I = V/R
If V increases and R is constant then I is higher.
I = V/R
You are increasing the voltage without changing anything else, why “current will have to drop”??, that’s not how it works.
Unfortunately, since R does not change, a higher V will result in higher P (V*I) hence doubling the V you are also increasing P and I
I doubles with double of voltage
If P is constant and V goes up, then I will have to drop.
P = IV
200 W = 100 V * 2 A
200 W = 200 V * 1 A
this is why 220 wires can be smaller for the same power motor. This is why when computer chips go from 5V to 3.3V volts they use more current.
You are correct though about ohms law and I was incorrect that if resistance is constant and V goes up, then I will go up for the same R.
A motor rated at up to 24V, so are we getting that RPM? Thinking about the g-code to test, but neither a straight down nor a horizontal cut. Have to dig back in the “pull tight” to get the max for this test I guess.
this old posts said motors/gearbox go faster and are warmer run at 24v for testing only, the first post above that says 41rpm at 24v, so no we are not getting that rpm .
You are taking R out of the equation.
12v * 0.2A = 2.4W
12V / 0.2A = 60R
now increasing V without changing R
24V / 60R = 0.4A (note double the I at the same R)
24v * 0.4A = 9.6W
I agree to disagree with this statement. Our motors have been able to break a lot of stuff and even where capable to make my bone marrow vibrate during debugging. 2 of them will take care of heavy sleds at ease.
Lol… didn’t mean to start a debate. I was trying to say that with a higher voltage, only half the current is needed for the same feedrate. Therefore, the motor controller has more headroom and maybe the feedrate could be increased.
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.
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.
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.