Re: [CAD_CAM_EDM_DRO] LARGE Gantry table: Geckos on large servos.

On Wed, 2004-03-10 at 11:53, Jon Elson wrote:
> >
> The Gecko doesn't have output filters, which causes 80 V (or whatever the
> DC supply voltage is) square waves to be applied to the motor's terminals
> at all times.

I think you mean whenever the motor is commanded to move by the drive.
You are saying that the drive controls the motor velocity/power with
pulse width modulation and that the voltage applied is the supply
voltage, ie 80V in this case.


If the motor had 80V across it all the time, it would be moving all the
time.

> This may cause high circulating currents at 25 KHz to flow
> through the transistors.

I think the only time current should be flowing through the <output>
transistors is when power is being supplied to the motors.

> It may also cause high currents to flow just to
> the windings if the capacitance to ground is low, or if there are EMI
> filters in the motor. It might be necessary to put an inductor in series
> with the motor.

I'm not following this argument.

> Finally, if the Gecko failed when stopping, the problem is due to the
> back EMF of the spinning motor PLUS the extra voltage created by
> currents in the motor inductance produces a DC supply voltage GREATER
> than the DC supply normally produces. In other words, during deceleration
> of large motors and heavy inertial loads, the bus voltage can rise to
> dangerous

I've had this argument with other people. Here is my understanding:

a) the terminal voltage of a DC <servo> motor is equal to the IR "drop"
of the armature, plus the KV product of the motor's speed.

VT = KV + IR.

b) with the software I am using, the motor accelerates at the same speed
as it decelerates.

Ie: Iaccel should equal Idecel


c) When we apply 80V to the motor while it is accelerating, the most the
KV product will get to is 80V. In actuality, it will be an IR drop away
from 80V, depending on the motor load.

ie: VTaccel = 80 = KV + IaccelR. IR will only be zero if I = 0, ie
there is no load on the motor. Therefore KV < 80V.

Or: KVaccel = VT -IaccelR.


d) When we decelerate the motor, the KV product will start at a max of
80V.


Ie: KVaccel = KVdecel at the start of deceleration.

Or: KVdecel = VT-IaccelR

During deceleration, the terminal voltage will be the sum of the KV and
IR, with IR being a voltage drop.

ie VTdecel = KVdec - IdecelR.

or VTdecel = VT -IaccelR - IdecelR

where VT is our 80V supply.

Therefore the terminal voltage during deceleration can never really be
greater than the supply voltage.

Think of the decelerating motor as a generator being turned at a
constant speed and the deceleration being a bigger and bigger load on
the motor. As the load increases, the terminal voltage drops.

So, I really don't see how the deceleration terminal voltage could be
higher than the acceleration voltage. The current could be larger, but
I don't see how the voltage could be.

Kim


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Kim Lux <lux@...>