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'[EE]: Step motor controllers'
2001\12\17@095344
by
David VanHorn
>
>David, there is little to be gained by using chopping
>with such a low supply voltage. What voltage and current
>motors are you using? Maybe if you explain more you
>might get more responses on this one? :o)
>-Roman
It's a 5 ohm motor, bipolar windings, running from a 6-8.5V supply.
The motor wants 0.5A.
I don't want to spend any more power than I must in driving it, since this
is a battery operated application.
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mitvma.mit.edu
2001\12\17@135240
by
steve
|
> It's a 5 ohm motor, bipolar windings, running from a 6-8.5V supply.
> The motor wants 0.5A. I don't want to spend any more power than I must
> in driving it, since this is a battery operated application.
You can't afford the losses of a darlington output device so your
best solution is probably to go semi-discrete with N-ch/P-ch half
bridge drivers or perhaps a low voltage DC motor driver device. Then
add your own current sense/chopper circuit.
I'd be tempted to have a play with MOS output controller IC's you've
already dismissed because of the minimum motor voltage.
AFAICT, the minimum would be dictated by the ability of the
charge pump to get a high enough Vgs on the top FET to turn it
hard on. I would expect that with a 6V supply you're going to be
close enough that even if the Rds on is twice what it is capable of,
you are still on the winning side. Try it. If it works OK for your
needs, flick an email to the chip supplier and ask for their opinion.
Steve.
======================================================
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TLA Microsystems Ltd Microcontroller Specialists
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Auckland, New Zealand ph +64 9 820-2221
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2001\12\19@060945
by
Vasile Surducan
On Mon, 17 Dec 2001, David VanHorn wrote:
>
> It's a 5 ohm motor, bipolar windings, running from a 6-8.5V supply.
> The motor wants 0.5A.
> I don't want to spend any more power than I must in driving it, since this
> is a battery operated application.
> --
I say a bridge half with bipolars, half with MOS will be the better
choice. But I don't see why not 3717, 3770 ; 8.5V x 0.5 = 4.25VA
10V x 0.5 = 5VA
No big difference but easy way.
Vasile
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2001\12\19@104703
by
David VanHorn
At 01:03 PM 12/19/01 +0200, Vasile Surducan wrote:
>On Mon, 17 Dec 2001, David VanHorn wrote:
>
> >
> > It's a 5 ohm motor, bipolar windings, running from a 6-8.5V supply.
> > The motor wants 0.5A.
> > I don't want to spend any more power than I must in driving it, since this
> > is a battery operated application.
> > --
> I say a bridge half with bipolars, half with MOS will be the better
>choice. But I don't see why not 3717, 3770 ; 8.5V x 0.5 = 4.25VA
>10V x 0.5 = 5VA
>No big difference but easy way.
>Vasile
Their spec says minimum motor voltage of 10V.
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2001\12\19@121131
by
Roman Black
|
> On Mon, 17 Dec 2001, David VanHorn wrote:
>
> >
> > It's a 5 ohm motor, bipolar windings, running from a 6-8.5V supply.
> > The motor wants 0.5A.
> > I don't want to spend any more power than I must in driving it, since this
> > is a battery operated application.
Sounds like a small 17 frame or pancake stepper,
2.5v 0.5A. You are unlikely to get better than
2:1 with any chopper IC so it will probably
cost you 200mA+ from the battery anyway.
Unless you need high speeds, I think the best
solution is to switch to a 12v 200mA motor,
these still have decent speeds and torques at
6v, and being unipolar you only need 4 tiny
fets to drive the motor, you can use low sat fets
as you only need 4 and the circuit is simple
enough to wire.
Another good point about the higher inductance
12v motor is that you can implement simple slow
pwm from the pic outputs, good for low power
"hold" modes and stuff. Now you will probably tell
me you have to use the 2.5v motor? ;o)
-Roman
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