Hi,
After reading A4988 datasheet as well as pololu description, I am still not able to clarify that what is the driver operating current for motor supply. Or how should I select the motor supply source depending on current. For example I have a 2A/phase stepper motor, I want to run it on 24Vmot. So is it compulsory to select a Vmot source 2A or higher or is there any calculation to setup. I have monitor power supply voltage which is always lower then the motor coil voltage (also read about current limit).
I want to deign my power source (with maximum current supply capacity, so right now My concern here is to ask power supply source, how it should be selected to run the stepper with A4988 driver. What is the maximum Vmot supply drain?
Our stepper motor drivers current limit by turning off the voltage applied to the motor when the set current threshold is reached, so they behave similar to step down regulators and effectively reduce the average voltage applied to the motor outputs. In such systems, the input current needed can be calculated with:
Pin = Pout
or
Iin x Vin = Iout x Vout
If you set the current limit of the driver to match the rated current of your motor, the output voltage you would use in the equation is the rated voltage of the motor. Of course there are some efficiency losses, but you can compensate for that by adding some margin to the result of your calculations.
All that being said, if you want to just be extra safe, you could get a supply that can handle double your current per phase rating (assuming your motor has two phases).
By the way, our A4988 stepper motor driver carrier is only rated for 1A per phase with without additional cooling. Driving the carrier up to 2A would require considerable cooling and would likely be very difficult to do. We do not have any stepper motor drivers that can handle 2A per phase without cooling, but we do have a few that can handle 1.5A or more. You can see all of them on our stepper motor driver category page.
Thank you for your answer. Well I am using a big heat-sink plate attached to the driver for cooling. What would you suggest ? I have used A4988, but now i see there is one more with digital current control option MP6500 driver. My application is linear guide (belt driven) load up to 4Kgs. So far i have tested DRV88s, Tic, A9s… driver but according to my application in machine which driver suits the best is able to limit current up to 2A/phase . I have about 48 similar motion profile in my machine design. So I want to select the best driver with more control option, with the fault detection for my software control. As well as I need to finalize PCB design for power supply of motor supply, that’s the reason of my concern for motor supply voltage consumption. SO I can design my PCB accordingly with sufficient power supplying capacity to my all 5 motors per motions profile. What I understand from your above statement is that if my motor is rated 2A/Phase, I shall use up to 4A of power supply. Is it correct. Is there any reference material or guide you have reading that how much the stepper motor consuming current. Because according to your A4988 current limiting video the power supply current shown is about 0.3 ampere, but motor is set to 1 ampere current limit. SO this is the confusing part now. Should i follow the power supply amperes to select my motor powering module or should i follow the phase current. Even is teh phase current is set to 1 ampere , why don’t the power supply showing this consumption. I hope you understand my point?
We do not have any specific recommendations for a heat sink (or fans), but if you have already tested with your large heatsink and it is working, then your heatsink would seem to be enough.
As Jim mentioned, the power supply displays the average current it outputs, but with a current limiting stepper motor driver like the ones on our boards, the voltage and current given to the motor are switched on and off (that is how the current limiting is achieved), so the maximum instantaneous current drawn by the motors will be higher than the average. For sizing your supply you would probably be fine just getting a supply rated for a little over the average input current, but like I mentioned before, if you want to be really safe, get a supply that can handle the full current of the motor.
So is it correct that if I say there is AC component in DC due to sine wave control of current supply to motor? , Then according to Power equation, i.e.
Iavg. = (2/π) x Ipeak
Ipeak-to-peak = 2xIpeak
If power regulated supply shows Iavg. = 0.3A, then
Ipeak = Iavg. x (π/2)
Ipeak = 0.3 x (π/2) = 0.4712A
Ipeak-to-peak = 2(0.4712) = 0.9424 A (this is the real current drawn from power supply)
Am I correct ? and if this is the case, dose it also mean the my regulated power supply Voltage is also considered Average Voltage ? Then I follow power equations to calculate Vp-p, because when stepper is running there is a constant up/down in my power supply voltage too. I am using Agilent Power Supply (20V, 4A)
If the power supply does its job, the output voltage will be constant.
Take a look at the A4988 data sheet for some typical motor current waveforms, which depend on the “decay” and full step/microstep settings of the driver.
The current through one coil changes as the motor steps, but there are two coils, and they are out of phase such that combined, they get constant power, and your power supply will deliver basically constant voltage and constant current. You can just simplify your math by using the peak current in one coil (at that step, the current in the other coil is zero). If it’s 1A on a motor that draws 1A at 4V, then you will need 4W, which is 1/3A at 12V or 1/6A at 24V. Those numbers don’t factor in efficiency, so something like 0.5A at 12V or 0.25A at 24V should cover you in this example. You should be able to confirm it by looking at what your supply is actually delivering. Any high-frequency deviation from the DC average is going to come from the details of how the stepper motor driver maintains current in the coils, not from your step position changing. The capacitance you have on your supply line will take care of that current variation.
