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Re: Guide 60HzMatch

Tesla List wrote:
> 
> Subscriber: tom_mcgahee-at-sigmais-dot-com Tue Jan 28 22:58:51 1997
> Date: Tue, 28 Jan 1997 15:40:47 -0500
> From: Thomas McGahee <tom_mcgahee-at-sigmais-dot-com>
> To: Tesla List <tesla-at-pupman-dot-com>
> Subject: Guide 60HzMatch
> 
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> 
> All,
> I need feedback on the following possible entry for the Guide. (It is no
> longer the Idiot's Guide. Just the Guide).
> 
> Critique style, tone, syntax, grammar, but especially home in on content,
> useability, etc. For all I know, this may be totally wrong.
> If you have ANY relevant material, please e-mail to me Re:Guide 60HzMatch.
> 
> *** Beginning of article
> MATCHING THE IMPEDANCE OF THE TRANSFORMER AND THE CAPACITOR
> 
<SNIP>
> The transformer secondary winding has a characteristic impedance at 50/60Hz
> that will limit the maximum current that the transformer can deliver. You
> can model the transformer as a perfect voltage source in series with an
> inductor and a resistor. The resistance is caused by
> the length and diameter of wire used in the construction of the
> transformer's secondary. The inductor represents the inductance of the
> transformer secondary, which causes its AC resistance at 50/60Hz. {Fig ??
> Eq ??}

This 50/60 Hz impedance can be estimated by Z = Vout/Iout where Vout and
Iout are the faceplate ratings. For example a 15 KV 60 MA neon will have
an effective impedance of:

    Z = 15000/0.060 = 250,000 Ohms 

The approximate secondary inductance can be estimated as:

    L = Z/(2*Pi*F)  Henries
        where F = either 50 or 60 Hertz

For our 15 KV 60 MA 60 Hertz transformer, the secondary inductance (L)
will be:

    L = 250000/(6.28*60) = 663 Henries (!)
 
> 
> Methods for measuring actual transformer secondary impedance are given in
> {Ref Transformer Sec ??}
> 
> The current available at any given instant from a given transformer is
> limited by the transformer's instantaneous voltage and total effective
> resistance at that instant.
> 
> If a capacitor and an inductor are connected in series, and if the
> inductive reactance and the capacitive reactances are made to be equal,
> then the effective total reactance equals ZERO. For a given inductance and
> capacitor, this will occur at only one particular frequency, known as the
> resonant frequency. {Eq ??}

For series resonance, the inductive and capacitive reactances cancel,
leaving only the winding resistance. For a given tank capacitance (C)
and the neon's secondary inductance (L), the resonant frequency can be
found as:

   Fr = 1/2*Pi*SQRT(LC)

For example, assume a tank capacitance of 0.01 uF:

   Fr = 1/(6.28*SQRT(663*.01*1e-6)) = 1/(6.28*SQRT(6.63e-6)
      = 1/(6.28*2.57*1e-3)
      = 1/0.01617 = 61.84 Hertz

This is VERY close to 60 Hz. The good news is that we can maximize power
transfer in this case. The bad news, is that the neon's secondary
voltage and the tank cap voltage will rise to many times the rated
voltage of the transformer. It turns out that if we didn't stop this
"resonant rise" by firing the gap, the voltage will rise to Q times
Vout, where Q can be anywhere from 3 to 10. A 15 KV transformer can
generate well over 45KV (untill either the cap or the transformer goes
"poof". This is why its CRITICAL that the gap NOT be openned to widely -
its the only thing preventing the voltage from rising to a destructive
level. 

<Major SNIP>


> 
> 2)      I need the actual formula for computing transformer impedance.
> 
>                 Someone on the list said Ztrans=Etrans/Itrans. Is that right?
>                 Does that mean that a Neon with a tag that says 12KV -at- 30ma
>                 has Ztrans=12000/.030 ? That seems AWFULLY High!
>                 That implies that a 12KV Neon produces 30ma when you short
>                 out the secondary and its voltage drops to ZERO.
> 

It's given above. And that's exactly what it does under short circuit
conditions. You can take a 100 ohm 2 watt wirewound resistor, connect it
across the output, and measure the output voltage Vx (carefully!!) with
an isolated VOM. The transformer current output is 10*Vx (in MA). This
is a very useful method for identifying warranty-return transformers
that have had their faceplates removed. 

A similar test can be used to determine the "faceplate" open circuit
output of a "label-less" neons. By applying a relatively low voltage Vin
(say 5 or 6.3 volts) to the primary of the neon and setting the VOM to
the 1000 Volt range and connecting it across the HV output, the
open-circuit voltage output of the transformer can be found. Assume we
measure Vx on the output while applying Vin:

      Vout = (Vx/Vin)*120
   
   Example: Vout = 625, Vin = 5 volts

      Vout = (625/5)*120 = 15,000 Volts. 

>                 That further implies that the Neon WASTES 360 Watts
>                 when the current is 30ma. Can't be, can it?
>                 Isn't the 30ma the rated continuous current.. and isn't
>                 12000 the OPEN circuit voltage with NO LOAD?
>                 (Or have I misunderstood the tag rating?)

Well... it means that the transformer is limiting the current reactively
- its pulling 360 Volt-Amperes of current, only a portion of which is
being dissipated as real power loss (Watts). This is where Power Factor
Correction comes in to bring the primary current more in-phase with the
applied voltage.

> 
>                 I am more familiar with regular transformers, where
>                 the power rating is the product of the voltage at
>                 rated current, and the rated current.
> 
>                 Shouldn't the circuit be loaded down
>                 until it produces the 30ma and then measure the voltage...
>                 Then the difference in voltage from 12000 would be the
>                 voltage being dropped ACROSS the impedance at a draw of 30ma
>                 and the formula would now be:
>                 Ztrans=(E1-E2)/I  where E1=open circuit voltage (12KV)
>                 E2=voltage at terminals when load draws rated current (Say 11KV)
>                 I=rated current (.030 Amps)

The Short-circuit current output IS 30 MA. The voltage actually declines
with increasing loading, until the transformer actually "looks" like an
AC current source.
> 
> ****
> 
> Thank you for reading this. I would appreciate any constructive criticism
> you might have regarding this particular article, or the Guide itself.
> 
> Fr. Tom McGahee

Keep up the effort - you are a good teacher, and an eloquent writer!

And... Safe coilin' to you!

-- Bert --