# Methods to reduce wall power current?

Oct 8, 2013
24
I have been working to modify an air popcorn popper so that the temperature of the heating element is a little lower so I can roast coffee. I have tried putting a few different things in-line with what is labeled "Thick white wire" on the schematic. I first tried a household light dimmer. Unfortunately, it didn't give me adjustable temperature. It was on at maximum "brightness" and turning it down a tenth of a millimeter (and lower) was off. Due to the price of potentiometers that can handle some wattage ($55+), I then tried a 0.2 ohm wirewound resistor. This worked, although it got very hot, but the temperature reduction was not enough. My latest attempt was with a 25W 0.47 ohm resistor, which seemed to be just about right, but it got way too hot (solder softened). So, now I am wondering about other methods of accomplishing my goal. I am aware that diodes can sometimes be used for this purpose, but I think I've only seen that mentioned in the context of low voltage DC circuits. Same for PWM circuits. In both cases, I haven't seen anything to indicate those are good approaches, or would work at all, for my 115V, 60Hz, 13 amp AC situation. (The Large Heating Coil is about 13 ohms, btw.) Can someone recommend a good approach that isn't too expensive? Ideally it wouldn't get hot at all. I have a general question which I haven't really looked into, which is, why are resistors used so much when their very purpose is to waste power. It's worth some money for learning experiences, even if it isn't for my solution, per se. #### Attachments • Popcorn-Popper.png 24.1 KB · Views: 87 #### Gryd3 Jun 25, 2014 4,098 I have been working to modify an air popcorn popper so that the temperature of the heating element is a little lower so I can roast coffee. I have tried putting a few different things in-line with what is labeled "Thick white wire" on the schematic. I first tried a household light dimmer. Unfortunately, it didn't give me adjustable temperature. It was on at maximum "brightness" and turning it down a tenth of a millimeter (and lower) was off. Due to the price of potentiometers that can handle some wattage ($55+), I then tried a 0.2 ohm wirewound resistor. This worked, although it got very hot, but the temperature reduction was not enough. My latest attempt was with a 25W 0.47 ohm resistor, which seemed to be just about right, but it got way too hot (solder softened). So, now I am wondering about other methods of accomplishing my goal.

I am aware that diodes can sometimes be used for this purpose, but I think I've only seen that mentioned in the context of low voltage DC circuits. Same for PWM circuits. In both cases, I haven't seen anything to indicate those are good approaches, or would work at all, for my 115V, 60Hz, 13 amp AC situation. (The Large Heating Coil is about 13 ohms, btw.)

Can someone recommend a good approach that isn't too expensive? Ideally it wouldn't get hot at all. I have a general question which I haven't really looked into, which is, why are resistors used so much when their very purpose is to waste power. It's worth some money for learning experiences, even if it isn't for my solution, per se.
Things to consider... The popcorn popper... is it digital? Or is it a stupid device with a simple on/off switch?

In order to reduce power to the popcorn maker you need to do one of two things:
-Increase the resistance in the path between wall and popcorn popper.
-Modify the sinewave to reduce the power. (Dimmers and Diodes)

You have noticed already that adding a resistor gets hot... this is no surprise... the heating element is essentially just a low value, high power resistor. So you essentially just added another heating element between the popper and the wall. It will always get hot, there is no way around this unless you increase the resistance mush more which will reduce the heat too much from the popper.

A diode will only allow half the waveform in, which will reduce power coming in. Use it just like the resistor.

A dimmer (depending on the type you used) will typically chop off the front part of the wave, or the back part of the wave so that only a partial wave reaches the popcorn maker.

If the device turned off suddenly, you may not be able to accomplish what you want without modifying the popcorn maker. You may need to go to a local big-box store and buy a simple cheap on/off switch type popcorn popper.

#### Arouse1973

Dec 18, 2013
5,177
I calculate the power dissipated by the resistor to be approx. 34 W so no wonder your resistor got hot. You will need a 50 W resistor mounted to a large heat sink to have a chance of using this approach.
Thanks

#### duke37

Jan 9, 2011
5,364
If the device has a thermostat, then reducing the power input will extend the ON time to compensate.

#### Greg J.

Oct 8, 2013
24
Thanks all for the responses.

This analog popper doesn't even have an on/off switch (ref. attached wiring diagram in the OP).

I forgot to mention that I attached a heat sink quite a bit bigger than the resistor itself, incl. thermal paste, for this latest try. Presumably the temperature wouldn't be any lower with a 50W resistor.

I think two diodes in parallel in opposite directions would pass the wall power mostly unchanged as long as they were within spec, right? Can a simple circuit instead be used to reduce the amount of power being passed through? Since this popper is basically two coil heating elements and a fan, hopefully a modified sine wave wouldn't cause any problems.

Adam, can you give me a link to a page that explains how this calculation is done for my scenario? I hunted all over and didn't find anything that made sense (presumably because I didn't have the background to understand what I was supposed to be calculating).

#### Arouse1973

Dec 18, 2013
5,177
Your correct the 50 W resistor will still get hot. What size heat sink did you use. I don't have a link to a web page sorry, what do you want to know?

#### Greg J.

Oct 8, 2013
24
Your correct the 50 W resistor will still get hot. What size heat sink did you use. I don't have a link to a web page sorry, what do you want to know?
Knowing that Power = VI and V = IR, I wasn't able to logically deduce what power rating the resistor would require. How would I have done that?

#### Arouse1973

Dec 18, 2013
5,177
Knowing that Power = VI and V = IR, I wasn't able to logically deduce what power rating the resistor would require. How would I have done that?

The voltage is 115 V a.c rms. This is the same as 115 V d.c for power loss in a resistor so you divide this by the total resistance 115/(0.47+13) to get the current. Then P = I^2*R = 8.85^2*0.47 = 34.25 W

#### Colin Mitchell

Aug 31, 2014
1,416
Use 4 x .47 ohm in series/parallel

#### Gryd3

Jun 25, 2014
4,098
Use 4 x .47 ohm in series/parallel
Haha. I was waiting for this kind of response.
WT serious F Colin?
You can waste your breath arguing with people, but when you choose to help you only say "Use X" without any explanations. I hope for your sake you don't get someone hurt with your lazy posts.

#### Colin Mitchell

Aug 31, 2014
1,416
"Haha. I was waiting for this kind of response.
WT serious F Colin?"

If you haven't done a 3 months electrical course and know what putting 4 resistors in series/parallel means, don't make a fool of yourself on this forum.

He already said: My latest attempt was with a 25W 0.47 ohm resistor, which seemed to be just about right but got too hot.

Rather than do any major calculations, simply putting 4 x 25 watt resistors in series/parallel will reduce the heat in each resistor considerably.

#### Gryd3

Jun 25, 2014
4,098
"Haha. I was waiting for this kind of response.
WT serious F Colin?"

If you haven't done a 3 months electrical course and know what putting 4 resistors in series/parallel means, don't make a fool of yourself on this forum.

He already said: My latest attempt was with a 25W 0.47 ohm resistor, which seemed to be just about right but got too hot.

Rather than do any major calculations, simply putting 4 x 25 watt resistors in series/parallel will reduce the heat in each resistor considerably.
So let's clarify that then. Do you expect everyone on here to have 3 months electrical courses under their belt to understand your 'helpful' comments?
I know what your post is getting at, but how do you expect everyone else to? Including that last line in this reply in your original post, or something like it in the one-liner replied you leave everyone would go a very long way to actually helping them.

#### AnalogKid

Jun 10, 2015
2,665
Don't knock the 3-months. In my first 2 weeks on this forum, his minimum required background to understand his wisdom was 40 years and 20 years (separate posts on two different threads).

Which raises a question - since quality is not a requirement, what does it take to be a VIP member?

ak

#### davenn

Moderator
Sep 5, 2009
14,043
Which raises a question - since quality is not a requirement, what does it take to be a VIP member?

it's a post count thing

Something I seriously consider needs changing to a different system

Dave

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