Re: L297 / L298 Help Required Pls!

Thanks for the info JJ...

Additional info!

The original controller board (not the stepper Drivers) has three
dual opto isolator's each one handle the Step and Dir.

Step is defaults to hi on the led side which sets hi transistor side
these then appear to go via two resister arrays which in turn connect
to a transistor (NPN) this then connects to a cable connector which
is a 16way (similar to an IDE cable but smaller) the ribbon cable
from this connector connects to a backplane that then connects to the
X,Y,Z stepper drivers.



------ ResArray Connector
Step X (H) -| o |- | |..|
Step X (H) -| p |- StepX(H)--|-(TR)-|..|-StepX(H)---StepCtlr-(H)
Dir X (H/L)-| t |- DirX(H/L)-|-(TR)-|..|-DirX(H/L)-- StepCtlr(L/H)
Dir X (H/L)-| o |- | |..|
------

You made need to copy the above into a text editor!

The Step pin on the Stepper Controller is HI by default when the
motors moves a LOW appears for as long as movement is requested.
This LOW is witnessed all the way back to the frontside of the
optoIsolator when the motor completes its task the Step go's HI again.

The question is how is this done? (regarding the circuit above)

TIA
Steve


--- In CAD_CAM_EDM_DRO@yahoogroups.com, John Johnson <johnatl@m...>
wrote:

> Give this a shot:
>
> http://homepage.mac.com/johnatl/PhotoAlbum21.html
>
> Regards,
> JJ
>
> On Tuesday, November 11, 2003, at 10:49 AM, paragon032003 wrote:
>
> > Sorry for not replying back sooner I have been away for a couple

of

> > days....
> >
> > Ok Ill try and explain more clearly ....
> >
> > The stepper controller is using 12v at the Step / Dir pins on the
> > euro connector.
> > The Step/Dir on the L297 is at 5v so this is OK.
> >
> > The voltage is always High (12V) on the Step/Dir of the euro
> > connector on the stepper card.
> > What I think I need to is sink the 12v to ground producing a LOW.
> >
> > I have seen some circuits on the web that look like the below....
> >
> > 12v
> > Step (from Eurocard)
> >
> > |
> > |
> > B | /C
> > D0 --------R1K------|/ TR1
> > |\
> > | \E
> > |
> > |
> > R
> > |
> > |
> > ------------------------ Gnd
> >
> >
> > Now from what I understand .. When D0 is High (5V) it basically
> > grounds the Collector input (Step 12V) thus sending it Low.
> > Now what I want is the reverse to this i.e. when D0 is low it the
> > Step (12V) goes Low and when D0 is High (5V) the Step (12V) is

High

> > i.e. at 12V.
> > Ideally an opto-coupler would be sitting between the Parallel port
> > and transistor.
> >
> > Hope this explains a little more.
> >
> > TIA
> > Steve
> >
> >
> >
> >
> > --- In CAD_CAM_EDM_DRO@yahoogroups.com, RichD <cmsteam@s...>

wrote:

> >> Steve,
> >> You have to fill in more detail.
> >> Where exactly are you checking this 0-12V?
> >> At the output of the L298? or the input of the L297?
> >> The L297's are 5V logic devices only.
> >> The input to the L298's are too, but the outputs (motor coil

drives)

> >> are whatever the motor supply voltage is. Is this the 12V you
> > mention?
> >> The outputs (motor coil drives) are phase drive and will seem

weird

> >> to see a correspondence to the step input.
> >> Otherwise, I see no problem so far.
> >> What IS the problem you have? Does the motor not step with step
> > input pulses?
> >> RichD
> >>
> >>
> >> paragon032003 wrote:
> >>>
> >>> John / Rich
> >>>
> >>> Thanks for your reply's!
> >>>
> >>> This is what I have found since the original post...
> >>>
> >>> The stepper driver cards are running at 12V for hi.
> >>> On the logic board there are three ILD74's (Dual opto

transistors)

> >>> these are accepting between 0-3.5V Lo/Hi respectively its set
> > high by
> >>> default then pulses down to low (step takes place) then back to
> > hi on
> >>> the output side a single pin reads 12V and pulsates down to 0
> > this is
> >>> connected to the stepper drives the DIR side is the same but
> > toggle
> >>> between high and low (high = Clockwise).
> >>>
> >>> Does this make any sense?
> >>>
> >>> I presume that when the logic side goes low it sinks the 12V

side

> > to
> >>> ground via a resister this then produces a low.
> >>>
> >>> Does this sound likely?
> >>>
> >>> TIA
> >>> Steve.
> >>>
> >
> >
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