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Link? For what driver? You can spin a stepper as fast as you want. People are (overly) worried about their stepper drivers.
By the way, what do you guys have your grbl Step idle delay ($1) set to for the X-Carve? (I don’t have one)
If you set it to 255 your motors will always be enabled when powered. If you try to back drive the motors while powered down, you can generate enough voltage to enable them and cause them to stop. If you’re forcing them while they’re enabled, well, there’s something wrong with you.
I know I can power the fan which is connected directly to the power supply when moving it by hand. Damage or no damage. It can’t be healthy for the life of your controller. It may not “blow it out” completely. But who’s to say what the cause of that “mysterious” lost step really is. In my opinion. If it was intended to be moved by hand. It would have a handle on it.
My point is that with $1=255, you should not be able to move your steppers when powered. If power is off, and you drive the steppers by hand, power will be applied to the circuit and enable the motors (also activate circuit protection).
I’m not convinced that the circuit protection would protect against manual movements. Overtemp protection can shut down a driver, and motor, controlled by the driver, but it can’t stop me from spinning my motor by hand. Making sure the motor is enabled when powered can stop me.
I burned out a stepper driver on the X-Controller doing this repeatedly over time. I’m an idiot. Fortunately Inventables didn’t force me to replace the whole unit, but they would have been within their rights to.
Nowadays if I want to move the gantry by hand — which is sometimes necessary for whatever reason — I power down the X-Controller and then physically unplug the motors from it.
This is not true. The driver chips have two power requirements, 5 volts for the logic and Motor Voltage for the motor windings.
The outputs (to the motor windings) do not apply any power to the Motor Voltage pins or the 5 volt logic, so even if you are moving the motor to generate back emf on the motor output pins you are not supplying any “power” to the chip protection circuitry.
A protection circuit against back EMF would absolutely protect against manual movements which produce the back EMF. That’s almost entirely their purpose of existing, so why wouldn’t they work?
Yes and No. It depends on the design and what the diode is expected to do.
Diodes can be biased in the “reverse” direction and unless you exceed the reverse junction voltage limit, no current will flow (if you exceed the voltage limit the diode is destroyed).
If you bias the diode in the “forward” direction once the voltage rises above the “threshold” voltage (typically 0.7 volts for silicon and 0.3 volts for germanium) current will flow and as long as you don’t exceed the current rating everything is golden.
Seems simple, yes?
NO.
What happens if we the have the same voltage on both of the diode pins? No current flow.
I’m going to simplify this for illustration purpose (reality is more complex) by saying that the FET is going to remain “off”.
So in the above design snippet if VM = 24 volts and AOUT2 = 24 volts no current flows through the diode (3rd FET/diode pair down the drawing).
If you remove the 24 volts from VM (remove power from the gShield), now you have a forward biased diode and current will flow from AOUT2 to VM.
There are hundreds of designs that use diodes this way so you can’t always assume the simple case.
H bridge designs challenge even experienced engineers when they do their first one.
When you throw in the FET and do the correct analysis you can have forward and reverse voltage and current through the AOUT2 pin without damaging anything and you can protect against back EMF when the driver is powered. Note that when the driver chip is not powered you have completely thrown out that analysis.
And if current flows from AOUT2 to Vm, and Vm is tied directly to Vin (not sure on X-Controller, but it is this way on others), one could power their entire controller if the stepper could provide the voltage. Does that sound right?
Here it is. The current is limited by the Vref input to the chip. Only works with power applied to the chip.
100% current settings (Current value)
100% current value is determined by Vref inputted from external part and the external resistance for detecting output current. Vref is doubled 1/3 inside IC.
Io (100%) = (1/3 × Vref) ÷ RNF
The average current is lower than the calculated value because this IC has the method of peak current detection.
Pleas use the IC under the conditions as follows;
0.11Ω ≤ RNF ≤ 0.5Ω, 0.3V ≤ Vref ≤ 1.95V
