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[moved] how to estimate copper plate connection resistance?

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Barry

Sat Oct 20 2007, 03:04PM

Registered Member #90 Joined: Thu Feb 09 2006, 02:44PM
Location: Seattle, Washington
Posts: 301

A high-power coilgun needs a high-power switch. In a single-stage coilgun we don't need the microsecond timing control provided by thyristors and igbt's. In theory, a mechanical switch will work just fine, as long as we take care to minimize contact bounce and arcing and resistance. I've traditionally used SCR switching, and I have some good IGBTs on hand, but I want to achieve a much lower voltage drop than even forward-biased diodes can provide. This will help maximize ringing (ie minimize decay rate) in a coilgun with nonpolarized capacitors.

I have a mechanical approach in mind that inherently reduces contact bounce and arcing. It can provide a simple, reliable, and of course non-polarized switch that scales well for very high current coilguns. It will probably need periodic contact cleaning but I'm willing to deal with that.

What I don't understand is the surface area requirements in the switch. How do I estimate the connection resistance between two copper surfaces? How much contact area will result in how many milliohms? Is there a standard formula that I can use? From this, I can determine the size needed for a desired resistance and current density.

Thanks, Barry
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likewhat

Sat Oct 20 2007, 03:30PM

Account deactivated by user request on 6/11/2009.
Registered Member #1071 Joined: Fri Oct 19 2007, 02:13AM
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Contact resistance is a pretty hard question to ask and electrical engineer that specializes in it. I know there is a relation between the area in contact and the pressure with which it is being pushed together, but I am not sure if constants exist to to use it generically for some given metal. You might just be able to measure it with a constant current source, slam them closed and look at the voltage drop or something.

Overclocked

Sat Oct 20 2007, 03:52PM

Registered Member #1056 Joined: Thu Oct 11 2007, 11:15PM
Location: CT, USA
Posts: 27

Taken from My Electronics Book; Introductory Circuit Analysis 10th ed By Boylestad (its from college). Assuming your Contacts are Square,

R=p* (L/A)
where,

p= resistivity of the material
L= length of the sample
A = Cross Sectional Area of the sample

If your using Copper, Its resistivity (p) is 10.37 @ 20 degrees C. Aluminum would be 17.0, and Iron would be 74.0 All at 20C.

H and A Must be in mils. Heres an example

L=3 ft
W=5 inches
Height = .5 inches
Material is Copper

Convert to mils
5 Inches = 5000 mils
.5 inches = 500 mils

A = 5000 mils*500 mils = W*H
=2.5 M sq mils

Now convert that into circular mils
2.5 M * (4/pi)
= 3.185 M CM (circular mils)

Now for the Actual Calc.
R= p*(l/A)
R=10.37 [(3ft)/(3.185 M Circular mils)]
R= 9.768 micro Ohms

Note: M = mega, I didnt want to keep typing out 10^6.

ADD: Even More Fun Stuff =D

Lets say you want to know the resistance if Temperature Goes up (which it will from switching..)

[abs (Ti) + T1] / R1 = [abs (Ti) + T2] / R2

where,
Ti = Inferred absolute temp of the material. This is a coefficient. For Copper its -234.5, Aluminum its -236 and Iron its -162.
R1 = Resistance at T1 temp
R2 = Resistance at T2 temp

Example: If a peice of copper has a resistance of 50 ohms at 20C, what will be its resistance at 100 C?

(234.5C + 20C ) / 50 ohms = (234.5C + 100 C) / R2

R2= (50 ohms*334.5C)/254.5C
R2 = 65.72 Ohms.

Barry

Sat Oct 20 2007, 04:16PM

Registered Member #90 Joined: Thu Feb 09 2006, 02:44PM
Location: Seattle, Washington
Posts: 301

Overclocked wrote ...
R=p* (l/A)

Thanks! That's the resistance of a solid block of material, right? This is Good Stuff for estimating the internal resistance of bus bars and the like. And for the resistance in the copper blocks in each part of the switch.

I wonder if your book also has some hints about the additional resistance due to two surfaces of the material contacting each other? I suspect that more factors come into play ... surface irregularities, slight corrosion due to exposure to air, and even the force pressing the plates together. I'll probably have to choose a limit on current density and then over-size the contact area.

Does anyone have an opinion on about safe current density through copper contacts?

This is all quite "practical engineering" stuff. <sigh> When was the last time anyone designed a switch, anyway? I'm not sure that it's done anymore, with all the good cheap products on the shelf. Unless of course, you want to handle 20,000 amps in a millisecond, which hmmm doesn't seem to be the first thing you see in Radio Shack.

Barry
PS - I had better go raid Boeing Surplus one more time before they close their doors next month!
PPS - If there's a sudden power outage due to parts vanishing from my nearby power substation ... it wasn't me!!

Simon

Sat Oct 20 2007, 11:49PM

Registered Member #32 Joined: Sat Feb 04 2006, 08:58AM
Location: Australia
Posts: 549

Overclocked: your calcs will give the resistance of a block of copper. The resistance of the contact surface of two sheets of copper will be much larger. I'd use those calcs and factor in a large number (for "effective contact area", since the contact will be imperfect).

TDU has some good experience with high power switching with spring loaded metal contacts, IIRC. I'll move this thread to General Science to give people like him a chance to add input.

Marko

Sun Oct 21 2007, 12:12AM

Registered Member #89 Joined: Thu Feb 09 2006, 02:40PM
Location: Zadar, Croatia
Posts: 3145

I wonder if your book also has some hints about the additional resistance due to two surfaces of the material contacting each other?

Contact resistance is very large compared to a block of metal of same shape.

Impossibility of creating exactly matched surfaces is first problem. Large area may help you nothing when first contact, and arcing, will happen at just one small spot melting and vaporising metal, which in turn deforms the surface further.

The arcing creates oxides which combined with charred dirt increases the resistance further greatly.

You are forcing enormous current through the contact in short time so contact surface will matter little, when majority of energy will be transferred through small first-contact spot and arcing in it.

In other words, you'l suffer massive ''turn on loss'' with such a kind of switch, because firing will be over long before the full contact actually forms.
This loss of energy will manifest in amazing shower of sparks from contacts and welded electrodes.

You can minimize this by making contacts slam together faster; but that's a difficult thing to calculate.

If you assume your first contact/arc forms at 0.1mm you would need impact speed of 100m/s to compare it to turn-on of 1us (some slow IGBT).

Even if you machine the electrodes better than that they will erode over time and create trouble.

Bjørn

Sun Oct 21 2007, 01:05AM

Registered Member #27 Joined: Fri Feb 03 2006, 02:20AM
Location: Hyperborea
Posts: 2058

It is even a bit more complicated. The magnetic fields will affect the formation of arcs. Some of the most powerful coil guns were switched by the projectile without arcs because the magnetic fields were shaped to suppress the arcs. So the design will greatly affect the performance.

Overclocked

Sun Oct 21 2007, 01:06AM

Registered Member #1056 Joined: Thu Oct 11 2007, 11:15PM
Location: CT, USA
Posts: 27

Barry wrote ...

Overclocked wrote ...
R=p* (l/A)

Unfortunately, No. However, I do know Copper Oxide is a bit of a semiconductor and when exposed to light produces a current (assuming its on copper in the first place) You Could just let H = 0 or some Very small Number and calculate it that way.

Marko

Sun Oct 21 2007, 01:20AM

Registered Member #89 Joined: Thu Feb 09 2006, 02:40PM
Location: Zadar, Croatia
Posts: 3145

It is even a bit more complicated. The magnetic fields will affect the formation of arcs. Some of the most powerful coil guns were switched by the projectile without arcs because the magnetic fields were shaped to suppress the arcs. So the design will greatly affect the performance.

Bjorn, isn't the magnetic field what is actually going to Z-pinch the initial arc and cause all the problems?

External magnetic field? I don't see how could one prevent arcing during closing of the switch.

Bjørn

Sun Oct 21 2007, 01:52AM

Registered Member #27 Joined: Fri Feb 03 2006, 02:20AM
Location: Hyperborea
Posts: 2058

Since there is no current before the arc it is not possible to affect the formation of the arc without an external system. When the arc has started it will be possible to affect it and it will be possible to avoid the arc restarting during the bounce.

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