## code to change the motor amp limit

### Re: code to change the motor amp limit

@benjamin... question: if motors are delta wound not wye, does one still need to double the resistance figure of the VESC detection to calculate lead to lead resistance?

### Re: code to change the motor amp limit

Benjamin seems to assume that most motors are Y wound. As far as I know, most motors (that we use) are delta-wound.

All this does not matter, if I allow you to take (only) external measurements of two motors, one Y wound, one delta-wound you cannot tell which is which.

So even when the motor is delta-wound, having a mental model of a Y-wound motor does not harm anything in any way.

Suppose I have a Y-wound motor with 1 ohm winding resistance. I measure the resistance from terminal to terminal and I measure two windings in series: 2 ohms. Now suppose there is a delta-wound motor with 3 ohm winding resistance. again from terminal to terminal I now there is one winding in parallel with the other two. 3 ohms parallel to 3+3, do the math, and again you get 2 ohms.

These two will behave 100% identical for measurements from the outside.

All this does not matter, if I allow you to take (only) external measurements of two motors, one Y wound, one delta-wound you cannot tell which is which.

So even when the motor is delta-wound, having a mental model of a Y-wound motor does not harm anything in any way.

Suppose I have a Y-wound motor with 1 ohm winding resistance. I measure the resistance from terminal to terminal and I measure two windings in series: 2 ohms. Now suppose there is a delta-wound motor with 3 ohm winding resistance. again from terminal to terminal I now there is one winding in parallel with the other two. 3 ohms parallel to 3+3, do the math, and again you get 2 ohms.

These two will behave 100% identical for measurements from the outside.

### Re: code to change the motor amp limit

rew wrote:Benjamin seems to assume that most motors are Y wound. As far as I know, most motors (that we use) are delta-wound.

All this does not matter, if I allow you to take (only) external measurements of two motors, one Y wound, one delta-wound you cannot tell which is which.

So even when the motor is delta-wound, having a mental model of a Y-wound motor does not harm anything in any way.

Suppose I have a Y-wound motor with 1 ohm winding resistance. I measure the resistance from terminal to terminal and I measure two windings in series: 2 ohms. Now suppose there is a delta-wound motor with 3 ohm winding resistance. again from terminal to terminal I now there is one winding in parallel with the other two. 3 ohms parallel to 3+3, do the math, and again you get 2 ohms.

These two will behave 100% identical for measurements from the outside.

rew, I'm not sure if I am right in this assertion, but could it be possible to detect which winding (delta or y) is used from outside measurements as follows:

first using one lead as a source and one lead as a drain, measure the resistance lead-to-lead.

now using one lead as a source, and the other 2 leads as drains, measure resistance from source using both drains.

if the 1 lead, 2 drain resistance is 1/2 the 1 lead, 1 drain resistance, then the motor is delta wound.

if the 1 lead, 2 drain resistance is 3/4ths of the 1 lead, 1 drain resistance, the motor is wye wound.

Anyway I'm not sure, but looking at the diagrams it seems that's how things might work out in determining delta vs wye from the outside... Just trying to learn.

The other reason I imagined Benjamin might have suggested doubling the resistance detecion measurement I think might possibly apply to delta windings but not wye... but I am not sure?

Even if the motor is delta wound, during "off" duty cycle time when the battery is disconnected from the coils but current is still flowing inductively... I am imaging the inductive current loop travels in series through both of the 2 out of 3 windings which I believe are "energized" at any given time... since the length of this energized path through 2 phases is double the length of an energized path from lead to lead with a delta wind, is this the reason ohm detection value would be doubled for my motor-amps-close-to-0rpm calculation, in particular with a delta wound motor, but not with a wye wound motor?

in simple terms, which of these 2 reasons are we doubling the detected resistance value:

reason A: we think the motor is wye (but it is in fact delta) and so we are doubling the detected resistance value because it is calculated to the virtual ground point, and with a wye wind, the resistance from lead to lead is double the resistance from lead to virtual ground point

or

reason B: we think the motor is delta, and because the duty off inductive part of the duty cycle creates an energized loop through not one, but 2 sets of windings in series, we are doubling the lead to lead resistance value because the inductive loop travels through 2 sets of windings in a delta winding, and this path through 2 sets of windings has double the resistance of the lead-to-lead resistance detection of a delta winding?

### Re: code to change the motor amp limit

i think my original guess was wrong about being able to distinguish delta and wye by comparing 1 to 1 and 1 to 2 lead resistance measurements because all 3 windings are energized at the same time with delta... and only 2 at a time with wye...

so recalculating...

if a lead-to-virtual-ground-point segment of a wye motor is 2 ohms

r=2ohms

1 to 1 wye = 4ohm

1 to 2 wye = 3ohm

3/4 = 0.75

1 to 1 delta = 1.333...ohm

1 to 2 delta = 1ohm

1 / 1.333... = 0.75

if a lead-to-virtual-ground-point segment of a wye motor is 3 ohms

r=3

1 to 1 wye = 6ohm

1 to 2 wye = 4.5ohm

4.5 / 6 = 0.75

1 to 1 delta = 2ohm

1 to 2 delta = 1.5ohm

1.5 / 2 = 0.75

in simple terms i was wrong, rew was right, you cant distinguish wye/delta from outside measurements, even by comparing the ratios between 1 to 1 and 1 to 2 lead resistance measurements.

interestingly it appears wye termination has 3 times greater resistance than delta, rather than 2 times greater as i'd believed previously...

so recalculating...

if a lead-to-virtual-ground-point segment of a wye motor is 2 ohms

r=2ohms

1 to 1 wye = 4ohm

1 to 2 wye = 3ohm

3/4 = 0.75

1 to 1 delta = 1.333...ohm

1 to 2 delta = 1ohm

1 / 1.333... = 0.75

if a lead-to-virtual-ground-point segment of a wye motor is 3 ohms

r=3

1 to 1 wye = 6ohm

1 to 2 wye = 4.5ohm

4.5 / 6 = 0.75

1 to 1 delta = 2ohm

1 to 2 delta = 1.5ohm

1.5 / 2 = 0.75

in simple terms i was wrong, rew was right, you cant distinguish wye/delta from outside measurements, even by comparing the ratios between 1 to 1 and 1 to 2 lead resistance measurements.

interestingly it appears wye termination has 3 times greater resistance than delta, rather than 2 times greater as i'd believed previously...

### Re: code to change the motor amp limit

Best way to learn is to start shouting "your're wrong!!", and finding out for yourself that instead you were wrong. Glad to have contributed to some understanding. (just for reference: This realization of Y<->delta came to me in exactly the same way).

### Re: code to change the motor amp limit

interesting...

with delta the use of 2 "underpowered" coils which cancel each other out effectively for motive purposes + 3rd coil which does the "work" results in more total "BEMF per RPM" and therefore higher kv and no load speed (delta) (delta is 1/3rd electrical resistance of wye) (at the cost of electrical to mechanical conversion efficiency compared to wye)

than

2 fully powered coils which have lower total BEMF per RPM than delta, with no work canceling and therefore lower kv & lower no load rpm (wye is 3x the electrical resistance of delta)

with delta the use of 2 "underpowered" coils which cancel each other out effectively for motive purposes + 3rd coil which does the "work" results in more total "BEMF per RPM" and therefore higher kv and no load speed (delta) (delta is 1/3rd electrical resistance of wye) (at the cost of electrical to mechanical conversion efficiency compared to wye)

than

2 fully powered coils which have lower total BEMF per RPM than delta, with no work canceling and therefore lower kv & lower no load rpm (wye is 3x the electrical resistance of delta)

### Re: code to change the motor amp limit

"cancel eachother" is not true. Each of the coils produces a magnetic vector. Put phase A at 0/360 degrees, B at 120 and C at 240, and you add the vectors they produce together, you'll get the same results. So for example, in Y you power A-to-B, you get positive current in A and negative current in B. So B becomes 120+180 = 300 degrees. So the vectors add to 330 degrees. Next phase, there will stil be A-positive, but C-negative instead of B. So we get A-C = 0 / 60 , or added to 30 degrees.

The same goes in Delta configuration, but the resulting vector will be a bit rotated. So we power A-B connection, but lets call that coil A. Coils B and C are powered at half the voltage, and if you add them up you get a magnetic vector. Next power the A-C connection, and you'll see the resulting vector rotate by 60 degrees.

The same goes in Delta configuration, but the resulting vector will be a bit rotated. So we power A-B connection, but lets call that coil A. Coils B and C are powered at half the voltage, and if you add them up you get a magnetic vector. Next power the A-C connection, and you'll see the resulting vector rotate by 60 degrees.

### Re: code to change the motor amp limit

so in simple terms are you saying there is no "wasted electrical to mechanical efficiency" with delta compared to wye considering there are 3 energized windings at a time with delta and 2 energized windings at a time with wye?

in simple terms are you saying all 3 windings are actively contributing to mechanical motion at all times with delta, and no electrical to mechanical conversion efficiency is lost compared to wye in the delta arrangement?

in simple terms are you saying all 3 windings are actively contributing to mechanical motion at all times with delta, and no electrical to mechanical conversion efficiency is lost compared to wye in the delta arrangement?

### Re: code to change the motor amp limit

Correct!

In principle, BLDC motors are a descendant of three phase motor. There you have three coils positions 120 degrees from each other, and three excitation voltages also 120 degrees shifted from each other. Add the three magnetic vectors from each of the coils together and the result is a ROTATING magnetic field. Now each of the coils is excited with a voltage between zero and one of the three phases or between two phases for each coil doesn't matter. In both cases, the excitation voltages are precisely those 120 degree shifted voltages.

In FOC Mode, the controller WILL try to enforce those sinusoidal 120-degree-shifted voltages on each of the three phases. No "unpowered leg" on the Y.

Motors specifically built for BLDC mode, or trapezoidal excitation, have some fancy stuff going on between the magnets and the coils so that not driving one of the coils is quite OK.

In principle, BLDC motors are a descendant of three phase motor. There you have three coils positions 120 degrees from each other, and three excitation voltages also 120 degrees shifted from each other. Add the three magnetic vectors from each of the coils together and the result is a ROTATING magnetic field. Now each of the coils is excited with a voltage between zero and one of the three phases or between two phases for each coil doesn't matter. In both cases, the excitation voltages are precisely those 120 degree shifted voltages.

In FOC Mode, the controller WILL try to enforce those sinusoidal 120-degree-shifted voltages on each of the three phases. No "unpowered leg" on the Y.

Motors specifically built for BLDC mode, or trapezoidal excitation, have some fancy stuff going on between the magnets and the coils so that not driving one of the coils is quite OK.

### Re: code to change the motor amp limit

if 100 turns = delta = 100kv

X turns = wye = 100kv

what is X?

if 100 turns = delta = 100kv

100 turns = wye = X kv

what is X?

X turns = wye = 100kv

what is X?

if 100 turns = delta = 100kv

100 turns = wye = X kv

what is X?

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