# Amps and Volts in Resistance Heating

Discussion in 'The Projects Forum' started by amspurge, Jan 19, 2016.

1. ### amspurge Thread Starter Member

Jan 15, 2016
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Hello Everyone,

I'm currently working on a resistance heating project that has been going on for some time now. We have yet to have a breakthrough but have not yet given up. I have a limited background in electronics and electricity in general, but I think I know just enough to be dangerous.

What we're trying to do is rapidly heat a metal tip (material TBD, but Nichrome and SS are both candidates), and use that hot tip with a concave "dome" shape to melt plastic. We would then rapidly cool it to solidify the plastic. The tips range in size from 1/8" up to about 1/2" (but the larger size is rare). Image of a tip is below. You can see the slit in the side that helps direct the power to the dome/tip area.

The test system that we currently have developed outputs a very high amperage with very low volts. It's been awhile since I've played with it, but last I remember we were getting 600A/1.5V out of the secondary with a tip rigged up. What I would like to know is this: How does my voltage affect the flow of the amperage and therefore the proper heating of the tip? My thoughts are that we have plenty of amperage, but not enough voltage to carry the current through the circuit and properly heat the area we need to heat. Any thoughts?

2. ### wayneh Expert

Sep 9, 2010
12,405
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You haven't said wether your power supply is regulated or not, or anything about the load properties. With a constant voltage supply, the current in the load will depend on the load and that voltage. The supply will have a maximum rated current. A constant current supply will vary its voltage in an attempt to drive current through the load. There will be some maximum voltage it cannot exceed.

3. ### amspurge Thread Starter Member

Jan 15, 2016
48
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We're driving our tip with AC power, so I believe we're getting a fixed amount from the secondary side of the transformer. Does this help?

The leads coming from the tip are connected to each leg of the secondary.

4. ### #12 Expert

Nov 30, 2010
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7,358
You obviously have enough voltage to drive 600 amps through the tip, you just said so, even if you don't know it.

Voltage affects amperage in accord with Ohm's Law.

ps, stainless steel expands surprisingly much when heated. You can use that to your advantage with the right shape, or find that you must avoid it. Depends on your application.

5. ### amspurge Thread Starter Member

Jan 15, 2016
48
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Would I find a better/faster heating result by increasing the voltage and lowering the amperage a bit? Would it help the power flow through the tip better?

6. ### ronv AAC Fanatic!

Nov 12, 2008
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1,496
You need to make some careful measurements. 1.5 volts at 600 amps is 900 watts. I have a 100 watt soldering gun that heats a copper tip, about the same size as your small tips, to about 400F in a few seconds.
@#12 is the resident thermal expert, but I would expect the thing to melt with 900 watts.

7. ### #12 Expert

Nov 30, 2010
16,705
7,358
That is impossible. Please learn about Ohm's Law. For any given resistance in the tip, the current will be the voltage divided by the resistance. E = IR
You can not increase the voltage in order to decrease the current unless you replace the tip with a different metal.

And, yes, that much power in a piece than small will melt most metals in less than a minute.
I can only assume you mean to apply power in short bursts. If you didn't, you would damage the wires too! They are certainly not built to carry 600 amps continuously.

Last edited: Jan 19, 2016
8. ### #12 Expert

Nov 30, 2010
16,705
7,358
Wow. I didn't realize I had shown that much expertise. Thank you for the compliment.

But, yes, thermodynamics is my day job.
I just don't use it on things this small. My usual job comprises about 10,000 to 20,000 cubic feet.

Last edited: Jan 19, 2016
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9. ### amspurge Thread Starter Member

Jan 15, 2016
48
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#12:

I appreciate your responses. I will read through your Ohm's Law post a few more times until I fully understand.

Thanks again!

10. ### amspurge Thread Starter Member

Jan 15, 2016
48
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Getting a better understanding of Ohm's Law! I had a very incorrect understanding of it before, so I'm glad it is a bit more clear.

I have analyzed one tip with a spectrometer, and its alloy came back as a 400 series stainless steel. The tip that I was just testing (that is a proven tip on a proven system) appears to also be stainless steel (I sent it out for analysis today).

Any thoughts on why one would use stainless steel for resistance heating? I was able to get the "dome" part of the tip to glow dimly (4mm diameter, 1-2mm thick area) in one second of heating and measured the power at about 136.53W (111A/1.23V). I've attached a picture of the tip I was testing. You can see that the tip is at an angle and the cut in the side appears to account for this.

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11. ### #12 Expert

Nov 30, 2010
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It's hard, resists corrosion, has more resistance than copper, melts at a higher temperature than copper, shrinks when it cools off, and is cheaper than tungsten. That's probably only part of the reasons.

12. ### ronv AAC Fanatic!

Nov 12, 2008
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I think the slot in this tip is functional. It allows for the concentration of the heat at the tip. Think of the tip as a filament in a light bulb and the rest of the tip as the wires that support the filament.
So you just need more power. Make sure to measure the voltage as close to the tip as you can, because the wires have resistance. So your voltage may not be as high as you think.

13. ### strantor AAC Fanatic!

Oct 3, 2010
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I can't see what size wires you have going to the tip, but be aware that voltage drop along the leads to the tip can greatly diminish heating ability. The more amps you have, the bigger your cables need to be. Your AC source probably puts out a lot more than 1.23V, and most of it is lost before it ever gets to the tip. For maximum power delivery to the tip, make the wire leads as short as possible and make them the highest wire gauge possible.

14. ### amspurge Thread Starter Member

Jan 15, 2016
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I think I'm getting a bit closer to understanding the process of resistance heating. I found an older tip that was designed by a previous engineer working on the project and gave it a shot. The video attached clearly shows that the current is "starting" in the right place, but flowing in the wrong direction. I have some ideas of how to overcome this, but I welcome any other ideas.

Video of test:

I was speaking with an engineer today about the potential to increase the AC frequency in order to increase the speed at which the tip is heated. Anyone know if this is true? If so, is it practical to increase the frequency of the current and is it safe and cost effective? He also mentioned the AC "Skin Effect." I'm having a little trouble finding good information on this, so if anyone can explain how it applies to my application I would really appreciate it!

15. ### #12 Expert

Nov 30, 2010
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1) only if you get into the megahertz region. Google Inductance heating. It is applied externally.
Cost effective? I don't think so.
Any frequency has a limited depth of utilization of the conductor. It's about 3/8 of an inch for 60 Hz, so there is no sense in using a wire more than 3/4 inch thick. At another frequency, a copper pipe is as good as a wire because the current is only using the outer 1/8th of an inch anyway.

The question remains: If you used a high frequency and tried to heat the outer 0.02 inches, the heat would immediately move toward the rest of the thickness, so what would you accomplish?

16. ### amspurge Thread Starter Member

Jan 15, 2016
48
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Your last point is great, #12. It really does come down to tip geometry apparently. Increasing the frequency would only serve to heat the upper portion faster. I have some new designs that should be ready for testing in a week or two. I will report back with my findings once complete.

17. ### strantor AAC Fanatic!

Oct 3, 2010
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Your tip is doing exactly what i would expect; heating up at the thinnest portions (the "bottle necks" or "choke points"): the outsides of the "T", which as I take it, isn't where you wanted. I believe you want the heat along the rim of the cup; if that's the case, then you need to make the rim of the cup the thinnest portion. That complicates things. It means you'll need an axial design; picture a rod inside a tube, separated all around by an air gap. The rod is connected to one wire lead and the tube to the other. The end of the rod is flared out (or the tube is flared in, or a combo of both ) at the end, so that the outer circumference of the rod touches the inner circumference of the tube, but only along a small band of area. The surface area of the contact circle should be less that the cross-sectional area of the tube and of the rod, to make that circle be the "bottle neck"/"choke point. "

18. ### amspurge Thread Starter Member

Jan 15, 2016
48
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Thanks for the feedback, guys. I have redesigned the tip in four different variants. I've attached all four with the hopes that you will all vote on which one you think will perform the best and heat the tip as the application requires. Take a look at my four designs and let me know which one you think will work. I'll be having these made this coming week and will have results the week following.

You'll notice that all of the designs have an extra step much closer to the tip. The video I posted shows how the current flows away from the tip, and I'm hoping less material at the tip (thanks to the step) will make it flow in the opposite direction.

Also notice that the step on 9505-T2A is much closer to the tip than the other three.

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19. ### ronv AAC Fanatic!

Nov 12, 2008
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My guess... That tip has only the one slot. If it had another slot opposite that one the hot spot would move to the tip.

20. ### amspurge Thread Starter Member

Jan 15, 2016
48
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Thanks for the feedback, ronv. What are your thoughts on the location of the last step? There are two versions of the tip that have a straight slot. Each has a different step location.