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« Reply #15 on: January 14, 2013, 12:07:52 am » |
For the final time, how long is the cat5 cable run? 
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« Reply #16 on: January 14, 2013, 07:41:27 am » |
8ft like it says in the wiring illustration and again a few replies back.
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« Reply #17 on: January 14, 2013, 09:42:21 am » |
Is this a repost of the same problem you asked about in 'How do you make the current draw from a device gradual instead of instant?', and again in 'How do I prevent video interference caused by electric motors', and also in 'Simple video interference problem. I'll pay to have it resolved.'? The people following this thread are in danger of wasting a lot of time and effort coming up with suggestions that have already been made.
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« Reply #18 on: January 14, 2013, 10:27:18 am » |
I'm trying to provide the information necessary for someone experienced enough to actually explain why the problem exists and exactly, with certainty, what will solve it.
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« Reply #19 on: January 14, 2013, 10:39:41 am » |
That sounds like a 'yes'.
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« Reply #20 on: January 14, 2013, 12:05:29 pm » |
Is the following analogy correct?:
If thought of like a Y connection in a water pipe, when the the high amp load is turned on all of the water (amperage) flows into it temporarily depriving the camera of current and causing the brief loss of video. Once the motor's on, the current flow stabilizes and the video is clear again while the motor is powered. Is that what's happening? if so, how do you make sure the camera will always have the current and voltage it needs to operate smoothly? Use an inductor?
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« Reply #21 on: January 14, 2013, 12:47:33 pm » |
Is that what's happening?
It's one possible explanation, which has already been suggested (several times) on the other threads you created about this issue.
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« Reply #22 on: January 14, 2013, 03:56:49 pm » |
If you can't help solve it, then there is no point in nagging about it. Please stop trying to hijack my thread.
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« Last Edit: January 14, 2013, 05:09:09 pm by David82 »
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« Reply #23 on: January 14, 2013, 06:17:39 pm » |
The camera draws 85mA. I have a 125mA, 10H inductor coming in the mail. The idea is that when the current gets sucked away by the 4A load, the inductor will be there to maintain current to the camera and hopefully prevent the momentary loss of video. If that doesn't work, I guess I can try a much larger 12v capacitor This is an image of how an inductor can smooth out current draw.  It seems that the camera can run on a range of voltages but it just doesn't like sudden changes in voltage.
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« Reply #24 on: January 14, 2013, 07:01:03 pm » |
just one is ground, one is 12v+ that powers the camera, one is the power for the 4amp load, one is +video, and the other four are going to two bi-directional motors which, by the way, have no effect at all on the video when used.
So whose idea was it to try to power a camera and a 4a motor thru a single conductor in a cat5 cable and not expect possible issues?
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« Reply #25 on: January 14, 2013, 07:33:03 pm » |
The load and camera are on separate + leads and share the same ground wire. The switch for the load is actually near the power supply and two separate wires power the camera and load.
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« Last Edit: January 14, 2013, 07:35:58 pm by David82 »
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« Reply #26 on: January 14, 2013, 07:53:19 pm » |
So just to confirm this, your using two completely separate power supplies that only share the same ground, correct?
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« Reply #27 on: January 14, 2013, 08:14:45 pm » |
nope, but I've tried that on a few different occasions with the same results.
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« Reply #28 on: January 14, 2013, 09:48:55 pm » |
The current going to the camera drops from a steady 85mA to about half, and sometimes zero, briefly, when the load is engaged. 1. Does that support the theory that the switching of the 4A load is momentarily diverting all of the current toward itself? 2. Does the lack of needed current to the camera explain why the voltage to the camera also drops? 3. If an inductor is used to maintain current, will it also maintain the voltage (assuming a diode and voltage regulator are used to filter out the voltage spikes)?
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« Reply #29 on: January 14, 2013, 11:00:02 pm » |
1. Does that support the theory that the switching of the 4A load is momentarily diverting all of the current toward itself? 2. Does the lack of needed current to the camera explain why the voltage to the camera also drops? Yes, I think this was described as a "brownout" days ago. I think the light is finally comming on! 3. If an inductor is used to maintain current, will it also maintain the voltage (assuming a diode and voltage regulator are used to filter out the voltage spikes)? I think you are wasting time on that approach.
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