View Full Version : How much torque would I lose?

04-06-2012, 05:04 AM
My unipolar stepper is rated at 6.48V and 1.2A.
Initially, I'd planned to power it with two ATX supplies connected in series at 24Vdc.
I calculated the current limiting resistors to be: (24-6.48) / 1.2 = 14.6R
with power rating at 1.2*1.2*14.6 = 21.02 W (the actual used are 15R at 25W)

Here's the thing...

one of the ATX blew out just before I was about to conduct tests to ensure my home made machine is working correctly. So I am like left with 12Vdc and the Z axis seems to lack the torque to move the axis properly. the x and y looks happy with a 12V supply.

the thing is...is there a way to find out how much torque I lose by supplying 12V to a circuit built for 24V?

04-06-2012, 06:21 AM
The motor torque should be proportional to the winding current (within limits).

04-06-2012, 11:53 AM
You're holding torque will be the same, but your maximum usable rpm will be about half.

04-06-2012, 03:47 PM
It sounds like the OP may be using a very simple driver - he is putting a resistor in series with the power supply to the motor in order to drop the voltage to the motor down to the nameplate voltage, instead of using a current-regulating driver that applies a higher-than-nameplate voltage to the motor.

04-08-2012, 02:50 AM
I connected another scavenged ATX in series and it obviously gives a much wider range of RPMs.
Yep...I am using a DIY tachus42 full-step driver. I have the 2M542 driver but am a bit clueless on how to connect to the breakout board.
For the power supply, is it better for me to get a 24Vdc SMPS or should I build one with a beefy transformer, bridge rectifier and reservoir caps?