Re: Info from Ah-ha! on BOSS's/Motors/HVdrivers

Richard,

An excellent and informative post! I would like to add these comments
though to some of your headings.

Microstep Driver Current Ratings:

If the currents are accurate sine and cosine then the RMS current has
to be .707 of peak current. The available torque is the same as what
you have on the "weak" step of a half-step drive.

A microstep drive will theoretically produce 63.66% ( 2/pi ) of the
holding torque of a full-step drive. This is the "area under the
curve" of a sine wave (0 to 90 degrees) when compared to a square
wave. At higher speeds where the benefits of microstepping fade (
above 4 RPS ), the motor should be presented with a full-step
waveform to extract maximum power.

These torque values are for holding torque. Dynamic torque (usually
specified at 5 full steps per second, full step sequence) is 60 to
65% of the motor's holding torque. The missing 35 to 40% is invested
in resonating the motor. A microstep drive invests 1% or less of
holding torque in resonating the motor.

Motor Speed:

This is really best looked at as a case of impedence matching. A high
current motor / low power supply voltage will produce the same power
as a low current motor / high power supply voltage. More
specifically, the relationship is "volts / square root of inductance".

Why use a size 42 motor? Two reasons:

1) High torque need at low speeds. This is a good reason. On a 5
pitch leadscrew and 2:1 reduction, transmission efficiency not taken
into account:

1100 oz-in = 4318 lbs of "push" at low speeds ( full-step drive, 9A )
Resonance not taken into account.
700 oz-in = 2748 lbs of "push" at low speeds ( microstep drive, 9A)
545 oz-in = 2138 lbs of "push" at low speeds ( microstep drive, 7A
into a 9A motor)

2) High power. This a more dubious reason. Let's say the motor stalls
at 200 IPM and the torque load (detent torque = 50 oz-in, load
friction = 20 oz-in ) is 70 oz-in. From that one can calculate the
motor output power is 104 watts mechanical, or less than 1/7 HP. This
is very small number for such a big motor; size 34 step motors can
easily output 150W and even some size 23 motors can output 100W.

If one thinks in terms of HP per cubic inch, then a size 42 motor
should be able to output about 1HP. Since motor power is proportional
to power supply voltage, 1HP would require a 336V power supply. That
is why 320V drivers exist and they can be leathal!

Motor Wiring:

There is less than a 2 to 3% "performance hit" running a motor in
half-winding instead of parallel. Let's take your 4.7A motor and 48V
power supply as an example. It is probably a 1.7 volt motor. If it
were wired in parallel the coil resistance would be halved and it
would drop .85 volts instead. Since motor power is proportional to
voltage, and this voltage subtracts from the power supply, what you
have is:

Half-winding: 46.30V ( 48 - 1.7 )
Parallel: 47.15V ( 48 - 0.85 )

Taking the ratio of the two, the half-winding motor "sees" 98% of the
voltage the parallel motor "sees", for a performance loss of 2%. At
higher speeds where inductive reactance dominates and phase currents
drop, this loss fades into insinificance ( << 1% ). Apart from I
squared R losses at low speeds, there is no advantage to parallel
connecting the motor.

Mariss

--- In CAD_CAM_EDM_DRO@y..., admin@a... wrote:
>
> howdy guys,
>
> Sorry for the delay with this 'tech post'.
> Business has been real hectic this week.
>
> Lots of interest in BOSS's here lately, so
> I'll start with that.
> (snip)
> Richard, KC0FVV