Stepper motor 101 was Re: Lini-stepper

UH...not quite.

When the BiPolar transistor is in a fully conducting state...the
resistance from Collector to emitter is low. And the transistor
effectively conducts as if it were a wire. Sending 1 amp of current
through at 30 volts does NOT dissipate 30 Watts in the transistor
body. (Do the thought experiment. Conducting 30volts/1Amp through
a wire doesn't dissiparte 30 Watts in the wire, does it?) The GREAT
BULK of that power is dissipated in the LOAD...that is being driven.

And the closer the transistors "turned on " resistance is to being
zero...the more close to true it is that the entire 30 Watts are
being disipated in the LOAD and not the transistor.

The reasons for choosing FETs over BiPolar (if you can find big
enough MOSFETs for the current you want to carry)[See Note 1] is
that they MIGHT have lower resistance (source to drain) than an
equivalent BiPolar transistor. That results in LESS power being
dissipated in the drive transistor..in its "fully on" state.

* * *

But really, the place to woory about power dissipation is NOT when
either sort of transistor is fully switched on (or off for that
matter) but when it is IN TRANSISTION. You will usually have two
transistors in a "totem pole" arrangement. And the idea is to
insure that you never have BOTH of them transistioning at the same
time...because that will cause a current surge (it is as if your
supplies plus was shorted to minus through the body of your
transistor.) So, what you do to avoid that is introduce "dead time"
where you switch the one that is on off before you switch the one
that is off on.

I have, of coruse, the picture of an H-bridge in my head as I am
writing this...so it is really a pair of totem poles. But each one
has the problem in itself.


Jim

[Note 1]: On some of my "projects"...I switch 312 volts at 200
Amps. No single MOSFET can handle that. (I drive an electric car.)

--- In CAD_CAM_EDM_DRO@yahoogroups.com, "Mariss Freimanis"
<mariss92705@...> wrote:

>
> I'm assuming the output transistors operate as current sinks. This
> means they are operating in the linear region (simultaneous

voltage

> across and current thru the transistor).
>
> For linear operation it makes no difference if bipolar NPN or n-
> channel MOSFET transistors are used. The power dissipation in the
> device will be exactly the same. 1A of current with 30V across the
> device is 30 Watts either way.
>
> Mariss
>
>
>
> --- In CAD_CAM_EDM_DRO@yahoogroups.com, "turbulatordude"
> <dave_mucha@> wrote:
> >
> >
> > > And the PIC makes an interesting and flexible way to drive

it.

> > >
> > > But OK, not linear, but an approximation of a sine wave,

similar

> to
> > micro
> > > stepping.
> > >
> > > Alan KM6VV
> > >
> >
> > > > IME Linisteppers work fine in their "designed-for"
> applications...
> > > > IMO they're great for learning about--and messing with--
> > > > PIC "stepper" code, and while the heatsink and power supply
> > > > requirements are bigger than for chopper drives; they are

still

> > > > inexpensive and user repair-able.
> > > >
> > > > ALSO, the Lini- is NOT linear stepping; but linear
> amperage/winding
> > >
> >
> > This is where the polishing of the program and circuit needs to

be

> > done.
> >
> > Switch to FET's to lower heat. a no brainer.
> >
> > Linear amperage does not equal linear steps, but then there has
> always
> > been the discussion of exactly where the step/armature is on a
> > microstep. is it exactly at 1/7th of a full step, or is the

power

> > ramping between steps smoother that offers better power

delivery ?

> > and it is 'somewhere' between 1/2 steps, but not exactly linear ?
> >
> > Also, there is that how close is good enough bit. does a screw
> with
> > 26,000 steps per inch suffer if the step if off by 1/10th of a
> step ?
> >
> > Dave
> >
>