Dumb question. How does the torque change with double voltage on the same motor?
the 2017 Etonm motors had bearings.
the redesigned 2018 etonm motors with the access “flap” got rid of the bearings. They didn’t even bother to tell anyone.
One of the reasons I don’t like Etonm. They change crap to save $1, but then demand $30K minimum orders.
1 Like
Upgrading XY motors
to the future is a good task to get more speed, I’m in. If this involves new motor shields , so be it.
This is exactly my point, since we are using the same motors, doubling the V won’t reduce the current, it will increase it instead. You don’t lower the current by increasing V
The amount of power needed to move the sled at a given feedrate is constant. The PID controller will reduce the duty cycle of the PWM signal to compensate for the greater amount of work done by each pulse.
Right, but power comes from V*I hence, doubling V you are actively increasing the power and current of the circuit.
You don’t lower current or power by increasing V.
On a side note, if the idea is to reduce the current in half, you will be reducing torque in half as well.
Do you know what’s the minimum duty cycle the firmware will go to compensate?. I would like to do some test and see if we’re not chopping the current and loosing torque there.
I hadn’t considered torque, and it’s been way too long since I’ve studied it. So is there no advantage to increasing voltage? I know what you are saving with respect to v=ir, and I’m not arguing against that… from an instantaneous perspective. But work is work… so something has to give or everything is converted to heat. What am I missing?
200 W = 100 V * 2 A
200 W = 200 V * 1 A
200 W = 400 V * 0.5A
200 W = 800 V * 0.25A
All the above is right… but that does not mean that 800v will push .25 A to a circuit, this is not taking in account the load (motors).
It’s the load and V what will determine the current of the circuit in the end (assuming power supply/shield are sufficient).
Increasing V will increase current, therefore motors will have more torque and RPM (if the motors winding can withstand the increased current)
But how fast can you go? How do you speed up gravity? the wood, router and bit being used at a given time will also be a factor.
You will never reach max 12 volt current at 24 volts without burning up the motor because it will exceed the power rating.
A look into industrial motor sizing explains that long distance motor installations go for higher voltage because at equivalent power, the wires are smaller because they take less current. This is why power lines run at 10000 volts over miles and miles very efficiently: there is so little current that less power is lost to heat allowing the wires to be smaller. It is all about equivalent power, not equivalent resistance. The same holds true for the maslow DC motor.
Right, maybe I’m being too simplistic in my thinking, but if the PID controller is running at a duty cycle of 50% for a 12v power feed, would it not have to run at a lower duty cycle for a 24v power feed to keep the feedrate of the router the same?
On these motor controllers, is the maximum current rating of the chip an instantaneous level (i.e., max current for each pulse) or a cumulative level (i.e., avg current over time)?
This I very much agree with. I’m very concerned about the stability of the sled without it being captured. A four motor design might be necessary.
HI 
This is exactly what i’m trying to say all the time (higher V will result in more current NOT less)
If they will burn or not need to be tested, per etonm they CAN be run at 24V
I believe is a misunderstanding here, how efficient is AC vs DC for long distances power transmission has NOTING to do with the power consumption of the motors or any circuit you want to power up. You will need to transform it down to whatever voltage your load (motors) need. The current of the circuit will be determined by the LOAD and increasing V will result in higher current, no way around it.
3 Likes
Right, and at 25% i believe we will end up with the same current as before not half (if we’re NOT chopping the current due to the frequency period being to short). We might need a speed constrain() to avoid this.
I’m not trying to be the Negative Nancy here, increasing to 24V could be a good idea, my initial point was that we can expect higher current running on 24V that’s all.
Depends on the definition of instantaneous, they don have clock to measure time and do avg, per TLE5206 datasheet it happens after the trip point is exceeded (8A at 25 °C) then outputs are turned off after 50 μs
I will again put out - data sheets don’t usually cover the efficiency drop as the temp goes up performance is not linear. A trip happens at 25c but efficiency will sag long before that.
Thank you
1 Like
I went back up to the post early on in the thread and looked at the 1280 (12V) and 2480 (24V) motors.
On the lower datasheet for the 19 RPM variants, at no load, the current draw on the 2480 is less than the current draw on the 1280 (precisely half). But if you click the link @metalmaslow posted you will find a datasheet where the current draw on the 2480 is higher than that of the 1280. I know they are different motors, but I have to believe one of them is incorrect. What is the physical difference between a 12V and 24V? just how its wound?
You can have two DIFFERENT motors that perform identically. A 24V motor can have the same power output as a 12V motor while drawing half of the current. BUT we are talking of increasing V on the SAME Maslow motors.
The trick is that the 24V motor would have twice as many turns of wire than the 12V motor therefore matching the magnetic field of the 12V motor (same torque). The wire used to wind the armature on the 24V will need to be longer to avoid an increase in current and can be thinner due to lower I.
Probably the motors on that table are winded different.
Hope it helps
Rather than spamming this thread, if you don’t mind helping me understand, I’m going to take this offline…
2 Likes
Sure. Just in case some one else is reading along. I believe I finally get the misunderstanding here.
Your statement that a 24V will draw half of a 12V current while keeping same torque is CORRECT. But for this to happen we need to have two different winded motors.
I am only referencing to the original idea of increasing the voltage on the current Maslow motors to from 12V to 24V (same motor/winding)
Hope it helps
P.S. feel free to PM me as many times as you want.
Well, I think I know where the issue is, let me know if this is correct… if the motors are powered by 24V, to make the motors turned at the same speed as if they were powered at 12V at 100% duty cycle (let’s say its 2A), the PWM would have to be 50% such that the same 12V is applied (and that was the thing I was missing) and therefore the same current (2A) gets applied. Not half, I stand corrected… I think it’s because I misunderstood your comment about current doubling. I thought you meant the current would double for the same speed but are you actually saying that you mean that the current would double if it ran at the same duty cycle (100%)… which would make the motor run faster. Sometimes you cant see the forest for the trees… or whatever that saying is.
2 Likes
You are correct
For this statement to be true we need a motor with different winding (longer). I am referring to the stock motors only.
Right, assuming current is linear (not chopping it with PWM).
if we have:
12v 2A at 100% duty cycle, therefore we could say we will have 1A at 50% duty cycle.
So, if we have
24V 4A at 100% cycle we could say we will have 2A at 50% cycle
Current will be double at any PWM duty cycle.
If the objective is having the same power output as the stock ones with half current, we need a different winded 24v motor.
Hopefully we can finally put this to bed
Cheers
Gab
1 Like