Re: [CAD_CAM_EDM_DRO] L297/298's

Ian Wright wrote:

> Hi,
>
> I've just built up a driver board based on the L297/298 chips and I'm
> a bit
> disappointed at the maximum motor speed I seem to be able to get from
> it - I
> managed 300 rpm with one motor and only 200 with another. The motors
> are, I
> believe, 3.2 volt, and 1.4 and 0.8 amp respectively (the 0.8 amp one
> went
> slowest) and I was driving them in half step mode from a 12 volt 3amp
> supply
> (L297 was set at motor's rated current output).
>
> On the machine I'm building with a 1:5 reduction driving a 1mm pitch
> leadscrew, the fastest equates to about 60mm or 2.5" per minute! - a
> bit
> slower than you all seem to talk about!! That, of course, is also the
>
> fastest I could expect - in G00 mode. Whilst I'm not too bothered
> about
> speed, I wondered what typical speeds you were getting from L297/298
> driver
> boards and consequently whether mine are performing as well as I might
>
> expect.
>
> In this test I was pulsing the step input with a 555-based oscillator
> and
> the motors ran really cleanly with plenly of torque up to the maximum
> speed
> I got and then suddenly stopped and sat there buzzing loudly.

Were you just turning on the pulse train at the test frequency, or were
you
ramping the pulse rate up and down? You will get much better results if

you gradually raise the pulse rate, to allow the motor to accelerate.

You may also need to give much higher voltage to the stepper driver.
I don't have any idea what the inductance of these motor windings are,
but they are usually quite large. The current in the windings must make
a full
cycle from max + current to max - current and then back to max + in 4
full steps. At 300 RPM, a 200 step/rev motor does 1000 steps/sec, or
250 full cycles of the current waves / sec. To do this, you need a
di/dt that
allows the current to go from zero to about rated current within about
one
step, here that allows about 1 Amp / mS, or 1000 A/Sec. With 12 V
available from the power supply, any inductance over V= L * di/dt
will not be able to get up to full current in the allowed time.
Solving, L = V / (di/dt) = 12 / 1000 = 12 mH
If the coil inductance is over 12 mH, which is not much, then you will
have much reduced current at this rate. Actually, due to transistor
voltage drop and the winding resistance, the drop in current will
actually ocurr at lower speeds.

More voltage will help, up to a point. But, doubling the voltage will
only double the speed, or double the inductance that can be run at
the same speed, at best.

Jon