# Thread: How much torque would I lose?

1. ## How much torque would I lose?

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?

2. The motor torque should be proportional to the winding current (within limits).

3. You're holding torque will be the same, but your maximum usable rpm will be about half.

4. 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.

5. 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?