RE: [CAD_CAM_EDM_DRO] Re: QUESTION ABOUT SERIES OR PARALLEL ON STEPPER MOTORS - ref post 52721

OK. Sounds like a reasonable thing to do.

Back to one of my original questions:

Since I can modify the power supply somewhat, and since the max drive
voltage is 80 volts, should I supply 62 volts, 68 volts (or some other
voltage which would mean I'd have to scrap the power supply and use
something else). Also, I've seen calculations on how to properly size a
capacitor for a DC power supply. There is a capacitor (unmarked of course)
about the size of a beer can with a 1K ohm resistor across it on the DC side
of the power supply. Can I assume that if it worked OK at 130 volts DC, it
will be OK at about 1/2 the voltage?


-----Original Message-----
From: Tony Jeffree [mailto:tony@...]
Sent: Thursday, February 13, 2003 5:21 PM
To: CAD_CAM_EDM_DRO@yahoogroups.com
Subject: RE: [CAD_CAM_EDM_DRO] Re: QUESTION ABOUT SERIES OR PARALLEL ON
STEPPER MOTORS - ref post 52721


Jeff -

Yes, in a simple L/R drive, you use a resistor in series with the motor
coil, forming a voltage divider, so you calculate the resistor value such
that the voltage drop across the motor coil is exactly what is needed to
cause the rated current to flow in the coil.

With the coil resistance and resistor values you are talking about, if it
were a simple L/R drive with a 130V supply, the rated current would be of
the order of 130/2.5, or approx 50 amps. I think this is highly unlikely!
So, my conclusion would be either that it is a simple L/R drive, and the
supply voltage across the motor/dropper resistor is actually somewhat less
than 130V, or your resistance measurements are in some way faulty, or you
are looking at some other (more complex) type of drive circuit altogether -
such as a "chopper" drive or a bi-level drive.

Either way, it might prove much simpler to go the "empirical" route I
suggested, and crank up the current on your CMD-260's till the motors start
to generate some heat.

Regards,
Tony

At 16:07 13/02/2003 -0500, you wrote:

>Thanks for the input. The NC controller has a bunch of large ceramic
>resistors in the bottom of the enclosure. The three axis circuit
>boards (labeled drive, so I guess that's what they are) are loaded with
>power transistors and smaller ceramic resistors if this helps with the
>identification of the type controller I have.
>
>If I understand you, the controller resisters are in series with the
>motor resistors and create a voltage divider?
>
>I measured the winding resistance (by wiring the four stepper coils in
>series, measuring, and dividing by four) and the per winding resistance
>comes to .3 ohm.
>
>I plugged the cables from the resistors in the controller into the axis
>drive board (in case my readings would be modified by some of the stuff
>on the board). Measuring across each of the four resistors I get 2.2,
>2.2, 2.7 and 2.7 ohms respectively.
>
>This is being fed by 130 volts A.C.
>
>I'm clueless about how the four controller resistors are combined to
>create the final resistance to use with the motor resistance to find
>out what the original motor voltage was.
>
>Any help would be greatly appreciated.
>
>Thanks and regards,
>Jeff
>
>
>-----Original Message-----
>From: Tony Jeffree [mailto:tony@...]
>Sent: Thursday, February 13, 2003 11:55 AM
>To: CAD_CAM_EDM_DRO@yahoogroups.com
>Subject: RE: [CAD_CAM_EDM_DRO] Re: QUESTION ABOUT SERIES OR PARALLEL ON
>STEPPER MOTORS - ref post 52721
>
>
>Jeff -
>
>An interesting problem - without knowledge of the motor's current
>rating, it is a little difficult to advise on a current setting for the
>controller cards.
>
> From your earlier posts, it sounds like the original drives may well
>have been simple L/R drives (using power resistors in series with the
>motor coils to limit the current). If this is so, then you should be
>able to work back to the motor current rating; divide the supply
>voltage by the ratio of the motor coil resistance to the series power
>resistor value to get the voltage needed to drive the max current. From
>I=R/E you can then work out what current that voltage will drive
>through the motor coil resistance. However, to do this successfully,
>you would need to know what the drive circuitry really is - whether it
>is a "simple" L/R drive, or a more sophisticated drive (a bi-level
>chopper, for example). If the latter, then it may not be so easy to
>determine by this method.
>
>An alternative "empirical" method might be to hook one motor up to a
>drive, and adjust the current up until the motor gets warm (giving
>enough time between current increments for the motor temperature to
>stabilize). Steppers are generally designed to run warm; if you set the
>current such that the motor gets warm (but not too hot to touch) you
>probably are not far off the right current setting.
>
>Regards,
>Tony
>
>At 11:04 13/02/2003 -0500, you wrote:
> >Guys:
> >
> >I really appreciate the input. I'm a newbie at this and your
> >discussion is very helpful. As a follow on question:
> >
> >I have the following challenges:
> >The DC power supply of the NC controller I'm cannibalizing puts out
> >130V. I don't know if this eventually reached the motors or was
> >reduced somewhere else in the old NC drive section before reaching
> >the drives. (There was a short circuit somewhere in the unit, so I
> >disconnected all the controls, drives etc so I could work on the
> >power supply). I can rewire the taps on the transformer to give me
> >either 62 or 68 volts DC, both of which fall below the range of the
> >80v max of the CMD260 drives I bought.
> >
> >Based on the input you guys gave (my NEMA 42 steppers are unmarked
> >and
> >undocumented) I'm going to try running them in parallel. I really
> >don't want to generate sparks and a puff of smoke. What would you
> >recommend the current setting on the drives to be placed at (they go as
> >high as 8 amps)? Should I drive them with 62 volts or 68 volts?
> >
> >Thanks in advance,
> >
> >Jeff
> >
> >
> >-----Original Message-----
> >From: turbulatordude <davemucha@...> [mailto:davemucha@...]
> >Sent: Thursday, February 13, 2003 8:33 AM
> >To: CAD_CAM_EDM_DRO@yahoogroups.com
> >Subject: [CAD_CAM_EDM_DRO] Re: QUESTION ABOUT SERIES OR PARALLEL ON
> >STEPPER MOTORS - ref post 52721
> >
> >
> >
> >The serial vs. parallel speed question was tossed around by the pros
> >from Dover a little while back.
> >
> >This is real close of an explanation. if you want more, read the
> >whole thread.
> >
> >
> >Message 52721 "jeffalanp" <xylotex@h. . . > Date: Tue Nov 19, 2002
> >12:56 am Subject: Re: Running Unipolar steppers with a Bi-polar
> >driver?
> >
> >"Mariss, OK, OK, I got it (I think). The inductive time constant for
> >both parallel and series motors will be the same (t=L/R, the time it
> >take to get to 62. 3%). But to overcome the increased inductance and
> >resistance in a series wound motor (compared to parallel), I would
> >need to double the voltage to achieve the same time as parallel. "
> >
> >There was more on the post, time to charge the coil, discharge it,
> >then reverse direction, that sort of stuff. but this is enough data
> >for most of us to understand the differences.
> >
> >Or the short and sweet, series goes thru a longer wire and that takes
> >more time. You can did deeper and get more data.
> >
> >Dave
> >
> >
> >
> >
> >
> >--- In CAD_CAM_EDM_DRO@yahoogroups.com, Alper YILDIRIM
> ><yildirimalper@y...> wrote:
> > > Hi,
> > > Take a look to the following site:
> > >
> > > http://www.microkinetics.com/contab.html
> > >
> > > For the bipolar series and parallel connections the
> > > time constant L/R is same. For parallel connection, inductance
> > > lower 4 times but resistance lowers 4 times too. So the time
> > > sonstant does not change. Can someone explain why parallel wiring
> > > allows higher speeds according to this approach.
> > >
> > > Best regards
> > > Alper
> > >
> > > Jon wrote:
> > >
> > > ""Yes. In almost all cases, the parallel connection
> > > is better. It
> > > provides lower inductance,
> > > and therefore, higher speeds, than the series
> > > connection. The only
> > > time
> > > to use series
> > > is if your stepper driver cannot supply the current needed for
> > > parallel connection, and
> > > if high speed is not important. (The fix is to get a driver that
> > > CAN handle full rated
> > > current in the parallel mode.)
> > > Jon""
> > >