Is it possible to design a heat block without cartridge heater?

My idea is to build a very small heat block to increase/decrease the heat as fast as possible. The resistance of the heat block will be used. The current to this block is 500mA and is set constant with a circuit. The voltage will be set with pwm. Is this possible with 500mA and 5V (2,5W)?

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    $\begingroup$ By the tone of your comments to the replies given insofar, is unclear what you are after: if the question is merely "is it possible to heat a piece of metal by passing current through it?" then the answer is "yes" (duh!). If the question is: "is it a design that makes sense on a FDM printer?" then the answer is "no". Yours seems like a bad solution looking for a non-existent problem. If this is not the case, please provide more context on why you would need a heat block that warms and cools faster. :) $\endgroup$ – mac Feb 23 '18 at 18:04
  • $\begingroup$ I want to make a usb powered printer. The power of usb is not much because of this i want it very small. I want only use 500mA and 5V for the heatblock. This is for me a existent problem. To heat a piece of metal fast enough to a given temperature with less power. Is this enough information? $\endgroup$ – user8886193 Feb 26 '18 at 10:54
  • $\begingroup$ I think a question on the lines of "Is it possible to use USB as a power source for the hot-end" could have been a better one than the one you asked, then. :) Anyway: I think you should go in the opposite direction than the one you chose: what you are after with a low-power source is an extremely well insulated hot-end with enough mass to hold the temperature ~steady when the extrusion speed spikes (as for example during infill). Up to a point, you can also compensate the limitations of such a low-power source by reducing drastically the print speed. $\endgroup$ – mac Feb 26 '18 at 22:56

The current to this block is 500mA and is set constant with a circuit. The voltage will be set with pwm. Is this possible with 500mA and 5V (2,5W)?

This means that the resistance of the heat block would have to be, by Ohm's law, 10 Ohms. You can't set voltage and current independently of each other, which it sounds like you're attempting to do.

2.5W is also rather low power, considering typical heating elements are 25W or even 40W. ABS plastic has a specific heat of ~1300 J/(kg K). Typical printing speeds are 20 grams/hour, so if you have to heat up the filament by 200 degrees C, you need a minimum of 1.44W. 2.5W leaves awfully little room (~1W) for losses due to convection or radiation.

  • $\begingroup$ The resistance is not important. The circuit (current source) will set the current to 500mA and the voltage is set with pwm to change the power. What do you think is it possible? $\endgroup$ – user8886193 Feb 23 '18 at 15:48
  • $\begingroup$ @user8886193 Unless I am misunderstanding your proposal, that is not how electricity works. You can't "fool" Ohm's law by combining a constant current source with PWM. $\endgroup$ – Tom van der Zanden Feb 23 '18 at 16:03
  • $\begingroup$ Fow example line Regulation (Variable Line Voltage) and before that a pwm and capacitor. $\endgroup$ – user8886193 Feb 23 '18 at 16:13

I'm afraid the idea itself is questionable. :)

Part of the reason for having a heat block in the first place is to leverage the volumetric heat capacity of the block to maintain the temperature constant even thought the extrusion speed (and thus the rate at which energy is used to make plastic warm and change state) is not.

A smaller block would probably:

  • require a lot more power to operate in order to compensate for the missing thermal inertia (most PSUs are already "stretched" when it comes to power requirements)
  • cause the temperature to fluctuate, with negative effects on the print quality

A second problem I see is that you would most probably electrify the whole printer:

  • creating a hazard
  • affecting the ground level and thus - potentially - the proper functioning of all electronics
  • $\begingroup$ There will be no connection (metal to metal). The heat block will have no connection. I will use a part in between to protect against voltage and heat. I think i will not need somethink to cool. $\endgroup$ – user8886193 Feb 23 '18 at 15:52

2.5W of electrical energy defines the heating rate for a specific mass (and thermal capacity of the material). It also determines the highest attainable temperature for a specific emissivity (clue, it won't get hot).

Any switching circuit to match the resistance of a block of metal to a 2.5W power source is a switch mode power supply in disguise. Yes, you can generate 150mV at 16A, but you need very thick wires to avoid loosing most of your generated power in the circuit.

  • $\begingroup$ If i transform the given 5V and 500mA to 150mV at 16A is the current the cause for the heat? Then it will be possible to just increase the current and set the voltage down to set the speed of heat generation and the opposite way. $\endgroup$ – user8886193 Feb 26 '18 at 16:31
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    $\begingroup$ no. it doesn't work how you want it to. $\endgroup$ – Sean Houlihane Feb 26 '18 at 16:32
  • $\begingroup$ Did you calculate it? My idea was to make the heatblock as small as possible (Alumimium) to save power because of less power and to indrease the temp as fast as possible. $\endgroup$ – user8886193 Feb 26 '18 at 16:54
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    $\begingroup$ @user8886193 An aluminium heat block will not have enough electrical resistance to make a practical heater. You can't control voltage and current independently - it's all determined by the resistance of the heat block and Ohm's Law. It simply will not work. If the heat block has the same resistance as the wires connecting it to the power supply, then those wires will heat up just as much as the heat block itself no matter what voltage and current you use. The resistance of the heater needs to be (much) higher than that of the wires. $\endgroup$ – Tom van der Zanden Feb 26 '18 at 21:34

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