Universal motor with commutator, recovering the back EMF?

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  • #1
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So while thinking about motors this suddenly struck me,
So as the universal series wound motor is spinning there is always some arcing going on around the place where the brushes contact the copper segments that slide past them, I assume this is at least partly because as each coil pair of the rotor is connected a current is built up in the coil , then as the segment for that coil pair passes by the brushes it disconnects electrically from the brushes at some point, this disconnection is rapid especially at higher RPM.

As much as I know from physics I have always remembered that an inductor always opposes changes in current , so upon sudden disconnection and circuit termination the inductor B field collapses inducing high voltage across it's terminals, this is the way car ignitions coils work I think.


So here as the coil slots disengage the brushes in theory on those slots there is a voltage buildup due to the stored flux collapsing ?
But this energy left in the disengaged coil gets wasted right?

So what would happen if I took a universal motor and positioned additional brushes at some distance from the current providing ones? I should pickup some voltage/current as each disengaged slot pair passed my new set of brushes by ?



PS. Now that I think of it , could it be that at higher speeds the rotation of the rotor is fast enough that each disengaged coil pair doesn't have the time to fully collapse it's field and as it completes a full turn and touches the brushes again there is still some voltage left over from the last cycle in that coil ?
 

Answers and Replies

  • #2
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Ok I just imagined one more thing so I might partly answer my own question,

isn't the rotor wound in such a way that each next coil overlaps a rather large armature area of the previous one, so as each previous coil disengages each next coil is already engaged so the overall flux through both of the coils changes minimally so this voltage buildup effect is minimized across the commutator ?
 
  • #3
sophiecentaur
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Where would any energy come from? You seem to be suggesting that you’ve found a PM machine????
PF has no interest in / bans such discussions, I’m afraid.
 
  • #4
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No, you can't recover that part of the energy. It is permanently lost due to constructive topology of the machine using commutator principles
 
  • #5
Baluncore
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An arc is drawn as the commutator bar moves away from the brush.
The arc continues so long as current continues to flow in one direction.
The negative voltage of the arc is in series with the inductance. V = L·di/dt.
A greater negative arc voltage reduces the current faster. di/dt = V/L.
The arc voltage is needed to stop the current.

You could catch the arc energy with a trailing brush.
It would have to provide the high negative voltage to catch, and stop the current.
For a Universal Motor running on AC, that “catcher” would keep reversing polarity.
It would be uneconomic.
 
  • #6
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@sophiecentaur , with all due respect I sometimes feel you just criticize my questions without even considering the point I am asking.

This has nothing to do with PM, and I am not a lunatic, thank you.

This energy that is as @zoki85 said perfectly in accordance with the laws of physics.

@Baluncore so the majority of the arcing in the commutator is indeed due to this collapsing flux in each coil pair that electrically disconnects from the brush contact?
I guess apart from the high voltage that develops there is little energy that could be recovered if after the power supplying brush there would be a as you said "trailing " brush that caught the leftover energy from the collapsing field?
 
  • #7
Baluncore
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Baluncore so the majority of the arcing in the commutator is indeed due to this collapsing flux in each coil pair that electrically disconnects from the brush contact?
How could it be otherwise. The brushes can straddle several commutator bars at one time so the field current in a UM continues to flow, while it takes many different parallel paths through the armature.

I guess apart from the high voltage that develops there is little energy that could be recovered if after the power supplying brush there would be a as you said "trailing " brush that caught the leftover energy from the collapsing field?
Your guesses are usually wild, "High voltage" is not "little energy". Your sentence appears to make no sense. Maybe you do not understand inductance; V = L·di/dt; and why a negative inductor voltage is required to reduce the current to zero.

When an arc forms from a commutator bar back to the brush, then the current in the arc is being passed through the brush to the commutator bars and windings more recently in contact with the brush, so the current is not wasted.
 
  • #8
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Combination "trailing brush-high voltage" would be technically (too) difficult and unpractical to realize.
Considering energy savings not worth an effort
 
  • #9
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@Baluncore , well as the coil pair in question disconnects from the brush supply the field in the coil starts to collapse as it does it generates a current in the opposite direction this is I think what you refer to "negative voltage" so the commutator slot that was say positive when it was connected to the brush now becomes negative and a PD develops across the slot and the brush and arc forms until the current is diminishes and the PD decreases, I think this should be right ?


well I said little energy because for a small coil even with high voltage the current might be low isn't it the case? and it's duration is short.

PS. I do agree this is not practical it was just a curiosity of mine.
 
  • #10
Baluncore
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Baluncore , well as the coil pair in question disconnects from the brush supply the field in the coil starts to collapse as it does it generates a current in the opposite direction this is I think what you refer to "negative voltage" so the commutator slot that was say positive when it was connected to the brush now becomes negative and a PD develops across the slot and the brush and arc forms until the current is diminishes and the PD decreases, I think this should be right ?
How can you communicate anything clearly in such a long sentence. If that is streamed consciousness, then inductance must be really confusing for you.

Are you really saying that when the gap opens and the field starts to collapse, the current reverses? Surely the voltage across the inductor must reverse in order for the inductor current to begin to fall back towards zero.

You appear to not understood the implications of V = L·di/dt.
di/dt is the rate of change of current.
A positive voltage causes the current to increase.
A negative voltage causes the current to fall.
Zero voltage maintains the original current, there is zero change.

well I said little energy because for a small coil even with high voltage the current might be low isn't it the case? and it's duration is short.
How little is little compared to high or low ?
The duration can be very short if the arc is permitted to form because the arc voltage makes for a faster collapse. Energy is the integral of V·I ·dt but you probably don't understand that.
 
  • #11
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@Baluncore I think I understand energy, VxI is voltage times current, which in a DC case would give the total instantaneous power.
Energy is then that power and how long it's there or power x time. Like the force I push on a spring and the length of time i'm doing it gives the energy I have used up to do so.


Ok I had a confusion with regards to the inductor, I had to do some searching because I could not entirely tell from your answer. So if current is flowing through an inductor , the moment the current gets interrupted in the circuit the inductor was part of, the inductor tries to maintain current flow but in order for this to happen the polarity across the inductor terminals must reverse. This reversed polarity then "brakes" the original current as the B field collapses and then when the field has fully collapsed the current has fully stopped and the reversed polarity voltage across it's terminals diminished to zero.


I take that the "negative voltage" across a inductor is at it's peak the moment the inductor is disconnected and then falls linearly with current ?


And now please be patient , one more possible confusion here for me, so the inductor reverses polarity but doesn't reverse current direction (the I just decreases to zero over time) as the field of it collapses, in a universal motor then that means that as the coil pair disconnects from the brush contacts it suddenly reverses polarity. But the current through the inductor isn't reversed so how can it give away it's stored energy back to the brush contacts?
 
  • #12
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one more possible confusion here for me, so the inductor reverses polarity but doesn't reverse current direction (the I just decreases to zero over time) as the field of it collapses, in a universal motor then that means that as the coil pair disconnects from the brush contacts it suddenly reverses polarity. But the current through the inductor isn't reversed so how can it give away it's stored energy back to the brush contacts?
Energy is indestructible. Inductivity won't disappear, current won't instantly change direction. Main point of application of a trailing brush with high impulsive voltage of opposite sign would be to suppress/eliminate formation of sparks when the main brush leaves/disconnects from the commutator's segment, and allow that part of current to flow through the trailing brush circuit. Easier said than done!
 
  • #13
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well I did some searching and I guess this is what I come up with.
so as the coil/inductor disconnects from the brush contacts it tries to maintain current , for this to happen it has to reverse polarity and develop a path/loop through which the current can close through the inductor, otherwise it cannot flow. So in a universal motor the decoupled inductor/coil simply develops a high voltage spike of opposite polarity which allows the current to be "dumped" back through the circuit (stator coils, power source) as the B field of the coil itself is collapsing.


In a transformer on the other hand , such abrupt disconnection of the primary would lead to a field collapse which would induce a current in the opposite direction in the secondary coil.

So would I be right in saying that
1) For a single coil/inductor switch opening means reversed polarity but current in the same direction
2) For a two or more coil coupled inductor/transformer , the disconnection of current in the primary coil would produce same field collapse which then creates currents in opposite direction for the rest of the coils ?
 
  • #14
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In a transformer on the other hand , such abrupt disconnection of the primary would lead to a field collapse which would induce a current in the opposite direction in the secondary coil.

So would I be right in saying that
1) For a single coil/inductor switch opening means reversed polarity but current in the same direction
2) For a two or more coil coupled inductor/transformer , the disconnection of current in the primary coil would produce same field collapse which then creates currents in opposite direction for the rest of the coils ?
1. Yes
2. Better to say it will produce coils' current in the direction that supports already existing mutual magnetic flux.
 
  • #15
Baluncore
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There has been too much focus on the inductance of the coil. Look at the way the armature segments are coupled by the windings, lap or wave. Each magnetic part of the armature is shared by several coils. When a commutator bar moves away from the brush there are other parallel coils available to support the bulk of the current and flux, so a complete collapse or dump of energy is not required. The analogy is a multiple phase transformer.

While each winding has some inductance, it also conducts the in-phase current necessary to do the real work of the motor. The inductive current necessary to provide the flux is a small part of the total current vector.

An arc strikes immediately on mechanical break, the air in the gap then provides an ionised conductive path that is rapidly stretched. The voltage dropped along that arc appears as a negative voltage across the inductor. The inductor adjusts it's terminal voltage to maintain it's current. The rate at which the winding current changes is therefore set by the voltage external to the inductance. If that includes the higher voltage of an arc, then the current falls more rapidly.

So long as inductive current flows it maintains magnetic flux, even if it is through a lengthening arc. Arc current is inefficient because it generates heat, so the sooner that inductive arc current can stop the better. That released current can then flow through other commutator bars, to more recently connected coils.
 
  • #16
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A pseudo similar thing happens in BLDC (trapezoidal BEMF) machines. At a sector change, a winding has to go from full phase current to zero current. The winding has inductance, and the energy stored in this inductance due to the previously flowing phase current has to be dissipated to bring the coil current to zero.

1585405003318.png


Note how Ia for example goes from full phase current to zero in essentially a step.

Its important to note that the induced voltage from the stator inductance experiencing a current change is distinctly different from the induced voltage of the moving rotor, more or less analogous to leakage inductance in a transformer.

This energy is typically dissipated in the body diodes of the switching bridge. I'm not sure if its practical to attempt to recover this energy!
 
  • #17
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Interestingly in a sine driven poly phase machine stator inductance is also charged and discharged at every half electrical cycle, but due to the switching architecture this energy is cycled to and from the bus, not burned up in a diode.
 
  • #18
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@essenmein Isn't it also easier to recover this energy because the waveform is sine and the current change in time is much slower than it would be for a square waveform?
 
  • #19
Baluncore
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There is a parallel here where the overlapping of the windings in a motor can make a transformer. The question comes as to how the energy can get back through the driver to the supply, or to the currently powered winding.


New Bitmap Image.jpg
 
  • #20
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@essenmein Isn't it also easier to recover this energy because the waveform is sine and the current change in time is much slower than it would be for a square waveform?

Recovery is perhaps the wrong word, the architecture of the drive and machine basically dictate whether or not the stator inductance constitutes an energy loss and phase angle (BLDC) or a voltage loss and phase angle (most other machines).

Its more to do with the switch pattern. In BLDC when a phase is active, ie voltage is applied to this phase coil via PWM to generate phase current, then when a sector changes, this PWM pattern is abruptly applied to a different set of switches (and therefore phases), the current in the previously active phase dissipates by freewheeling via what ever diode it feels like. BLDC machines are essentially replacing the commutator and brushes with hall sensors and a 3ph bridge, vs a "real" 3 phase machine which runs with more or less continuous phase current. I'm not sure if its reasonable to say the phase current discontinuity in BDLC is the cause here...

In a sine PWM inverter all the phases always have some PWM applied, so when the inductance is being charged with increasing current, the inverter bucks current into the coil, when this current has to reduce, the inverter breifly boosts this energy difference back to the bus, in reality this looks like a little additional phase angle on the applied voltage vs the current, and this phase angle slightly changes the angle of the energy flow in and out. Since this inductance is in series with the machine, another way to look at it is the voltage vector iωL is produced at 90deg to the phase current, this voltage amplitude is speed dependent (ω = rad/s) and current dependent (i) and reduces the maximum useful voltage applied to the machine, especially as speed increases (assuming your inverter bus voltage is fixed).
 
  • #21
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There is a parallel here where the overlapping of the windings in a motor can make a transformer. The question comes as to how the energy can get back through the driver to the supply, or to the currently powered winding.


View attachment 259577

This transformer action is a PITA, most machine guys I've worked with only think in terms of machine fundamental, not the 10-20kHz voltage actually applied. So now do some distributed windings where coils from different phase share slots, voila, one inverter phase interferes with the PWM applied to the other at the inverter frequency not the machine frequency, many control problems as a result...
 

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