Re: [CAD_CAM_EDM_DRO] Re: Power Supply for 4-axis CNC stepper driver

In a message dated 9/9/2005 10:22:53 A.M. Mountain Daylight Time,
johnd@... writes:

I wrote up a description for a friend on why the higher power rating is
required. I'll paste that below. His motor is slightly different from
yours.
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This will take some time....

Motor is rated at 2.3V and 5.5A. That's 12.65W per winding or 25.3W; say
25W. Since you are microstepping you never have both windings totally on
hence the derating of the power supply from 5.5A per phase to say 0.707 * A
per phase. That means about 18W total.

So if you were running an LR circuit (Remember high voltage is high pressure
to get those pesky electrons flowing into the winding faster so you get
rated torque faster) you'd run 25 * 2.3V or about 60V into the motor. But
60V across a 0.418 Ohm winding (2.3V / 5.5A) is 143 Amps. Yikes. Burnt
Marshmallow time.

So the resistor in series with the supply needs to drop 60 - 2.3V == 57.7
volts at 5.5A which is 10.5 Ohms. And it needs to be a 320W. Can you spell
space heater. And as the current climbs the voltage drop increases so the
pressure is reduced so you never really get full current before it's time to
step again.

Alright. Let's use a comparator across a very low resistance like 0.05 Ohm
in series with the motor windings. The voltage drop across this resistor is
0.275V when 5.5 Amps are flowing. When the comparator sees this voltage the
output changes to turn off the current drive.

It doesn't really matter if the applied voltage is 6V or 60V. But (for
example) with 60V you'll get up to your 5.5A ten times faster than with 6V
because of the coil inductance. So say you want a step rate of one step
every 0.1 seconds. But it takes 1 second at 6V to get 5.5A because of
inductance. That means at full torque loading on the motor you can only
step once ever 1.0 seconds. But use 60V and you get full torque after 0.1
seconds. True the controller then shuts off the current until it decays
below 5.5A but you don't care, you are stepping again anyway.

If you stop stepping and just hold at full torque the chopper circuit will
hold the current at 5.5A or the LR circuit will drop 57.7V across the
resistor and 2.3V across the winding. Let's see.... 57.7V + 2.3V == 60V *
5.5A is 330VA. The chopper circuit needs to supply 5.5A too but from a 60V
supply.

Now back to the coil inductance. Once the current has reached 5.5A the
chopper turns it off. It decays based on the same RL time constant with the
back emf being forced either through the windings with the back emf diode or
in the case of an H-Bridge back into the power supply. As soon as the
current reaches the hysterisis of the 0.275V comparator it turns on the
power again, and current starts flowing. This happens at about 20kHz
depending on the circuit.

So then the question is, where is the 5.5A coming from? Remember, the
voltage across the winding is maintained at 2.3V for our 5.5A. At this
point, the transformer could indeed be about 4V at 5.5A or 22W assuming
small losses over the FET and bridge rectifier.

Can you see that it's a dynamic situation? For LR circuits the transformer
does indeed need to be the full 5.5A at 60V. For chopper circuits it needs
to be 60V at 5.5A until 5.5A is reached and then it really only needs to be
about 4V at 5.5A.

What if we used a bigger cap and lower power transformer? Well, we're back
into the LR time constant and CR time constants. Start with 60V charge in
big capacitor. Apply this to motor. We need 5.5A in 0.1 seconds. This
means we could discharge the cap down to 4V and then hold it there with our
22W transformer driving 5.5A. OK. Now step one step.

Oops. We need to get our power back up to 60V before we step. Alright,
how long does it take to charge our capacitor back up to 60V while the
chopper is drawing 5.5A. This next part is a bit complicated but bear with
me.

0. Motor step and 0.1 seconds later current reaches 5.5A & 60V power supply
almost shorted out.
1. 2.3V reached and 5.5A reached so chopper off.
2. 0.0005 seconds go by.
3. voltage below 2.3V so chopper on.
4. 0.0005 seconds go by. Goto to step 1.

So you have 0.0005 seconds to charge up the power supply capacitor back to
60V during each chopper off period. Trouble is, at maximum step rate, it
took 0.1 seconds to get to 5.5A with the 60V applied voltage using energy
stored in the cap. So if we want to step again, how do we fill up the cap?

The transformer is effectively almost shorted out because you're applying a
60V transformer to a 2.3V load every 0.0005 seconds (20kHz chop rate).
-------------

Hope the above description makes sense. I've generalized a bit in places to
simplify what is happening.

John Dammeyer



The above should be printed out by most, as a future reference as what
happens and why a chopper type of driver should be used...
bill


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