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I am making a winch that will spool out line under tension. I've had success using an old brushed motor. I short the windings, so that when I unspool the line, the motor provides braking torque and line tension. The old motor is not powerful enough, so I upgraded to a stronger, brushless motor with 5000w power rating.

Here's the problem: as I unspool the line from this motor, it builds torque until it reaches a certain value (130lb in my case), and then the current and torque suddenly drop to zero and then build up again, then drop again.

The motor has 3 phases (which are shorted for the generator mode) and a 5-pin hall sensor connector (unused here). Here's the exact motor: http://www.aliexpress.com/item/Electric-bike-brushless-dc-hub-motor-5000w-for-electric-bicycle-5kw-ebike-hub-motor/32234157960.html

Why is it dropping the current in generator mode? Do these motors have some kind of a built-in internal protection? I thought that there was nothing inside but windings and magnets (and halls), as this one uses an external 100Amp controller.

Any thoughts would be appreciated. PS. Does this 3-phase brushless generate AC or pulse DC?

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  • \$\begingroup\$ I am not expert on motor control. But here are some basic questions that anyone might want answered, to help you. Are you using a commercial controller with the motor? Do you disconnect the controller when spooling out wire? Is the motor wired as a "Wye" or "delta" configuration. I am guessing "Wye." When you say you shorted the windings together for generator mode are the three windings also shorted to ground, or just to each other? Is there any load to ground on the generator at all? \$\endgroup\$
    – user57037
    Commented Feb 2, 2015 at 1:45
  • \$\begingroup\$ When you take away the controller from a brushless DC motor, it is essentially a permanent magnet synchronous motor (AC). If you hook up the phases to different channels of an oscilloscope and spin the motor shaft somehow (maybe with a power drill or something) you will see a three phase voltage waveform. Might be sinusoidal or kind of trapezoidal. In generator mode, I believe you can achieve constant drag during spooling by putting resistors from each phase to ground. Because the sum of power delivered to all resistors is virtually constant under constant speed. \$\endgroup\$
    – user57037
    Commented Feb 2, 2015 at 1:56
  • \$\begingroup\$ MKeith, thanks for the informative response. You indeed got me thinking. The motor is likely a Wye, but I don't know for sure. Yes, the controller is disconnected when I unspool line. There are three wires from the phases and they are just shorted to each other. The most glaring thing from your post is that I don't have a ground. I don' have any ground wire and have no idea how I would introduce ground into the system here! \$\endgroup\$
    – Greg
    Commented Feb 2, 2015 at 2:21
  • \$\begingroup\$ The reason I don't use resistors is because they would cut the current and cut the torque. The phase to phase resistance is 0.3 ohm. If I add even another 0.3 ohm, it will cut the current and the turque in half. Even if I get such resistors (I was gonna use heating elements), how would I connect them to "ground"? The motor does not appear to have any... \$\endgroup\$
    – Greg
    Commented Feb 2, 2015 at 2:24
  • \$\begingroup\$ Exactly how are you shorting the phases to each other? What diameter is the winch drum? \$\endgroup\$ Commented Feb 2, 2015 at 2:44

4 Answers 4

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The load angle

Your BLDC behaves like a synchronous machine, the only difference is that you give it block voltages instead of sinusoidal voltages as input. That means that the principles of the synchronous machine can be used on your machine.

The load angle is a measure (as the name says) of the loading of the machine. As you can see from the figure below it is the relative displacement of the rotor in regard to the field axis. Load angle in relation to flux lines

The second figure shows you how the torque(or power) of the machine depends on the load angle. You can see that there is a maximum at \$90^\circ\$. After this points the flux lines tear apart, and the torque reduces. That means if you pass this point your motor will loose torque and slow down.

Torque dependence in regard to load angle

The simplified equation describing this behavior is: $$ T=\frac{3UE}{X_s\omega}\sin\delta $$

For references and a short explanation look this link.

EDIT: In regard to the comment

The synchronicity is irrelevant, that is true. The thing that is worth thinking about is the stability. Let's do a little calculation, the induced voltage is divided between the short circuit resistance and the reactance of the machine(we will neglect \$R_{SC}\$ as needed):

\begin{equation} \underline{E}={{jX_s}\underline{I}+R_{SC}\underline{I}}\nonumber\\ E\angle\delta={{jX_s}I\angle\varphi+R_{SC}I\angle\varphi}\\ I\angle\varphi=\frac{E\angle\delta}{{jX_s}+R_{SC}}; R_{SC}<<X_s\\ I\angle\varphi=\frac{E\angle(\delta-90^\circ)}{{X_s}}\\ I\angle-\varphi=\frac{E\angle(\delta-90^\circ-2\varphi)}{{X_s}} \end{equation}

Let's multiply by the current to get the losses (multiply by\$\underline{I}^*\$): $$ IE\angle(\delta-\varphi)={{jX_s}I^2+R_{SC}I^2} $$

The losses are the last term \$P_{loss}=R_{SC}I^2\$, we can equalize them with the real part of the rest of the equation: $$ P_{loss}=\mathfrak{R}\{IE\angle(\delta-\varphi)-{jX_s}I^2\}\\ P_{loss}=\mathfrak{R}\{\frac{E\angle(\delta)E\angle(\delta-90^\circ-2\varphi)}{{X_s}}\}\\ P_{loss}=\frac{E^2\cos(2\delta-2\phi-90)}{{X_s}} $$

These are the losses, not really comparable to the previos torque equation. Let's look what happens to the load angle in the meantime: $$ 1\angle\delta=\frac{{jX_s}I\angle\varphi+R_{SC}I\angle\varphi}{E}; R_{SC}<<X_s\\ 1\angle\delta=\frac{{X_s}I\angle(\varphi+90)}{E}\\ \delta=\varphi+90 $$ For an ideal short, it will be unstable as you can see from the last equation. Put some resistors instead of a short, it should work better. Capacitors would help too, just calculate it first with the given equations. Watch out for three phases, this is the one phase calculation!

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  • \$\begingroup\$ In synchronous machines, the rotation is in sync with the supplied AC. But in a generator mode, there is no supplied AC, so I would think synchronicity becomes irrelevant. Yes, the torque would vary FOR EACH COIL as presented in the figure above, but the motor has at least 6 coils, and that pulsating effect is not felt. I am having more than just torque oscillation caused by passing magnets. I am getting a compete drop in current for 1-2 seconds at a time. \$\endgroup\$
    – Greg
    Commented Feb 3, 2015 at 17:38
  • \$\begingroup\$ @Greg Edited the question, added some calculations. \$\endgroup\$
    – WalyKu
    Commented Feb 3, 2015 at 22:54
  • \$\begingroup\$ I don't really understand this. I mean, I haven't taken the trouble to try. But intuitively, it makes sense that capacitors could help. Because we do want the current to be out of phase with the back EMF in order to produce drag. As I said before, it is a power transfer problem. We want to maximize the transfer of power to some load. If the source impedance of the synchronous machine is reactive in generator mode, then a matching network could help. \$\endgroup\$
    – user57037
    Commented Feb 5, 2015 at 5:50
  • \$\begingroup\$ @Kurtovic - how does your calculation explain why the same phenomenon does not happen with a brushed motor (see my latest comment at bottom of page)? Also, what place and role do capacitors have in the system? Are they hooked up parallel to the braking resistors? \$\endgroup\$
    – Greg
    Commented Feb 12, 2015 at 16:01
  • 1
    \$\begingroup\$ I finally tested a different set of resistors. Three 0.25 Ohms in parallel per phase. My ohm-meter is showing such resistance (0.25/3) as zero. I couldn't measure the current or voltage because of this, but I got a steady torque of 120lb ! Still don't know all the theory, but somehow resisters prevent the current drops I was seeing before. The drops happened at 130lb, so it remains to be seen if I can take this thing to torques above 130lb and still have steady torque without the sudden drops. \$\endgroup\$
    – Greg
    Commented Mar 18, 2015 at 1:34
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I don't believe you're stalling (stopping) the rotor, there's a car pulling it...

However, remember the generated current generates its own field, opposing the PM field. At some point there will be (partial) field cancellation, and thus reduced voltage generation, reduced stall current, reduced torque, at a point where the field from one pole is falling off anyway and before the next pole fully engages. It's a little more complex than this, when a rotor pole is between two stator poles, the phase of the voltage it generates in each will be different. I don't have a good enough understanding to describe it in more detail.

So there will be a point between poles at which the motor "lets go", before the field increases again at the next pole.

With suitable resistances instead of a short you will be able to pull it round by hand and feel the cogging as each pole engages.

I think that imposes an upper limit on the braking you can achieve from the motor : to get more, you need a bigger motor, or a smaller diameter winch drum, or gearing.

EDIT : given the information that the pulsation is at 1-2second intervals, it's clearly not the poles cogging.

I suspect the start/stop motion may be from the cable acting as a spring, storing energy while the motor brakes efficiently, but it's not clear why the brushed DC motor acted differently, sorry.

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  • \$\begingroup\$ The "certain point between poles" corresponds to "the middle of a pole" for another phase. There is no point that a BLDC cannot develop torque, just points of reduced torque due to cogging \$\endgroup\$
    – user16222
    Commented Feb 2, 2015 at 12:36
  • \$\begingroup\$ @JonRB - Not exactly - you are correct about the exact midpoint between poles, but that's not where the torque troughs occur. \$\endgroup\$
    – user16324
    Commented Feb 2, 2015 at 13:41
  • \$\begingroup\$ They do actually.If you look at the line-line backEMF of a BLDC machine, especially "rectified" in relation to absolute rotor position, there are troughs that correspond to the points of commutation as a rotor pole crosses between two teeth. This however does not result in a point of zero torque otherwise you would have breakout issues and specific starting & parking requirements when dealing with BLDC machine. These troughs are compounded by motor-drives as they commutate the stator current & produce higher torque ripple. Thus the statement that the "motor "lets go" isn't correct \$\endgroup\$
    – user16222
    Commented Feb 2, 2015 at 14:51
  • \$\begingroup\$ I can pull it by hand even when it's shorted. It creates steady torque when I spin in by hand slowly. When I spin it with a car much faster than by hand, the torque starts to "pulsate" (abruptly drops and gradually rises) in 1-2 second intervals (much slower than the phase rotation speed). \$\endgroup\$
    – Greg
    Commented Feb 2, 2015 at 18:40
  • \$\begingroup\$ Just a quick comment. If you imagine a BLDC with three phase sinusoidal back emf being used as a generator, the total instantaneous voltage output is actually constant at constant speed. For a trapezoidal back emf, I guess this is not true, or doesn't have to be true, but is still an approximation of the truth. Anyway, constant voltage + resistive load implies smooth torque characteristic. So a PMSM used as a generator would have drag but zero torque ripple with a resistive load. The trapezoidal EMF might give rise to some torque ripple, but not create detents. \$\endgroup\$
    – user57037
    Commented Feb 2, 2015 at 23:44
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Not sure if I am supposed to post as a separate answer to my own question or just use comments. I've done a bunch of comments (see above), but this one was too long.

"What is occurring here is the generated torque is higher that your load so the rotor stalls." Interesting... The motor is not driving anything. It is disconnected from any power supply. The "generated torque is higher" than what load? - There's no load, the motor is not driving anything, it's disconnected from power. The torque only opposes the mechanical rotation. The mechanical rotation never stops, as the car unspooling the line is way stronger that this motor. What stops is the torque, at which point the rotation accelerates suddenly (b/c there's no opposing torque at that moment), and then slows down again when the torque is back. It didn't happen on a burshed motor that I used to try this first. Maybe the current gets so high that it messes up the next phase's magnetic field... - Your second explanation seems to make a lot more sense. I had no idea the flux direction changed at higher current. Again, didn't happen with the brushed motor - that sucker produced torque even as it would start smoking at high current (35A while rated at 25A).

I'll try the limiting resistors, but I have now had two BLDC motors that don't produce consistent torque as generators (I just tried another BLDC that I have). Interesting phenomenon. The torque builds up and is continuous to certain RPMs of external mechanical rotation, and then starts to disappear for 1-2 seconds and only comes back in short spurts after that. The manufacturer just told me the motor has no internal over-current protection. I can't think of any theoretical reason why BLDCs would behave this way. Since they produce AC, I'll try a 3-phase diode rectifier and a low-resistance heating element as a load.

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  • \$\begingroup\$ You can edit your own question to add the new information and clarifications, rather than long comments or an "answer" \$\endgroup\$
    – user16324
    Commented Feb 2, 2015 at 18:07
  • \$\begingroup\$ Permanent magnet motors and generators are very efficient. Here is what I propose. Power = force * speed. You know what force you want. You know what speed you want it to happen at. Calculate that power in Watts. Call it P. Spin up the motor to the target speed with no electric load on the phases. Measure or estimate the RMS voltage from any one wire to any other. Call that V. The resistor (or heater element) you need can be calculated using P/3 = V^2/R. Now, when you load the generator, the voltage will sag, but hopefully not enough to throw off the calculation. \$\endgroup\$
    – user57037
    Commented Feb 2, 2015 at 22:30
  • \$\begingroup\$ Don't rectify an AC signal just to attach a resistive load. Just apply the resistive load directly. Actually three resistive loads. Conceptually, arrange the resistors in a delta configuration (triangle) and connect one wire to each corner. \$\endgroup\$
    – user57037
    Commented Feb 2, 2015 at 22:35
  • \$\begingroup\$ I think you are overheating the core. When you do, flux density crashes, reducing torque. Then it cools and comes back up. Short-circuiting the coils is not going to produce the maximum torque, nor maximum total heat, but it will produce the maximum current and heat in the windings. So far you have been assuming that current in the windings is good because it creates torque, and that is why you are hesitant to add any resistance. I am looking at it from conservation of energy. For a motor at a given speed, maximum drag will occur when load is optimized for maximum total heating. \$\endgroup\$
    – user57037
    Commented Feb 2, 2015 at 22:47
  • \$\begingroup\$ Productive discussion:) Yes, P=F*v. F=130lbf=60kgf=600N (g=9.8 ~10). v=10mph=4.5m/s. P=600*4.5=2700 Watts. The motor is rated for 5000w, and I am trying to get it to pull at 150lbf by speeding the car (and thus linear line speed) past 10mph. When I do that, the torque simply drops to zero and then re-builds back to about 60lb and then drops again, such cycles being 1-2 sec. The car never exceeds 12-15mph. So even at max, P=700N*5.5m/s=3850w ~4000w. \$\endgroup\$
    – Greg
    Commented Feb 3, 2015 at 19:35
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In short your generation capability is a lot higher than your load.

You are essentially "plugging" the machine, shorting the 3phases together.

The backEMF constant is proportional to speed. As the rotor starts to turn, voltage is induced. Since the phases are shorted a current will flow which in turn generates a torque, a torque that will try to oppose your load.

The faster the rotor goes, the more torque is generated ...

Depending on specifics of the electrical machine, you could be experiencing field weakening as an increase in speed will result in increase in current which will shift the stator flux vector with respect to the rotor flux vector, which results in a drop in developed torque -> current.

Solutions? Some series resistance would limit the current and thus the torque but that is a waste. Some gearing would help. Smaller cog on the load, larger on the rotor. Some frictional component equally would help but again that's loss

Also a BLDC machine will generate an AC backEMF, it will not be a pure sine wave (unless the emag designer was ... Quick). It will have a slightly flatter top

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  • \$\begingroup\$ "In short your generation capability is a lot higher than your load." So you are saying that the shorted wires are not strong enough load to absorb the generated current? If I add resistors across the windings, they will limit the current and will dissipate the generated energy as heat. Even though it is a waste of sorts, I'd be fine with it. In frichtion-based systems, such dissipation is not channeled anywhere specific and results in brakes overheating, etc. \$\endgroup\$
    – Greg
    Commented Feb 2, 2015 at 17:44
  • \$\begingroup\$ "You are essentially "plugging" the machine, shorting the 3phases together." I thought this would only be applicable if the motor was running "direct", connected to a power source. I am running it in "reverse", by spinning it mechnically and generating current. \$\endgroup\$
    – Greg
    Commented Feb 2, 2015 at 17:48
  • \$\begingroup\$ "shorted wires not strong enough" - thats dependent on what you pick but that wasn't what I was alluding to. Thinking a bit more about your situation, without knowing specifics of the machine (especially terminal inductance) you are quite likely experiencing field weakening at the top speed where the stator inductance is shifting the current w.r.t. backEMF and thus are no longer in phase. If this carries on then the zero torque point will be reached and current will drop off \$\endgroup\$
    – user16222
    Commented Feb 3, 2015 at 10:30
  • \$\begingroup\$ yeah, it looks like either field weakening or resistance growing/current weakening. This doesn't happen with a similarly designed brushed motor (ebay.com/itm/…) as opposed to this brushless (ebay.com/itm/…) \$\endgroup\$
    – Greg
    Commented Feb 3, 2015 at 20:02

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