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Old Nov 30, 2012, 12:35 AM
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Question
how does no-load current affect motor?

background: my focus is on vtol multirotor


when i was playing with ecalc multirotor portion.. i found an interesting thing

when everything else stays the same...
increase no-load current.
the hovering time and flight time decrease

so i wonder how does it affect the motor's performance?


from my knowledge.. .. i think the no load current is (Vin - backEMF)/resistance

mechanically
the volt diff comes from the difference between ideal rpm and actual rpm....
so literally the smaller no-load current means there are less friction in the system...


but how about on the electrical side of the motor?


thx
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Old Nov 30, 2012, 03:06 AM
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Quote:
Originally Posted by jy0933 View Post
from my knowledge.. .. i think the no load current is (Vin - backEMF)/resistance
True, but not relevant. Resistance could be zero and there would still be a no-load current. It is caused by magnetic (hysteresis, eddy current) and frictional (bearings, windage) losses. A (small) resistive loss also occurs as the no-load current passes through the windings.

No-load current is not available for torque production, so it simply wastes battery capacity, as well as heating the motor up a bit and slightly reducing output rpm. High no-load current reduces motor efficiency, but this is more noticeable at low power levels where Io is is a significant proportion of the total current draw. Having a low Io is important for getting efficient operation at low throttle.

NOTE: the no-load current you measure with a wattmeter also includes current used by the ESC for its internal operation, and possibly BEC current going to the receiver etc. This current doesn't even get to the motor, but is still an excess that the battery has to supply. The main effect of this is the same - a reduction in run time.
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Old Nov 30, 2012, 10:31 AM
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You can see it this way :

voltage "goes through" the winding resistance and produces current,
current produces torque,
torque produces movement --> increasing rpm,
rpm produces bEMF,
bEMF comes decrease the voltage through the winding,
less current produces less torque,
rpm stop increase.

Then no-load current is just the current needed so that torque is enough to make the motor rotate with no load at the rpm allowed by the applied voltage.
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Old Nov 30, 2012, 12:18 PM
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Originally Posted by yomgui View Post
Then no-load current is just the current needed so that torque is enough to make the motor rotate with no load at the rpm allowed by the applied voltage.
There are actually two main components to motor current.

In an ideal motor, no torque is required to turn the armature under no load.

There will still be no load current due to magnetic and other losses.

It may be seen in the simple motor model below that from a theoretical point of view used for analysis, the no load current does not pass through the armature but shunts it and does not contribute to torque.
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Old Dec 01, 2012, 08:46 PM
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Then no-load current is just the current needed so that torque is enough to make the motor rotate with no load at the rpm allowed by the applied voltage.
If you consider 'internal' torque required to spin the rotor then yes, (some of) the no-load current will be contributing to this.

However, normally we are only interested in shaft output torque, and the torque produced by Io is 'wasted' before it gets there. What the no-load current is doing inside the motor is not of interest to us. For all we know a large part of it could be going through a big resistor wired in parallel with the motor terminals (which is effectively the case for a shunt-wound non-PM DC motor).
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Old Dec 02, 2012, 03:54 AM
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Hi
No load losses are mainly coming from eddy currents and magnetic losses
A second order polynomial is a good fit for the losses vs rpm
L = A + B x rpm +C x (rpm)^2
Example on S5030-220 F3A
Louis
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Old Dec 03, 2012, 08:02 PM
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thanks for the info guys..


so it seems.. the lower no-load current.. the less loss.. so that the motor is better? (regard less of balance. just purely from stats aspect)

i'm kind of wonder why some motors have so high no-load current.. but still consider good?
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Old Dec 04, 2012, 10:44 AM
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Quote:
Originally Posted by jy0933 View Post
thanks for the info guys..


so it seems.. the lower no-load current.. the less loss.. so that the motor is better? (regard less of balance. just purely from stats aspect)

i'm kind of wonder why some motors have so high no-load current.. but still consider good?
That's why I think it's useful to see it as "rotor torque", the torque needed to rotate itself. Bigger motor = motor torque.

Torque = amps / kv , with the good units (also most of the time people use Kt = 1/Kv, but that's the same logic)

You'll find that identical motors of different Kv have different no-load current (higher Kv = Higher no-load current), but in fact probably the same "rotor torque"
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Old Dec 04, 2012, 10:55 AM
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Originally Posted by yomgui View Post
That's why I think it's useful to see it as "rotor torque", the torque needed to rotate itself. Bigger motor = motor torque.

Torque = amps / kv , with the good units (also most of the time people use Kt = 1/Kv, but that's the same logic)

You'll find that identical motors of different Kv have different no-load current (higher Kv = Higher no-load current), but in fact probably the same "rotor torque"
Hi
Yes the graph "no load losses versus rpm " (see post # 6) is the same for the same motor kit- different turns (different Kv)
Louis
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Old Dec 04, 2012, 11:06 AM
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Originally Posted by jy0933 View Post
thanks for the info guys..


so it seems.. the lower no-load current.. the less loss.. so that the motor is better? (regard less of balance. just purely from stats aspect)

i'm kind of wonder why some motors have so high no-load current.. but still consider good?
No load current is only half the picture - resistance is a bigger contributor to losses at high power levels. Usually, larger motors have lower resistance and higher no load current since they have more stuff to move.

The balancing act is to choose a prop that draws enough current so that no load current becomes a fairly small % of power consumed, yet not so much current that resistive losses become too large of a % of power consumed. And then do so for the range of speeds you think you will be using.

By way of example, an extremely large motor might have miniscule resistance but if you run a 3 inch prop on it drawing only a few amps, while it still draws a few amps just to turn itself, then you're not gonna have a very efficient system. Alternately, running a massive prop at high speed on a very small motor will make the running of the motor (the no load current) negligible compared to the running of the prop, but you will be drawing too much current through too large a resistance and most of your energy will be lost to heat (and the motor will probably burn up).
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Old Dec 04, 2012, 11:37 AM
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Originally Posted by Nereth View Post

The balancing act is to choose a prop that draws enough current so that no load current becomes a fairly small % of power consumed, yet not so much current that resistive losses become too large of a % of power consumed. And then do so for the range of speeds you think you will be using.

.
Hi
Not exactly.
It is better to try to use enough voltage and the right prop and rpm, to try to balance:
no load losses =~ copper losses.
That is the way to work close to Peakeff
or to manage a good efficiency compromise
** Climbing or take off, full gaz (WOT) overcurrent / Peakeff with a reasonable percentage
** Level cruise just under Peakeff
The only solution is to choose the right voltage (high enough)
Louis
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Old Dec 04, 2012, 11:44 AM
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Hi
Not exactly.
It is better to try to use enough voltage and the right prop and rpm, to try to balance:
no load losses =~ copper losses.
That is the way to work close to Peakeff
or to manage a good efficiency compromise
** Climbing or take off, full gaz (WOT) overcurrent / Peakeff with a reasonable percentage
** Level cruise just under Peakeff
The only solution is to choose the right voltage (high enough)
Louis
Indeed modifying voltage can be superior, but most people don't have as much control over voltage I think, they try to use the same cell count they are already using etc.
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Old Dec 04, 2012, 01:24 PM
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Quote:
Originally Posted by jy0933 View Post
i'm kind of wonder why some motors have so high no-load current.. but still consider good?
It depends on what application the motor is designed for. Obviously a larger motor will tend to have a higher no-load loss, simply because it has more 'stuff', and a motor with higher Kv will have higher Io at the same voltage. However you have probably noticed that even motors of the same size and Kv can have quite different no-load currents.

So why is it that a 'high' no-load current can be still acceptable in a 'small' motor? It depends on what application the motor is designed for. All motors run at best efficiency when drawing enough current to double the loss, ie. when iron loss and copper loss are equal. So if you want the highest possible peak efficiency then you need to keep the no-load current down. However, Io becomes less important as current increases. If you want to optimize efficiency at high current then it may be better to concentrate on reducing copper loss, and accept higher iron loss.

Why can't you have both low iron loss and low copper loss? In motor design the principle of TANSTAAFL applies - if you change the configuration to improve one parameter, another one will suffer.

Iron loss can be reduced by using fewer magnets. However to keep Kv down you then need more turns, which increases winding resistance (and therefore increases copper loss at high current). The result is a motor which is optimized for lower current. Or you can keep the same number of turns, and get a motor which is optimized for higher rpm. If you want to improve efficiency at high current then you can increase magnet coverage so that less turns are required, but magnetic losses will increase (resulting in higher Io).

Other ways to reduce iron loss include using thinner stator laminations (raises cost), better stator material (also raises cost), reducing thickness of the stator arms (lowers saturation current), weaker magnets (raises Kv), increasing air-gap (raises Kv), going slot-less or eliminating the iron altogether (raises Kv dramatically).

Cheap motors tend to have thicker laminations and poorer quality steel, so their no-load current is higher. However Io doesn't have much effect at high current, so a cheap motor can still work well at high power if it has good copper fill. If you aren't concerned about getting the highest possible peak efficiency at low power and/or don't mind having a higher Kv, then that cheap motor with higher Io may still be considered 'good'.
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Old Feb 19, 2013, 08:50 AM
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Interesting discussion.

A little different application:

RC Car with variable motor timing

I am trying to find the the "sweet spot" of advancing the motor timing.

Motors are all sensored brushless 13.5T through 25.5T

If I advance the timing to much I just generate heat.
Can I measure the peak current draw from the battery (7.4V) with no load using a digital meter?
How would I know what what current value provides the best power and isn't just loss in heat? I guess a I am looking for max effiency (power)
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Old Feb 19, 2013, 09:28 AM
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Best efficiency will be at the lowest advance and noload current. As you increase the advance the noload current goes up as does the noload RPM. Generally it is best to run at the lowest advance and change gearing to get more speed. Once you are close to the performance you want then adjusting the advance can be tweaked a little to fine tune the setup.

Steve
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