Tv reception

wrote:



My electric fence was interfering with my TV reception.

Lucifer Morningstar wrote:

If I had an electric fence that would be one of the first suspects but  
who would think turning a light on would cause it, if I lived with  
anyone I may have figured it out earlier.

Why would you have figured it out earlier if you lived with someone?  

Once upon a time on usenet Max wrote:


Less time with pants around ankles? <g>

Max wrote:

Because two are in more places than one.
If one was watching the tele and the other turned on the light there  
would be more chance to work out a relation for instance.

wrote:



My electric fence was affecting the signal quality to where the
picture and sound dropped out completely but not in pulses
like the electric fence.

They're not entirely dissimilar to automotive CDI - they have a transistor  
inverter generating about 350V for the capacitor discharge section.  

Its been a possible problem with CFLs for years.

Some of the LED replacement bulbs use a wattless dropper instead of SMPSU  
circuitry. You might be able to adapt it, but don't forget to include a  
surge limiting resistor - the current is very high around the zero crossing  
portion of the AC waveform.

All the LED bulbs I've pulled apart had SMD resistors, so I didn't bother  
tracing the schematic - there was a very handy CFL blog, maybe LED bulbs  
have caught up.  

That's mad, the current will be very low around the zero crossing of the AC  
waveform.


  

Three words - "rate of change".  

It has nothing to do with rate of change, the voltage by definition
is close to zero, so the current is in fact very low. Ohms law.
  

Geez - and I thought I wasn't great at AC theory..........................

A coupling capacitor (wattless dropper) blocks DC, the AC waveform is almost  
that at the peaks. The capacitor only passes a rate of change - that's at  
its greatest around the zero crossing points.

Maybe you should lay off the booze (or whatever)........  

On 07-Feb-17 5:15 AM, Benderthe.evilrobot wrote:


I have put such a circuit in a simulator, and except for the inrush  
current I don't see anything like you said.
Try it yourself, like on: http://www.falstad.com/circuit/
Select 'File', 'import from text', and paste this text, a simple example:

$ 1 0.000005 3.452441195350251 45 5 43
s 688 96 752 96 0 0 false
v 752 368 752 96 0 1 50 330 0 0 0.5
d 480 352 480 240 1 0.805904783
d 576 112 576 208 1 0.805904783
w 688 96 576 96 0
w 576 96 576 112 0
w 480 96 576 96 0
w 480 352 480 368 0
d 416 144 416 96 1 0.805904783
d 416 272 416 352 1 0.805904783
w 480 96 416 96 0
w 416 352 416 368 0
w 432 368 480 368 0
370 688 368 752 368 1 0
162 480 224 480 160 1 2.1024259 1 0 0 0.01
162 512 224 512 160 1 2.1024259 1 0 0 0.01
162 544 224 544 160 1 2.1024259 1 0 0 0.01
w 480 144 480 160 0
w 480 160 512 160 0
w 512 160 544 160 0
w 544 224 512 224 0
w 512 224 480 224 0
w 432 144 416 144 0
c 560 368 624 368 0 5e-7 -9.1361757825205
w 688 368 624 368 0
w 560 368 480 368 0
w 416 144 416 160 0
w 416 160 416 272 0
w 432 368 416 368 0
w 432 144 480 144 0
w 480 240 480 224 0
r 560 320 624 320 0 220000
w 560 320 560 368 0
w 624 320 624 368 0
w 576 208 576 224 0
w 480 240 576 240 0
w 576 240 576 224 0
o 1 64 0 2083 640 0.1 0 -1 0
o 30 64 0 2083 640 0.1 1 -1 0

You have to reset the simulation and start again do get the current  
range to normal after the inrush.
AFAIK The only important thing to remember is to use at least an X3  
rated capacitor and a discharge resistor across the cap, so yo don't get  
zapped when touching the open pins.

If you have to reset it after the inrush - that's probably why you're not  
seeing the cycle by cycle current peaks around the point of maximum rate of  
change.  

On 08-Feb-17 3:42 AM, Benderthe.evilrobot wrote:


No, you have to reset it because the range of the current display has  
expanded so much, by the inrush, that you don't see the normal current  
curve. There is no such thing as a current peak once it is on. There is  
a switch on the top right of the schematics. You can click on it to  
switch on and off and see what the circuit does, as well.

I have built and used such circuits often.
Using an X rated capacitor or/and a small "sacrificial" resistor in  
series will make sure that it won't catch fire due to a short in the  
capacitor.

"normal current curve"? - its only the sinusoidal drive that takes the edge  
off the capacitor current that peaks around the zero crossing point.

You could switch the circuit on *ANYWHERE* on the cycle - it could be at the  
peak and cause a big current pulse, or at zero where it starts at zero and  
rises pretty much the same as continuous operation.

The AC current in a theoretically ideal capacitor leads voltage by 90  
degrees - its not some great mystery from an alternate universe with  
different laws of physics.  

On 09-Feb-17 3:18 AM, Benderthe.evilrobot wrote:


It's you that need to get real. The current does not "get very high'  
around the zero crossing. It just reaches its maximum there because of  
the phase shift of the capacitor. The resistor you were talking about is  
only there to protect the circuit from melt down in case the capacitor  
shorts out. Full stop. Unless you have a non-linear load that's just the  
end of the discussion.

Your twisting what I said doesn't make it any less real.  

On Thu, 9 Feb 2017 21:51:41 -0000, "Benderthe.evilrobot"



Would like to know how old "F Murtz" aerial is?
Is it a old Analogue or new digital?

Live in almost same area and my old Analogue was causing loss of
reception when turning almost anything on.  

Decided to remove  Analogue get Terracotta Tile roof washed new
gutters and fascia.
Then paid $90 for TV guy to put in new digital aerial problem gone.