60 vs. 30 ma
From: John H. Couture[SMTP:couturejh-at-worldnet.att-dot-net]
Sent: Friday, June 20, 1997 4:04 PM
To: Tesla List
Subject: Re: 60 vs. 30 ma
At 02:14 PM 6/18/97 +0000, you wrote:
>From: Gary Lau 18-Jun-1997 0810[SMTP:lau-at-hdecad.ENET.dec-dot-com]
>Sent: Wednesday, June 18, 1997 8:10 AM
>Subject: Re: 60 vs. 30 ma
What you have brought up regarding the pri capacitor charging vs the power
transformer wattage is important. Most coilers have never considered this
From the "charge time" equation you show it is obvious that the time is
dependent on the voltage. However, with a neon transformer the voltage
varies with the load. In fact the voltage is almost zero when the cap is
first connected to the neon. It is obvious that the problem is not simple
because the voltage is varying as the natural log plus the load on the neon.
With a pole transformer this may be less of a problem if voltage regulation
of the transformer and circuit is good.
Assuming you start with a correct size cap, the "C" in the equation will
not vary but the R may vary. This would be because the impedance of the neon
would vary with the varying current. As you can see modeling this
transformer would be difficult. I do not think the cap would deliver twice
the power as the smaller transformer because of the non linear operating
conditions. The breaks per second would also have to be increased.
The optimum rate to dump energy pulses (BKS/SEC) into the tank circuit is
well established by the equation
BKS = 2 Wp EFF/(Cp Vp^2)
This is actually an energy equation and indicates the necessary wattage
(secs) to properly charge the cap of a certain size for a certain (optimum?)
number of breaks. The quenching time (spark duration?) and tank ring-down
(Q?) would be additional considerations.
The solution of the above looks difficult but it should be possible using
a simple computer iterative program. All the variables are either known or
can be estimated and the equations are easy to implement. Note that this is
a method of digging deeper into the operation of Tesla coils without the
cost in money and time of building, testing, and rebuilding coils. This is
the perfect approach for coilers who do not have the money to build
expensive coils but have the skills to do the programs. Of course, building
and testing small coils should not be ignored.
>>>If you add larger transformers, you can certainly also increase
>>>the capacitance to take advantage of the higher charging current.
>>>All I'm saying here is you do not have to increase the capacitor
>>>value when higher current transformers are added. Some folks
>>>seem to think one requires the other.
>>Actually you do have to increase the capacitor size if you increase
>>the charging current otherwise you will not be using the additional
>>current. Realize of course this assumes that you were already using
>>the largest practical size capacitor for the charging current. The
>>capacitor is really the controlling factor as to how much power we
>>can cram into a given Tesla coil.
>It would seem to me that if one switches to a transformer with
>twice the current rating, using the same capacitor, that the
>capacitor would simply charge up to the spark gap voltage twice
>Modeling the transformer as a voltage source Vt and series R.
>In a series V-R-C circuit, solving for the time to charge the cap
>to a certain voltage Vc, t = -RC ln((Vt-Vc)/Vt). Doubling the
>transformer current rating would halve the R value. From
>equation above, this would also halve the time for the cap to
>charge to the spark gap voltage, Vc. Thus the same capacitor
>would dump it's load into the primary twice as often, delivering
>twice the power as with the smaller transformer.
>What I have not considered is, is there an optimum rate at which
>to dump energy pulses into the tank circuit? I'm sure that this
>is related to one's gap quenching time and tank ring-down time.
>But then the conventional wisdom of matching one's capacitor
>reactance to the transformer's impedance doesn't consider this