MY CNC sessions are now like laundry day....

[Jonathan]

Just taken delivery of a chinese TB6560 based CNC stepper driver board.

Oh dear, see below for why...

When I connect up my steppers, the sound from them is akin to water rushing through pipes (the same motors connected to another Allegro A3977 based driver board .... silent as can be)....so what's the noise all about - it makes me want to run to my missus & ask for some Daz.

If the switching frequency of the current limiting circuit in the driver is in the audible range, i.e. 20-20kHz (depending on your age!), then you will hear the sound you describe. The Allegro must switch at above this frequency, so you can't hear it. I wouldn't worry as there's not a lot you can do about it.

(The above is a bit of a guess, so treat it accordingly)

I'm struggling here to make sense of the rated voltage spec. Just about every driver on the market takes at least 24V DC as an input ...so where does the 3V come into it (I can hazard a guess the resistance is 2 ohm, therefore 3V across the coil would yield 1.25A ...but where do I set the 3V - would this be a setting on the driver board?)

You are correct in saying as V=IR, V=2.15*2=4.3V ... yes the numbers you've quoted are a bit fishy but nevermind.
The important point is the inductance of the motors limits the rate of change of current, so you need a high voltage to overcome that. It's common to drive them at 20-30 times the rated voltage. As a rough guide V=32*L^0.5 [Edit: Had wrong number]

Each phase of the motor can be modelled as a resistor and inductor in series. So you can write down the differential equation in terms of the voltage across each component. For a constant applied voltage, V:

V=Ri + Ldi/dt + idL/dØ*dØ/dt

Where Ø is the displacement.
The last term is the back-emf, for the purposes of this discussion that can be neglected. So solving the remaining first order linear differential equation yields:

i(t)=V/R-(V/R-I)*e^(-t/τ)

Where I is the initial current in the motor phase when the phase it turned on, and τ is the time constant, given by:

τ=L/R

We want the current to rise as quickly as possible, so that it has reached the rated value ideally well before the next step is commanded. If this is not the case the motor is effectively being run on less than the rated current, so the torque drops dramatically.
It's clear from the equation for i(t) if you increase V (or decrease R), i(t) increases, achieving our objective.
For the rate of change of current to be greater the time constant of the exponential must be decreased and if L or R are decreased τ is less, so the current rises faster.

So in conclusion you should run the motors at a high voltage, and select motors with a low inductance and resistance to obtain the most torque at high speed.

Now it is obvious why running the motors in parallel (though your driver can't take it) is optimal. The inductance in parallel is a quarter of what it is in series.

(Before someone complains :whistling: - the above is a simplified explanation. I have neglected certain factors for clarity such as initial conditions and the fact the motor inductance can vary with displacement)