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Author Topic: Are neon sign transformers already OU ?  (Read 5416 times)
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@ All.
I have in my hand   a Hansen EVG 20-3 Neon sign transformer.

Reading input    230v     0.20 A  50 hz
Output     3000v 20 mA    25 kHz

14 W OU ? What's the deal ?
E??
   
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Well... the very first thing to do would be to check the Data Sheet.

It can be very... enlightening.

 >:-)

   

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Well... the very first thing to do would be to check the Data Sheet.

It can be very... enlightening.

 >:-)


I just checked the data sheet you linked TK,and here is the info i see on there.

Input
230 V, +/- 10 %, 50 / 60 Hz
Depends on the connected tube load;
max. 0.35 A
cos phi 0.95<--- power factor seems pretty good here
Taking into account the power factor,we have an input of 76.47 watts

Output
3,000 V with 20 mA constant current,
Here the output seems to show 60 watt's.
So the efficiency is around 78.46%

Bugga--missed another OU device.


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As you say looking at specs can be very revealing  ;D


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VAR is just an angle on a scope. Nothing to see here -  move on.
   

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Stunned silence??

OK  let's do the math as the US would say.

120x 2.36  =    283.2  watts in..  Correct?

12,000 x 0.042  =  504 watts out.  Correct?

So the device gives out    505 -283.2   = 220.8  watts more than it consumes.

Is that correct??

WHY THE STUNNED SILENCE??


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VAR is just an angle on a scope. Nothing to see here -  move on.
   
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I heard it's measured in different situations. Input power is in closed circuit , output in spark gap. The difference can be used only in Tesla method of conversion circuit ;-)
   

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Stunned silence??

OK  let's do the math as the US would say.

120x 2.36  =    283.2  watts in..  Correct?

12,000 x 0.042  =  504 watts out.  Correct?

So the device gives out    505 -283.2   = 220.8  watts more than it consumes.

Is that correct??

WHY THE STUNNED SILENCE??

The "Stunned silence" is because "some people" look at Data Plates and claim OU from what the data plate says.

We actually see this quite frequently, as when Steve Spisak measures the input power to an unloaded motor-generator using a DMM and then looks at the Data Plate for an "output" measurement, even though there is no load attached to the output.

It's just too bad you can't run a home on a Data Plate. Maybe if you collected enough of them and connected them in series.....
 :D


(ETA: I can't find a Data Sheet for that particular transformer but the _truth_ will be something like Tinman's calculation using the numbers from the other transformer's data sheet as compared to its Data Plate.)
   
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Let us remember that the transformer, sat on a table, is no more capable of being OU than the common London fletton.
   

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Stunned silence??

OK  let's do the math as the US would say.

120x 2.36  =    283.2  watts in..  Correct?

12,000 x 0.042  =  504 watts out.  Correct?

So the device gives out    505 -283.2   = 220.8  watts more than it consumes.

Is that correct??

WHY THE STUNNED SILENCE??
Lol-no it dosnt give out 220 watts more.

12000v is P/P voltage.
Vrms is 0.3535 x Vpp
vrms =4242
4242 x .042=178.164 watts
178.164 / 283.2 x 100 =62.91% efficiency.

Not a very efficient OU transformer that one O0


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The spec plate gives the input, (maybe rated average current), but for the output the voltage is open circuit voltage and the current is short circuit current so it gives max values for both V and A for the output, however that never happens in reality, in reality they will be both less I would say. The neon uses the HV to "strike" and conduction lowers the applied voltage. Similar to a fluro. there is a voltage drop across the powered tube.

There is no OU neon sign transformers nor any other type of OU transformer. There is also no neon sign transformers that output more real power than is input. Just to say it two ways.

And the fact that Don Smith swore they were OU tells me he was absolutely full of it. And all his stuff was faked.
..
   

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So let us do some more mathematics.

Assuming the output is peak to peak
we have:

504x 0.707  =  356.328  watts  r.m.s.

so  283.2 watts in   356.328 watts out  =   73.128  watts  over unity

correct?

Or are we going to put our heads in the sand and say maths doesn't work with electronics " except when it gives the results
I want." >:(


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VAR is just an angle on a scope. Nothing to see here -  move on.
   
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So let us do some more mathematics.

Assuming the output is peak to peak
we have:

504x 0.707  =  356.328  watts  r.m.s.

so  283.2 watts in   356.328 watts out  =   73.128  watts  over unity

correct?

Or are we going to put our heads in the sand and say maths doesn't work with electronics " except when it gives the results
I want." >:(


You can't assume peak to peak for the power waveform you get from AC current and voltage.  Nor can you talk about rms watts, that is gibberish.  The power waveform obtained by multiplying sine-wave voltage by sine-wave current is a sine-squared function that is all positive and is a sinewave at twice the frequency sitting on zero, mathematically 1/2 + sin(2wt)/2.  It is the average value of this power waveform that you should use.  You are in good company in getting it wrong, Bearden's original paper on the MEG shows the math channel of a tek scope being incorrectly used to compute the rms value of the power waveform.  Tinman got the math right.

Smudge
« Last Edit: 2015-04-13, 14:02:13 by Smudge »
   

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So let us do some more mathematics.

Assuming the output is peak to peak
we have:

504x 0.707  =  356.328  watts  r.m.s.

so  283.2 watts in   356.328 watts out  =   73.128  watts  over unity

correct?

Or are we going to put our heads in the sand and say maths doesn't work with electronics " except when it gives the results
I want." >:(


Quote
504x 0.707  =  356.328  watts  r.m.s.

How did you come up with this from 12000 vpp ?.


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not a sine wave --- nothing like!


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Ok thanks for the info. Much appreciated.


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VAR is just an angle on a scope. Nothing to see here -  move on.
   
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