Chosing resistors for heating

I've been messing around with resistors alot and made some discoveries that are probably obvious, but just thought I'd throw them up for people to know who have absolutely no training in electronics.

-The big ceramic resistors tend to just get longer and longer for an increase in their wattage. Given it's generally found that longer melt regions are worse, it's generally a good idea to try to arrange things so that you use lower wattage resistors.
-A 5 watt resistor will not always dump 5 watts, thats just the ceiling value where the resistor will break down if you go above. A 5 watt resistor that has only .1ohm might only throw off half a watt or something.
-Lower resistance resistors do not throw off as much heat as a higher ohm resistor, even if it's rated as a 5 watt resistor a 3.3 watt will give off much more energy than a .1 ohm.
-A basic minimum cutoff to the temperatures that we get to are about the 3.3ohm resistors. A little bit above this is where they get practical for us.
-Low ohm resistors, even in series will trip out the circuit breakers in the curcuits.
-As a rough estimation, by just throwing the resistors into the heater driver in the gen 3 electronics in series yields: A 3.3 ohm resistor takes about 30 seconds to get to 100C, while a .1 ohm simply doesn't get there within 5 minutes.

Edited 1 time(s). Last edit at 07/24/2009 03:26AM by letsburn00.

Hi, just thought I'd point out that some of those points are only true for some setups. With a constant voltage across it, the lower the resistance the greater the heat output. Your observation to the contrary shows that your drive electronics are not providing a constant voltage source. As the resistance decreases, more power is radiated from the transistor (I assume that's what you're using?), and less from the resistor. However, with better transistors, until you get down to small fractions of an ohm, heat output will increase with reduced resistance. Your point about a 3.3 watt resistor giving off more energy than a .1 ohm resistor does not make sense - resistance and rated wattage are two completely unrelated properties. Incidentally, rating resistors by maximum wattage is fairly pointless, it is the high temperatures which occur at these wattages which do the damage.

I've been using two 12ohm resistors
(http://uk.farnell.com/jsp/search/productdetail.jsp?sku=1109309)
in parallel to make a single 6ohm heater. Both resistors are embedded into a single block of mild steel, wrapped in foil and forced into 7.5mm holes.

[picasaweb.google.co.uk]

This takes a few minutes to warm up to 220C, but once there is much easier to maintain at that temp because of the thermal inertia of a big block of metal.

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I agree on the nonsensicalness of the rating, but the are sold as 5 watt resistors, and I figured I'd note that given if people are buying them then thats the basic way it'll go.

Lower rating resistors giving more heat out doesn't really make sense to me. Given I figured that the resistance that the resistor is creating has to go somewhere. And that is given off as heat. So more resistance for a constant amps and voltage would give out more energy. But I don't really know about this stuff (I'm mostly taking these guesses from my knowledge in heat transfer)

"Given I figured that the resistance that the resistor is creating has to go somewhere. And that is given off as heat. So more resistance for a constant amps and voltage would give out more energy."

There is a simple relation between voltage, current (Amps) and resistance (Ohms).

Volts = Amps * Ohms
(Corrected. I originally wrote Amps/Ohms which is wrong. Sorry)

There is also a simple relation between voltage, current and power (Watts):

Watts = Volts * Amps

From that you can see that for a given voltage, the current goes up as the resistance goes down and the wattage goes up as the current goes up. This means that the power dissipation (heat production) of a resistor goes UP as the resistance goes down FOR A CONSTANT VOLTAGE.

You can use the formulas to determine how low a resistance your power source can stand. For example if you have a 10 Volt source capable of 5 Amps you can produce 50 Watts of heat and you can use a resistor as low as 2 Ohms.
If you use a 4 Ohm resistor it will only let 2.5 Amps of current through and will only put out 25 Watts (10V*2.5A) of heat.

The power RATING of a resistor has nothing to do with how much heat it produces at a given voltage. The rating just tells you how much heat the package is guaranteed to stand. It IS true that higher power rated resistors put out less CONCENTRATED heat. That's how they stay intact at the higher power levels, by spreading out the heat.

Edited 1 time(s). Last edit at 08/25/2009 09:35AM by JohnWasser.

Well put, I would just like to add a couple more points:

Power is the rate of flow of energy, and energy into a system and out of a system must be the same, therefore the electrical power in must equal the heat power out.

The temperature of the resistor will however vary on the magnitude of the input power, and the rate at which the resistor can dissipate heat. The only difference with 'higher power' resistors is that they are able to dissipate heat more quickly, hence the highest power ones are enclosed in aluminium heatsinks.

The power rating is only the power rateing this becomes invalid as soon as we put the device in a lump of metal. The rating will have been made based on normal use.

We can only use the rateing as a rough guide as to the devices garantted maximum power capability. The way its mounted in the block the block material etc will significantly change its effective power rating I guess mounting in a block will increase its power rating. ~Hower wrapping insulation around t extruder and block he will have the oppiset effect and be effectvely de
decreasing its maximum power capability.

So it better to only use the rating as general guide.

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JohnWasser Wrote:
-------------------------------------------------------
> There is a simple relation between voltage,
> current (Amps) and resistance (Ohms).
>
> Volts = Amps / Ohms
>
>
> From that you can see that for a given voltage,
> the current goes up as the resistance goes down
> and the wattage goes up as the current goes up.

PLEASE edit your post with the correct formula...
It should be Volts = Amps * Ohms!
Or V = R*I in short notation.

We don't want to redefine the ohm's law here, do we?
[en.wikipedia.org]

Protonias Wrote:
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> PLEASE edit your post with the correct formula...
> It should be Volts = Amps * Ohms!

OOPS! Thanks for pointing that out. I have made the correction.

This post is from almost 5 years ago. But for those who can help I made reference to this post in another thread: Help replace dead hotend heater (Ohm's Law and Rated Wattage)