# Run 15A 120VAC HeatPress on 15A outlet, limit current to 12A (limiter boxe???)

I want to limite the amount of current a appliance could draw to 12A in order to use it on a 15A outlet without tripping the breaker.

I have a HeatPress (heating element(coil) with a PID controller) working on 120VAC rated 15A with a T-Brace plug for 20A outlet. Link to Heatpress website spec

The previous owner told me he add a electrician make him a adapter cable with a box in the middle to be able to plug in it a 15A outlet without tripping the breaker.(but it was nowhere to be found)

I think I must keep the voltage to 120V to make the PID and maybe other electronic component happy so I dont think a variac can be the answer.

Common theorical sense told me to raise the resistance from 8Ω to 10Ω

120V 15A 8Ω 1800W / 120V 12A 10Ω 1440W

But adding in serial a 2Ω 1440W if I'm right sound in my head like a joke.

So here I need your help, what was in that boxe? Is it possible to limit the current the appliance can draw? I know the heating element will take more time to heat and wont be able to be as hot but that's not a problem for me.

Thank You for you time

Sam

• Sounds like you should adjust the PID controller so that it doesn't use 100% power. That's the first place I'd look. – jonk Nov 22 '16 at 7:46
• Measure the actual resistance. If it's rated 15 A your chances of it consuming 15.0 A ar very low. Moreover, how long do you use it for? Say it consumes 13 A, then your 12 A breaker will take tens of minutes if not hours before tripping. – winny Nov 22 '16 at 8:02
• @jonk. It is a heating system. The PID controller switches ON/OFF. The current during the on state remains the nominal value. The consumed power over time (energy) can be adjusted with the PID. – Decapod Nov 22 '16 at 8:51
• @Decapod I developed PID controllers for heating systems for well over a decade. I don't recall any of them being used as on/off control systems (they can be, but I think rarely are.) We allowed for a max power setting and I seem to recall all of my competition did, as well. (Power output from PID generally isn't on/off. So I'm sincerely flummoxed by your comment on this point.) – jonk Nov 22 '16 at 9:31
• @jonk I'm no expert on PID but in my field(screen printing shop) all the heat press and way bigger conveyor dryer I have open and work on used a PID powered by 120VAC with a thermocouple swiching a internal DC low voltage Relay connected to a SSR DC low voltage input swiching the other end AC high voltage on and off with the current remaining at the nominal value. More like Decapod said and very similar to the way a baseboard heater work in a home. I'm no expert in PID trough but have work on many all wired this way. Any link or information about other way to use a PID will be welcome. – Samuel P Nov 22 '16 at 16:00

The easy way is to find a 24VAC transformer able to handle 12A at the output. Therefore 288VA AC or a little bit more. Connect the primary of the transformer in parallel with the 120V mains and the secondary side (24VAC) in series with the heating element. Beware to connect the secondary side of the transformer in such a way that the combined output voltage of the series circuit becomes 120VAC - 24VAC = 96VAC. This way your PID keeps the 120VAC and the heater can get 96VAC thereby reducing the current to the desired 12A.

You might try to keep the transformer outside the press and feed the same with only 96VAC. It is very well possible that the PID can work properly on 96VAC only

• Nice! Here's a decent candidate with a price (USD 51.83 from Digi-Key) which won't break the bank . – EM Fields Nov 22 '16 at 10:51
• This really look like it could work. I will also look at the other component and PID to see if they can work on lower voltage if yes it might have been a Transformer or a variac that was inside the boxe. It will be more easy to keep it outside and run everything on say 96VAC if posible and also been able to unplug the adapter to come back to full power would be great. Thanks – Samuel P Nov 22 '16 at 16:17

If you lower the current into the press to 12 amperes by connecting a 2 ohm resistor in series with the press's 8 ohm heating element, then the power dissipated by the resistor will be: $P=I^2R = 144\times2\Omega = 288\text{ watts.}$

Something like this would work:

The entire datasheet is available here, and it's in stock at Digi-Key for USD 15.29.

• This look like a possible thing. Will it be possible to put it in serial with the whole unit, dropping the voltage, if the other componnent cant handle lower voltage? – Samuel P Nov 22 '16 at 16:21
• In a word, "no". – EM Fields Nov 22 '16 at 16:27
• I will open the unit tonight and look for a place for this resistor. The first thing that came to my head in the begenning was if it was possible to have a higher resistance element and this look like it. Thank You – Samuel P Nov 22 '16 at 16:35
• @Samuel P: Part 1: Be aware that since the machine's heating element's resistance is 8 ohms, it'll dissipate 1800 watts with 120 volts across it, but if you put 2 ohms in series with the heating element , the element will only dissipate 768 watts, which is about 43% of what it would dissipate under full power. That means it might take the heater a long time to get to the temperature needed to get your process to work, or maybe never... – EM Fields Nov 22 '16 at 17:35
• @Samuel P: Part 2: Also, be aware that the resistor will get very hot, so you should house it in a cage which will itself stay moderately warm and won't restrict airflow, but will make it impossible for anyone to contact the resistor. As for mounting the resistor safely, I've never had a problem with the holders which slip into the ends of the resistors, suspend the resistors in air and provide a secure footing for the resistors. The part numbers are given on the full data sheet. – EM Fields Nov 22 '16 at 17:37