Re: A3977 Calculating ITrip

And I didn't express the idea correctly either. Its not that *I* am
quartering the average power in any way, its that 4x the voltage
takes 1/4 the time to reach 1 amp.

So my Vref voltage should be 2 volts (1amp=2volts/8*.25ohms). After
work I'll figure out the heat dissapation by the A3977 and sense
resistors. Odd as it sounds, I actually like the math - its hard
sometimes but very satisfying when I (eventually) get the correct
results :D

--- In CAD_CAM_EDM_DRO@yahoogroups.com, "Mariss Freimanis"
<mariss92705@y...> wrote:

> 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!