Is it not amazing that the hillbillies from backwoods Missouri with a 3rd grade education can make a rotary phase converter without all the math...
Interesting discussion. John Kirk Wallace wrote: > On Mon, 2011-03-14 at 00:04 -0400, Przemek Klosowski wrote: > >> On Wed, Mar 9, 2011 at 12:35 AM, Kirk Wallace >> <kwall...@wallacecompany.com>wrote: >> >> >>> Let me try to provide more details on my understanding of the phase >>> timing of DIY converters. Attached is a schematic of a common rotary >>> converter. The source is here: >>> http://www.metalwebnews.com/howto/ph-conv/ph-conv.html >>> http://www.metalwebnews.com/howto/ph-conv/fig1.html >>> >>> I used this to make my converter and is the only design I have some >>> understanding of. As the schematic shows L1 and L2 go straight to the >>> output and are unaffected. Since L1 and L2 are single phase, L1 is a >>> mirror of L2, therefore 180 degrees out of 360 apart. >>> >>> >>> >> But, but, you are committing a tautology. If you reference L1 to L2 and vice >> versa, of course L1 is a mirror of L2. In your scheme there is no 'ground' >> or reference point, so you only have one phase coming in, >> > I agree if you combine any two sine waves of the same period, the result > is a single wave of the same period. I think what does change is the the > amplitude of the resulting wave as one changes the phase between the two > original waves. But I'll also agree that the ground or neutral is a red > herring because it isn't connected to any part of the motor circuit > (other than 120 VAC for a relay). > > So what we have at this point is a 240 Volt sine wave across the idler > motor connection at L1 and L2. If nothing else is connected, voltage and > current is alternating between L1 and L2. L3 looks like a center tap, so > V-L1L3 should be 120 Volts and 0 degrees in reference to V-L1L2, and > V-L2L3 is 120 Volts and 180 degrees in reference to V-L1L2, just as a > neutral would be. If the rotor had no mass, my guess is that it would > vibrate between 0 and 120 degrees. When the idler L1 and L2 is first > connected to the mains, a start capacitor is switched on between L1 and > L3, which for an instant, will look like a temporary short circuit of L1 > and L3, as it charges, the field will rotate giving the rotor a kick in > one direction (?). The rotor mass should carry the rotor through > subsequent cycles and the start capacitor is disconnected. So now we > have the idler motor with 240 VAC across L1 and L2 with the rotor > spinning and nothing connected to L3. Now I'm thinking, what's going on > at L3? What comes to mind is that any motor is just as much a generator > as a motor, so there must be a very significant three phase signal > coming off of L1, L2 and L3 and because of the physical layout of the > stator windings, each generated leg is 120 degrees apart. Now I'm > beginning to think of the idler as a three phase generator that is being > back driven on one phase (L1L2), or as a motor that has two dead phases. > The only difference between a 240 Volt three phase motor and a 240 Volt > three phase generator is the current direction, so the idler has motor > like currents for 120 shaft degrees and generator currents for 240. Also > the rotor is decelerating or converting inertia into electrical energy > for 240 degrees and storing inertial energy for 120. I suppose the > capacitors between L1 and L3, and L2 and L3 are another means to store > energy to help the rotor get through the generator phase. Uugg, my brain > hurts but I think I'm getting a better understanding. I haven't even > begun to think about the math that follows. But, getting the correct > starting point helps the thought process. Thanks > > >> and the motor >> generates the second (and third) phase L3, shifted with respect to L1/L2. >> One way to see it comes from the trigonometric identity holding that sum of >> three sines shifted by 0, 120 and 240 degrees is zero: >> sin(x)+sin(x+2*%pi/3)+sin(x+4*%pi/3)=0 >> >> Let's say we take some arbitrary reference point, so that L1(t) is the >> voltage on the first leg. Assume L2(t)-L1(t) = V sin(2 pi f t). If the >> converter generates L3(t)=L1(t)-V sin(2 pi f t + 2/3 pi), then L1-L3 is V >> sin(2 pi f t + 2/3 pi), i.e. shifted by 120 degrees, and L3-L2 is -V sin(2 >> pi f t + 2/3pi) - V sin(2 pi f t) which reduces to V sin(x+4*%pi/3) i.e. >> the same AC shifted 240 degrees. Note that it doesn't matter what the >> reference point is, because its potential L1(t) drops out --- all that >> matters are voltage differences between the legs of the circuit, not the >> potentials of the legs themselves. >> > > > ------------------------------------------------------------------------------ > Colocation vs. Managed Hosting > A question and answer guide to determining the best fit > for your organization - today and in the future. > http://p.sf.net/sfu/internap-sfd2d > _______________________________________________ > Emc-users mailing list > Emc-users@lists.sourceforge.net > https://lists.sourceforge.net/lists/listinfo/emc-users > ------------------------------------------------------------------------------ Colocation vs. Managed Hosting A question and answer guide to determining the best fit for your organization - today and in the future. http://p.sf.net/sfu/internap-sfd2d _______________________________________________ Emc-users mailing list Emc-users@lists.sourceforge.net https://lists.sourceforge.net/lists/listinfo/emc-users
