19v to Rambo?

I have the RAMBo v1.4, and it came with a 12v power brick. Can I just swap it for a 19v brick? Has anyone done this?
At the very least I may need to reduce motor current to prevent too much power from going through them.

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Well, I know you can use 24V. I believe that you could use a 19V.

You will not need to adjust current. That’s what the drivers do, is to adjust the current going to the motors to be appropriate.

The internal electronics all run on lower voltage, already regulated down from the 12V, so they ought to be able to handle the higher voltage. As I said, the board will take 24V

Unless you are having trouble with the 12V supply and you have the 19V supply ready to hand, there’s little reason to do this. Higher voltage will allow the motors to move a little faster, but won’t give you more power at the speeds we use them.

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What is the procedure to adjust it if I wish? I noticed even on 12v they get very hot. Is it done with a M906, M907, M908, or in Configuration_adv.h?

For the RAMBo, I believe that this is in Configuration_adv.h

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Once you flash it, the current will not change by recompiling. On the rambos, it should be M907.

They are constant current drivers. We are used to constant voltage stuff. So normally, we worry that we have a 5V supply for a pi, with at least 1A of current capacity.

On a constant current system, the power (and heat and torque) is all based on the current setting and the voltage just needs to be high enough to reach the set current. The only time a 12V will be any different than a 19V is when the motor is spinning fast, and the motor resists the voltage enough that 12V is not enough.

So, if your motors or drivers are too hot, reduce the current setting in the firmware. The source voltage, 12V or 19V doesn’t have an effect.


For a wire the efficiency conversation is all about the amps. But I believe for a stepper driver, it’s about the watts.
12v 1A = 12 watts, 19v 1A = 19 watts.
So assuming 90% efficiency in both cases, there is an extra 0.7 watts of heat generated when running the higher voltage/power.

The driver chips set constant current, which for a given load, means a constant voltage (up to the supply voltage.)

If my drivers are set to deliver 1A current, they will adjust the voltage until it is enough to supply 1A current, provided that there is enough source voltage from my supply to do this. I think that if I were to scope my driver output with the motors that I have, this would end up being less than 3V whether I supply them with 12V or with 24V.

Where higher supply voltage comes in is at high speed. Because of the stepper motors design, at speed they generate a back voltage caused by moving permanent magnets past the coils. This is called “back EMF” or backwards electromotive force. The higher the speed of the motors, the greater this generated power is, and your power supply has to overcome it. The driver chips compensate for it automatically, as much as they are able.

So once at speed, the net voltage that I have to supply is STILL 3V, however there might be 6V of back EMF, so if my supply does not have at least 9V available, I’ll start to lose torque. Go fast enough, and I might get 9V of back EMF, and now my 12V supply gets marginal. It might start to lose torque because the driver chips regulator isn’t 100% efficient, and it can no longer supply the entire 1A of current.

Back EMF can be considered like drag. It can never fully overcome the driving voltage, but it can approach it, until the differential can no longer overcome other friction or drag on the motor.

So the higher voltage does not result in more motor power, except at high speeds where back EMF reduces the effective supply voltage. My printer has the motors turning at very high speeds, (It’s all using leadscrews so about 4X the RPM that we’re using with 16T pulleys, plus print speeds of up to 90mm/s, rapid moves up to 120mm/s) We aren’t approaching those kinds of speeds with the MPCNC, and so it’s doubtful that you’d see any difference whatsoever with the same defined current to your motors.

The source voltage will have zero effect on the motor heat, that’s all about dissipated power through the motors. The driver chips themselves dissipate power, too, but it depends on the driver, and the regulation transistors. Their efficiency curves more based on switching duty cycle rather than actual current. Hard to predict the increase in TDP for a voltage change

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It’s the stepper drivers that I am worried about, I have no concerns about the motors. It’s the voltage difference across the stepper drivers that worries me. Perhaps what I am saying makes more sense if we just think of them like power regulators. Take the humble LM7805 for example. Feed it 6 volt with a modest load, and it runs cool. Swap to 24 volts input, and it cooks.

In this case, should be fine though, since the board is rated to run everything on 24V. I get that there is a change in TDP through the stepper drivers, but that’s still more closely related to current than voltage.

For the power calculation, you’re looking at a voltage delta rather than an absolute, and input current drops with a more advanced regulator for the same output current, since it tries to match input and output wattage. Therefore the voltage delta across the regulator is always kept to a minimum for this configuration, and the major factor in TDP is amperage. Try to push 1.4A through an LM7805 fed with 6V, and it’ll still do some pretty terrible things. Run the same thing with a simple buck-boost, and it’ll dissipate a few mW of heat, and maybe a few more at 24V. and only take a few hundred mA of source current. You lose more power to internal resistance pushing the output current than you do to the actual regulation.

Thanks Dan. I agree, it should be fine, I will give it a go. And thanks to Jeffeb3 for the M907 tip. I will research how to use M907 if it runs noticeably hotter than it does on 12v.

This isn’t right, because the way the driver achieves 1A is by turning the source voltage on and off. If the duty cycle is 40% at 12V, then it will be closer to 25% with 19V. The resistance doesn’t change (on a stationary motor) and the power will still be IIR in both cases.

This is because the LM constant voltage supplies just dissipate the extra voltage as heat. This is closer to a buck converter.

Doubling the voltage will ensure the power is connected at half the duty cycle. The average current is constant, and independent of the source voltage.

The one place where a 19V will make any difference is if the stepper is spinning fast. In that case, it is effectively creating a reverse voltage. So the effective source voltage may be low, like 2V. In that case, the driver may want to output 1A, but it can’t at 12V, even with 100% duty cycle. At that point, the 19V will have 7V more headroom, and can achieve 1A with a duty cycle of say, 80%.

The average current is constant, through the wires and the coils, the voltage varies with motor speed.