What is the power consumption of the heatbed and hotend?
What is the power consumption of your heatbed (size) and hotend (model)?
I want to verify that it is possible to use a battery to power them.
With a sufficiently large battery you can power anything! :)
I have printed using a massive car battery when my original PSU blew. I wanted to print so bad so I just connected it and went with it. It worked fine but do keep a charger on it because it drains rather quickly.
I have a Kill-A-Watt meter so I got a pretty good measurement for you with my Anet A6. Like Petar said each model is different but this should give you a idea. When heating both the nozzle and heat bed the printer consumes 160 W of power, once to temp it backs down to 9 W (it also uses 9 W when just "sitting doing nothing and is on"). When the nozzle and bed get down in temp it hits back up to 160 W. Basically it is never a consistent heating, it is on and off. Like a refrigerator.
When it comes to heating only the nozzle the printer uses 60 W (so 51 W is going to the nozzle for heating).
When it comes to heating only the bed the printer uses 142 W (133 W to the bed).
This is interesting because it would make sense the printer needs more than 160 W when 51 W is going for the nozzle and 142 W going to the bed, that makes 193 W. I make mention of this because that may suggest my power supply is not big enough and the printer could really use around 200 W.
As a little bonus when the printer is moving around (stepper motors are active) I find it using 35-40 W (or 26-31 W) to power the steppers.
So with all the said, is it possible to use a battery? Yes, you could. And to give a example a car battery should have 80 Amp-hours (or something like that, but we will go with it). With that battery you can get 960 Wh (Watt-hours) from the battery before it dies. Going with my printer using 160 W I will get 6 hours of printing time. But keep in mind as the battery is used the voltage will drop, so in the end the printer will be getting something like 10 V which I am sure will affect heating and overall performance.
Last thing I feel that needs to be said. If using a inverter to convert the 12 V battery to 110 V (or whatever voltage you use) a cheap one will not be healthy for the printer. Cheap inverters put out square waves instead of sines waves. Basically it will hurt the printer. You can learn more at this WEBSITE
"Update" on March 4
I read a comment that mentioned running right off the battery without a battery and then I thought of something that I did not think of before. And that would be protecting the battery itselfSo I said you can run the printer off the battery. There was one issue that I had not thought of. And that was the voltage drop and the battery discharged. A battery usually does not have voltage-cut off to keep the battery from being overly discharged, and a printer does not have anything to measure voltage (why should it). So a simple hook up of a 3D printer to a battery is prone to drain the battery much lower than 10 V, which will greatly shorten a battery life-span. This can be prevented two ways.
A circuit between battery and 3D printer. There is plenty of circuits that can be bought as long as they cut power to printer at 10 V or something (for lead acid anyway) and can handle the amperage draw.
An inverter can also be used because this voltage cut off is already in them. But remember that square waves are bad for the printer.
At least for Cura output and my kit's firmware, the bed and the hotend are never both in the "full heating" mode. This limits the max current draw for teh system as a whole, which I see as a good thing.
@CarlWitthoft I use Craftware and it does heat the bed and nozzle at the same time (but can be disabled). I have noticed it simply takes longer when doing both. But once one is done, being the nozzle or bed, it focuses all the power to the other component that needs to heat. And I do agree that it is good to limit the max draw to the printer, I will should disable the heat simultaneously option.
I really want to take a printer out to my off-the-grid shed and see if I can sustain it. If I assume a 50% duty cycle of heat on/off, that'd be about 8-10 amps/hour. Hmmm. (no inverter necessary)
@tedder42 You are correct with 8-10 amps per hour. And I recommended a inverted because the printer does not know when to cut off, so the battery voltage will go well below what is recommended, which is around 10v or something. A inverted does have a voltage cut off and will stop the printer from 'wrecking' or shorting the life span of the battery. But because the person was asking if he could run it off a battery that is why I answer "yes". I should has put more detail about why a inverter is recommended, which I may do yet.
thanks. Seems it'd be a lot more efficient to do a low voltage switch/sag protector than to invert. Depends on the circumstance, though.
@tedder42 It would be. But I honestly didn't put time into thinking of the best way to protect the battery. It will be interesting how it goes running the printer off-grid.
Note that "watt" is a measure of "energy per second" and "amp" is a measure of "charge per second". Saying "amps/hour" makes no sense, because "amp", in and of itself, already contains the notion of "per time unit". The printer simply draws 8-10 Amps (not "per hour" or anything like that). Battery capacity is measured in Amp-hours (an 80 Amp-hour battery can supply 80A for 1 hour, or 40A for 2, or 1260A for 30 minutes), and the total amount of energy a battery can deliver is measured in Watt-hours.
@TomvanderZanden Thank you for the correction, I should have been thinking about that when I was adding but that slipped through my mind. But it is a really good thing you noticed :)
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Carl Witthoft 5 years ago
At the very least, identify your printer, your intended bed temperature, extruder temperature, and length of print time. A big enough battery can power an entire town; a small one can't keep a LED lit.