Re: A3977 Calculating ITrip

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!