richard versus Gerald_D argument for the unedited posts

Did you catch the part that voltage is the force that "forces" current into the coils. Inductance is the resistance to that happening. The lower the inductance, the easier it is to force current into the cols.

Heat is generated as the product of voltage X current. The greater the voltage, at a given current, the greater the heat.

The goal is to get adequate performance without frying things. Some motor manufacturers allow an 80-degree 'C' rise in temperature with a 100-degree 'C' maximum. edited

If you need slightly more torque than your current motor/driver/gearbox can give you, get a bigger motor and stay within the recommendations. Paying for four larger motors is much less expensive than paying for a burned up shop.

Last edited by Gerald D; Fri 12 July 2013 at 20:40..
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  #19  
Old Thu 11 July 2013, 23:14
IPenev
Just call me: Ivan
 
Varna
Bulgaria
Hi
I started to build MM recently, sorry about the boring discussion, just would like to plan carefully, but I got the idea.
So I am getting a 35V (AC) which will become 35Vac X 1,4 = 49Vdc and 13A
my PS will have 49 x 13 = 637W
my motors are: 86HS9802 (5A and 2.4mH )and the controllers areM856 (20V to 80V)

By the way I spoke with a professional and learned more about that 1.4 constant increase of voltage. This factor comes up at the capacitor, it moves the voltage to the peak voltage, there is a formula but I will not go into it, but the voltage will drop as soon as we put load on the circuit. I will find out how much does it drop and keep you posted
Regards
Ivan
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  #20  
Old Fri 12 July 2013, 05:11
Gerald D
Just call me: Gerald (retired)
 
Cape Town
South Africa
Ivan, your calculations are exactly right, and they are within the rules. However, our experience over a large number of MechMates tells us that you can go up to 40V (AC) for the transformer and you can reduce its size to 300 Watts (VA). It depends what is easily available to you.

The rules were written for places where the motors have a constant, permanent load load, such as driving a conveyor belt, or for a blood pump in an operating theatre, etc. With our CNC machines we have seen that a single motor seldom has full load for a long time. The load is shared between the 3 axes at different times.
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  #21  
Old Fri 12 July 2013, 09:15
IPenev
Just call me: Ivan
 
Varna
Bulgaria
Hi
So, about the 1.4 factor, it comes after the capacitor but after we put load on the PS the voltage drops. The amount it drops depends on the load, but for the woest part it should not drop below that 1.4 factor, what I mean is that if we have a 35V AC and we convert it to DC after the capacitor it comes as a peak current to 35X1.4= 49VDC, after the load it should drop no more then 35VDC. It is getting an initial bust and then drops, so we are protecting the controllers from that initial bust.
I heard that SIEMENS is making very good PS ready to use, and relatively to the same cost, I will look into it

Regards
Ivan
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  #22  
Old Fri 12 July 2013, 11:35
danilom
Just call me: Danilo #64
 
Novi Sad
Serbia
If you look at this motor datasheet

http://www.jat-gmbh.de/pdf_engl/mot/2_3_34S_e.pdf

it's one that I own, it was designed with an encoder and run as closed loop.
If you look at the characteristics curve diagram there are curves for three voltages

highest is 150 V !!! and the motor has a inductance from 1.2 to 4mH
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  #23  
Old Fri 12 July 2013, 21:15
Gerald D
Just call me: Gerald (retired)
 
Cape Town
South Africa
The posts deleted from the above thread can be found at: VOLTAGE: the Mike Richard versus Gerald_D argument
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  #24  
Old Mon 15 July 2013, 12:18
digit
Just call me: Dominic
 
Quebec
Canada
Voltage rating for stepper are related to inherent resistance value of coil. When you direct connect your coil to source you have

V = r * I

r = internal impedance of coil
I = max value of (heat, winding wire gauge)

To get rated current (I) you apply (V) voltage. In fact, (V) is not a rating but a result of (r) and (I).

With good drive, current is controlled (limited) by the chopper drive. You can apply higher voltage and drive will make sure current is limited to set value nevertheless source voltage.

Higher voltage allow current to overcome inductance and reach spec current faster = reaching max torque faster = better performance.

Please also note that motor specs are rated for steady situation. 1A steady or 2A 50% duty cycle is same.

The only real (V) rating is coil winding isolation. Too much voltage and voltage will spark to other coil or to casing. Heat should also be consider.

Last edited by digit; Mon 15 July 2013 at 12:44..
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  #25  
Old Mon 15 July 2013, 15:28
paulus
Just call me: paul
 
jhb
South Africa
Hi guys.
I was wondering if there is a worry of the motor overheating why not just put force cooling on it. Just wondering if it will help. I have lost you guys on the second post with all the electrical terms. To prevent the motor from getting to hot you must get rid of the heat so cool it with a fan.

Hopeee this puts a smile on every bodys faceses
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  #26  
Old Mon 15 July 2013, 22:12
KenC
Just call me: Ken
 
Klang
Malaysia
Paulus,
force cooling is a good idea, but why not design something which doesn't need force cooling or other auxiliary equipment, minimum parts count, lightest, smallest, cheapest...
Hence the argument...
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  #27  
Old Mon 15 July 2013, 23:11
MetalHead
Just call me: Mike
 
Columbiana AL
United States of America
You should design your machine to be within the tolerances that you can accept and support. If you choose to push your hardware, you will need to pay close attention to ensure specifications are followed and aligned. Motor, Wire, Power Supplies etc. Safety is each persons own responsibility as well as the risk of damaging electrical componets and having to invest in their replacement.

If you do not want to be the type to worry so much about those tolerances, then a solution for that approach is available as well and has been tested on many MechMates.

Both methods have proven to work. How well will be more specific to what your trying to achieve.

Sometimes it is better to spend a little more up front on better technology (So you do not need fans for example) than have to spend the money twice by trying to buy low cost solutions. But again this is balance and tolerance to down time that drives this typically.

The costs of getting as cheap as possible is typically not directly and only related to the hardware costs. Inevitably the point in time that you determine that you went to far in stretching the technology edge will be right in the middle of a high volume hi profit job.
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  #28  
Old Mon 15 July 2013, 23:51
KenC
Just call me: Ken
 
Klang
Malaysia
Engineering design is an art of compromise.

Like investment, risk tolerance is a big factor in how far one will go.
Most importantly, don't mistaken guide lines like they are carved in stone.
I'm bias toward Gerald, I too will push my equipment into no man's land provided that I think I'd learn enough to know what is dangerous. BUT I'll be as cautious as Richard when I really ignorant with stuff eg, when I still need to ask why I need to multiply 1.414 for ratification...
I agree with Mike as its false economy to look purely on $$ sign. but than again, expensive will not guarantee the best solution.
To come up with a rational solution, one must acquire enough knowledge, not just engineering, it must include everything around it.
My best advise in MM build is, COPY to the teeth if you still need to ask fundamental questions such as rms & the difference between AC & DC.
BECAUSE WE HAD PROOF THEY WORKED!!! OVER 100 recorded MM built, isn't that enuf??? I built 2 for god sake!!!
If you still worry, forget about building your own CNC Router.

With all that said, there are no design which will satisfy everyone.

Richard, you are right about safety, but we should not stop anyone from pushing equipment limits... honestly, if Gerald took your advise on liabilities, this forum might not had existed...
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  #29  
Old Tue 16 July 2013, 07:03
darren salyer
Just call me: Darren #101
 
Wentzville mo
United States of America
Well said, Ken.
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  #30  
Old Thu 18 July 2013, 18:54
Richards
Just call me: Mike
 
South Jordan, UT
United States of America
Let's look at a very popular motor, the Oriental Motor PK296A2A-SG7.2. It is advertised as having 704 oz*in of holding torque. With a 1.24" pitch diameter spur gear, it moves 0.000273" inches per step with a geckodrive stepper driver.

A much larger PK299-F4.5A motor has 880 oz*in of holding torque when wired bipolar parallel, but without a gearbox or belt-drive, it moves 0.001964" per step. Quality would suffer.

The smaller PK296-F4.5A, when wired parallel, has 440 oz*in of holding torque. With a 4:1 belt-drive, it would have 1,760 oz*in holding torque and it would move 0.000491" per step. That's not bad. Very good torque and more than adequate quality.

The often-overlooked PK268-E2.0A size 23 motor has 240 oz*in holding torque (parallel). When used with a 6:1 belt drive (10 tooth to 60 tooth pulley) and a 1.25 spur gear, it has 1,220 oz*in of torque and it moves 0.000327" per step. If that little motor is used with a 10-tooth pulley, you would need two idler rollers pushing against the belt so that the belt wrapped around the little pulley properly.

There are a lot of options. The ability of a motor to push hard is based on torque. Everyone wants speed. I've always thought that a 4:1 belt-drive was the best compromise between high torque and high speed, but there are many ways to solve the motor "problem".

Read the motor's data sheet. Do the math. Read the forum to see comparisons between a non-geared motor, a geared motor and a motor with a belt-drive. Get out your sketch pad and draw up several designs for a belt-drive to see if it would physically fit your machine. Contact Mike (Metalhead) about his belt-drive kit. When you've done that, you'll have a very good idea of which way to go to get the speed, the torque and the price that's acceptable.
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