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I want to make a DIY spot welder

I am planning to rebuild a non-rebuildable heating coil (Eleaf HW-N) and I need to weld two non-resistive (low resistance) wires. A stainless-steel net to a nickel-chromium alloy to protect the rubber(red on video) at the bottom of the assembly from melting

Coil specs: I want to use the "net" version of the coil but I didn't find a video for that The original: Dim:37.5*12.6mm Mat:Kanthal Res:0.2Ω≈ 1.6 Ω/ft

My Found: Dim:∞*13mm Mat:Kanthal Res:0.2Ω≈ 1.2 Ω/ft with this as the legs

enter image description here

I can find a lot of DIY builds for "wire zappers" but they are all for connecting two wires (around 0.5 max) I don't know what is my current/voltage need to connect the wire to the mesh

Click: Video of the coil disassembly

The circuit is quite simple I found one here Wire zapper Schematic

R1: 3.3 OHM 10W

R2: 33 KOHM 1/4W

Fuse1 : 3A

D1: 1N5400

C1: 2200 uF 50V

I'd use an LM2596 board(35V output) I know that this works for .5 wires.

What kind of capacitor should I use to make a good weld?

What would be the approximate current that is enough to melt the 13*1mm area?

And what voltage would be needed? I saw others with 60V, 45V, 35V

Also, I'm not sure if I can weld SS-Ni as they have such different melting points.

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  • \$\begingroup\$ Maybe with a bigg L-Ion battery array and an MOT transformer with low turns in series, but they probably are using an induction step-down welder like a big soldering iron or ultrasonic welder to fuse the materials together \$\endgroup\$ – Tony Stewart EE75 Jun 23 '18 at 21:27
  • \$\begingroup\$ You mean the whole capacitor solution is not gonna work? Why? I thought if one capacitor is not enough I may use more as a charge pump \$\endgroup\$ – Roland Jakubik Jun 23 '18 at 21:42
  • \$\begingroup\$ Not for welding . No way. Max power must match impedance of the joint. \$\endgroup\$ – Tony Stewart EE75 Jun 24 '18 at 3:49
  • \$\begingroup\$ According to Amada, even commercial spot welders use "Caacitive discharge" as a Power Storage-Source \$\endgroup\$ – Roland Jakubik Jun 24 '18 at 22:13
  • \$\begingroup\$ Thanks Roland yes Amada/Miyachi make really nice spot welders for $1.7k \$\endgroup\$ – Tony Stewart EE75 Jun 24 '18 at 22:36
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Imagine a loose contact resistance to be welded (switch closing after a spot weld). The 1st weld is easy, Each subsequent weld current gets shared by each previous weld as the gap gets closer.

schematic

The Ni-Chrome wire does make it a little easier to make the resistant-weld significantly smaller than the inter-weld resistance.

Assumptions:

  • Ni-Chrome end-to-end R= 200 mOhms, height = 1/4 Diameter or 50 mOhms
  • weld contact surface area over height must be 100% otherwise there will be hot spots where current during operation has more resistance and less path to share.

This means the weld must be continuous along the height of the foil which involves a 90 deg wrap around the SS terminal wire.

Imagine only 10 spots, instead of 20 spots or 40 spots. In order for each weld to be lower resistance than the heater, it must be less than 50 mOhms/10 or 5 mOhms which needs to be achieved in 1mm² spot.

So what energy does it take to raise a spot of Ni-Chrome per mm³ well above 1400°C melting point to diffuse into SS with a similar melting point?

10 Joules? ( tiny spot weld) 100 Joules? 300 Joules?

Since Energy of a Cap is \$E=½CV²= I²ESR\$ and max current must be much greater than the average bond current has ESR which is falling with V so it is not simple I=V/R and is limited by Cap Array ESR. Rather the cap array must be far less than the spot weld ESR so as not to absorb much of the stored energy lost in heat.

Let's assume the caps must be explosion proof and must be plastic self-healing Y cap rated or better. What what is better high V or high C? That depends on ESR of each part of the system ( caps, SS, nichrome and existing spot welds)

Start with 100J or C = 2 x 100[J] / V² and assume CAP ESR< 5mΩ and T=ESR*C minimum

At 1V , C = 200 Farads with T = 1 second and I = 200A max = 1V/5mΩ

At 10V , C = 2 Farads with T = 10ms and I = 2,000A max

At 100V, C = 20mF with T = 100 us and I = 20,000A max ( no way on a small array)

If you can get tungsten electrodes and an array of L-ion cells in parallel to supply 200A for a few seconds, you might get some practical results. But ideally you want to maximize current yet have enough voltage to strike an arc. ( at 1kV/mm you need to have surface roughness < 1um to arc at 1V and not covered with carbon from 1st attempt.) 1um is unrealistic.

Therefore you want slightly higher voltage around 4V to get > 100A for > 1 second.

Others can contribute and compute volume of 1mm² metal and the energy to raise the temperature 1600'C.

Or decide if the resistance rise in the SS to 800% or so causes the heat to be localized for a continuous seam weld.

Anecdotal

But I still think this one is done with Ultrasonic welders. For 1 point Cap discharge, is easy. But seam weld, not so easy. In '79 I designed an instrument to monitor difussion bonding of a Zirconium shim inside a two Monel Steel tubes at ~10kA * 4V. It was water-jet cooled as the tube rolled around and power had to increase as R reduiced around the circumference.

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  • \$\begingroup\$ Tony, I'm terribly sorry for, forgetting to mark your answer, It is a really well taught, and written answer, thank you for your time, and effort \$\endgroup\$ – Roland Jakubik Aug 4 '18 at 19:09

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