A4983 and A4988 Getting Started Guide

And I have a new question.

First - I was able to run my motor for the very first time and thanks to the instructions on this forum.

Then, in the Test instructions, it looks like the MS1 is being fed 5 volts. While this I imagined would come from one of the PWM pins of arduino, it doesn’t look like it comes from any particular part. Can it be powered (HIGH) using the 5V output of the arduino?

Also, is it the intent to set the MS1, MS2, MS3 to H,L,L to configure the A4988 into half steps?

Thanks.

Farzad

If you are using the A4988, the MS1, MS2, and MS3 pins are all pulled-down internally. Assuming MS2 and MS3 are tied to ground or left disconnected, tying MS1 high will put the driver in half-step mode and leaving it disconnected or tying it to ground will put the driver in full-step mode. You can connect MS1 to a digital pin on your Arduino to control the step-mode, or tie it to either the 5V pin or GND pin (depending on what microstep mode you want) if you don’t need control over it. I am not sure the intent of the original poster, but tying 5V to the MS1 pin with MS2 and MS3 tied to ground will put the A4988 in half-step mode.

-Brandon

Thanks, Brandon.

Farzad

Hello, I have a A4988 stepper motor driver and trying to control small stepper motor with these specs - 3.75 degree, 24V, 60 ohm. The stepper motor is unipolar, 6 wires and I set them as “bipolar half winding” ie I’m using one end and the centre tap of each phase.My calculations was 24V \ 60 Ohm > 0.4A so I set the current limit of the A4988 to 0.4A, the power suply is 19V and everything looks fine. The motor run smooth, full step, half step, direction reverse is OK. The problem - when the motor stop and stall, it start overheating.I tested the motor runing about 10 min. and the temperature of the stepper motor is no more 40 - 50 degree Celsius. In IDLE with power up it start heating, and after 5 min. it is hard to tuch with finger. To be clear, I use an optoisolated breakout bord and control the driver with Mach3 via PC parralel port.
I also tested to reverse the directions of windings with no result. The motor start overheating after stop running. Any advices would be very helpfull.

Hello.

When the stepper motor is holding its position, is when it will draw the most current. If you tell me what you measured for Vref, I can verify what your current limit is set to.

The way you have your stepper motor connected to the driver with one side to the center of each coil, means that the resistance of the portion of the coils you are using is half, so it is possible that the stepper motor is drawing twice the rated current. We recommend connecting the stepper motors as shown in the FAQs on our stepper motor product pages. If you connect it like this does the motor run cooler?

- Grant

Thank you Grant.
The resistance between one end and center of the coil is 60 Ohm, the measured Vref is 0.16V x 2.5 = 0.4 A
Here is my stepper motor

The measured resistance between ends of each coil is 120 Ohm, so if I use both ends , my calculation is 24 V / 120 Ohm = 0.2 A?

I did some math based on the number you gave to find the power that motor would use:

P = I^2 × R = (0.4 A)^2 × 60 ohm = 9.6 Watts

When the stepper motor is holding its position, the power used by the motor is going to be dissipated as heat, so the heat you are getting from the motor might not be unusual. Since you are controlling it with a bipolar stepper motor driver, it is also possible that the stepper motor is drawing twice as much power because the stepper motor driver is driving two separate coils at once.

If the motor is fine when you are moving it, you might try just disabling the driver when the motor is not moving. You might also try using the whole coil on each side since the added resistance should cause the current to decrease (as your calculation shows) and halve the power dissipation.

- Grant

Thank you for the answer Grant. I was qurious why this motor generate too much heat driving it with A4988 as bipolar.When it run with its original unipolar driver with MTD1120F chip, the teperaure is no more 50 C". Is there any other parameter, may be the inductance of the coil, wich is in relation with heating? What will be happen if I lower the V mot.?

(1) I assume you are setting the motor to the same current limit in both cases. If not, that would be the main reason it is heating.

(2) Given (1), the bipolar mode will run a given current through twice as much copper. The unipolar mode will run a given current through only half the coils at a given time, so it will produce about half as much heat loss (and half as much torque.)

Bipolar also has a little more (probably insignificant) induction/ACripple/someECEconceptIdontReallyUnderstand losses because the direction is changing more.

I agree with Tomek and expect what he said to probably be the case for the additional heat you are observing.

In your case, the current limit of the driver is probably not really limiting the current of the motor since you are using the stepper motor at its rated voltage, so lowering the supply voltage will probably cause the motor to dissipate less power, but you will also get less torque from the motor.

- Grant

Thanks for the answers.

For the record, I am not sure why grant said what he said about the current limit probably not doing anything.

For a given motor, say rated V=12V, I = 1A, a current limit of 0.5A will still be useful to limit the current, even if you are only running it at 12V.

It should be noted however that at 12V for a 12V motor, you have a very low top speed before the current starts limiting itself by induction.

This is also an issue when you go from unipolar to bipolar. You may find for a given voltage your top speed drops off, even if your low-speed torque improves. That’s because the inductance of the bipolar motor is more than twice that of the unipolar motor.

I suspect he said what he said because of the speed limit, but you may be using your motor at low RPM, in which case what he said does not apply.

Generally the way you want to reduce power is not by reducing the motor driver voltage input, but by reducing the current limit. Reducing the input voltage is a very mediocre solution that should be reserved for the rare situations it is the only solution (in which case it is no longer a mediocre solution.)

**The reason why it’s normally a mediocre solution is because it reduces your top speed and as a consequence can result in torque problems that take a little while to diagnose.

Actually this motor is rated 24V 60 Ohm. I am running it at 19V and the current limit is set to 0.4A. The main issue is not speed but overheating. I tried to set a minimum possible current limit with the trimer, and even at 0.2A it generate too much heat.

Weird, is it a very small stepper motor?

At 19V it should never pull more than its rated current (it should pull~0.32A [19V/60Ohms]) (I guess this must be what Grant was talking about…Not that a current limit wont do anything, but that it shouldn’t be necessary.)

Have you used an Anmeter/Current meter to test the current going through the windings?

I would do some digging, since it’s unusual for a stepper to be rated at something they can’t continuously deal with. You may also be overestimating the concern for heat, as steppers are designed to tolerate somewhat substantial continuous temperatures.

That said, certainly, a 0.2A current limit (0.1V on Vref) should be fine for a 0.4A rated motor…

It seems like Tomek already figured out what I meant in my last post, but just in case, I was trying to say that because color was using the stepper motor at the rated voltage (which for stepper motors is just the voltage at which the motor will draw the rated current), setting a current limit of the rated current was probably not doing anything. I agree with Tomek though, it is generally better to adjust the current limit to reduce the current through your motor, rather than reducing the applied voltage.

- Grant

I have connected a 100micro farad capacitor across Vmot, Ground of A4988 Driver. but capacitors are blowing within 1 minute after switching ON 13V smps power supply. Capacitor +,- terminals are connected in Vmot,gnd respectively. What may be the problem?. i have changed new capacitor too… new capacitor also blowing out.

Note: Same circuit was perfectly gave output the previous day.

What voltage rating are your capacitors? For a 13V PS you want something like 20V rated caps (V*1.5). Not as you might expect, 16V

What does “The same circuit was working perfectly yesterday” mean? Do you mean the same type capacitor on said circuit?
Have you measured Vout on the PSU?

I have used 25V 100microfarad capacitor and the problem is smps supply, by changing smps power supply, capacitor working fine. thanks for your ideas…

I was wondering if the user Color is progressing in driving his CKD stepper motor.

No, the problem with motor heating still persist. I measured the current with ammeter, it is 200mA per coil and when running the motor is cool, 15 min. in IDLE and temperature jump to 60* C and may be more. No matter how the coils are wired - parallel or series. I think the only way to run these motors in bipolar is to use a driver which is capable to limit the current in IDLE.