Re: Success Re[TCML] Measuring secondary current

[mddeming@xxxxxxx]

I think part of the problem is trying to mix DC theory with steady-state AC theory and applying it to narrow pulsed RF phenomena. Reactance isn't resistance (eg no heat generated by reactance). 1MW at a 0.01% duty cycle is only 100W continuous power etc.etc. For example, a couple of millinia back when I was in H.S. we were able to get 1KW peak pulses from an old 6L6 tube by making the pulses narrow enough that the heat dissipation was less than 10W.

Matt D


-----Original Message-----
From: Andreas Hahn <andreas.hahn@xxxxxxxxxxx>
To: Tesla Coil Mailing List <tesla@xxxxxxxxxx>
Sent: Mon, Sep 3, 2012 9:05 pm
Subject: Re: Success Re[TCML] Measuring secondary current


Fascinating, thanks! This goes a long way towards explaining why this 
thing behaves the way it does... I'm somewhat surprised there isn't much 
more heat dissipated by the secondary, though. Those 440uJ have to get 
dissipated at some point, and the resistive dissipation (JavaTC gives 
53 ohms for the secondary DC resistance, which I should check when I next 
get a chance) ought to be significant  (V^2 / R = very high).

Come to think of it, that can't possibly be right. V^2 / R = 8300V^2 * 53 
ohms = 1.3MW of dissipation in the secondary...


On Mon, 3 Sep 2012, paul wrote:

> Andreas wrote:
>
>> For a 48kOhm secondary reactance, that means secondary
>> voltage output due to resonant rise should be about 8.3kV
>> (48kOhm * .170mA) ...
>
>> ... 18V * 2A = 36W in to the primary.
>
>> ... 8.3kV * 170mA means 1400W sloshing about in
>> the secondary circuit...
>
> Seems reasonable to have 1400W of reactive power in the secondary,
> this is the stored energy exchanging back and forth between L and C
> on each RF cycle.
>
> The stored energy is
>
> Xl * I_rms^2 / (2 * pi * F)
>
> = 48k * 170mA^2 / (2 * pi * 500kHz)
>
> = .00044 Joules
>
> The storage efficiency is measured by the Q factor:
>
> Q = 2 * pi * F * stored_energy / power_input
>
>   = reactive_power / power_input
>
> where power_input is equal to the power dissipated once the
> oscillation has 'rung up' to a steady state.
>
> Some of your 36W DC input is replacing the energy dissipated
> each second.   If the oscillator and primary were 100% efficient
> so that all the 36W goes into the secondary, then the Q would
> be around 1400 / 36 ~= 40
>
> If the efficiency was 50%, then Q would be 80, and so on.
> Typically secondary Q factors are in the range of 50 to 200 or so.
> A good oscillator/primary efficiency might be 70%, and 90% or more
> would be excellent.
>
> Measure Q independently by measuring the 3dB bandwidth,
> then back-calculate efficiency.
>
> --
> Paul Nicholson
> --
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