AC synchronous motor power factor

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SUMMARY

AC synchronous motors equipped with DC slip rings and excitation current can achieve a power factor of 1.0 when operating in synchronous mode. This occurs because the rotor's DC current counterbalances the rotating AC magnetic field, allowing the stator winding to behave as an almost purely resistive load. The power factor can be adjusted by modifying the exciter voltage, enabling a wide range of power factors. In modern applications, synchronous generators are often overexcited to compensate for the lagging power factor present in the bulk power grid.

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  • Understanding of AC synchronous motor operation
  • Knowledge of DC excitation systems
  • Familiarity with power factor concepts
  • Basic principles of electrical transmission systems
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  • Explore the characteristics of synchronous generators in power grids
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Electrical engineers, power system analysts, and anyone involved in the design or optimization of synchronous motors and generators.

girts
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Is it true that AC synchronous motors (the ones with the DC slip rings and excitation current running into rotor poles) have a power factor of 1.0 when they run in synchronous mode?
Would that be because with sufficient rotor DC current he rotating AC magnetic field is precisely counterbalanced so there is almost no induction and the incoming stator winding AC "sees" an almost resistive load instead of an inductive load like in a transformer or an induction motor?
can I say that in a sense the synchronous motor is like a magnetic amplifier (saturable reactor)thanks
 
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The power factor is adjustable by changing the exciter voltage. So you can have almost any power factor you want.

The picture shows the capability curve for a synchronous generator. It's the same for a synchronous motor. The point where VARs are zero and power factor is one is neither overexited or underexcited. It doesn't have a name but you can call it "just right" or maybe "excited to perfection". :biggrin:
loss-of-excitation-22-638.jpg
 

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so I guess due to all the capacitive filters in power supplies and smps , the generators of today are rather overexcited then underexcited? to make the sine wave with a steeper climb in order to balance out the lagging current right?
 
You need to overexcite enough to compensate for the typical lagging PF on the grid (total load) as a whole.
 
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girts said:
so I guess due to all the capacitive filters in power supplies and smps , the generators of today are rather overexcited then underexcited? to make the sine wave with a steeper climb in order to balance out the lagging current right?

For the bulk power grid (think continental scale), the inductance and capacitance of the transmission lines are more significant than the L or C components of the loads.
 
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Most likely this can only be answered by an "old timer". I am making measurements on an uA709 op amp (metal can). I would like to calculate the frequency rolloff curves (I can measure them). I assume the compensation is via the miller effect. To do the calculations I would need to know the gain of the transistors and the effective resistance seen at the compensation terminals, not including the values I put there. Anyone know those values?

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