# Making sense of Generator wiring configuration Parallel Zig Zag

Discussion in 'General Electronics Chat' started by ben22, Jul 10, 2013.

1. ### ben22 Thread Starter New Member

Jul 10, 2013
25
0
Dear all,

I'm looking at a 3-phase, 12-wire generator for single phase operation, and trying to make sense of how the potential wiring configurations differ and the math behind how they produce the "split-phase" voltage. As I see it, I have two options that will produce 120/240 volts (assuming each stator is producing 120V):

1) double delta
2) parallel zig zag

Attached to this post are two pictures, one which show the 6 stators and 12 wires on the generator (each wire labeled from 1 through 12), and the other which shows the wiring configurations for both options mentioned above. These are both taken from the generator user manual.

I understand how the double delta produces the 120/240 volts because line 1 (wire #4) and line 2 (wire #1) come from stators on the same phase (same as they would be in one side of a series delta configuration) so adding the output from two 120V stators, both on the same phase, gives 240V. And placing the neutral wire in between them means 120V from line to neutral for each.

But the parallel zig zag leaves me perplexed...I understand the 120V from the line on the right side (wires #1 and #3) to neutral, but not A) how you get 240V from line to line, or B) how you get 120V from the line on the left side (wires #7 and #5) to neutral.

For A) it seems that you would get the same output as line to line in a parallel star configuration: 208V, since they are 120 degrees out of phase. For B) it seems you might get double that, or 416V, since you have two different phases but also opposite "ends" of the stators.

I think its pretty safe to assume that the generator manufacturer (Mecc Alte) knows what they are talking about, and it seems to be well-established that parallel zig zag will produce the same 120/240V output as double delta, but I can't seem to make heads or tails of WHY.

Thanks!
Ben

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2. ### t_n_k AAC Fanatic!

Mar 6, 2009
5,448
784
I'm puzzled that you can understand how the double delta works but not the parallel zig-zag. I suspect you don't really understand the former. Otherwise how do you account (say) for the delta connection branch comprising series connection of windings [Line]7-8,11-12[Neutral] producing the same emf as its parallel connected winding [Line]4-3[Neutral]. If winding [Line]4-3[Neutral] produces 120V at angle 180° then so must the connection [Line]7-8,11-12[Neutral].

One has to first adopt some phase convention for the various winding emfs. Suppose we take

V(1,2) = 120 @ 0° [Volts]
V(3,4) = 120 @ 0° [Volts]
V(5,6) = 120 @ -120° [Volts]
V(7,8) = 120 @ -120° [Volts]
V(9,10) = 120 @ -240° [Volts]
V(11,12) = 120 @ -240° [Volts]

If winding 3-4 produces 120V @ 0° relative to the other windings then with terminal 3 tied to neutral the line-to-neutral voltage at terminal 4 will be 120V @180° with respect to the other line-to-neutral voltage created by the connection of winding 1-2 with terminal 2 tied to neutral.

If winding 11-12 produces 120V @ -240° relative to the other stator windings and winding 7-8 produces 120V @ -120° relative to the other windings then the total emf created by the series addition of windings 7-8 & 11-12 as shown in the diagrams would be

V(7,12) = V(7,8) + V(11,12) = 120@-120° + 120@-240° = 120 @180° [V]

So V(7,12) will equal V(4,3) in both magnitude & phase and the two may be safely connected in parallel.

Similar reasoning may be applied to the zig-zag winding topology.

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3. ### ben22 Thread Starter New Member

Jul 10, 2013
25
0
t n k,

Thank you very much for the explanation. The way you explained it is very clear. Its true that I did not fully understand how the double delta configuration works, and that, as you say, once you understand double delta voltage you can pretty easily also understand parallel zig zag voltage and vice versa.

Given the phase references you set and a double delta configuration, I get that the voltage difference between wire 4 (120@180°) and wire 1 (120@0°) = 240@0°. This makes sense since, put another way, you're adding up two 120V coils that are both @0°, giving you 240 at the same phase angle.

I guess where I ran into problems was just looking at the wires terminals in isolation and trying to look at voltage difference between them in the parallel zig zag. For instance, given the references you set, the voltage difference between wire 5 (120@-120°) and wire 1 (120@0°) would be 208@30° ...but I see that you cant do the math that way, you have to take into account a full coil, not just one side of it, and also account for the connections between coils.

Just to make sure I got it, would the top of the other (righthand) triangle of the double delta look like this:

V(10,5) = V(10,9)+V(6,5) = 120@-60° + 120@-300° = 120@0° = V(1,2)

Is that right?

Final questions: For parallel zig zag, since there are two sets of parallel coils between line and neutral on the left side, and only one on the right side, does that affect the potential max current output - would it be different for each L-N, or the same? And would double delta have the same potential max current output? The AAC online book addresses current output for three phase generators here, but it doesn't seem to really explain how you calculate it, just that, for delta configuration, you use sqrt of 3 to calculate line current given a phase current. So this makes me think that zig zag would have a higher potential current output than double delta, given that it has coils in parallel. Can you clarify?

Thanks again,

Ben

4. ### t_n_k AAC Fanatic!

Mar 6, 2009
5,448
784

If you look carefully at each topology every effective winding between any line and neutral terminal has two paths in parallel through which current may flow. The zig zag left side has only two effective current paths even though there are four windings comprising that section.

When one is considering 3-phase balanced load Delta systems the current at a delta node will be a factor of √3 times the individual phase currents. This is true because of the relative phase displacements between line & phase voltages & currents. In this 2-phase situation we know the "designer" has configured the windings such that the there is no phase difference between the overall emf's generated by the parallel connected windings. So the current flow though either part will be in phase and the magnitudes simply add algebraically.

Last edited: Jul 11, 2013
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5. ### ben22 Thread Starter New Member

Jul 10, 2013
25
0
Ok, so since current does not add in series, the left side of the zig zag can provide as much current as the right side.

So looks like the same reasoning means that max current is the same for double delta as it is for zig zag. In effect, the the top two sides of either triangle in a double delta (5-6-9-10 or 7-8-11-12) function the same way that, say, coil 1-2 or 3-4 functions on the right side of the zig zag, and the same as either of the two coil combinations 7-8-9-10 or 5-6-11-12 on the left side of the zig zag.

Thanks again t n k for responding, you have been extremely helpful!

Last edited: Jul 11, 2013
6. ### lazzaparks New Member

Sep 17, 2013
1
0

I hope it's ok to reply to an old thread like this- basically I have the same issue where I need to reconnect a generator in a double-delta single phase set up. What I dont understand is the vector maths.

I thought that windings 1-2 and 3-4 would be totally in phase and not 180º out... as these are connected in series in a normal Wye-type set up. Can someone explain this'

Many thanks
Larry

7. ### tinkerman Member

Jul 22, 2012
136
34
As shown by Parks, whether 1-2 and 3-4 are connected in series or parallel they ARE in phase. In series you get twice the voltage, in parallel you get twice the current. The other windings (5-12)combine vectorily and are 180 deg out of phase with 1-2 and 3-4.