# Stepper Motor Calculation

I'm using this Stepper Motor from TI.

This is the first time I am working with Stepper Motor.

I need to calculate the load current or the current that the Motor takes.

The Stepper is rated at 0.6A/phase.

The application of the stepper is to move up and down an object. So, in this case, what are the factors that determine the current through a stepper motor?

I feel my question is very incomplete. Since, this is my first attempt at working with Stepper Motors, please tell me what should I look for and what are the parameters that I should focus on while working with stepper motors.

• You might start by selecting a stepper motor driver IC. Steppers of the type found in 3D printers etc are often driven from 24V supplies and the motor would draw far too much current if connected directly to the supply for any length of time. The relatively high voltage allows the motor’s inductance to be overcome quickly and provide sufficient torque for the application. The IC limits the current both when the motor is turning and when it’s ‘holding’ (actively preventing rotation). The required current depends on the load and the required speed. – Frog May 9 at 5:20
• I have been playing with stepper motors in the last couple of weeks. You might like to skim my old answers, and the references there, to get a vague idea how to play with step motors. (1) electronics.stackexchange.com/questions/558386/… (2) electronics.stackexchange.com/questions/559986/… / to continue, ... – tlfong01 May 9 at 6:00
• – tlfong01 May 9 at 6:00
• This answer with tutorials is good for newbies to start off. (5) raspberrypi.stackexchange.com/questions/97975/… – tlfong01 May 9 at 6:02
• It's weird how you suggest/promote some of your own hugely-downvoted answers to the OP. @tlfong01 – Mitu Raj May 9 at 6:06

The motor spec for L and R and V will definitely e your currents.

Holding current: I= V/R. For V=Vavg if PWM modulated.

Moving current is reduced by L initially as dI/dt starts 0A at t=0 then rises (exponentially) with a 63% I time constant of T=L/R which tends to be in the 1ms range for small steppers.

This means you can use a higher voltage for a period shorter than this time constant then with PWM reduce the average voltage by sensing the current limit or step to the next voltage =0 for full steps before it reaches a current higher than rated (continuous conduction mode)

If the friction and inertial load is smaller than the force and current required to accelerate (f=ma) and stop in the time to reach the next step is achieved, then you have a faster seek time. But this is difficult to achieve.

Otherwise one uses a PWM modulated active current limiter with current sensing and a higher voltage than the I=V/R would indicate.

These are your options for the design of a stepper, current motor driver or any servo motor design.

The limiting factor is the temp rise in the motor from $$\Rth * I^2*DCR\$$ for a thermal resistance Rth (with or without cooling fans) and DC resistance , DCR. This is why stepper motors are rated in current.