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My project involves powering 7 motors from a wall outlet, using switching power supply units. I am a little over-budget on possible output power, but it is very unlikely that all motors will run max power at the same time. Below is the architecture: enter image description here

EDIT - Sorry folks, typo in my "infographic". Should be 1700W from the PSUs.

My questions are below. I'm a bit of a newbie so please excuse :)

1) Can the wall outlet, plus a generic extension cord, really transmit up to ~1800W ? Like is it safe, would anything melt, etc. I've only used extension cords for small electronic devices like laptops.

2) The amperage requested by the motors may be in the 30-50 A range in total, but the wall outlet provides 15A. Just to confirm, the PSUs take care of the power conversion to the amps/volts required, right?

3) What will happen when the system requests more than 1800W ? E.g. if all motors are on and at their highest torque requirements.

4) Similar to above, but for an individual PSU with 400W. What happens if both the 288W motors are trying to request their max power, will power be split evenly at 200W per motor? What if one starts 'working hard' first (taking the max 288W), then after a few seconds the other starts requesting as much power as possible, does it get stuck with only 400-288 = 112W ?

EDIT - one more important question:

My motors are using motor drivers with adjustable current settings (so I can control max #s) and overcurrent protection. Can I leverage these current settings to operate at lower power and ensure I generally don't go over e.g. 1700W? example product: http://www.omc-stepperonline.com/24-phase-nema-23-stepper-motor-driver-2450vdc-15a45a-256-microstep-m542t-p-293.html (I am using Stepper Motors)

Thanks for any insights, as well feel free to suggest improvements to the setup. It's based mostly on available parts and physical space constraints, but could change. What's fixed is using the wall outlet as a source, and having those particular motors.

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    \$\begingroup\$ 1800W at 110V outlet is a bit higher than the allowed 15A. Also I don't see how do you intend to power the 528W motor with 500W PSU. \$\endgroup\$ – Eugene Sh. Jun 29 '17 at 19:07
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    \$\begingroup\$ The power supplies are not 100% efficient, so they will waste some additional power. If the maximum current draw of the PSUs is not specified, you can assume they take 10-20% more power in than they give out. \$\endgroup\$ – Oskar Skog Jun 29 '17 at 19:15
  • \$\begingroup\$ Sorry folks, typo in my "infographic". Should be 1700W from the PSUs. \$\endgroup\$ – JDS Jun 29 '17 at 20:03
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    \$\begingroup\$ It would be preferable to plug the power supplies directly into wall outlets -preferably outlets on at least two different circuit breakers. If you must use an extension, make sure you have a heavy-duty one (or two), with #12 or #14 wire. \$\endgroup\$ – Peter Bennett Jun 29 '17 at 20:05
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    \$\begingroup\$ Also note that a motor rated for X watt steady state power may draw a more than X watt when powering up, especially under heavy mechanic load. If the motor is just switched from off to on instantly (no soft start) the instantaneous power draw can be many times larger. \$\endgroup\$ – marcelm Jun 29 '17 at 20:15
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I am a little over-budget on possible output power…

No, you are way over budget on power.

  • You've added something up wrong. Your four power supplies sum to 1700W, not 1900W. They are completely inadequate for the aggregate load of your motors.

  • You cannot assume that it is "unlikely that all motors will run max power at the same time". Electric motors draw the most power under stall torque, which will occur when the motor first starts turning from a dead stop.

    In the case of PSU1, this means that it will probably overload the power supply every time motor 1 starts up. You need a larger power supply for that motor.

    In the case of PSU2/3, the only way this configuration would be safe would be if you had interlocks in place to make it impossible for both motors to be powered at the same time. 288W x 2 is well in excess of the capacity of those power supplies.

  • Power supplies are not 100% efficient. The wattage ratings you are quoting are for the output of the power supplies; the input will require more power. Read the specifications on your power supplies for details.

  • With regard to the extension cord… read its specifications. Any reputable cord will have its power ratings marked on it. Do not chain extension cords under this type of load.

  • With regard to the outlet, you absolutely cannot draw more power from the outlet than it's rated for. At best, you will trip a circuit breaker. At worst, you may cause an electrical fire.

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  • \$\begingroup\$ Thanks for the answer, some quick q's and comments. 1) Does stall torque == highest current and occur when starting from deadstop even under minimum load? I'm using steppers. 2) I've used a smaller than PSU1 to drive MOTOR1 with no problems, just less torque available. 3) I think I can limit total power in PSUs 2 and 3 with the motor drivers, does this make sense? \$\endgroup\$ – JDS Jun 29 '17 at 20:14
  • \$\begingroup\$ And 4) with regards to the power outlet, since I am drawing from PSUs with a total of 1700W (<1800W) that should be fine right? \$\endgroup\$ – JDS Jun 29 '17 at 20:15
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    \$\begingroup\$ 1) Difficult to say without knowing much more about your situation, but probably yes. 2) Doesn't matter. You're still working outside specs. 3) I wouldn't count on it. 4) No. Your power supplies are probably less than 95% efficient, so they will draw over 1800W at the wall when fully loaded. \$\endgroup\$ – duskwuff Jun 29 '17 at 20:16
  • \$\begingroup\$ Thanks :) I posted a link to a motor driver in the original question for clarification. What's important for my project is being able to use the wall outlet safely, even if motors could potentially draw more power than available. I didn't know efficiency meant the PSUs will attempt to draw more, good tip! \$\endgroup\$ – JDS Jun 29 '17 at 20:21
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1) A 15A wall outlet at 120v can deliver 1800W. A 15A extension cord can deliver 1800W, a generic one might not be able to. The difference is in the cable thickness. There should be a current or power rating on anything you buy.

2) Yes, a low voltage output SMPS delivers more current than it draws. Note that it delivers less power than it draws, the efficiency will never be 100%, 90% ought to be achievable, but you still might only see 80%, depending on the brand and operating power. That means your max total power of 2062W may need a wall power input of possibly 2500W, assuming 85% efficiency.

3) If you try to draw 16A from your 15A socket, you will probably get away with it, at least for a while. The fuses or breakers are probably not set that close. The temperature rise in the cables behind the socket will be slow, and with luck there will be some margin of error built into their design. At sufficiently high overcurrent, the fuses or breakers will disconnect you before you damage the socket or the wiring.

That's what happens physically, if a socket should happen to get overloaded. Please note that if you plan to overload a socket, that violates safety regulations, building codes, it could invalidate buildings insurance etc. You should not plan to do it as part of a project, and anybody responsible for you will be delinquent if they let you proceed with your plan.

4) It depends on the specific power supplies. They may trip out at their power budget, or they may reduce output voltage, it depends what they've been designed to do. The book of words that comes with them should tell you what they do when overloaded. Unlike a wall socket which is probably not protected very accurately to the stated current, the power supplies might give you no margin for overload.

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  • \$\begingroup\$ Thanks for your response. Any insight on my newly edited-in question, on whether I can use motor drivers to safely limit power drawn by my motors? \$\endgroup\$ – JDS Jun 29 '17 at 20:26
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    \$\begingroup\$ @JDS I can't even begin to think why you should think that a motor controller that advertises it can limit the power draw should not be able to limit the power draw. Or are you asking me whether you have the intellect to figure out how to set the power draw limit? Or are you asking me whether the datasheet is wrong, or you've read it correctly? The 'Current Settings' table looks fairly unambiguous to me. \$\endgroup\$ – Neil_UK Jun 30 '17 at 5:36
  • \$\begingroup\$ Because someone else was who answered was implying that I shouldn't count on it... Sorry sometimes I just like getting some confirmation before I do things with electronics :) Thanks again for your answer. \$\endgroup\$ – JDS Jun 30 '17 at 16:11
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I deny any liability, consult an electrician for actual installation.

Wall outlet

The outlet could carry even more than 15A, it will depend your installed wiring and circuit breaker. Single pole residential circuit breakers are pretty cheap, as is the wiring associated with them. In regards to the extension cord, depending on your location, inspectors may not approve permanent extension cords. You are likely going to need to run 'permanent' cable through an armored conduit.

Power Conversion Amps volts Watts

Power converts consist of power electronic circuits capable of changing the voltage/amperage waveforms. However, power out < power in, though most modern power supplies are ~90% efficient.

Overload

Well the circuit breaker for the wall outlet should trip, dropping your system abruptly. You probably should consider fusing individual motors, this will protect them and prevent a single motor from tripping off the whole system if it has a fault. Further depending on your resources, some microprocessor, PLC, or controls could be good.

Individual power supply

In general the speed of the motor rotation is the balance of electrical power in vs mechanical power out. Generic motors will have a speed vs available torque curves when nominal voltage is applied. As voltage is reduced (voltage sags during current rushes), the motors will slow. The exact interplay between your two motors will depend on the actual motors and PSU. The motors themselves are greedy, and will compete for power from the power supply. A microprocessor or other technology would be required to get them to 'cooperate'.

Overall Thoughts

You might need to talk to your facilities or building management folks about the 15A wall outlet. Rewiring this outlet and installing a larger breaker should be easy compared to getting the 7 motors to do what you want. I think you'll want/need a larger circuit, and a 20A breaker and wiring should be < $100. Though you know your constraints better than I...

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I'm surprised no one else has noted that in the United States most, if not all electrical codes,specify that the maximum current drawn from a 15 amp circuit should not exceed 12 amps (in general the maximum current should not exceed 80% of the circuit rating). Thus you only have about 1440 watts available, not 1800 watts.

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You appear from your comments to be driving stepper motors ....in addition you are micro-stepping the motors which makes things less predictable. These motors have fixed (maximum) current rates (which defines the hold torque) and if your drivers are not reducing the current when stationary (unlikely with micro-stepping) they are always at the current defined for each microstep by the controller.

Stepper motors do not draw larger currents when they start. Stepper motors do not provide higher torque when starting.

Since you clearly have a motion control system, the motors could (unlikely though it might be) all be at a maximum current step and stationary so your power calculations are way over the top.

A more sensible way to arrange your motion control is to gear down (typically toothed belts) which allows you to reduce the holding torque required and hence the maximum motor current.

Added after comment Based on the controller you have it looks like you have a maximum capability of 4.5 A per phase at a maximum of 50 V and probably have a simple 2 phase bipolar stepper.

Since you are micro-stepping, the current is the peak (ignore any references to RMS current). The absolute peak power in each phase is 4.5^2 * Phase R .....I don't see the steppers specified, but likely you have not more than 1 Ohms ...so about 20 W peak per phase (and with micro-stepping you never have both at peak current) and the voltage on each phase is no more than 4.5 V.

Here is a link to a good quality Nema 23 stepper. It has 4.2 A peak and with a 0.33 Ohm phase R would only dissipate 4.2^2 * 0.33 = 6 W per phase winding. The worst case dissipation (no micro-stepping would be 12 W per motor.

Note: I think you have by mistakenly taken your per phase current and multiplied by your power supply voltage ...which would give something of the order of 216 W per phase ...two phases --> 432 W ...Very incorrect....

Given this sort of calculation mistake ...I think your stepper motors in total are probably no more than 100-200 W Total power including all losses in drivers and power supplies.....and you need to revisit your calculations.

Driving stepper motors ....if you use a site such as http://www.orientalmotor.com/, you will find a set of tools that may help you understand the problem better.

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  • \$\begingroup\$ Thank you! I should have mentioned that I'm driving stepper motors. And they are already geared down to what my application requires :) Though you are right that I can probably gear down even more (sacrificing speed) for meeting the power specs. \$\endgroup\$ – JDS Jun 29 '17 at 21:34
  • \$\begingroup\$ So then do you think the drivers, such as the one I linked, can reliably limit the power that my motors draw? Via their current settings. \$\endgroup\$ – JDS Jun 29 '17 at 21:35
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    \$\begingroup\$ Added to answer ...hopefully this will help you. I don't think you are overloading the power plug at all ...I think you have incorrectly calculated the power you are consuming. \$\endgroup\$ – Jack Creasey Jun 30 '17 at 0:08
  • \$\begingroup\$ Regarding power - that seems really weird that your numbers are so low. Yes, I calculated based on e.g. 4A/phase * 2phases * 36V supplied = 288W. Maybe that's wrong. But 20W? Seems very low. Power is also = Torque * Angular velocity. From a sample Nema23 torque-speed curve, I pick a point on the curve and multiply e.g. 1.2Nm * 720rpm (~ 75 rad/s) and get 90W. Still lower than what I calculate though. Sample curve: oceancontrols.com.au/datasheet/lea/MOT-181-ST.png \$\endgroup\$ – JDS Jun 30 '17 at 16:22
  • \$\begingroup\$ I guess I need to make a new question for how to really calculate the power required with these stepper motors! \$\endgroup\$ – JDS Jun 30 '17 at 16:22

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