Re: A3977 Calculating ITrip

Actually a "font of knowlege" that makes stupid math mistakes.

The "attack" or increasing current slope should have been:

Supply voltage - motor voltage; (20V - 5V) / 10mH = 1,500 Amps/sec

This changes the "attack" time to 16.67uS, (.025A * 10mH / 15V). The
decay time is fixed at 50uS. Now the numbers come out right; on for
16.67uS out of a period of 66.67uS (50uS +16.67uS) for a duty cycle
of 25%. This meshes with 5V motor / 20V supply = 1:4 rato or 25%.

Mariss


--- In CAD_CAM_EDM_DRO@yahoogroups.com, "ghidera2000"
<ghidera2000@y...> wrote:

> Oh ok, I get it now. So I was correct in thinking I had to quarter
> the *power* used by the motor, just not in how that quartering is
> accomplished.
>
> A font of knowledge that Mariss is. Thanks once again!
>
> --- In CAD_CAM_EDM_DRO@yahoogroups.com, "Mariss Freimanis"
> <mariss92705@y...> wrote:
> > You want to set "ITrip" equal to the motor's rated current which
> is 1
> > Amp. This is because the motor's rated torque is proportional to
> the
> > winding current. If you set it at .25A then you will get only 25%
> of
> > the rated torque.
> >
> > Where I think your confusion comes from how switching drives
> operate.
> > Here's how things would work with a trip current of 1A and a 20V
> > supply:
> >
> > 1) The drive applies 20V across your 5V motor causing winding
> current
> > to increase. The rate of current increase (slope) is V/L amps per
> > second. If your motor inductance is 10mH, the slope would be

20V /

> > 10mH or 2,000A per second.
> >
> > 2) When current reaches 1A (1/2,000 of a second later), the drive
> > shorts the winding for a fixed period of time (50uS). The current
> now
> > decays at a rate of V/L, where V is the motor voltage of 5V.
> >
> > 3) The motor current decays 25mA in 50uS to .975A (5V/10mH * 50uS
> =
> > 25mA).
> >
> > 4) Supply voltage is again applied to the winding. It will take
> > 12.5uS for the current to rise from 0.975A to 1A,( .025A * 10mH /
> 20V
> > = 12.5uS). The process then repeats.
> >
> > The motor current bounces between 0.975A and 1A, the power supply
> is
> > connected to the motor for 1/4 as long as it is disconnected.
> >
> > Mariss
> >
> >
> >
> >
> >
> > The motor will have the supply voltage across it
> >
> > --- In CAD_CAM_EDM_DRO@yahoogroups.com, "ghidera2000"
> > <ghidera2000@y...> wrote:
> > > I'm missing a vital bit of information about setting up an

A3977

> > > based unit. The A3977 explains how to set ITrip (the current at
> > > which you want to cut off the motor supply) but I can't seem to
> > find
> > > anything that specifically says how to determine what ITrip
> value
> > > you'd want to use.
> > >
> > > Lets say I use a 5V 1A motors. I'm going to be supplying them
> with
> > > 20 volts which means the maximum current I want to allow
> is .25A.
> > > Is .25A where I want to set ITrip? Seems a tad high, supplying
> > > maximum current all the time...
> > >
> > > Is there a formula to determine the ideal current or even just
> > > a "increase until X happens" method?
> > >
> > > Thanks in advance!