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COMMUTATION OF DC MACHINE 1

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electricalkannan

Electrical
Mar 16, 2002
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I am fresh in electricalfield.I don't about commutation of dc machine,how does it convert ac into dc or dc into ac,i want deep discussion.could you please write here about commutation of dc machines
thanks
electricalkannan
 
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hi my friends,thanks to electrageek .i want to know about commutation of dc machines.in dc machine,commutator is there,how does it do magneticflux reserval.actually commutator convert ac to dc for electrical genaration and dc to ac for mechanical genaration.could you please write me
thanks
electricalkannan
 
electricalkannan

I am not sure I understand your question. You need to read the description of the DC motor theory on the website? A DC motor only uses DC voltage and DC current at its terminals. A DC Generator only produces DC voltage and DC current at its terminals. The principle that makes a DC motor turn, is shown on the Website. The coil, that represents the copper armature on the DC motor rotor has DC current flowing through it, and because it is in a magnetic field, supplied by the field coils mounted around the outside on the stator, it tries to move to a horizontal position. But because the DC current is passing through the carbon brushes as soon as the coil moves a little ways a new coil is now under the brush and it tries to move to the horizontal position and so on and so on. The coil actually has to make an entire 180 deg revolution before the same coil has the DC current reverse through it. (this is for a two brush motor) Its the orenintation of the brushes and the field coils that generate the torque to turn the rotor.

I use the following equations to help me understand the characteristic of a DC motor.

Terminal Voltage Equation:

Vt = Vcemf + Ia*Ra + BD

Where Vt = terminal volts (what you can measure)
Vcemf = volts counter EMF (used by motor)
Ia = armature current
Ra = armature resistance
Ia*Ra = internal motor voltage loss
BD = brush drop (consider 2volts and ignored)

Speed equation

n = Vcemf/(N*If)*K1

Where n = speed
N = Number of turns in field winding
If = field current
N*If = field flux
K1 = constant

Torque Equation

T = Ia * (N*If) * K2

Where T = Torque
K2 = Constant

Horsepower Equation

HP = (T * n)/746

Where HP = horsepower
746 = constant KW to HP conversion

From the equations you can see that speed is directly proportional to Vcemf and inversely proportional to If. Double the armature voltage and you double the speed. Cut the field current in half and you also double the speed. However you notice that the Torque equation is not related to speed or voltage. As you weaken the field in a constant torque application the speed will go up but the armature current will have to also increase to keep the torque at the same value. So you can only weaken the field until you either reach top speed or the armature current reaches 100% then your out of Torque and HP.
 
Hi,
The voltage in the armature of any D.C. machine is A.C. As a coil passes through the flux of a field coil a voltage is generated. If the coil passes a north field, the voltage is generated as a sine wave in one direction (let's say positive, in this instance). As the coil leaves the north field and passes through a south field, a voltage is generated as a negative sine wave. This is the A.C. voltage-a full 360° sine wave. The system repeats depending on how many pairs of poles are available and the number of revolutions made.
The point on the sine wave where the voltage reaches 0 is the 'commutation zone'. This is where the brushes are located, either collecting or applying (+) or (-) current and voltage (generator or motor).
Hope this helps.
 
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