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Re: code to change the motor amp limit

Posted: 10 Jun 2017, 07:42
by devin
rew wrote:Who is paying me to check your homework?

Your end result is wrong, there must be something wrong with your reasoning.

I still think you're confusing the electrical power that goes into the motor, at stall being 100% losses, with something going on with the magnetic fields. You are throwing with formulas that I am unfamiliar with. I would have to study to get to know them and their limitations.


@rew Can you provide any formulas supporting the assertion that following statement is always true, even when wire cross section changes, & not only with constant copper volume, without using the statement itself as supporting evidence:

"kv x turns = constant"

Re: code to change the motor amp limit

Posted: 10 Jun 2017, 10:41
by rew
KV is essentially the voltage induced when using the motor as a generator. The voltage in each turn depends on the change in magnetic flux. This is measured at zero current, so there is no drop due to I*R. So the voltage over the motor is Nturns * <stuff that depends on the motor> * rotation speed.

Re: code to change the motor amp limit

Posted: 10 Jun 2017, 17:10
by devin
rew wrote:KV is essentially the voltage induced when using the motor as a generator. The voltage in each turn depends on the change in magnetic flux. This is measured at zero current, so there is no drop due to I*R. So the voltage over the motor is Nturns * <stuff that depends on the motor> * rotation speed.


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Let's imagine a virtual motor where:

A = Mechanical Output Shaft
B = Rotor
C = Stator - lead-to-lead winding resistance 0.00353ohms
D = Ground
E = Rotor Kinetic Energy in Joules at no load RPM @ 4.2V DC (relative to C = Stator)

KV testing is performed with a 4.2V DC source, commutation timing is achieved via optical sensors, and effective voltage to motor is 4.2V directly from the DC source with no PWM.

When we measure the kinetic energy of the unpowered rotor at the same rpm as no load rpm 4.2V, we find the rotor's kinetic energy is 5000 joules.

Therefore:

Example #1

A = Mechanical Output Shaft
B = Rotor
C = Stator - lead-to-lead winding resistance 0.00353ohms
D = Ground
E = 5000 Joules = Rotor Kinetic Energy in Joules at no load RPM @ 4.2V DC (relative to C = Stator)

5000 Joules of kinetic energy is 5000 watt seconds of energy.

In other words, an average of 5000 watts of mechanical power could be extracted from the kinetic energy of the rotor for a duration of 1 second.

Assuming an air core motor, frictionless bearings, dielectric rotor and motor case construction and operation in a vacuum-- the only place the rotor CAN transfer its kinetic energy is to the mobile charge carriers in the copper windings. Therefore 100% of the rotor's kinetic energy relative to the stator WILL transfer first to the mobile charge carriers, then mechanically to the windings themselves through resistance, then to the ground.

Therefore if the transfer of 100 percent of the kinetic energy of the rotor to mobile charge carriers in the 0.00353ohm winding takes one second, the average wattage during that second would be 5000w and the average amperage in the windings during the second would be 1190.47amps.

5000 watts of power for 1 second from a 4.2V battery is also 1190.47 amps.

If we apply 4.2V to the 0.00353ohm winding with the motor stalled for 1 second, this is also 5000 watts of electrical power and 5000J of energy.

Simply at no load rpm 4.2V, the rotor has 5000j of kinetic energy it can transfer to the mobile charge carriers in the windings.

With motor stalled at 4.2V for 1 second, 5000j of energy will transfer to the mobile charge carriers in the windings from the battery. (ohm's law: 4.2V - 1190.12A - 0.00353ohm - 5000W)

Therefore in this case kinetic energy in the rotor at no load rpm is equal to kinetic energy applied to the mobile charge carriers in the winding in one second with the motor stalled.

Example #2

If we triple the cross section of the wire, we lower winding resistance to 0.00117 ohms. (ohm's law: 4.2V - 3589.74A - 0.00117ohm - 15076.92W)

In one second with motor stalled the energy transferred to the moving charge carriers in the windings is 15076.92 Joules.

The magnetic moment of the stator has tripled but the magnetic moment of the rotor has stayed the same.

At the same RPM which was no load rpm in the first example, the rotor still has 5000J of kinetic energy available to transfer to the moving charge carriers.

But in the new motor with 3 times the cross section winding there are 3 times as many moving charge carriers and they are moving at same voltage as in the first example, meaning the total cumulative power of the mobile charge carriers is 3 times larger than in first example.

Since at the same rpm which was no load rpm in the first example the kinetic energy of the rotor is still 5000J, only 5000J is available at this rpm to act in opposition to the now 3 times as many mobile charge carriers, each having the same voltage as before.

In order for the rotor to have the same completely-opposing effect on the total kinetic energy of the moving charge carriers in the windings as in the first example, since there are 3 times as many moving charge carriers in a 3 times thicker winding, and the 3 times larger number of mobile charge carriers has the same voltage as before, the rotor needs 3 times as much kinetic energy to completely counteract the 3 times greater kinetic energy of the mobile charge carriers in the new winding at the same voltage. (as in example 1 -- Back EMF V works in opposition to the battery voltage).

Simply since at stall there are 3 times as many moving charge carriers, each moving at the same voltage as in the first example, and cumulatively the charge carriers have 3 times the kinetic energy as the first example, the rotor would necessarily require 3 times as much kinetic energy @ the same total rotor magnetic moment to act in equal opposition to them (compared with example #1).

Since kinetic energy of the rotor increases proportionally to its velocity squared (V^2), only 1.73205 (sqrt(3)) times as much rotor velocity is needed to produce 3 times as much kinetic energy and counteract the tripled kinetic energy in the windings with tripled cross section and 1/3 resistance.

In a retermination of any motor from wye to delta, the resistance lessens by 1x(1/3), which leads to triple the number of moving charge carriers at the same applied voltage, which leads to triple the total kinetic energy of these moving charge carriers at the same applied voltage. To completely counteract the kinetic energy of this tripled quantity of moving charge carriers at the same applied voltage in the wye->delta retermination, the rotor necessarily requires triple the kinetic energy to act in equal opposition to the tripled total kinetic energy of the moving charge carriers. Triple the kinetic energy of the rotor only requires a 1x1.73205 [sqrt(3)] increase in rotor velocity since kinetic energy increases proportionally to velocity squared (V^2).

^This is why I THINK in any standard wye->delta re-termination the "Max RPM per volt" (KV) increases by a factor of 1x(1.73205) [sqrt(3)].

:mrgreen:

Re: code to change the motor amp limit

Posted: 15 Jun 2017, 22:33
by devin
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Re: code to change the motor amp limit

Posted: 27 Jun 2017, 19:05
by devin
[post removed by author]

Re: code to change the motor amp limit

Posted: 28 Jun 2017, 17:37
by devin
I made a mistake in the drawing of the wye motor stator magnetic field vectors yesterday....

If one looks closely at yesterday's overlay of the rotor and stator fields it becomes apparent that not all of the energized stator phases would contribute to rotor torque in the same direction.

So I re-drew & I believe this stator magnetic field vector drawing and stator/rotor magnetic field overlay more accurately represent the proper field vectors at one point in the rotor's rotation:

Where:

Red Area = Phase "A" -- Energized
Green Area = Phase "B" -- Energized
Blue Area = Phase "C" -- Not Energized
Black Line = 12 Pole Wye Stator Magnetic Field Vectors w/ 2 Energized Phases
White Line = 14 Pole Rotor Magnetic Field Vectors

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Commutation Sequence Source: http://m.eet.com/media/1179124/bldc2tab1.jpg

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Wye Diagram Source: the internet

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Re: code to change the motor amp limit

Posted: 29 Jun 2017, 00:10
by devin
Just finished the Delta version of the stator magnetic field vector diagram...

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Delta Diagram Source: the internet

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...notice all 12 stator teeth are energized and generating rotor torque in the same (clockwise) direction.

Re: code to change the motor amp limit

Posted: 29 Jun 2017, 07:50
by rew
devin wrote:Blue Area = Phase "C" -- Not Energized
The C phase is, in BLDC mode only, not energized. It is the least efficient in generating torque. Suppose we ground the star-point and are able to apply a positive or negative voltage to each A, B or C terminal. In the situation you describe the A and B windings generate more torque-per-amp than the C winding. And the winding scheme is designed such that the two winners will always be one positive, one negative.

Now you might say: Why bother exciting the C winding at all when in FOC Mode? Well: because the losses are proportional to current-squared, you still win some efficiency by pushing a small enough current through the C winding.

For Y, if I understand your drawing correctly, you've drawn the blue triangles exactly in the position of zero torque (if there would have been current) and in FOC mode zero-current. You've rotated things a bit in your delta drawing so that torque is not exactly zero. At least not in the blue region.

Re: code to change the motor amp limit

Posted: 04 Jul 2017, 03:12
by devin
rew wrote:The C phase is, in BLDC mode only, not energized. It is the least efficient in generating torque. Suppose we ground the star-point and are able to apply a positive or negative voltage to each A, B or C terminal. In the situation you describe the A and B windings generate more torque-per-amp than the C winding. And the winding scheme is designed such that the two winners will always be one positive, one negative.

Now you might say: Why bother exciting the C winding at all when in F
OC Mode? Well: because the losses are proportional to current-squared, you still win some efficiency by pushing a small enough current through the C winding.

For Y, if I understand your drawing correctly, you've drawn the blue triangles exactly in the position of zero torque (if there would have been current) and in FOC mode zero-current. You've rotated things a bit in your delta drawing so that torque is not exactly zero. At least not in the blue region.


@rew If it's so that phase C is energized simultaneously with phase A & B in a wye-wound motor in FOC mode (but not in BLDC mode)... in the following diagram, exactly which mosfets would be ON to permit this all-3-phase-wye-winding energized state?

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Re: code to change the motor amp limit

Posted: 04 Jul 2017, 07:10
by rew
A full sine cycle is 2*pi. So a third of that is 2/3 PI

Phase A is energized with the voltage: Va = Vs * sin (f * t + 0*2/3PI);
Phase B is energized with the voltage: Vb = Vs * sin (f * t + 1*2/3PI);
Phase C is energized with the voltage: Vc = Vs * sin (f * t + 2*2/3PI);

here Vs is the "duty cycle" dependent voltage scale. and f is the RPM (in the proper units).

Now the question is: what does it mean that a phase is energized with a voltage: We can only drive either to V+ or to V-?

Well, when the target voltage is 0, we alternate the highside and the lowside switch in a 50/50 fashion. 50% PWM. So now by varying the PWM percentage we can apply (effective) voltages anywhere between -Vbat/2 to +Vbat/2.