DC generator armature winding power factor - Page 1
Forum section: electronics / projects

author: Circlotron
author_rank: Super Contributor
timestamp: October 30, 2022, 12:21:15 am
content: It is commonly known that if a transformer feeds a full wave rectifier and then a large capacitor and thence to the DC load, depending on the peak to peak value of the ripple voltage on the capacitor, the transformer will see a power factor of about 0.6. That is to say, a 100VA transformer in this situation will only be able to deliver about 60 watts before it is fully loaded. Part of the problem is the transformer only delivers current near the voltage peaks, not over the entire AC cycle.Now picture a DC generator that has many segments on the commutator. As each winding on the armature moves into position the commutator connects the winding to the load via the brushes. This connection lasts only a relatively few degrees of rotation and so the winding can only deliver current during this time, when presumably the winding voltage is near it's maximum. The rest of the time the winding can't deliver current.So... The question is - does the DC generator suffer from a utilisation figure of only 60%, similar to the transformer situation? If so, are there any traditional work-arounds for this? Generators and motors may look simple, but the details are often far more complicated than might first appear!

author: moffy
author_rank: Super Contributor
timestamp: October 30, 2022, 06:32:20 am
content: The answer is basically, no. DC machines are rated by their losses and temperature rise i.e. how hot they can get before the insulation of the windings melts. The losses are comprised of I2R for the windings and eddy current losses of the magnetic material which is related to the speed of rotation and magnetic field. Each segment of the armature winding caries the full load current i.e. if you have a resistor across the output of a generator you get that constant current being switched from one armature winding to the next by the commutator.

author: Circlotron
author_rank: Super Contributor
timestamp: October 30, 2022, 11:20:49 am
content: Quote from: moffy on October 30, 2022, 06:32:20 amYes, but the load current of a given segment of the armature winding occurs only while it's matching commutator segment is in contact with the brushes. The rest of the time there is (presumably) no current in that winding so it's current and induced voltage waveshapes are different. That sounds like a power factor of less than 1. Why would that situation be different to the transformer and rectifier mentioned?

author: Wolfram
author_rank: Frequent Contributor
timestamp: October 30, 2022, 06:16:21 pm
content: The commutator can be considered as a synchronous rectifier. This is unlike a normal diode rectifier that can only transfer power in one direction. The generator will have an output ripple that depends on the commutation angle per rotor bar and the waveform of the induced voltage in the rotor coils. You can estimate this by counting the number of rotor bars, assuming the induced voltage is sinusoidal.The generator will se a ripple current that's determined by the impedance of the generator itself and the load connected to it. I can imagine there could be some ripple current if you connect a large capacitor directly to the generator terminals, best is to measure it under real world conditions.

author: moffy
author_rank: Super Contributor
timestamp: October 30, 2022, 10:48:01 pm
content: Quote from: Wolfram on October 30, 2022, 06:16:21 pmFollowing on from Wolfram's explanation, the commutator/windings determine the number of "phases" being rectified, hence the "ripple" is small, because the effective number of phases is high. Treat it as a black box, the output current and voltage are in phase, for a resistor, so it has a PF close to 1.

author: bdunham7
author_rank: Super Contributor
timestamp: October 30, 2022, 11:17:48 pm
content: Quote from: Circlotron on October 30, 2022, 11:20:49 amWhile you could calculate a power factor like that, your analysis would need to be confined to the element that it applies to, which is that particular segment of the armature. And the losses in that segment are indeed higher than you would calculate using the RMS current and voltage measuredacross that segmentover time. But where does that get you?