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4hv.org :: Forums :: General Science and Electronics
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rogowski coil integration

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rp181
Sun Feb 01 2009, 05:28PM Print
rp181 Registered Member #1062 Joined: Tue Oct 16 2007, 02:01AM
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Posts: 1529
So i recently wound a rogowski coil around a coax cable. It works great, and now i am trying to add integration. I am taking a passive approach first. I put two 510k ohm resistors in parallel, and then in series with the coil. I then put a .01uf ceramic capacitor in parallel with it. My laptop sound card is no picking anything up from the coil, but it works fine with no integration. Any idea's on why this is not working? I would like to go to a active integrator later, but it is too expensive right now.

EDIT: I added one more 510k resistor in parallel, and i get results! this time, the peak is not nearly as high. To see the waveform, the time scale needs to be on the 1millisecond setting, rather then 2 seconds without integration. Do you think the wave looks good for a 3900uf capacitor charged to 160v hard shorted through a SCR?
Th
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likewhat
Sun Feb 01 2009, 05:49PM
likewhat Account deactivated by user request on 6/11/2009.
Registered Member #1071 Joined: Fri Oct 19 2007, 02:13AM
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Posts: 44
Why not just numerically integrate the signal? You have the data on the computer already.
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rp181
Sun Feb 01 2009, 06:19PM
rp181 Registered Member #1062 Joined: Tue Oct 16 2007, 02:01AM
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Posts: 1529
How do i do that? I put the integration because earlier with just 1 capacitor, the peak was cut off. I also dont know how to use it to calculate the peak current and such yet, as i have no RLC meter to run a sim against.
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likewhat
Sun Feb 01 2009, 07:15PM
likewhat Account deactivated by user request on 6/11/2009.
Registered Member #1071 Joined: Fri Oct 19 2007, 02:13AM
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Posts: 44
The easiest way is to put your data trace into excel. You should end up with column A for the time and column B for the voltage.

In column C on the second timestep you can type in this formula

C2 = B2*(A2-A1)+C1

What this is going to do is take the area of a box of size timestep by voltage. This way the area under each point in your derivative signal is added to the previous areas, and therefore you are getting the area under the curve.

Just copy that equation down to all cells where data is present and then make an XY scatter plot of it and you should have you derivative signal and your integrated signal.

If you want to get more fancy there are advanced numerical integration algorithms that can be used for a more accurate integration. Now that it is so easy to get data onto a computer and do whatever you want with it there is not really any reason to use a hardware integrator.
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rp181
Sun Feb 01 2009, 07:17PM
rp181 Registered Member #1062 Joined: Tue Oct 16 2007, 02:01AM
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how can you read voltage on a soundcard scope?
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likewhat
Sun Feb 01 2009, 07:24PM
likewhat Account deactivated by user request on 6/11/2009.
Registered Member #1071 Joined: Fri Oct 19 2007, 02:13AM
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Posts: 44
The vertical axis is voltage. That voltage somehow corresponds to the current flowing through your rogowski coil. After you integrate the signal you will have what the actual current looks like. If you run a known current source into your rogowski you can find what number you need to multiply by to get the graph to show the actual current.
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rp181
Sun Feb 01 2009, 09:01PM
rp181 Registered Member #1062 Joined: Tue Oct 16 2007, 02:01AM
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Posts: 1529
I housed the integration in one of those satelite combiner things. It works great, but now it stopped working =( when a coil says 50 ohm impendance, does that mean the resistance is 50 ohm?

EDIT: does it matter if the capacitor is before or after the resistor?
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