Student question on DC motors

I'm reading a book on DC motors design.
It shows the schematic of a motor and it follows that a brush short- circuits 2 wires of the armature, those that go on the same part of
the commutator. The book says that if those wires were under voltage then a large current would flow through the brush and this would wear out the brushes and waste a lot of energy.
So it says that the brushes should be positioned in such a way to always short-circuit wires that are under zero voltage, those that are on the neutral zones of the magnetic field.
My question is, if the brushes always make contact with zero voltage wires, then how can we have voltage in the output of the generator?
And another point. As the commutator rotates the brush will make contact with two parts of the commutator which means it will short circuit four wires, which complicates matter even more. Isn't this taken into account?
Please excuse some terminology errors as the book is not in English and I'm translating.
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Maybe I'm dumb, really, but I still can't understand anything.
Aren't the coils shorted by the commutator anyway?
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On 5/9/07 3:51 AM, in article Xns992B8D10471DEp3ifw90nsdek@193.92.150.76,

No. There are insulating spacers between commutator segments. Moreover, although not spelled out in an earlier post,Kelly I believe, when the brush is aligned along the neutral axis, the coils connected to the commutator bars that get shorted get little emf induced. That is because those coils are moving almost parallel to the magnetic field and not cutting flux.
Bill -- Fermez le Bush--about two years to go.
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remove the X to answer ---------------------------- Aren't the coils shorted by the commutator anyway?
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Here is a common type of winding, the lap winding:
http://img513.imageshack.us/img513/614/mtdfig14uq2.gif
At the moment that the picture is showing, can you tell me which conductors carry what voltage? and which conductors are shorted?
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On 5/10/07 4:39 AM, in article Xns992C952DB572Bp3ifw90nsdek@193.92.150.76,

If I were drawing this illustration I would have the magnetic poles shifted just a bit to the left. As shown, turn two is between poles and not cutting flux. Turn 3 is near the pole edge. And not producing its maximum emf. If drawn the way I would prefer, the brush would be sitting smack on top of segment 2. Then, as the armature moves to the left, segments 2 and 3 would short out through the brushes but only while one of the coils is just beginning to generate emf while the other is stopping to generate emf.
Bill -- Fermez le Bush--about two years to go.
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Because your gif has four poles and four brushes, things are repeated. But I'll just talk about the left most pair of brushes, the ones touching commutator segments 11/12 and 2/3.
You're right this is a 'lap' winding. It is 'progressive' in that the coils are connected to the segments in a certain way as opposed to 'regressive'. There are also 'wave' windings.
In this 'lap' winding we see that coil number 11 is connected to bars 11 and 12. Coil number 12 is connected to bars 12 and 1, and so on across.
When bars number 11 and 12 slid under a brush, you want there to be zero voltage between the bars. That way there will be no sparking as the brush shorts the bars together. To get this, we find the point in rotation where coil number 11 (the one connecting bars 11 and 12) has its two sides located midway between poles. At that point, the voltage induced into the 'left' side of the coil will exactly equal the voltage induced into the 'right' side of the coil, *BUT* with opposite polarity. So the net voltage between bars 11 and 12 is zero, just like we want. The same holds true for the 'left' and 'right' sides of coil number 2 which is connecting bars 2 and 3.
To see how the current flows, let's assume this is a generator and current is returning into the positive brush from the external load. In a simplistic/perfect world, the load current will first split in half and half will enter the left most brush (a '+' brush) and the brush contacting bars 5 and 6. This split happens in the circuit of the brush rigging and can't be seen in your drawing as those external connections aren't shown.
Then the current that enters the left '+' brush splits again and half of the brush current flows into bar 11 while half flows into bar 12.
The current that enters bar 12 flows through coil 12 to bar 1. Then from bar 1 through coil 1 to bar 2 and out the '-' brush. The 'left' side of coil 12 and the left side of coil 1 are both under an S pole, while the right half of coil 12 and the right half of coil 1 are under an N pole. So the conductor has an EMF induced of one polarity as the coil goes down the length of the rotor from the commutator end towards the opposite end, and the polarity is the same on the return trip as the coil comes back up from the opposite end towards the commutator again (traveling opposite direction, but under opposite polarity magnetic pole keeps the induced EMF the same).
The current that enters bar 11 flows through coil 10 to bar 10. Then from bar 10 through coil 9 to bar 9 and out the '-' brush that contacts it. A similarly thing happens as before with the 'left' and 'right' sides of these coils. But the 'left' side of the coils is under an N pole and the right sides are under an S pole. This works out okay since we're traveling around these coils in a counter-clockwise direction where as in the previous paragraph we were traveling around the coils clockwise. So we still travel 'away from the commutator' under an S pole and 'toward the commutator' under an N pole.
Notice that none of the current travels through coils 11, 2, 5, or 8 at this moment in the rotation. But in a very few degrees of rotation, the bars associated with these coils will no longer be shorted, and the sides of these coils will no longer be mid-way between N and S poles. So these coils will carry current when the rotor has rotated slightly, but another set of coils will be in position midway between poles and have their bars shorted.
So the current in the coils is AC. When the bars of a coil are between two brushes with the positive brush on the left side the current flows in one direction throught the coil. As the rotor moves and the coil ends up between two brushes with the negative brush on the left side, the current flows in the opposite direction through the coil. And when the coil is exactly where its two bars are shorted, ideally there is no current flow through the coil at all.
As you might see, if the bars are sliding to the right, then the left edge of each brush is where contact is first made with each new bar as they approach the brush, and the right edge of each brush is where contact is finally broken as each bar travels past. In a perfect design, the coil would have zero EMF induced into it during the entire time the two bars are shorted out. But this isn't really possible in most machines. So there are some compromises made. Longer commutators bars (axially speaking) with multiple brushes riding on the same bars that aren't too 'thick' to short out too many bars at once are used when heavy current is needed.
Careful positioning of the brushes can get just enough EMF induced into the shorted coil as its bar leaves the trailing edge will reduce the current flow from brush to bar just in time so there is very little current to interrupt as the bar leaves the brush.
And as Don mentioned in an earlier post, the magnetic field is distorted by the MMF of the current in the rotor windings so the 'ideal' position of the brushes changes with the amount of current flow. There are some techniques used to mitigate this problem as well.
daestrom
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snip
Thanks for the great reply. Saved for future reference.
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