very new to cncs and electrical stuff and basically learning as i go
i am building an xyz gantry running 3 NEMA 17 stepper motors, x and y are belt driven and z with a leadscrew. i have 3 1.5A stepper motors and one of those little 9g servos and running tmc2208 drivers on them with an arduino cnc shield (i have some a4988 too)
the only power supply i have is a 12V 2A AC/DC wall adapter plus the arduino. Is this enough power?
in the future how would one go about figuring this out?
When using steppers with a current-controlled driver (such as the TMC2208), you must select the power supply unit by calculating the required power. To do this, you need the rated current of the coils (the rated current in the data sheet is almost always per coil) and the voltage or resistance of the coils. Then you can calculate the required power per coil. With 3 steppers you have to multiply this by 6 (2 coils per stepper). Then use at least a factor of two to select the power of your power supply unit.
The coil current alone is not sufficient to select a suitable power supply unit with a current driver.
[EDIT] So we need more info about your steppers. At least the rated voltage ( coil voltage at rated current ) or the coil resistance. Provide a link to the datasheet of your steppers.
No, this is not obvious - you often can read incorrect advices about this in the forum. But your PSU is surely too weak.
I second this. 12V 2A powersupply is surely too weak.
You shoud use one that has more than 4.5A. There should be still some "air" to maximum rating
I suggest a Powersupply with 12V minimum 6A. Which is 12V * 6A = 72Watt
For the stepper-motors you should not go below 12V.
So just as an example, suppose the coil resistance was 3 ohms. At 1.5A each coil would need 6.75 watts and there are 6 of them, so total is 40.5W. Motors work better at higher voltage, 24v is a good choice, so you need 1.6875A at 24v , but some safety factor too so a 3A 24v supply is probably a minimum choice.
A few points to note here. The combined current of the motors at the 24v supply is not 9A for two reasons. First, the 2209 drivers are acting as switch mode power supplies, so what's important is the total power not coil current. Second, the drivers never drive both coils at full current while microstepping, so the max current per motor is more like 1.414 times the single coil current. But always nice to have some safety factor.
You may not need 1.5A max current, it depends what torque you need. I have a small CNC mill with NEMA23s on X and Y and NEMA34 on Z and I only use 1.5A on those (at about 36v supply) and get plenty of torque. So you may have some scope to reduce current, not to use a smaller supply but to let your drivers run cooler. The motors may get rather hot, but they are designed to.
No they are not.
The drivers maintain the desired coil current by using PWM to control the voltage across the coil.
They don't act as a buck circuit in order to reduce the voltage. If you look at any driver design, you won't see any inductors that a buck circuit requires.
Are you saying the the driver somehow use the motor coil in some kind of buck configuration in order to reduce the voltage across the coil itself? How is that possible?
Yes, they are. The driver chip itself is only part of the system. The bulk capacitor at Vmot of the driver, the driver itself and the coils are the complete system. Every part ist needed for the system to work correctly.
The voltage across the coils is not reduced - the current through the coils is controlled. It works more like a short-circuited step-down converter, where the current limitation comes into play. This is only possible because the coils are an inductive load.
Some years ago I wrote an explanation in this forum and it was hosted by user Tommy56 - but unfortunately it is in german. Maybe google translate can help
( Perhaps I should write an english version, as a stepper driver is very often misunderstood ... ).
No, things are not that easy. It's not only ohms law when capacitors and coils come into play. Coils and capacitors are energy storage devices.
With a buck converter the current on the low voltage side is also higher than the current on the primary side. But as you know it works! With your simple assumption that would be impossible too.
And you forgot, that all those driver ICs need a big capacitor at the input side.
The VM pin is a bidirectional pin. Current flows into and out of that pin. That's why the capacitor is mandatory. It stores the energy when current flows out of the pin. And a big part from this energy is used, when current flows into the pin again.
This is a picture from the DRV8825 datasheet where you can see all possible current flow through the H-bridge. Its not only simply from input to output:
As you can see, the current through the coil is always the same direction ( with no stepping ), but the current throug the H-bridge is not. Switching between modes 1-2-3 is done very fast to control the coil current.
[EDIT] Because of losses in the coils and mechancal work the stepper has to do, the energy flowing out of the VM pin is less then the energy flowing into the pin. Only this difference has to be provided by the PSU.
