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Re: 8 kHz Tesla Coil

Original poster: "Gerry  Reynolds" <gerryreynolds@xxxxxxxxxxxxx>


Hi Malcolm,

I'm not sure what it is you "dont think so". Is seems you are saying the same thing regard to AC_resistance/DC_resistance as a function of wire_size/skin_depth. Certainly DC resistance goes up as wire size goes down. Also, you were only commenting about my second paragraph only??? See comments below:

Gerry R

Original poster: "Malcolm Watts" <m.j.watts@xxxxxxxxxxxx>

Hi Gerry,

On 20 Sep 2005, at 10:33, Tesla list wrote:

> Original poster: "Gerry  Reynolds" <gerryreynolds@xxxxxxxxxxxxx>
>
> Since I dont have a complete design for a huge 8KHz coil (and I
> really dont want to invest too much time in it), I use my 8x36 coil as
> an example.  It uses a 15Kx120ma power source and a synchronous gap
> with 55nf Cp.  Ip_peak is 700 amps (with a 8x32 toroid and tapped at
> 12 turns).  Lp=70uh and Ls=80mh.  If I use the energy equations and
> assume no loss from primary to secondary, then Is_peak=Ip_peak *
> sqrt(Lp/Ls) or about 21amps.  The energy transfer time is 50us and I
> run at 120bps.  If I assume that Is_peak does not decay and stays
> constant for the full 50us then I calculate the
> Is_rms=0.707*Is_peak*(50us*120bps) or about 90ma.  Since my coil is 61
> ohms DC, the power loss in the windings  is about 1/2 watt.  However,
> as you rightfully mention, the skin depth for my operating frequency
> is 9 mils and I use 23 awg with a radius of about 11 mils.  So my wire
> accomodates only a little more than one skin depth.  Therefore, the
> current density at the center is about 1/e or 40% of the current
> density at the surface.  The AC resistance will therefore be larger
> and I will guess that the 61 ohms is no more than 120 ohms at
> resonance.  Even with the larger effective resistance, the power loss
> in the coil would be 1 watt.  I've tried to make my simplifying
> assumptions to error on the high side so the actual power loss should
> be less.
>
> If the skin depth is very large compared to the wire size, it seems
> like the resistance approaches the DC resistance (current density in
> the center is about the same as at the surface).  If the wire size is
> very large compared to the skin depth, most of the current flows on
> the surface and the resistance of the wire become inversely
> proportional to the wire diameter instead wire cross sectional area.
> In other words, a wire with too large of a diameter is a waste of
> copper.

I don't think so. I struggled with understanding this for quite a
while. If the diameter of the wire is equivalent to a skin depth or
so, the AC resistance (ignoring proximity effect) becomes
commensurate with the DC resistance of the wire (which of course also
climbs as the wire gets smaller).


Yes, this is essentially what I was saying above.

Even in a large wire, the current
distribution remains according the skin depth so that as the wire
gets larger, the effective cross-sectional area increases and the
resistance goes down in both quarters.

Absolutely. However, if you double the wire diameter and the cross sectional area is increased by 4, the resistance is only reduced by 2 instead of 4 because the conducting cross sectional area is porportional to the circumference instead the wire cross sectional area.