Simple heater question SOLVED
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Yup that correct - but as you say - make sure to have a proper power rating for the resistor -
Alternativ, change the voltage - Run a 24V catridge at 12V will require no extra resistor, and will run on half the power.
If you need to use 24V on the printer, use a step down converter (buck converter) - -
@martin1454 said in Non standard heater question - quick check on my sanity:
Run a 24V catridge at 12V will require no extra resistor, and will run on half the power.
It would run at one quarter of the power (half the voltage and half the current).
Duet WiFi hardware designer and firmware engineer
Please do not ask me for Duet support via PM or email, use the forum
http://www.escher3d.com, https://miscsolutions.wordpress.com -
on e3d you get 30 and 40w heaters for 12/24v - ca. 5,-
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@dc42 said in Non standard heater question - quick check on my sanity:
It would run at one quarter of the power (half the voltage and half the current).
Oh, yeah true -
But since you are here @dc42 - It would be possible to use a buck converter, since the buck converter puts 24V down to 12V, but it is GND that is switched on the duet heater channel, so that would be no problem right?
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@martin1454 said in Non standard heater question - quick check on my sanity:
@dc42 said in Non standard heater question - quick check on my sanity:
It would run at one quarter of the power (half the voltage and half the current).
Oh, yeah true -
But since you are here @dc42 - It would be possible to use a buck converter, since the buck converter puts 24V down to 12V, but it is GND that is switched on the duet heater channel, so that would be no problem right?
You could use a buck converter to drive the positive heater wire, instead of connecting that wire to the heater output terminal on the Duet. Do not connect the heater output of the Duet to the input of a buck converter.
If you are looking to reduce heater power, you could also consider turning the VIN voltage down. A 20% drop in voltage causes about 36% drop in power. But of course all heaters and fans running from that power voltage will be affected.
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Biqu sells 30W 12V/24V as well as 50W ones
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As per my OP, "standard" heater cartridges are cheap as chips and readily available - I already know this.
I also have a supplier lined up that can make me non-standard cartridges of pretty much any wattage (at a price). So first I need to evaluate what the heating requirements are and for that I'd need a range of heaters between say 15 and 40 watts in 5 watt increments.
So, if anyone can point me to a link where I can buy 15, 20, 25, and 35 watt @24V heater cartridges for about £1.00 to £2.00 each (the cost of a 50 Watt power resistor) , then please do. Otherwise I'll stick to plan A and use series resistors.
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To halve the power, a 14ohm resistor would do the trick, but will also waste half the power at the resistor. The result would be 10W of heat at the heat block and 10W of heat at the series resistor.
Using a voltage divider network, you could reduce the voltage "seen" by the heater to about 17V (24 * √2) which will halve the dissipated power in the heat block, which is more likely what you want to do. the resistance that the current flows through can be much lower, therefore it will dissipate less power.
A buck/boost supply which will support the appropriate current (about 1.25A) should be useable to supply the positive rail, so long as it's fed a different ground, and is probably the cheapest option.
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Addendum: If you want to make it adjustable power output, then a power regulator is really the only way.
So if you don't mind throwing away the power (Have a high wattage resistor strapped to a heat sink in series) then your 14 ohm one will give you 10W at the heater. You figure the dissipated power for the overall circuit, then calculate the voltage across each device in the circuit, like a standard voltage divider, then work out the power from there.
For example a 10 ohm resistor in series would give a total series resistance of 24 ohm, TPd is 24W with 14W at the heater cartridge and 10W at the resistor. (simple case) this is probably close enough to your 15W case (9.85 ohms gives you close to 15W, if I've done my math right.)
3 Ohm in series gives you 17 ohm total series resistance, TPd is 34W, with 28W at the heater cartridge and 6W at the resistor.
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Seriously guys, all I want to do is a quick and cheap evaluation. Basically just a heater tune with various wattage heaters.
Bunging a series resistor in, is both quick and cheap. It'll cost me about £1.00 to £2.00 per resistor. I don't care about wasting half the power for a quick 5 minute test. I don't want to spend money on a buck/boost supply that I'll never use again. Nor do I want to build an adjustable power regulator. Nor do I want to buy heater cartridges from China that I'll only use for one quick test at £10 a pop.
Not wishing to be too blunt and thanks anyway, but my question was answered by @Martin1454 at 9.30 this morning.
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One last solution (based on @SupraGuy's ideas) which is both cheap and allows you to modify the wattage of the heater would be to buy a 10-pack of LM2596 based buck-down converters together with a 10-pack of small heatsinks.
They come from China and take a while but cost less than 10 bucks altogether (I assumed you wanted to test all the heaters (you talk in plural in your OP) at the same time. The advantage is that you can adjust the power by regulating the voltage instead of swapping out resistors.
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All I was doing there is reminding you that the half the power you waste doesn't go into the heater, so the answer @Martin1454 gave you wasn't correct -- at least not completely. I was assuming that you wanted to simulate the various wattages at the heater block.
High power resistors seem to be cheaper where you are. Around here, as soon as you get past 5W, they start getting about the same price as the odd wattage heater cartridges, and they aren't available in the same kinds of values.
If I were doing this, I'd probably end up using a few 5W devices and series/parallel them to get the resistance that I want. High wattage devices aren't available in better than 5% precision, and if you're lucky, you can get them in E12 series values. (Or completely screwball values that are obviously for specific purposes.) As such, it's going to be somewhat tricky to get the specific values for 5W increments
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Why do you care what power it is? It's all PID controlled anyway, by using a lower power heater it's just going to take longer to heat up and have a harder time holding temperature.
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@surgikill As he said in his OP he needs to evaluate something.
Manuel
Duet 3 6HC (v0.6) with RPi 4B on a custom Cartesian
with probably always latest firmware/DWC (incl. betas or self-compiled)
My Tool Collection -
@wilriker Would help if we knew exactly what was going on.
Also, if all is he looking to do is change power at the heater block he can change the PWM duty cycle and get the same results. If he's looking to increase power draw on a FET or something then he would need some power resistors.
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@surgikill I cannot disclose any details but he needs to find an appropriate amount of heating power that suffices his project.
But you are right. PWM of the heater connectors could really work in this case easiest. Since this is purely about evaluation and not something trying to use a 12V heater on a 24V
VINthis will probably work without any additional hardware.Manuel
Duet 3 6HC (v0.6) with RPi 4B on a custom Cartesian
with probably always latest firmware/DWC (incl. betas or self-compiled)
My Tool Collection -
@wilriker Well, I have the equations here if he's interested. It's cheaper than going out and buying power resistors too.
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@surgikill said in Simple heater question SOLVED:
Why do you care what power it is? It's all PID controlled anyway, by using a lower power heater it's just going to take longer to heat up and have a harder time holding temperature.
How do you know that when I haven't given any details about the application? You know nothing about the thermal masses involved because I haven't given any details. You know nothing about the desired temperatures and desired rise times because again, I haven't given any details. You know nothing about the flow rate and temperature differential of any incoming fluids because I haven't given any details. You know nothing about how many heaters are involved and what their configurations and orientations are, with respect to each other and with respect to the design as a whole.
So how do you know that lower powered heaters are going to have a hard time time holding temperature? Or I just use a high powered heaters and rely on PID? And what happens when a FET fails and the PID gets stuck on full power and burns someone's house down?
And I'm not going to provide that information because this is a prototype design of something that may end up as a marketable product and I don't want to give out any details at this time. The only thing that is relevant to this thread is my first sentence, quote " I have a need to evaluate some heater power requirements." And my question was answered in the first reply.
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