Re: current set resistor

Hi,

The new drive has two, overlapping current ranges; 0.3A to 2A range
and the original 1A to 7A range.

For the 0.3A to 2A range use: R = (47,000 * I) / (2 - I)

For the 1A to 7A range use: R = (47,000 * I) / (7 - I)

The low range will yield 47K while the high range gives 8.2K for a 1A
current set. Hope that clears things up.

Mariss

--- In CAD_CAM_EDM_DRO@y..., cadcamcenter@y... wrote:
> Mariss wrote: "The current set resistor tells the drive what
current
> to send to the motors. Use a 1/4W 47K resistor in the low current
> jumper setting or an 8.2K if the jumpers are set for the 1A to 7A
> range."
>
> Do you mean to ignore the results from the above equation and to
use
> these (47k or 8.2k)?
>
> Example, 5.1V 1A stepper
>
> R = 47,000*1/(7-1) = 7.833k Use 8.2k ?
>
> Example. 3.2V 2A stepper
>
> R = 47,000*2/(7-2) = 18.8k Use 8.2k ?
>
>
> I R
> 0.5 3.6
> 1.0 7.8
> 1.5 12.8
> 2.0 18.8
> 2.5 26.1
> 3.0 35.3
> 3.5 47.0
> 4.0 62.7
> 4.5 84.6
> 5.0 117.5
> 5.5 172.3
> 6.0 282.0
> 6.5 611.0
> 7.0 #DIV/0!
>
> 47k: low current jumper setting, 8.2k 1-7A (higher current?)
>
> It appear that on should use higher resistance for higer current?
>
> Don't understand.
>
> Peter
>
> > What is the purpose of the current set resistor in the
> > Gecko stepper drives?
> >
> > (I am asking because I am trying to understand the
> > formula R = 47,000*I/(2-I)