[SOLVED] Reducing heat on heating element

[Mattman4494]

Here is my question: currently I have a circuit with a 30 watt heating element and a dimmer switch that plugs into a 110v wall outlet. With no resistance from the dimmer, the heating element can reach around 700 degrees Fahrenheit. I am wanting to limit the max temp of the heating element (with no resistance from the dimmer) to around 450 degrees Fahrenheit by adding resistors. Unfortunately I have very little knowledge on this matter, so I am turning to you all for some help ;-)


Please let me know if I am leaving out any important info, and thanks for any help!!

 

[chuckey]

So 30 W gives 700 F, 0W gives 60 F (room temperature). So for 30 W gives increase of 640 F, you want increase of 390 F, so you need 390/640 X 30 W = 18 W. At 110V, 18 W gives R = 110 X110/18 = 672 ohms. You have 110 X 110 /30 = 403, so you need to add 269 ohms to the circuit.
Frank

 

[keith1200rs]

Using a series diode might work but without the power loss.

Keith

 

[Mattman4494]

Thanks Frank for the info, that is extremely helpful! Would the resistor need to be rated for 18W then? Keith, could you possibly point me site that sells series diodes that would be applicable to my senario?

Matt

 

[Syncopator]

Using a series diode might work but without the power loss.

That would reduce the applied voltage to half its normal value, and therefore the power to a quarter. That's probably going to be too little to achieve the desired temperature.

 

[keith1200rs]

If you want to try a a diodes, something like the 1N4007 would be fine. The effect will depend on your "system" as heat loss is not necessarily linearly proportional to temperature difference. Any series resistor could be similarly affected and the value might need adjusting.

Keith

 

[chuckey]

Sorry, Syncopator, but what will happen is the power will be halved as alternate half cycles of the sine wave will be missing. So it would give 15 W, a bit low, but it certainly will not dissipate any power so its very efficient. A sneaky way to increase the power would be to put a resistor in parallel with the diode to provide the extra 3W during the non conducting part of the cycle. So the extra 3W would be equivalent to 6W for a complete sinewave. so 6W into 403 ohms is a current of (6/403)-1/2 = .122A, so volt drop across heater V = 403 X .122 = 49.1 V, so voltdrop across extra resistor = 60.9V @ .122 A = 500 ohms. So with a 269 ohm resistor it would dissipate about 20W thats a BIG resistor. With the diode +resistor , the resistor would dissipate 3.5 W and you should get your 18W.
Frank

 

[keith1200rs]

but what will happen is the power will be halved as alternate half cycles of the sine wave will be missing.

Frank,

You are correct. I have used a diode before in a couple of cases where I needed to reduce the power of a heating element. Quick and cheap.

Keith

 

[Mattman4494]

I think 15W will be just fine. One last question, will I be able to wire up the said diode straight into my circuit without any other components? Once again thank you for all the info guys... I really appreciate it

 

[kripacharya]

I think 15W will be just fine. One last question, will I be able to wire up the said diode straight into my circuit without any other components? Once again thank you for all the info guys... I really appreciate it

yes of course. just connect in series on any wire to your heater. Its a no-brainer

 

[andy93]

Sorry, Syncopator, but what will happen is the power will be halved as alternate half cycles of the sine wave will be missing. So it would give 15 W, a bit low, but it certainly will not dissipate any power so its very efficient. A sneaky way to increase the power would be to put a resistor in parallel with the diode to provide the extra 3W during the non conducting part of the cycle. So the extra 3W would be equivalent to 6W for a complete sinewave. so 6W into 403 ohms is a current of (6/403)-1/2 = .122A, so volt drop across heater V = 403 X .122 = 49.1 V, so voltdrop across extra resistor = 60.9V @ .122 A = 500 ohms. So with a 269 ohm resistor it would dissipate about 20W thats a BIG resistor. With the diode +resistor , the resistor would dissipate 3.5 W and you should get your 18W.
Frank

well spotted ... but don't forget the power verses temp of this element won't be linear , I would guess you would only need about half the power to keep it at 450 as opposed to 700 . Another factor no one has mentioned is drop in resistance at lower temp. if iron wire the resistance will about halved , nichrome about 8%drop , so with a diode must get more than 17w, I would think just right ......... Mattman please put a diode in there and tell us the temperature.

 

[Mattman4494]

well spotted ... but don't forget the power verses temp of this element won't be linear , I would guess you would only need about half the power to keep it at 450 as opposed to 700 . Another factor no one has mentioned is drop in resistance at lower temp. if iron wire the resistance will about halved , nichrome about 8%drop , so with a diode must get more than 17w, I would think just right ......... Mattman please put a diode in there and tell us the temperature.

Ok I ordered some diodes and I will let you know how it turns out

 

[andy93]

So 30 W gives 700 F, 0W gives 60 F (room temperature). So for 30 W gives increase of 640 F, you want increase of 390 F, so you need 390/640 X 30 W = 18 W. At 110V, 18 W gives R = 110 X110/18 = 672 ohms. You have 110 X 110 /30 = 403, so you need to add 269 ohms to the circuit.
Frank

Greetings Frank ... I hope you won't take offense at a newbie respectfully pointing out 3 errors in your post. First the temp/power curve of this element won't be linear , that's to say the power needed to raise the temp from 639 to 640 is much greater than from 60 to 61..... I would guess about 12w might be enough. ..... Your element and resistor in your calculation will draw 18w , but 7.2w of this is heating up the resistor so the element is only getting 10.8w . Lastly the resistance of the element will be less at the lower temperature, if made from iron wire about 50% less , nichrome about 8% less so that needs to be accounted for......Andy