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Thread: Determining Minimum Surface Area for a Piece of Steel in a Point Contact System

  1. #1
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    Determining Minimum Surface Area for a Piece of Steel in a Point Contact System

    I have a piece of steel that I am looking to supply ~12VDC to. My goal is to have the voltage driven power supply, send ~12VDC and 2W of power to the piece of steel through a point-contact system (the piece of steel touching the power supply terminal).

    My question is, what is the best way to determine the minimum surface area of the piece of steel needed at the point of contact?

  2. #2
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    I believe that is a very tough question to answer.

    Experiment.

    After dinner I will provide some additional comments.

    .

  3. #3
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    Quote Originally Posted by Elektrotechnik View Post
    My question is, what is the best way to determine the minimum surface area of the piece of steel needed at the point of contact?
    The formula for the bulk resistance from a surface connection (electrode) on a chunk of metal of uniform resistivity is pretty simple as long as the area is small compared to the dimensions of the piece. But the contact resistance and the resistance on the power supply terminal side of the connection need to be considered too.
    How do you intend to insure that the area remains unchanged? Will you be bolting it down or just applying spring pressure?
    And finally, the area requirement ultimately depends on how much voltage drop you will accept and whether that will cause a heating problem as well as a voltage drop problem.

    Your description of 10V and 2 watts implies that something will be limiting the current. Do you expect the limiting factor to be the bolted connection or the resistivity of the steel or some other series resistance?

  4. #4
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    Elektrotechnik:

    Your description implies that you want to dissipate 2 W at the interface of the point contact with the bulk piece of steel, and that both the point contact, and bulk material are steel, and not stainless.

    This is a lot of power in a very small area, and that means a lot of heat that will change the character of the interface contact.

    Without doing a controlled experiment, and only a two terminal ohmmeter instead of a 4 terminal test, I get about possibly 0.1 ohms with square shank screwdriver, and a scribe with a tip diameter of about 0.01".

    For 2 W at 12 V you need a resistance of 72 ohms. You won't get 2 W dissipation at the interface without an extremely small contact area and/or controlled contact force. If you just want 0.167 A of current, then use external current limiting. Power dissipation of 0.167 A in 0.1 ohms is about 0.003 W.

    If the contact resistance is 0.2 ohms, then to get 2 W dissipation in 0.2 ohms requires a current of 3.16 A. And 0.1 ohm requires 4.47 A.

    What is your goal and do you really want 2 W at the interface? The surface of a 2 W resistor dissipating 2 W will burn your finger. A 2 W resistor is fairly large compared to a point contact. I suspect 2 W dissipation at the tip end of my scriber would probably fuse it to the other piece (weld).

    .

  5. #5
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    Conductivity of iron is approximately 1/6 that of copper.
    I didn't think OP said it was the connection causing the resistance (load) but maybe I read it wrong.
    You make the lights come on and we make them go off.

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