-=> Quoting Tesla-at-grendel.objinc-dot-com to Tesla List <=-

 Te> * Original msg to: Esondrmn-at-aol-dot-com

 Te> Quoting Ed Sonderman:
 ES> Richard,
 ES> In a recent communication, you said a 12kv 120ma 60 cycle n  
 ES> power supply will effectively charge a .026mfd capacitor.    
 ES> I've been curious about this and was doing some math and I   
 ES> came up with a different answer. Maybe others will check     
 ES> this out and comment.

 Te> Oh lord Shaitan help me! 

 Te> You guys know that math and I don't get along. Just the thought
 Te> of meandering through this makes my head start to hurt. Who is
 Te> the resident MATH expert anyway? 
 ES> I calculate as follows:  The available time to charge the    
 ES> capacitor is one quarter of the 60 cycle waveform (peak      
 ES> voltage).  

 Te> My understanding was that current was delivered from zero voltage
 Te> point to zero voltage point on the 60 cycle waveform. That covers
 Te> one-half of the waveform as opposed to 1/4 of the waveform that
 Te> you calculated from. Again, I could be wrong, but I believe that
 Te> this is another transmission line problem that Mark Graalman
 Te> would be the resident expert consultant on. I figured that the
 Te> voltage peak lagged behind the current peak in the line charging
 Te> the capacitor. 
 MG> Ed, you are correct. actually, a 1 milliamp source will charge any size
     capacitor, just as a small battery charger will charge a large 
     automotive battery. It will just take longer, if the source can't 
     charge in the required time frame you simply wind up with a average
     voltage level that is much less than would be possible had the source
     charged at a faster rate. 
 ES> This is .0042 seconds. To achieve 97% charge on a capacitor  
 ES> we need 3 time constants. The thevenan equivalent of a 12kv  
 ES> 120 ma transformer must look like an infinite current 12kv   
 ES> supply with a 100k ohm resister in series.  We know we must  
 ES> achieve full charge (97%) in .0042 sec and it takes 3 time   
 ES> constants to do this so we have .0042 / 3 = .0014 sec for
 ES> one time constant. We know one time constant = RC and C =    
 ES> one time constant / R.  Solving for C we have .0014 sec /    
 ES> 100k ohms = .014mfd. Thus, a 12kv 120ma transformer will     
 ES> fully (97%) charge up to a .014 mfd capacitor. 

 Te> Oh boy. You lost me after "This is .0042 seconds."

 Te> My tried and true (in practical applications and circuits) Tesla
 Te> coil calculator states simply that at 60 cycles a 12kv 120ma
 Te> power supply will fully charge a .0265 mfd capacitor. This is not
 Te> to say that this is a 100% correct answer, but it works OK.
 MG> The formula works by accident, the imped. of a transformer is a 
     CONSTANTLY CHANGING attribute, the current level and voltage level are
     different at every point on the 60 hz. sinewave. At the begining of the
     capacitor charging curve (0 degrees) the current is maximum and the
     voltage level is minimum, at the top of the sinewave (90 degrees) the
     voltage is maximum and the charging current is minimum. So for the sake
     of exhample, at the bottom of the curve you may have a neon delivering
     60 ma. at 1 volt and at the creast of the curve it may be delivering
     15000 volts at 1 ma. So what do call a match!?!? The reactance of the
     capacitor never changed, the transformer voltage and current was all
     over the place!!
 ES> If this math is correct, I calculate that we need a 12kv     
 ES> 171ma transformer to fully charge a .026mfd capacitor.

 Te> My calculator says this power supply will charge a .0377 mfd
 Te> capacitor at 60 cycles. I do not have the latest version of Mark
 Te> Graalman's calculator program, so if you do not get a response
 Te> from him, forward your post directly to his attention.
 MG> In TESLAC I use the same accidental formula that every one else does
     because is gives a workable answer. But, to insure there is a safety
     margin in the calculation I convert the RMS voltage level of the
     source to the peak value, this results in a capacitor calculation that
     appears too small, but it insures the source has ample reserve to 
     charge the capacitor. It is better to have a capacitor thats a bit too
     small, than a bit too big.
 ES> What happens when you have a power supply large enough to    
 ES> charge the tank capacitor in say 1/8 of a cycle?  Then we    
 ES> need a rotary spark gap to take advantage of the extra power 
 ES> -- do I have this correct?        Ed Sonderman

 Te> This is correct.

 Te> Richard Quick

 MG> Mark Graalman

... Alias, Mark the spark
___ Blue Wave/QWK v2.12