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General Mach Discussion / Re: Mach III motor tuning with servos and G320X
« on: June 29, 2018, 05:24:12 PM »
Hi,
the bottom line is to get the fastest and most powerful response from your servo you will need maximum current. Current is proportional
to torque which is proportional to acceleration. Thus you need to set the drive to deliver maximum current it can without fault.
The second thing you need to do to screw up the maximum Proportional gain you can within the limits of stability. You can't avoid this, it is absolutely
basic to any feedback control system, the higher the feedback gain the better. In order for you to do this you will need to turn the following error window
as wide as possible or even 'off' if you could.
In Machs motor tuning page specify an effectively infinite speed and acceleration, this will mean that the speed and acceleration that you see at the servo are the
most the servo can produce with this drive. You do not want the command to the servo limited in such a manner that you can't find the servos natural limits.
So in Machs tuning page put some ridiculously high numbers for max speed and acceleration. Then try tuning.
Next mistake you are making is that the current going into the drive from the power supply IS NOT THE SAME as the servo current. The output of the drive is PWM,
that is to say the supply voltage switched on and off in proportion to the deemed required output voltage. Thus the output is in effect the output of a buck regulator,
it output voltage will always be somewhat less than the supply voltage but its output current can be GREATER than the supply current. How is this so?
Lets put some numbers in to illustrate the idea.
Vsupply=100V
Vout(commanded)=10V
Thus the duty cycle for a 10V output is 10%.
The charge balance equation is:
100 X I(input)=10 X I(output)
I(output)/I(input) = 100/10
=10
So the driver is in effect behaving like a transformer, its input voltage is being transformed downwards and so its output current is being transformed upwards.
It makes no sense then to measure the supply input current because the output current could differ markedly.
Craig
the bottom line is to get the fastest and most powerful response from your servo you will need maximum current. Current is proportional
to torque which is proportional to acceleration. Thus you need to set the drive to deliver maximum current it can without fault.
The second thing you need to do to screw up the maximum Proportional gain you can within the limits of stability. You can't avoid this, it is absolutely
basic to any feedback control system, the higher the feedback gain the better. In order for you to do this you will need to turn the following error window
as wide as possible or even 'off' if you could.
In Machs motor tuning page specify an effectively infinite speed and acceleration, this will mean that the speed and acceleration that you see at the servo are the
most the servo can produce with this drive. You do not want the command to the servo limited in such a manner that you can't find the servos natural limits.
So in Machs tuning page put some ridiculously high numbers for max speed and acceleration. Then try tuning.
Next mistake you are making is that the current going into the drive from the power supply IS NOT THE SAME as the servo current. The output of the drive is PWM,
that is to say the supply voltage switched on and off in proportion to the deemed required output voltage. Thus the output is in effect the output of a buck regulator,
it output voltage will always be somewhat less than the supply voltage but its output current can be GREATER than the supply current. How is this so?
Lets put some numbers in to illustrate the idea.
Vsupply=100V
Vout(commanded)=10V
Thus the duty cycle for a 10V output is 10%.
The charge balance equation is:
100 X I(input)=10 X I(output)
I(output)/I(input) = 100/10
=10
So the driver is in effect behaving like a transformer, its input voltage is being transformed downwards and so its output current is being transformed upwards.
It makes no sense then to measure the supply input current because the output current could differ markedly.
Craig