Re: A3977 Calculating ITrip

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!